ELECTROCHEMISTRY CHAPTER 2
Chapter Outline
2.1 Electrical Conductivity
2.2 Electrolytic Conductance
2.3 Kohlrausch Law
2.4 Electrolysis
2.5 Electrochemical Cells
2.6 Electrochemical Series
2.7 Nernst Equation
2.8 Batteries
2.9 Corrosion
Electrical current flows in metals without changes occurring in metals. Solutions of acids, alkalies and salts conduct electricity, but material changes occur during conduction. Electrochemistry is mainly concerned with the relation between electrical energy and chemical energy.
2.1 ELECTRICAL CONDUCTIVITY
The substance that does not allow the flow of electricity through it is called an insulator.
S.No
1. Conductivity is due to mobility of free electrons from negative to positive end
Diamond is a familiar example of an insulator. Rubber, glass, plastics, etc., are other examples of insulators.
Electric current is considered in general as the flow of electric charges through the conducting medium. This medium is termed as an electrical conductor. Electrical conductor is defined as the substance that allows electric current to pass through it. Electrical conductors may be solid metals, fused metals, molten salts, or aqueous solutions of acids, bases, and salts. Electrical conductors are broadly classified into two types.
Electronic conductors: Metals, alloys, graphite, and salts like cadmium sulphide, copper sulphide, etc., are examples of electronic conductors.
Electrolytic conductors: Fused salts and aqueous solutions of acids, alkalies, salts, etc., are examples of electrolytic conductors. The main difference between the two types of electrical conductors are listed in Table 2.1
Conductivity is due to mobility of free ions towards oppositely charged electrodes
2. During conduction, matter is not transferred During conduction, matter is transferred in the form of ions
3. Passage of current may bring only physical changes.
4. The conductivity power is high
5. An increase in temperature decreases conductivity due to increased thermal vibration of particles or resistance increases
Passage of current brings physical as well as chemical changes.
The conductivity power is relatively less.
With an increase in temperature, conductivity increases due to the increase in the degree of ionisation and increase in the mobility of ions.
In electrolytic conductors, the flow of current is due to the movement of ions in the solution under the influence of an applied field. The mobility of ions under applied voltage is called electrical migration or mobility or conductivity of ions.
Electronic conduction is also called metallic conduction. It depends upon nature, structure, density, temperature, and valency of the metal. It also depends upon the composition of the alloy formed between metals.
2.1.1 Electrolytic Dissociation
Electrolyte is defined as a substance that conducts electric current in molten or in solution state and is simultaneously decomposed. There are substances that do not conduct electricity even in the solution state. Such substances are called non-electrolytes. Cane sugar, urea, glucose, etc., are examples of non-electrolytes. The important points of the theory of electrolytes are:
1. An electrolyte in aqueous solution breaks up into two oppositely charged ions and the ions are in equilibrium with the unionised electrolyte, BA B+ + A–
2. The process of splitting of the electrolyte into ions is called ionisation or dissociation. The fraction of total number of molecules present as ions in the solution is called degree of dissociation. It is den oted by ‘ a ’.
Numberofmoles ofelectrolyte asions
Totalnumberofmoles of electrolyte α=
3. The degree of dissociation depends upon nature of the electrolyte, concentration of the electrolyte and temperature.
4. The equilibrium constant for the dissociation of electrolyte (K) is related to the concentration (C) as K = Ca2/(1–a).
5. When an electric current is passed through the solution of electrolyte, the ions move
towards electrodes of opposite charge and are discharged.
6. The degree of dissociation of electrolyte increases with decrease in concentration. It approaches unity at infinite dilution. So, at infinite dilution, all electrolytes are said to be completely ionised.
7. Degree of ionisation increases with increase in temperature.
8. Theory of electrolytes is applicable for weak electrolytes only.
2.2 ELECTROLYTIC CONDUCTANCE
Conductivity of electrolytic solution depends on the nature of eletrolyte, concentration, and temperature of the solution.
Electrolytic solutions also obey Ohm’s law in a manner similar to electronic conductors. A conductivity cell is shown in Fig.2.1.} Platinum wires Black Platinum plates
Electrolyte Area, a Distance, l
Fig. 2.1 Conductivity cell
The resistance ‘R’ to the flow of current offered by the solution of electrolyte in the cell is inversely proportional to the area of cross section (a) of the electrode, and it is directly proportional to the distance of separation (l).
Resistance offered by the solution, () or ∝=ρ l RR a l a
Here, ρ is the specific resistance, also called resistivity. The ratio of length ( l) and area of
cross section (a) is called cell constant. Unit of cell constant is cm –1 or m–1.
Once the cell constant is determined, we can use it for measuring the resistance or conductivity of any solution. The set up for the measurement of the resistance is shown in Fig.2.2 .
Conductivity cell containing electrolyte
Fig. 2.2 Arrangement for measurement of resistance of a solution of an electrolyte
It consists of two resistances R 3 and R 4 , a variable resistance R 1 and the conductivity cell having the unknown resistance R 2. The Wheatstone bridge is fed by an oscillator O (a source of A.C. power in the audio frequency range 550 to 5000 cycles per second). P is a suitable detector (a headphone or other electronic device) and the bridge is balanced when no current passes through the detector. Under these conditions, unknown resistance
14 3 2 RR R R =
These days, inexpensive conductivity meters are available that can directly read the conductance or resistance of the solution in the conductivity cell. Once the cell constant and the resistance of the solution in the cell is determined, the conductivity of the solution is given by the equation * Cellconstant G k= = RR
The reciprocal of resistance is known as conductance (C).
1 11 / C Rla ==× ρ 1 Ck(or)kC (/a) a ==× l l
Here, k (kappa) is specific conductance, also called conductivity. It is the reciprocal of specific resistance. Resistance is expressed in ohm. The symbol of ohm is ‘Ω’. Conductance is expressed in ohm–1 or mho. In the SI system, ohm –1 is siemens. The symbol of siemen is ‘S’. Units of specific conductance are ohm –1cm–1 in CGS system and S m–1 in SI system. Specific conductance is the conductance of the electrolyte in unit volume of the solution. Conductance and specific conductance are one and the same if cell constant is unity.
The conductance of an electrolytic solution can be expressed as:
1. Specific conductance ( k)
2. Molar conductance ( Λm)
3. Equivalent condutance ( Λeq)
2.2.1 Specific Conductance or Conductivity (k)
The specific conductance of an electrolyte is defined as the conductance of the solution enclosed between two parallel electrodes of unit area of cross section separated by a unit distance or the conductance of the electrolyte in the solution of volume of 1 cm3 (CGS units) or the conductance of the electrolyte in the solution of volume of 1 m 3 (SI units).
In practice, the concentration of solutions of acids or bases or salts is expressed in molarity or normality. Conductance of electrolytic solution with reference to the concentration is of two types: molar conductance ( Λm) and equivalent conductance ( Λeq).
Molar Conductivity (Λm)
Molar conductivity of a solution at a dilution V is the conductance of all the ions produced from one mole of the electrolyte dissolved in V cm3 of the solution when the electrodes are
one cm apart and the area of the electrodes is so large that the whole of the solution is contained between them. It is usually represented by Λ m .
Relationship between Molar Conductivity and Conductivity
The molar conductivity of a solution is usually not found directly but is calculated from the specific conductivity. The relationship between molar conductivity and specific conductivity may be obtained as follows:
Consider a rectangular vessel with its two opposite walls one cm apart and made of some metal sheet so that these act as the electrodes.
Case-I : Suppose 1 cm 3 of the solution containing 1 mole of the electrolyte is taken in the vessel, the conductance of this solution will then be its conductivity. Further, if 1 cm3 of the solution taken contains one mole of the electrolyte, the conductance of the solution will be its molar conductivity. Thus, when 1 cm 3 of the solution containing one mole of the electrolyte is considered, the molar conductivity is equal to its conductivity.
Case-II : Suppose that 4 cm 3 of the solution containing one mole of the electrolyte is taken. The conductance of the solution will be still equal to its molar conductivity at this dilution but, now, there will be four cubes each of volume 1 cm3, as shown in Fig.2.3
molar conductivity, is four times the specific conductivity.
Therefore, the molar conductivity is related to the specific conductivity as follows: 10001000 Molarity mVmcc kVorkk c Λ=×Λ=×=×
where k is the conductivity and V is the volume of the solution containing one mole of the electrolyte and c is the molar concentration, i.e., mol L–1 (or mol dm–3).
Unit: The units of Λ m will be ohm–1 cm2 mol–1 or S cm2 mol–1 or Ω–1 cm2 mol–1 or S m2 mol–1 in SI units.
1 S m2 mol–1 = 104 S cm2 mol–1
Therefore, the formula, along with units of each quantity, may be written as
() 1 31 21 m 1 kScm1000cmL Scmmol MolaritymolL × Λ=
In terms of SI units, the formula becomes
1 21 m 3 kSm Smmol Molarity(molm) Λ=
However, if molarity is expressed in mol L–1, the above formula becomes ()() 1 21 m 31 Sm Smmol 1000LmMolarity(molL) κ ∧= ×
Equivalent Conductivity (Λeq)
The conductance of each 1 cm 3 of the solution is equal to its conductivity so that the total conductance of the solution, i.e.,
Equivalent conductivity of solution at a dilution V is defined as the conductance of all the ions produced from one gram equivalent of the electrolyte dissolved in V cm3 of the solution when the distance between the electrodes is one cm and the area of the electrodes is so large that the whole solution is contained between them. Its symbol is Λ eq Equivalent conductivity = specific conductance × V Λ eq = k× V
If the solution has a concentration of c gram equivalents per litre, i.e., c gram equivalents are present in 1000 cm3 of the solution, then the volume of solution containing one gram
equivalent will be 1000 , c i.e., 1000 V c = .
Variation of Conductance with Concentration
As the ions are current conducting species, conductivity of an electrolyte depends on the concentration or number of ions present in a unit volume of the solution. Therefore, conductivity of solution increases with increase in concentration of solution. Equivalent conductance or molar conductance decreases with increase in concentration, as shown in Fig.2.4
increases. The increase in the conductivity of a weak electrolyte at lower concentrations is attributed to increase in the extent of dissociation. The variation in equivalent conductance with concentration is not significant in the case of strong electrolytes because strong electrolytes are almost completely ionised at all concentrations, and increase in eq Λ with dilution is only due to decrease in the inter-ionic forces. In case of weak electrolytes, the decrease in eq Λ is significant with increase in concentration.
Λ eq increases on dilution and reaches a maximum value at infinite dilution because degree of dissociation of weak electrolyte becomes nearly equal to 1.
The variation of equivalent conductance of an electrolyte with concentration depends on the type (valencies of ions) of electrolyte than on its nature.
■ For strong univalent electrolytes (KCl), the decrease in equivalent conductance () eq Λ with increase in concentration is not very large.
■ The decrease in equivalent conductance () eq Λ with concentration is more as the valency of the ions increases.
■ Weak electrolytes like CH3COOH, NH4OH exhibit an apparently different behaviour due to incomplete ionisation.
Fig. 2.4 Variations of equivalent conductivity with concentration
As the concentration decreases for any electrolyte, the equivalent conductance
The ratio of equivalent conductance at any concentration () C Λ and at infinite dilution () 0 Λ is called conductance ratio.
Degree of dissociation (or) conductance ratio c m o m Λ α= Λ or c m m . α Λ Λ For a weak electrolyte, the conductance ratio is equal to de gree of dissociation.
a is higher for 0.01 M CH 3COOH when compared to that for 0.1 M CH 3COOH.
For weak electrolytes, limiting equivalent conductance cannot be calculated from eq Λ vs graph, because, for weak electrolytes, at low concentrations, eq Λ increases sharply with decrease in the concentration.
For weak electrolytes, Λ 0 is indirectly calculated from the Kohlrausch law.
On the basis of k (specific conductance) value, electrolytes are of two types.
Equivalent conductance of an electrolytic solution at very low concentration is known as equivalent conductance at infinite dilution or at zero concentration or limiting equivalent conductance () o eq Λ . Direct determination of conductance at zero concentration of solution is not practically possible. The work of Debye and Huckel on the quantitative treatment of equivalent conductance marked the commencement of a new era in electrochemistry.
For strong electrolytes, Λ m vs c graph is a straight line with a negative slope as shown in Fig.2.5.
Intercept is Δ0 Slope=-veStrongelectrolyteWeakelectrolyte
c
Fig. 2.5 Variation of equivalent conductance with square root of concentration
For strong electrolytes, limiting equivalent conductance can be known by extrapolating m Λ vs c graph to zero concentration.
m0 Kc Λ=Λ−
In this equation, K is a constant. It depends on the type of electrolyte (1 : 1 or MX etc.)
1) Weak electrolytes: These electrolytes have low k value, e.g., carboxylic acids.
ClCH 2 COOH has higher k value than CH3COOH.
2) Strong electrolytes : These electrolytes have high k value. Example: salts, mineral acids, alkalies.
The conductance of an electrolyte is due to its ionisation. The extent of ionisation is maximum for weak electrolytes as dilution reaches to maximum.
Equivalent conductance of an electrolyte can be determined experimentally by the following method:
i. Determination of conductance by using conductance meter
ii. Calculation of specific conductance from conductance l ck a ×=
iii. Calculation of equivalent conductance
1000 × Λ= eq k N
1. Specific conductance is 0.0382 ohm–1 cm–1 for a solution that is 0.1 M in KCl and 0.2 M in MCl (here, M is a metal radical). If the ionic conductances of K+ and Cl– ions in the solution are 74 and 76 S cm2 mol–1, find the ionic conductance of M+ ion.
Sol. () o m KCl Λ = 74 + 76 = 150 S cm2 mol–1 = m 1000 0.1 × Λ= k 3 21 KCl
1500.1 15100.015Scmmol 1000 × ==×= k
Ksolution = 0.0382 = KKCl + KMCl
Therefore, KMC = 0.0382 – KKCl = 0.0382 – 0.015 = 0.0232
o MCl 10000.02321000 116 0.20.2 ×× Λ=== k
ooo mmm MMClCl +−λ=∧−λ = 116 – 76
()()()
() o 21 m M40Scmmol +−λ=
Try yourself:
1. Water with conductivity, K = 7.5 × 10–8 Ω–1 cm –1 is used to make 0.01 M solution of acetic acid. Calculate the molar conductivity of acetic acid, when its conductivity is 1.465 × 10–4 Ω–1 cm–1 at this concentration. Ans: 14.64 S 2cm –1mol
TEST YOURSELF
1. Which of the following is a good conductor of electricity?
(1) Diamond (2) Graphite (3) Solid NaCl (4) Wood
2. Which of the following solutions of NaCl has the higher specific conductance?
(1) 0.001 N (2) 0.01 N (3) 0.1 N (4) 0.5 N
3. The specific conductance of 0.1 N KCl solution at 23ºC is 0.012 ohm–1 cm –1. The resistance of cell containing the solution at the same temperature was found to be 55 ohm. The cell constant will be (1) 0.142 cm–1 (2) 0.66 cm–1 (3) 0.918 cm–1 (4) 1.12 cm–1
4. The conductance of 0.1 M HCl solution is greater than that of 0.1 M NaCl. This is because
(1) HCl is more ionised than NaCl
(2) HCl is an acid whereas NaCl solution is neutral
(3) H+ ions have greater mobility than Na + ions
(4) H+ ions have lesser mobility than Na + ions
Answer Key
(1) 2 (2) 4 (3) 2 (4) 3
2.3 KOHLRAUSCH LAW
Kohlrausch studied the molar conductivities at infinite dilution Λ o m for a number of pairs of strong electrolytes, each pair having a common cation or anion. Then, he calculated the difference of these Λ o m values for each pair. It is observed that the difference between Λ o m values for each pair of sodium and potassium salts having a common anion was same, irrespective of what this anion was. Similarly, the difference in the Λ o m values for each pair of salts having the different anions and a common cation are the same, irrespective of what this cation was.
Table 2.3 Conductances of some ions at infinite dilution and 25°C
Kohlrausch concluded that each ion makes a definite contribution to the total molar conductivity of an electrolyte at infinite dilution, irrespective of the nature of the other ion of the electrolyte. This individual contribution of an ion towards the total molar conductivity of the electrolyte is called molar ionic conductivity. As a result of these studies, Kohlrausch, in 1876, put forward a generalisation, known after him as ‘Kohlrausch’s law’.
The limiting molar conductivity of an electrolyte (i.e., molar conductivity at infinite dilution) is the sum of the limiting ionic conductivities of the cation and the anion each multiplied with the number of ions present in one formula unit of the electrolyte.
Mathematically, o oo m forAB +−Λ=λ+λ xyxy
where, Λ o m is the limiting molar conductivity of the electrolyte, ο + λ and ο λ are the limiting molar conductivities of the cation (A y+) and the anion (Bx–), respectively.
For example,
ο Λ m for NaCl = +− οολ+λ NaCl
ο Λ m for BaCl2 = 2 2 +− οολ+λ BaCl
ο Λ m for Al2(SO4)3 = 32 4 23+− οολ+λ AlSO
In terms of equivalent conductivities, Kohlrausch’s law is defined as follows:-
ο Λ=λ+λoo eqca
where, ο λ c and λ o a are called the limiting ionic conductivities for the cation and the anion, respectively.
Relationship between molar conductivity and equivalent conductivity of electrolytes: Since normality = nf × molarity,
eq m f 1 n Λ=Λ ()()() o o3o2 m24m m4 3 AlSO2Al3SO Λ+− =λ+λ ()() o m4 3 o eq24 3 f AlSO AlSO n Λ
() o m24 3 AlSO 6 Λ =
( n-factor of Al2SO4=6)
o m λ = molar ionic conductivity of the ion at infinite dilution or at zero concentration.
o o3o2 eq24 eq eq4 3 AlSOAlSO Λ+− =λ+λ
()()()
o eq λ = equivalent ionic conductivity of the ion at zero concentration or at infinite dilution.
2. The molar conductivities at infinite dilution of Ba(OH) 2 , BaCl 2 , and NH 4 Cl solutions are 457.6, 240.6, and 129.8 S cm 2 mol –1 , respectively. Find the value of ο Λ eq and ο Λ m of NH4OH solution.
Sol. (()) 2 m BaOH 2 BaOH2+− οοο Λ=λ+λ . . . . .(1)
() 2 m2 BaCl BaCl2+− οοο Λ=λ+λ . . . . .
() 4 m4 NHCl NHCl +− οοο Λ=λ+λ
οο Λ=×Λ
(2)
(3) 1 2 × equation (1) + equation (3) – 1 2 equation (2) gives () 4 m4 NHOH NHOH +− οοο λ+λ=Λ Therefore, ()()()m4 11 NHOH457.6129.8240.6 22 ο Λ=+− = 238.3 S cm2 mol–1 eq m f 1 n
() eq4 m4 1 ofNHOHNHOH238.3 1 οο Λ=×Λ= (n-factor of NH4OH = 1)
Try yourself:
2. The equivalent conductivities at infinite dilution of Ba(CN)2, HCl, and BaCl2 are 100, 400, and 120 S cm2 eq –1. If the equivalent conductivity of 0.1 M HCN solution is 3.8 S cm2 eq–1, find the degree of dissociation of HCN at this concentration.
Ans: 0.001
TEST YOURSELF
1. The molar conductivities NaOAc Λ and HCl Λ at infinite dilution in water at 25°C are 91.0 and 426.2 S cm2/mol, respectively. To calculate HOAc Λ , the additional value required is
(1) 2 HO Λ (2) KCl Λ
(3) NaOH Λ (4) NaCl Λ
2. Specific conductivity of 0.01 M solution of HX is 10-8 mho cm–1. Molar conductance of the solution is equal to X mho cm2 mole–1. Value of ‘X’ equal to
(1) 0.1 (2) 10
(3) 0.001 (4) 1
3. Conductance of Al+3 is ‘x’ Sm2 geq–1 and that -2 4 SO is ‘y’ Sm2 geq–1 Molar conductance of aluminium sulphate is ___ Sm 2 mol–1
(1) 6 + xy (2) x+y
(3) 6(x+y) (4) 2 x + 3 y
4. The conductivity of a 0.8 M solution of an electrolyte is 200 Ω –1 cm –1 . Its molar conductivity (in Ω–1 cm2 mol–1) is
(1) 2.5 × 104 (2) 2.5 × 105 (3) 0.25 (4) 2.5
5. Resistance of 0.2 M solution of an electrolyte is 50 Ω. The specific conductance of the solution is 1.4 m–1. The resistance of 0.5 M solution of the same electrolyte is 280 Ω. The molar conductivity of 0.5 M solution of the electrolyte in S m 2 mol–1 is (1) 5 × 103 (2) 5 × 102 (3) 5 × 10–4 (4) 5 × 10–3
6. Molar conductance of 0.01M weak monoprotic acid (HA) is 40 mho cm 2 mol–1. pH of the solution is (limiting molar conductance of HA is 400 mho cm 2 mol–1).
(1) 2 (2) 4 (3) 2.3 (4) 3
7. At 300 K, the conductivity of 0.01 mol dm-3 aqueous solution of acetic acid is 19.5 × 10–5 mho cm–1 and limiting molar conductivity of acetic acid at the same temperature is 390 mho cm2 mol–1. The degree of dissociation of acetic acid is
(1) 5.0 × 10–5 (2) 5.0 × 10–2 (3) 2.5 × 10–5 (4) 7.5 × 10–2
8. Molar ionic conductance of Ca2+ and 3 4 PO at infinite dilution are ‘x ’ mho cm 2 mol –1 and ‘y’ mho cm2 mol–1 respectively. Molar conductance of calcium phosphate at infinite dilution is ___ mho cm2 mol–1
(1) (x + y) (2) 3 x + 2 y
(3) 6 (x + y) (4) 32 6 xy +
Answer Key
(1) 4 (2) 3 (3) 3 (4) 2 (5) 3 (6) 4 (7) 2 (8) 2
2.4 ELECTROLYSIS
Sodium combines with chlorine directly to form sodium chloride. The reaction involves oxidation of Na to Na+ and reduction of Cl to Cl–. Thus, the reaction between a metal and a non-metal is described as a redox reaction.
The formation of sodium chloride from its elements is an example of spontaneous change. The process of getting back the elements from an electrolyte involves a non-spontaneous redox reaction. Such a process is made possible using electrical energy.
Electrolyte: A substance that is in the molten state or in the dissolved state, containing ions and functioning as an electrically conducting medium, is called an electrolyte.
Electrolytic conductors are called electrolytes. Salts in molten state, like NaCl, KCl, fused hydrides and halides of Ag, Ba, Pb. a-form of Ag2S, solutions of NaCl, KCl, K2SO4, CuSO4, HCl, H2SO4, HNO3, NaOH, KOH etc. in water are electrolytic conductors. In general,
all fused salts or that which contain free ions are electrolytes.
Non-electrolyte: A substance that does not conduct electric current either in the molten state or in aqueous solutions is called as a non-electrolyte.
Examples: Urea solution, aqueous solution of sugar, non polar covalent carbon compounds
Electrolysis: Electrolysis is a non-spontaneous redox reaction which occurs under the influence of an applied electromotive force (EMF)
Electrolysis is performed in a device called the electrolytic cell. So, in this cell, electrical energy is used to carry out a redox reaction. Electrons enter into the cell at cathode and, hence, it is negatively charged. Electrons flow in the external circuit from anode to cathode.
Electrons Anode Cathode Electrolyte
Fig. 2.6 Experimental setup of electrolysis
Positively charged ions migrate towards the negative electrode and negatively charged ions migrate towards the positive electrode. A common setup of electrolysis is shown in Fig.2.6.
During electrolysis, the cations move towards the negative electrode (cathode) and the anions move towards the positive electrode (anode). They get discharged and get deposited or liberated.
Let BA be an electrolyte, containing B + and A – ions. During electrolysis, B + ions move towards cathode, whereas A– ions move towards anode.
At cathode: BeB +−+→ (reduction)
At cathode, gain of electrons or electronation or reduction takes place.
At anode: AAe →+ (oxidation)
At anode, loss of electrons or deelectronation or oxidation takes place.
During electrolys the electrons enter into the electrolyte at cathode, and they are taken up by the cations. The electrons lost by the anions leave the electrolyte at anode. As a result, ions are converted into atoms or molecules and get deposited or liberated at respective electrodes.
The minimum potential that is required for performing electrolysis and subsequent discharge of ions at their respective electrodes is called discharge potential. It varies from one ion to another ion.
When a solution contains two or more anions and cations, then the ion with the lower discharge potential will get discharged first.
Discharge potentials of anion:
Order of deposition of anions:
Decreasing order of discharge potentials for various cations:
Li+ > K+ > Na+ > Ca2+ > Mg2+ > Al3+ > Mn2+ > Zn2+ > Cr3+ > Fe2+ > H+ > Cu2+ > Ag+ > Hg+2 > Ag+ > Pt2+ > Au3+
Order of deposition of cations: Au3+ > Pt2+ > Ag+ > Hg2+ > Cu2+ > H+ > Fe2+ > Cr3+ > Zn2+ > Mn2+ > Al3+ > Mg2+ > Ca2+ > Na+ > K+> Li+
The process is illustrated by taking the electrolysis of molten sodium chloride as an example. When electric current is passed through molten sodium chloride, the Na+ ions move towards the cathode and the Cl– ions move towards the anode.
2NaCl2Na2Cl+−→+
At cathode (reduction): 2Na2e2Na +−+→
Sodium ions are reduced and sodium metal (Na) is deposited at cathode.
At anode (oxidation): 2 2ClCl2e →+
Chloride ions are oxidised and Cl 2 gas is liberated at anode.
Examples and Applications
Electrolysis of fused and aqueous solutions with inert electrodes is used for producing generally metal at cathode and non-metal at anode. Though hydrogen is a non-metal, it is electrolytically collected at cathode from aqueous solutions. However, hydrogen is liberated at anode during the electrolysis of fused metal hydrides.
The electrolysis of aqueous sodium chloride solution differs from that of molten sodium chloride. Ionisation of aqueous salt solution produces Na+ and Cl– ions. Na+ ions travel to the cathode but cannot undergo reduction to metal at platinum cathode in aqueous solutions. Water preferentially undergoes reduction to give hydrogen gas.
22 2HO2eH2OH +→+
(Reduction or electronation at cathode)
Chloride ions travel to the anode and undergo oxidation to give chlorine gas.
2 2Cl2eCl →+
(Oxidation or de-electronation at anode)
∴ In electrolysis, concentrated aqueous NaCl, H2 gas is obtained at cathode and Cl2 is obtained at anode. But Na+ and OH– ions remain in the solution. So, the pH increases during electrolysis.
However, on prolonged electrolysis, when most of the chloride ions are oxidised, water is oxidised finally to liberate oxygen gas at anode.
Electrolysis of water, using platinum electrodes liberates hydrogen at cathode and oxygen at anode. Water is a very
poor electrolyte. In order to increase the conductivity, a small amount of an electrolyte is dissolved in water during electrolysis.
At cathode: 22 4HO4e2H4OH +→+
At anode: 2H2O → 4e– + O2 + 4H+
6H2O→ 2H2 + O2 + 4H2O
The overall reactions is:
→+
Thus, for the electrolysis of 2 moles of water,
4 faradays of electricity are required.
The products of electrolysis at inert electrodes are the same using dilute sulphuric acid solution, using dilute caustic soda solution or using an aqueous potassium sulphate solution as an electrolyte.
Electrodes that do not involve in chemical reaction during electrolysis are called inert electrodes.
Example: Graphite, platinum, etc.
Active electrodes involve in chemical reaction during electrolysis. In general, anode metal dissolves in solution. Example: Cu anode in aqueous CuSO4
When the aqueous solutions of salts of alkali metals, alkaline earth metals, Al +3, Zn+2, Fe+2 and Ni+2 etc., are electrolysed by using inert electrodes, at cathode, H2 is obtained by the reduction of either H + or H 2O molecules in preference to metal ions.
22 2HO2eH2OH +→+ or 2(g) 2H2eH +−+→
When the aqueous solutions of SO 4 –2 , NO 3 – , and PO 4 –3 are electrolysed by using inert electrodes at anode, H 2 O molecules undergo oxidation in preference to oxoanions to give O2.
Pt 22 2HOO4H4e+−→++
2.4.1 Faraday’s Laws
NOnooxidation
Pt 3
Pt 2 4
SOnooxidation
Faraday put forward two laws regarding the quantity of electricity passed through electrolytes and the amount of substance deposited or liberated at electrodes.
POnooxidation
Pt 3 4
A metal salt solution is electrolysed by using the same metal electrodes.
At cathode, metal is deposited due to reduction of metal ion and at anode metal ions are obtained due to the oxidation of anodic metal.
Example: Electrolysis of CuSO 4 with Cu electrodes
() 22 4 4aq CuSOCuSO+−→+
At cathode: 2 Cu2eCu +−+→
At anode: 2 CuCu2e+−→+
Some examples of electrolysis reactions and the products obtained at the respective electrodes are summarised in Table 2.3.
These laws are also known as fundamental laws of electrochemistry and they give quantitative relationships of electrolysis. These laws are applicable to all electrolytes at all the temperatures, pressures, and other conditions.
First Law
The amount of substance deposited or liberated or dissolved at an electrode during electrolysis is directly proportional to the quantity of electricity passing through the electrolyte.
The first law is expressed as m ∝ q m=eq . . . . . (1) where m = weight of the substance in grams q = quantity of charge in coulombs q = c × t
2HO2eH2OH +→+ Fused NaOH Pt 22 4OH2HOO4e →++
NaeNa +−+→ Aq. NaOH Pt 22 4OH2HOO4e →++
22 2HO2eH2OH +→+
Aq. HCl Pt 2 2ClCl2e →+ Pt 2 2H2eH +−+→
Aq. H2SO4 Pt 22 2HOO4e4H−+→++ Pt 2 2H2eH +−+→
50% H2SO4 Pt 4228 2HSOHSO2e →+ Pb 2 2H2eH +−+→
Aq. K2SO4 Pt 22 2HOO4e4H−+→++
Fused CuCl2 Pt 2 2ClCl2e →+
Aq.CuCl2 Pt 2 2ClCl2e →+
Aq.CuSO4 Pt 22 2HOO4e4H−+→++
Fused NaH Pt 2 2HH2e →+
Aq.AgNO3 Pt 22 2HOO4e4H−+→++
Pt 22 2HO2eH2OH +→+
Pt 2 Cu2eCu +−+→
Pt 2 Cu2eCu +−+→
Pt 2 Cu2eCu +−+→
Pt NaeNa +−+→
Pt AgeAg +−+→
c = current strength in amperes
t = time in seconds
m ∝ c×t (or) m = e × c × t ......(2)
where ‘e’ is a constant known as electrochemical equivalent, and it is the characteristic of the substance deposited. If q = 1 coulomb, then e = m.
Electrochemical equivalent may be defined as the mass of substance deposited by passing one coulomb of electricity through the electrolyte. It is also defined as the mass of substance deposited by passing a current strength of one ampere for one second. The units of electrochemical equivalent are gram/ amperes or gram /coulomb. The SI unit of electrochemical equivalent is kg coulomb –1 .
Second Law
If the same quantity of electricity is passed through different electrolytes, then the weight of substances deposited at the respective electrodes are in the ratio of their chemical equivalents. The second law is mathematically expressed as, w ∝E, where ‘E’ is the equivalent weight of the substance.
If the same quantity of electricity is passed through two or more different electrolytes, the second law is applied. Let the weight of the substance ‘A’ be wA and equivalent weight of ‘A’ be EA. Let the weight of the substance ‘B’ be wB and equivalent weight of ‘B’ be EB. Then, according to the second law,
(3)
All the above 1, 2 and 3 equations are valid only when current efficiency is 100% i.e., η = 1
current efficiency
Actual mass 100 Theoreticalmass η=×
Faraday is defined as the quantity of electricity required to deposit or dissolve one gram equivalent weight of a substance.
From first law m e q =
From second law E e F =
Therefore, mE e qF ==
The total charge present on one mole of electrons is known as faraday.
Both coulomb and faraday are the units of quantity of charge. Coulomb represents a smaller quantity of charge and faraday represents a larger quantity of charge.
One faraday = Charge of one mole of electrons
= Charge of 6.023 × 10 23 electrons = 6.023 × 1023 × 1.602 × 10–19 coulomb = 96488 coulomb.
One faraday is usually taken as 96500 coulomb. Units of faraday are coulomb/mole.
Electrochemical equivalent (e) is
Equivalentweight e 96,500 = (or)
Atomicweight e n96,500 = ×
Here, n is the valency of the element. Faraday’s laws of electrolysis are useful for the determination of the amounts of substances deposited, the quantity of electricity, the number of electrons passing through the electrolyte, equivalent weights, atomic weights and valency.
3. Electrolysis of dilute aqueous solution of NaCl was carried out by passing 0.01 Amperes of current. How much times is required to liberate 0.01 moles of H 2 gas at cathode in the electrolysis.
Sol. –22H2eH ++→
One mole of H2 gas is liberated with 2F charge. The charge required for 0.01 moles H2 = 0.01 × 2F
q = 0.01 × 2 × 96,500 C
q = i × t = 0.01 × 2 × 96,500 C
0.01296500C t 296500C 0.01 ×× ==×
Try yourself:
3. For converting 54 g of water into H 2 and O2 gases, how long a steady current of 96.5 amperes should be passed into the water? (Report the answer in minutes)
Ans: 100 minutes
TEST YOURSELF
1. When an electric current is passed through acidified water, 112 mL of hydrogen gas at N.T.P was collected at the cathode in 965 seconds. The current passed in ampere is (1) 1.0 (2) 0.5 (3) 0.1 (4) 2.0
2. Which among the following has highest electrochemical equivalent?
(1) Ca (2) Na (3) Mg (4) Al
3. The electric charge for electrode deposition of one equivalent of the substance is (1) one ampere per second
(2) 965000 C per second
(3) one ampere per hour
(4) charge on 1 mole of electrons
4. How many electrons are delivered at the cathode during electrolysis by a current of 1 amp in 60 seconds?
(1) 3.74 × 1020 (2) 6.0 × 1023
(3) 7.48 × 1021 (4) 6.0 × 10020
5. Electric charge on 1 g ion of N 3– is (1) 4.8 ×10–19 C
(2) 10 × 1.6 × 10–19 C (3) 1.6 × 10–19 C (4) 2.89 × 105 C
6. A silver cup is plated with silver by passing 965 coulombs of electricity. The amount of silver deposited is (1) 9.89 g (2) 107.87 g (3) 1.0787 g (4) 1.002 g
7. One coulomb of charge passes through solutions of AgNO3 and CuSO4. The ratio of the amounts of silver and copper deposited on platinum electrodes used for electrolysis is
(1) 108 : 63.5 (2) 54 : 31.75 (3) 108 : 31.5 (4) 215.8 : 31.5
8. Dilute nitric acid on electrolysis using platinum electrodes yields (1) both oxygen and hydrogen at cathode (2) both oxygen and hydrogen at anode (3) H2 at cathode and O2 at anode (4) oxygen at cathode and H2 at anode
9. When 0.1 mole 2 4 MnO is oxidised, the quantity of electricity required to completely oxidize 2 4 MnO and 4 MnO is (1) 96500 C (2) 2 × 96500 C (3) 9650 C (4) 96.50 C
10. One faraday of electricity will liberate one gram atom of the metal from the solution of (1) CuCl2 (2) CuSO4 (3) AgNO3 (4) AuCl3
Answer Key (1) 1 (2) 2 (3) 4 (4) 1 (5) 4 (6) 3 (7) 3 (8) 3 (9) 3 (10) 3
2.5 ELECTROCHEMICAL CELLS
Electrochemical cell is a device in which electrical energy is produced at the expense of chemical energy.
2.5.1 Voltaic Cell
When plates of two dissimilar metals are placed in a conducting liquid, such as an aqueous solution of a salt, the resulting system becomes a source of electricity. When a piece of zinc is kept in copper sulphate solution for some time, the zinc piece turns red. This is due to the deposition of copper on zinc. The reaction is given as
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
In this redox reaction, two electrons are transferred from zinc atom to cupric ion. However, these electrons cannot be trapped to obtain any small quantity of electrical energy because they are directly transfering from zinc to Cu+2 within the vessel. When the reactants are separated from each other and given an indirect contact, the electrical energy can be trapped.
A device in which electrical energy is generated by performing a redox reaction is called galvanic cell. this cell is also called voltaic cell or electrochemical cell.
So, chemical energy is converted into electrical energy in a galvanic cell. A galvanic cell can be reversed to function as an electrolytic cell. A typical galvanic cell is shown in Fig.2.7
Zinc anode
Copper cathode
Zinc sulphate solution Coper sulphate solution
Fig. 2.7 Galvanic cell
Comparison between the two types of cells are given in the Table 2.4.
Electrolytic cell
A galvanic cell consists of two half cells or electrodes. One electrode contains zinc metal plate dipped in ZnSO4 solution and the other is copper plate in CuSO 4 solution. The two half cells are joined by a salt bridge. In Daniell cell, salt bridge is replaced by a porous pot.
When the Zn and Cu electrodes are joined externally by a wire, the following observations are made:
1. There is a flow of electric current through the external circuit.
2. ‘Zn’ rod loses its mass, while ‘Cu’ rod gains the mass.
3. The concentration of ZnSO 4 solution increases, while the concentration of CuSO4 solution decreases.
4. The solutions in both the compartments remain electrically neutral.
The standard emf of a Daniell cell is 1.1 V. If an external potential of 1.1 V is applied on this cell, there will be no chemical reaction and no flow of electrons. If the external potential is less than 1.1 V, the spontaneous redox reaction is continuous. If the external potential is more than 1.1 V, the cell functions as an electrolytic cell and the electrical energy is used to carry out non-spontaneous chemical reactions. Functioning of Daniell cell with applied external voltage is shown Fig.2.8(a) Fig.2.8(b)
Electrochemical cell
1. Electrical energy is converted into chemical energy. Chemical energy is converted to electrical energy
2. Anode is positive electrode. Cathode is a positive electrode.
3. Cathode is negative electrode. Anode is a negative electrode.
4. Oxidation occurs at anode. Oxidation occurs at anode.
5. Reduction occurs at cathode. Reduction occurs at cathode.
6. Electricity flows from the circuit to the cell. Electricity flows from the cell to the circuit.
7. Irreversible and non spontaneous. Reversible and spontaneous at irreversible.
8. Ions are discharged at both the electrodes. Ions are discharged only on cathode.
Fig. 2.8 Daniell cell with applied external voltage (a) less than 1.1 V and (b) more than 1.1 V
Chemical Reactions in a Galvanic Cell
Zinc goes into solution as Zn2+ ions in the half cell in the left hand side.
2 Zn(s)Zn(aq)2e +−→+
Since the reaction involves the loss of electrons, it is known as oxidation half reaction. This electrode is known as anode. The electrons from the Zn electrode flow through the external wire to the Cu electrode. Cu+2 ions of the CuSO4 solution take up these electrons to form Cu atoms.
Cu2+(aq) + 2e– → Cu(s)
Since this reaction involves gain of electrons, it is known as reduction half reaction. This electrode is known as cathode.
At anode: 2 ZnZn2e+−→+ (oxidation)
At cathode: 2 Cu2eCu +− +→ (reduction)
The net cell reactions is Zn + Cu+2 → Zn+2 + Cu
Voltmeter is used to measure the potential difference between the two electrodes. In the Daniel cell, the salt bridge is replaced with a porous pot.
The electrical neutrality of the Daniell cell is maintained by the porous vessel.
Salt Bridge
Salt bridge is a U-shaped table. It is filled with a strong electrolyte, such as KCl, NaCl, KNO3 etc.
It must be inert towards the chemicals used in the electrochemical cells. The cations and anions of the electrolyte of the salt bridge must have same migratory speed.
In the salt bridge the electrolytes are held as a gel like agar-agar.
In the absence of salt bridge, ions of electrolytes of galvanic cell are accumulated in the half cells. Then, electrons flow is stopped and the cell stops functioning.
When salts bridge is connected, it supplies its ions to the half cells of galvanic cell and maintain electrical neutrality.
Important functions of salts bridge are: (1) It completes the electrical circuit (2) It maintains electrical neutrality (3) It minimises the liquid – liquid junction potential.
Line Notation for Electrochemical Cells
1. The line notation for the anodic half cell is written on the left and that for the cathodic half cell on the right.
Representation of Daniell cell is Zn(s)/Zn2+(C1)||Cu+2(C2)|Cu(s) C1 and C2 are the molarities of the Zn +2 and Cu+2 in the electrolytic solutions.
2. A single vertical line or comma indicates a change in state or phase.
3. A double vertical line is used to indicate the elimination of liquid junction potential (using salt bridge) between the half cells.
4. When gases are involved, their pressures are written in parentheses.
Representation of hydrogen – oxygen cells is
Pt, H2 (P1) |HCl(C)| O2 (P2), Pt.
This indicates that HCl solution with concentration, C acts as an electrolytic in both H2-half cell and O2-half cell. P1 and P2 are pressures of H2 and O2 respectively.
4. What is the line representation of the cell for the cell reaction? (The molarity of the solution is 1.0 M and pressure of the gas is 1.0 gas)
2AgClH2HCl2Ag +→+
()()()() s2g aq s
Sol. In the given cell H2 is oxidised to H+ ions. AgCl is reduced to Ag(s) and Cl–. Platinum is required for electrical constant.
The cell notation is
Pt(s), H2(g) (1.0 bar) | HCl (1.0 M) | AgCl(s), Ag(s)
Try yourself:
4. What are the half cell reactions possible for the electrochemical cell: Pt (s)| () Br2 |Br–(0.01M)||H+(0.03M|H2(g)(1.0 bar)|Pt(s)|H2(g) (1.0 bar)?
Cell reaction : (aq)+2H + 2 (aq)–Br → +2(g)H )( 2Br
Reduction half reaction (aq)+:2H + 2e → 2(g)H
Oxidation half reaction →(aq)–:2Br )( 2Br +2e
Ans:
2.5.2 Single Electrode or Half Cell
Single electrodes are of two types (i) metal electrode (ii) non-metal electrode or gas electrode.
Types of Electrodes or Half Cells
1. Metal–metal electrodes: Metal with intermediate reactivity can be used for making these half cells.
Example: Zn|Zn+2, Cu|Cu+2
2. Non-metal Electrode or Gas Electrode:
Gas bubbled into the solution containing its anion in presence of platinum electrode at atmospheric pressure is called a nonmetal electrode.
Examples: ()()() aq2g H/H1atm,Pt + . . . . . . Hydrogen electrode
()()2g aq
Pt,O(1atm)/OH . . . . . . Oxygen electrode
()() 2 aq Pt,Cl1atm/Cl . . . . . . . . Chlorine electrode
()() 2 aq Pt,Br1atm/Br . . . . . . . . Bromine electrode
Generally platinum or Gold is used as inert electrode
Electrode reaction: O2 + 2H 2O + 4e – 4OH–
3. Metal–metal insoluble salt–salt anion electrode
Examples: i. ()()() 2 Hg|HgCl|KClsaq
Electrode reaction: Hg 2 Cl 2 + 2e – 2Hg(ℓ)+2Cl–(aq)
ii. Ag(s)|AgCl(s)|KCl(aq)
electrode reaction: AgCl(s) + e Ag(s) + Cl–(aq)
4. Oxidation – reduction electrodes
In this half cell, inert electrode is dipped in the solution containing ions of a species having two different oxidation states.
Examples: i. Pt/Fe+3, Fe+2
Electrode reaction: Fe +3 Fe+2
ii. Quin hydrone electrode, Pt/Q, QH 2
Electrode reaction:
Q + 2H+ 2e QH2
Here,
Q = Quinone, QH2 = hydroquinone
Single Electrode Potential
In each half cell, there is a tendency of a metal to go into the solution as ions, leaving behind electrons at the electrode. At the same time, metal ions of the solution show tendency to deposit on the metal electrode. At equilibrium, there is a separation of charges between metal and the solution. As a result, a potential difference exists between them. The electrical potential difference set up between the metal and its ions in the solution is called electrode potential (i.e., single electrode potential).
If the potential is developed due to the tendency of electrode to undergo oxidation, it is called its oxidation potential. If the potential is developed due to the tendency of the of electrode to undergo reduction, it is called its reduction potential. For a given electrode, if oxidation potential is (V) volts, its reduction potential will be (–V) volts. Electrode potential is an intensive property.
Electrode potential depends upon the
1. nature of the metal (or non-metal)
2. concentration of the electrolyte
3. temperature of the electrolyte solution
4. pressure in case of gas electrode
5. number of electrons involved per atom.
2.5.3 Standard Electrode Potential (E°)
The potential of an electrode measured at standard conditions of 25°C temperature and 1 bar pressure, and using ions at unit concentration, is known as standard electrode potential. It is denoted by ‘E°’.
For any half cell or single electrode, standard reduction potential = – (standard oxidation potential) For example, 2 Zn2eZn; +−+→ the standard reduction potential is – 0.76 V. 2 ZnZn2e; +−→+ the standard oxidation potential is + 0.76 V.
2.5.4 Measurement of Electrode Potentials
The absolute value of the electrode potential of a single electrode (i.e., single electrode potential or half cell potential) cannot be determined. Because no net reaction occurs in a half cell. It can be measured using some electrode as a reference electrode.
The hydrogen electrode is primary reference electrode for measuring single electrode potential. The construction of standard hydrogen electrode is shown in Fig.2.9.
The standard hydrogen electrode (SHE) consists of a platinum plate dipped into a solution of protons with one molar concentration at 25°C. Hydrogen gas is passed over the Pt plate at one atm. The half cell reaction is written as 2H + (aq) + 2e– → H2(g)
H2(g) (at 1 atm)
Pt black electrode
H3O+(aq) (1M)
2.9 Standard hydrogen electrode
The half cell is represented as, Pt, H +(1M)/ H2(1atm), E°= 0.0 volts. The potential of SHE or NHE is arbitrarily taken as zero.
It is a reversible electrode with respect to H+ ion, i.e., it can act like a cathode or anode.
1 2 2 HHe+− →+− when acting like anode
2 1 HeH 2 +−+→− when acting like cathode
Secondary reference electrodes like quinhydrone electrode and calomel electrode are also used now-a-days. Single electrode potentials are measured in a potentiometer using reference electrodes (for SHE, [H+]=1M).
For the determinations of electrode potential of any electrode, an electrochemical cell is set up using the electrode or the half cell under study and standard hydrogen electrode. The emf of the cell is measured. It gives electrode potential of the electrode under study.
For example, the determination of standard electrode potential of Zn|Zn+2(1M) and SHE is set up. Then, voltage is recorded (it comes out to be 0.76 volts). In this cell, one can observe that the concentration of Zn+2 increases, while that of H+ decreases. Thus, the cell reaction is Zn + 2H+ → Zn+2 + H2. Hence, zinc electrode acts as anode.
EMF of cell = (standard reduction potential of cathode)
—(Standard reduction potential of anode)
0.76 = 2 H/HZn/Zn EE++ οο
0.76 = 2 ZnZn 0E + ο
Therefore, 2 Zn/Zn E + ο = – 0.76 volts.
Similarly, when standard potential of copper electrode is to be measured, a cell is set up with copper half cell and SHE. Then, the measured voltage is expected to be 0.34 volt. In the cell, one can observe that the concentration of Cu+2 decreases while the concentration of H+ increases. Thus, the cell reaction is
Cu+2 + H2 → Cu + 2H+
Hence, the coper electrode acts a cathode.
EMF of the cell, 0.34 V = 2 Cu/CuH/H2 EE++ οο
Therefore, 2 Cu/Cu E0 + ο = + 0.34 V
In electrode – SHE couple, if the electrode undergoes oxidation, the standard reduction potential of the electrode is negative.
In this manner, the standard electrode potentials of a large number of electrodes were determined using SHE as a reference electrode. These electrode potentials are relative standard half cell potentials.
TEST YOURSELF
1. In Daniell cell, the process that occurs at anode is
(1) ()() 2 aq s Zn2eZn +−+→
(2) ()() 2 CusCuaq2e +−→+
(3) ()() 22 ZnCuaqZnaqCu ++ +→+
(4) ()() 2 ZnsZnaq2e +−→+
2. At 298 K, reduction potential of H-electrode is –0.118 V. Hydroxyl ion concentration of the solution in which the H-electrode is immersed is ____M
(1) 10–10 (2) 10–9 (3) 10–12 (4) 10–11
3. Saturated solution of KNO 3 is used to construct ‘salt – bridge’ because (1) velocity of K + is greater than that of NO3
(2) velocity of NO3 is greater than that of K+
(3) velocities of both K+ and NO3 are nearly the same
(4) KNO3 is highly soluble in water
4. Which of the following statements is correct w.r.t. both electrolytic cell and galvanic cell?
(1) In both cells, anode is shown by +ve sign.
(2) In both cells, cathode is shown by –ve sign.
(3) In both cells, reduction reaction takes place at the cathode.
(4) In both cells, oxidation reaction takes place at the cathode.
Answer Key
(1) 4 (2) 3 (3) 3 (4) 3
2.6 ELECTROCHEMICAL SERIES
The arrangement of electrodes in the order of increasing standard reduction potentials is called electrochemical series. This is also called activity series.
The negative sign of standard reduction potential indicates that when the electrode is connected with standard hydrogen electrode, it acts as the anode and oxidation takes place. A positive sign of standard reduction potential indicates that the electrode acts as cathode when connected with a standard hydrogen electrode. The activity series containing important half-cells is listed in Table 2.5.
The values of standard electrode potentials for some selected half-cell reduction reactions are given. It can be seen that the standard electrode potentials for fluorine is the highest, indicating that fluorine gas (F 2 ) has the maximum tendency to get reduced to fluoride ions (F–). Therefore, fluorine is the strongest oxidising agent and fluoride ion is the weakest reducing agent.
Lithium has the lowest electrode potential, indicating that lithium ion is the weakest oxidising agent, while lithium metal is the most powerful reducing agent in an aqueous solution. It may be seen that as we go from top to bottom in Table 2.5, the standard electrode potential increase.
Thus, the tendency to undergo the corresponding reaction (reduction) decreases, the reducing power of the particles decreases.
2.6.1 Applications of Electrochemical Series
With the help of electrochemical series, one can understand the spontaneity of certain reactions and one can predict the occurrence of metal-salt solution reactions. Using these series reducing and oxidising abilities of elements can be estimated.
Chemical Reactivity of Metals
The metal that has high negative value of standard reduction potential readily loses electrons and it is converted into cations. Such a metal is said to be chemically very active. The chemical reactivity of metals decreases from top to bottom in the activity series.
Relative Strength of Oxidising and Reducing Agents
The reducing ability decreases from top to bottom in the electrochemical series. The oxidation ability increases with increase in the reduction potential.
Oxidising power ∝ reduction potential
Reducing power ∝ oxidation potential
Fluorine, with highest reduction potential, is the powerful oxidising agent. Lithium, with least reduction potential, is the powerful reducing agent in aqueous medium based on the potentials.
A given metal in activity series can reduce all the metal ions present below it. A given metal ion can oxidise all metal atoms present above it. A metal can reduce all the non-metals present below it in the activity series.
For storing a salt solution a metal container with smaller SRP value should not be used. Instead a metal container with higher SRP can be used.
Examples: FeSO4 solution () 2 o Fe/Fe E0.44v + =− cannot be stored in Zn container () 2 o Zn/Zn E0.76v + =− but it can be stored in a Cu container () 2 o Cu/Cu E0.34v +=+
A non-metal can oxidise all metal atoms present above it in the activity series. It can also oxidise non-metal anions present above it.
Table 2.5 Comparison between the two types of cells
S.No.
1. Li(aq)eLi(s) +−+→
2. K(aq)eK(s) +−+→ K+ / K – 2.93
3. 2 Ca(aq)2eCa(s) +−+→
4. Na(aq)eNa(s) +−+→
5. 2 Mg(aq)2eMg(s) +−+→ Mg2+/Mg – 2.37
6. 3 Al(aq)3eAl(s) +−+→ Al3+/Al – 1.66
7. 2 Zn(aq)2eZn(s) +−+→ Zn2+/Zn – 0.76
8. 2 Fe(aq)2eFe(s) +−+→
9. 2 Co(aq)2eCo(s) +−+→
– 0.28
10. 2 Ni(aq)2eNi(s) +−+→ Ni2+/Ni – 0.25
11. 2 Sn(aq)2eSn(s) +−+→ Sn2+/Sn – 0.14
12. 2 2H(aq)2eH(g) +−+→ Pt, H+/H2 0.00
13. 2 Cu(aq)2eCu(s) +−+→ Cu2+/Cu + 0.34
14. 2 I(s)2e2I(aq) +→ Pt, I2/I– + 0.54
15. Ag(aq)eAg(s) +−+→ Ag+/Ag + 0.80
16. 2 2 Hg(aq)2e2Hg(l) +−+→ Hg+/Hg + 0.85
17. 2 Br(l)2e2Br(aq) +→
18. 2 Cl(g)2e2Cl(aq) +→
19. 3 22 O(g)2H(aq)2eO(g)HO(l) +− ++→+
20. 2 F(g)2e2F(aq) +→
Displacement of Hydrogen from Dilute Acids by Metals
A metal with negative SRP can displace H+ from dilute acids to liberate H 2 gas.
A metal with positive SRP like Cu, Ag, Pt etc., cannot displace H+ from dilute acids to liberate H2.
Standard EMF of Cell
Br2/Br– + 1.07
Cl2/Cl– + 1.36
The sum of standard oxidation potential of anode and reduction potential of cathode is called standard cell emf or the difference in the reduction potentials of cathode and anode is also called cell standard emf of the cell, cell E ο ο=+ celloxred (anode)(cathode) EEE
But oxidation potential = –reduction potential οο =− o EEEcellcathodeanode
ooo EEEcellrightleft
In a similar way, the cell potential or emf of cells (Ecell) = Ecathode – Eanode (i.e., difference of reduction potentials)
5. Can PbO2 oxides Mn+2 according to the reaction: PbO2(s) + Mn+2 + 4H2O → Pb+2 + MnO4– + H+ Assume that all the reactants and products are in their standard states
(Given 2 Mn/MnO4 E1.51V +− ο =− and
2 2 PbO/Pb1.47V E + ο =+
Sol. For the given reaction, cellca EEE οοο =−
22 24 PbO/PbMnO/Mn EE+−+ οο =−
(difference of reduction potentials)
= + 1.47 – (+1.51) = – 0.04 V
Since, the potential of the cell reaction is negative, the given reaction is not spontaneous. In other words, the PbO2 cannot oxidise Mn+2
Try yourself:
5. Find the standard emf of the cell reaction, 22 34 27 2 CrO3Sn14H2Cr3Sn7HO −++++ ++→++
Given: 2 27 CrO/Cr E33V, −+ ο=+ Sn/Sn42 E0.15V ++ ο=+
Ans: 1.18 V
TEST YOURSELF
1. The standard reduction potentials for the following half-reactions are given against each at 298 K.
()() 2 aq s Zn2eZn;0.762 +−+=−
()() 3 aq s Cr3eCr;0.74V +−+=−
2H2eH;0.0V +−+=
()2g
() 32 maq aq FeeFe;0.77V +−++=+
Which is the strongest oxidising agent?
(1) Zn(s) (2) Cr(s) (3) H2(g) (4) Fe3+(aq)
2. The standard electrode potentials M/M E + ο of four metals A, B, C and D are –1.2 V, 0.6 V, 0.85 V, and –0.76 V, respectively. The sequence of deposition of metals on applying potential is
(1) A,B,C,D (2) B,D,C,A (3) C,B,D,A (4) D,A,B,C
3. Which of the following has least tendency to liberate H2 from mineral acids
(1) Cu (2) Mn
(3) Ni (4) Zn
4. Which of the following is true for electrochemical cell made with SHE and Cu electrode?
(Given 2 Cu/Cu E0.36V + ο=+ )
(1) H2 is cathode and Cu is anode
(2) H2 is anode and Cu is cathode
(3) Reduction occurs at H2 electrode
(4) Oxidation occurs at Cu electrode
5. Regarding the standard hydrogen electrode, the incorrect statement is (1) H2 gas is bubbled at 1 bar
(2) 2 M HCl is electrolyte
(3) SRP is arbitrarily taken as Zero. (4) It is a primary reference electrode.
6. Zn gives H2 gas with H2SO4 and HCl but not with HNO3 because (1) Zn acts as oxidising agent when reacts with HNO3.
(2) HNO3 is weaker acid than H2SO4 and HCl.
(3) In electrochemical series Zn is above H2. (4) NO3 is reduced in preference to H3O+
7. The standard reduction potentials of Cu +2 , Ag + , Hg +2 , and Mg +2 are +0.34 V, +0.80 V + 0.79 V, and –2.37 V respectively. With increasing voltage, the sequence of deposition of metals on the cathode from a molten mixture containing all those ions is (1) Ag, Hg, Mg, Cu (2) Cu, Hg, Ag, Mg (3) Ag, Hg, Cu, Mg (4) Cu, Hg, Mg, Ag
8. The chemical reaction
()()()() s2g aq s 2AgClH2HCl2Ag +→+ taking place in a galvanic cell is represented by the notation.
(1) ()()()()() s2g aq PtH1bar1MKClAgClAgss ,
PtH1bar1MHC1 , lMAgAg +
(2) ()()()()() s2g aq aqs
(3) ()()()()() s2g aq PtH1bar1MHClAgClAgss ,
(4) ()()()()() s2g aqs s PtH1bar1MHClAgAgCl , Answer Key
(1) 4 (2) 3 (3) 1 (4) 2
(5) 2 (6) 4 (7) 3 (8) 3
2.7 NERNST EQUATION (FOR ELECTRODE POTENTIALS)
Effect of concentration and temperature on electrode potentials is explained by Nernst equation.
The electrode potential required under different conditions (i.e., if the electrolyte ion is not with 1.0 M concentration or not with 1.0 bar pressure and temperature is not 298 K), can be obtained using Nernst equation.
2.7.1 For a Metal Electrode (or Half Cell)
The Nernst equation for the electrode reaction,
()() n aq s MneM +−+→
The electrode potential, ++
Here, n M/M E + ο = Standard reduction potential
R = gas constant (8.314 J K –1 mol–1)
T = Absolute temperature
n = number of electrons involved in the reaction
F = one Faraday = 96,500 C
For pure liquids or solids or gases at 1.0 atmospheric pressure, the molar concentration is taken as unity.
Therefore, [M] = 1 M/MM/M M/M RT1 EE F M 2.303RT1 E log F M ++ + ο + ο
nn n n n n n n at 298 K M/MM/M 0.0591 EE log M ++ ο + =−
For a gas electrode
If the electrode reaction is X2(g) + 2e – →2X–(aq)
Electrode potential is 22 2 // 2 2.303RT EE log F ο
6. What is the half cell potential of the half cell, Pt, Cl2 (2 atm)//Cl– (0.1 M). standard reduction potential of chlorine electrode is 1.36 volts.
Sol. For the given half cell, the reduction half reaction is ()()2g aq
Cl2e2Cl +
Since, temperature is not mentioned, the potential can be estimated at 298 K.
For the reaction
0.1 0.01 Q P22
0.059 1.3620.301 2
2 Cl/Cl E1.428Volts∴=+
Try yourself:
6. Estimate the reduction potential value of magnesium half cell containing 0.1 M solution of Mg+2 () 2 Mg/Mg E0.2363V + ο =−
Ans: –0.266 volts
2.7.2 Nernst Equation for a Galvanic Cell
For a cell reaction: ne aAbBcCdD +→+ cd
CD 2.303RT EE log nF AB
cellcell ab
at 298 K
cellcell ab CD 0.059 EElog nF AB
For a Daniell cell reaction:
2e 22 ZnCuZnCu ++ +→+ 2 cellcell 2 Zn 2.303RT EE log 2F Cu + ο +
(here cellcathodeanode EEE οοο =− )
7. Calculate the emf of the cell at 25°C, ()() 32 Cr/Cr0.1M//Fe0.01MFe ++
Given : 3 Cr/Cr E0.74V + ο =− and 2 Fe E0.44V + ο =−
Sol. In the given cell, the cell reaction is 6e 23 2Cr3Fe2Cr3Fe ++ ++
Q = 104
cellCa EEE οοο =− = (–0.44) – (–0.74 = 0.3 Volts
EMF of Cell, Ecell = cell 0.059 ElogQ n ο 4 cell 0.059 Elog10 6 ο () 0.059 0.34 6 =− cell E0.26volts∴=
Try yourself:
7. Estimate the cell potential at 298 K of the cell
Ag/Ag+(0.01 M)//Cl–(0.2M)/Cl2 (0.5 atm)/Pt
Given : Ag/Ag E0.8V + ο=+ and 2 Cl/Cl E1.36V ο=+ Ans: 0.17 volts
2.7.3 Equilibrium Constant from Nernst Equation
The cell reaction of the Daniell cell is ()()() 22 saq aq ZnCuZnCu ++ +→+
If the cell is functioning, the concentration of Zn +2 keeps on increasing, while the concentration Cu+2 keeps on decreasing. At the same time voltage of the cell slowly decreases. After some time, voltmeter gives zero reading, as there is no change in concentrations of both Zn +2 and Cu +2. This indicates that the cell reaction attains equation state.
2 cellcell 2 Zn 2.303RT EE log nF Cu + ο + =−
at equilibrium Ecell = 0 and 2 cc 2 Zn QK Cu + +
cell C 2.303RT 0E logK 2 ο =− (or)
cell C 2.303RT E logK 2 ο =
at 298 K, cell C 0.059 ElogK 2 ο =
In general, for any cell at equilibrium, cell C 2.303RT E logK nF ο = at 298 K cell C 0.059 ElogK n ο = or cell C nE logK 0.059 ο =
8. Find the log K value (here K = equilibrium constant) for the disproportionation reaction:
()()() 2 aq saq 2CuCuCu ++ + at 298 K.
(Given: 2 Cu/Cu Cu/Cu E0.52V,E0.15V +++ οο=+=+ )
Sol. For the reaction at equilibrium, E cell = 0
2 C Cu/CuCu/Cu 0.059 EE logK 1 +++ οο =−−
C 0.059 0.520.15logK 1 =−− = 0.37 – 0.059 log KC C 0.37 logK6.27 0.59 ==
Try yourself:
8. Calculate log KC for the reaction, Fe+2 + Ce+4 Fe+3 + Ce+3
(Given: 43 32 Ce/Ce Fe/Fe E1.44V,E0.68V ++++ οο == ) Ans: 1.28
2.7.4 Electrochemical Cell and Gibbs Energy
The maximum reversible work done by a galvanic cell is given by
ΔG = – nF Ecell
Here, Ecell = emf of the cell
nF = charge of n moles of electrons passed
For the cell reaction: ()()() 2e 22 (aq) s aqs ZnCuZnCu ++ +→+
ΔG = – 2F Ecell
When coefficients are changed in the reaction
()()()() 4e 22 s aq aq s 2Zn2Cu2Zn2Cu ++ +→+
ΔG = – 4FEcell
If the concentration of all the reaction species is unity at 298 K,
ΔG° = – n F E° cell and ΔG° = – 2.303 RT log K
(K = equilibrium constant)
Thermodynamic Data of Galvanic Cell
1. For enthalpy change of reaction (ΔH): Gibbs equation: ΔG = ΔH – TΔS …… (1) or ΔG = ΔH + () P G T T ∂∆ ∂ ………… (2)
nn ( ΔG = –nFE)
(3)
2. Temperature coefficient of emf cell:
EHFE TFT ∂∆+ = ∂ n n . . . . . . . (4)
3. Entropy change of reaction:
(from equations (1) and (2))
(or)
Hence for each of the above half reactions, ΔG is calculated.
Hence,
2.7.5 Electrode Potential and Solubility Product
Let us consider silver electrode dipped in a solution of AgNO3. If excess of NaI is added to this half cell, the equilibrium established is AgI + e Ag + I–. This is corresponding to the electrode, I–/AgI/Ag. This is metal –insoluble salt–soluble anion electrode.
Nernst equations for the half cell potentials are as follows: o Ag/AgAg/Ag 0.0591 EE log 1 Ag ++ + =−
(for the reaction Ag+ + e Ag)
I/AgI/AgI/AgI/Ag
(For the reaction AgI + e Ag + I–)
Since, Ag/Ag
I/AgI/Ag
EE0.059logAgI −+ οο+− =+
I/AgI/AgAg/Ag
οοο −=−−− 332211 nFEnFEnFE
()() 3 20.340210.522 E 1 ο +− =
3 E0.1584volts ο=+
Try yourself:
9. Calculate the half cell potentials
3 Au3eAu,E1.5V ++→°=
u AeAu,E1.7V ++→°=
Then find the standard half cell potential for the reaction, Au+3 + 2e → Au+
Ans: 1.4 volts
TEST YOURSELF
1. Reduction potential of iron electrode is minimum when it is in contact with ________Fe2+ solution.
(1) 10–3 M (2) 10–3 M (3) 10–2 M (4) 1 M
or sp
EE0.059logK −+ οο=+
I/AgI/AgAg/Ag
9. If standard reduction potentials of Cu +/Cu, Cu +2/Cu are 0.522 volts and 0.3402 volts, respectively, calculate the standard half cell potential for the electrodes, Cu +2/Cu+
Sol. n1
CueCu,E0.522V +=+→°= (1) n2
2 Cu2eCu,E0.3402V +=+→°=+ ...(2)
2 CueCu,E? +++→°= .
(3)
Electrode potentials are not additive, but ΔG values are additive.
2. Find the emf of the cell in which the following reaction takes place at 298 K. ()()()()()() 0.001M 0.001M 2 s aq aq s Ni2AgNi2Ag ++ +→+ (Given that 0 cell 2.303RT E1.05V,0.059 at 298K ) F == (1) 1.0385 V (2) 1.385 V (3) 0.9615 V (4) 1.05 V
3. Give that zn/Zn22Cd/Cd E0.763VandE0.403V ++ °=−°=− the emf of the cell () () 22 ZnZn//CdCd a0.004a0.2 ++ == will be given by (1) E0.36log0.0590.004 22
=−+
4. Consider the following four electrodes:
A = Cu2+ (0.0001M) / Cu(s)
B = Cu2+(0.1 M) / Cu(s)
C = Cu2+ (0.01 M) / Cu(s)
D = Cu2+ (0.001 M) / Cu(s)
If the standard reduction potential of Cu +2/ Cu is +0.34 V, the reduction potentials (in volts) of the above electrodes follow the order
(1) A > D > C > B
(2) B > C > D > A
(3) C > D > B > A
(4) A > B > C > D
5. The standard electrode potential for the following reaction is +1.33 V. What would be its potential when the pH is 2.0?
(1) +1.820 V (2) +1.990 V (3) +1.608 V (4) + 1.0542 V
6. For a cell reaction involving two electron transfer the standard emf of the cell is found to be 0.295 V at 25°C. At the same temperature, equilibrium constant will be (1) 10 (2) 1 × 1010 (3) 1 × 10–10 (4) 29.5 × 10–2
7. If the cell E ο for a given reaction has a negative value, then which of the following gives the correct relationships for the values of ΔG° and K eq ?
(1) eq G0;K1∆>< (2) eq G0;K1∆>>
(3) eq 1 0; G K ∆<> (4) eq G0;K1∆<<
8. The standard electrode potential of a Daniell cell is 1.1 V. Find the standard Gibbs energy for the reaction
()()()() aq aq 22 ss ZnCuZnCu ++ +→+
(1) –212.3 kJ / mol (2) –2.123 kJ / mol (3) –21.23 kJ / mol (4) –2123 kJ / mol
9. ()()()21 22 PtClPHCl0.1MPtClP; cell reaction will be spontaneous if (1) P1 = P2 (2) P1 > P2 (3) P2 > P1 (4) P1 = P2 = 1 atm
Answer Key
(1) 2 (2) 3 (3) 4 (4) 2 (5) 4 (6) 2 (7) 1 (8) 1 (9) 3
2.8 BATTERIES
Battery is an example of an electrochemical c ell. It consists of two or more voltaic cells connected in series. The voltage of a battery should not vary much during its use. Mainly two types of cells are in use: primary cells and secondary cells.
2.8.1
Primary Cells
Primary cell is a galvanic cell that cannot be recharged once the reactants are consumed. This is because the electrode reactions cannot reversed to regenerate reactants from the products of the cell reaction.
The most familiar example of primary cell is the dry cell, commonly called Leclanche cell, which is used in clocks, transistors, and torches.
Dry Cell
Dry cell consists of a zinc container that acts as anode and graphite as cathode rod, surrounded by powdered manganese dioxide and carbon, as shown in Fig.2.10.
Carbon rod (cathode)
Fig. 2.10 Primary Cells: Common Dry Cell
The zinc electrode is in contact with paste of ammonium chloride and zinc chloride. The paste, C+MnO2 and NH4Cl + ZnCl2 are separated by a porous sheet. The zinc vessel is surrounded by a cardboard. The zinc vessel is sealed with pitch. The reactions that occur in dry cell are complex, but approximately shown as,
At cathode (reduction):
3 MnONHeMnO(OH)NH +− ++→+
24
(or) 22 MnOHOeMnO(OH)OH ++→+
MnO2 acts as cathodic depolariser.
At anode (oxidation): 2 Zn(s)Zn2e+−→+
The secondary reactions taking place in dry cell are
432 NHClOHNHClHO +→++
2 3 Zn2NH2ClZnCl.2NH23 +−++→
322 or[Zn(NH)Cl]
Ammonia produced in the reaction forms a complex, 2 34 [Zn(NH)] + . The potential of the cell is nearly 1.5 V.
Mercury Cell
Mercury cell is another dry cell suitable for low current devices. It consists of a paste of potash and zinc oxide acting as electrolyte. A paste of carbon and mercuric oxide acts as cathode and zinc amalgam as anode. The electrode reactions are,
At cathode (reduction):
2 HgOHO2eHg(l)2OH ++→+
At anode (oxidation): (s)2
ZnHg2OHZnOHOHg2e −+→+++
The overall reaction is, (s) ZnHgHgOZnO2Hg(s)() −+→+ l
This cell does not involve any ion, changing its concentration. That is why this cell can give almost constant voltage.
The potential of mercury dry cell is approximately 1.35 V.
Anode Cap
Anode
2.8.2
Secondary Cells
The secondary cells are rechargeable. Their chemical reactions can be reversed by supplying electrical energy to the cells. These are also known as storage cells.
Most common reversible storage cell is lead accumulator, used in inverters and automobile vehicles. It is called acid storage cell. Alkali storage cell is called Edison battery.
Anode
Cathode
Negative plates: lead grids filled with spongy lead
Positive plates: lcad grids filled with PbO2 38% sulphuric acid solution
Fig. 2.12 Lead Storage Battery
Lead Storage Battery
Lead storage battery consists of two lead electrodes, and both are reversible. Sponge lead is used as anode and a grid of lead packed with plumbic oxide is used as cathode. A 38% aqueous solution of sulphuric acid is used as electrolyte.
It is represented as:
Pb|PbSO4(s), H2SO4(aq) PbSO4(s), PbO2(s)/ Pb.
In a simple way, it can be represented as Pb/ H2SO4. PbO2
When the battery is in use, i.e., during discharge, the cell reactions are:
The reaction at cathode is 2 2(s)4(aq)(aq) PbOSO4H2e −+− +++→
PbSO2HO4(s)2(l) +
The reaction at anode is ()()() 2 44 s aq s PbSOPbSO2e +→+
The overall cell reaction is: discharge
PbPbO2HSO(s)2(s)24(aq)++→
2PbSO2HO4(s)2(l) +
During the charging of the battery, the electrode reactions are reversed as follows:
At anode: PbSO4(s)+2H2O(l) →
PbO2(s) + SO4–2(aq) + 4H+(aq) + 2e–
At cathode
PbSO4(s) + 2e– → Pb(s) + SO4–2(aq)
Lead sulphate is accumulated at cathode and anode is converted into lead and lead dioxide, respectively.
Charging and discharging of lead storage battery is known as double sulphation. The voltage increases with the concentration of electrolyte. The voltage varies from 1.88 V (5% H2SO4) to 2.15 V (40% H2SO4).
Current is generated when lead accumulator undergoes the process of discharging. During this process, sulphuric acid is consumed and its concentration decreases. When the concentration of sulphuric acid decreases to a certain low value, the accumulator should be charged. During charging, reverse reaction takes place and H2SO4 is generated resulting in an increase in its concentration or specific gravity. Hence, the cell is tested by determining the specific gravity of H 2SO4
Nickel–Cadmium Cell
Another important secondary cell is nickel–cadmium cell.
Positive plate
Separator
Negative plate
2.13 Rechargeable Ni–Cd Cell
The Ni-Cd cell is more expensive to manufacturer, but it has longer life than lead storage battery. The overall reaction during discharge of nickel–cadmium cell is,
Cd(s) + 2Ni(OH)3(s) → 2Ni(OH)2(s) + CdO(s) + H2O(l)
2.8.3 Fuel Cell
A galvanic cell that directly converts chemical energy into electrical energy is called a fuel cell.
The fuel cell also has two electrodes and one electrolyte. The fuel is oxidised at anode and oxidant is reduced at cathode. Fuel and oxidant are continuously and separately supplied towards the two electrodes of the cell at which they undergo reaction. Fuel cells are primary cells and are capable of supplying
2: Electrochemistry
current as long as they are provided with fuel and oxidant.
A fuel cell is represented as:
Fuel / Anode / Electrolyte / Cathode / Oxidant + Cathode Aqueous electrolysis Anode – H2O
Fuel cells are now used in automobiles on an experimental basis. Fuel cells have silent operation and high efficiency, and they do not cause pollution.
The absence of moving parts in fuel cells eliminates wear and tear problems. They were used as power sources in spacecrafts Gemini and Apollo. The water formed in a hydrogen–oxygen cell is pure, and hence it is used for drinking purpose in spaceships.
TEST YOURSELF
1. The overall reaction of a hydrogen–oxygen fuel cell is
(1) ()()() 22 g eq O2HO4e4OH ++→
Fig. 2.14 Hydrogen–oxygen fuel cell
Hydrogen–oxygen, (as shown in Fig.2.14.) fuel cell consists of porous electrodes suspended in concentrated caustic soda solution.
Fuel and oxidant gases are bubbled at the surface of porous carbon electrodes embedded with suitable catalysts. Catalysts like finely divided platinum or palladium are incorporated into the electrode for increasing the rate of electrode reactions.
The electrode reactions are,
At cathode (reduction):
O2HO4e4OH ++→
2(g)2() (aq)
At anode (oxidation):
H2OH2e2HO2() +→+
2(g) (aq)
The overall reaction is given as
2(g)2(g)2() 2HO2HO +→
The heat of combustion is directly converted into electrical energy. The efficiency of energy conversion in fuel cells is generally more than 70%.
Methane–oxygen fuel cell consists of platinum electrodes suspended in phosphoric acid, which acts as an electrolyte.
(2) ()()() 22 g aq 2H4OH4HO4e +→+
(3) ()()() 222 gg 2HO2HO +→
(4) ()() 2 aq 4OH4e2HO +→
2. Select the incorrect statement.
(1) Primary batteries are non-rechargeable, so can be used only once
(2) Secondary batteries are rechargeable
(3) Primary batteries may act as electrolytic cells
(4) Secondary batteries may act as electrolytic cells
3. During the discharge of lead storage battery 1F of electricity is generated. Amount of PbSO 4 formed at cathode would be (MW of PbSO4 = 304)
(1) 304 g (2) 152 g
(3) 608 g (4) 456 g
4. During the charging of lead storage battery, the reaction occurring at cathode is represented by
(1) 2 PbPb2e+−→+
(2) 2 Pb2ePb +−+→
(3) 22 44 PbSOPbSO ++→
(4) 2 4224 PbSO2HOPbO4HSO2e +−−+→+++ 2 4224 PbSO2HOPbO4HSO2e +−−+→+++
5. A depolariser used in dry cell batteries is (1) manganese dioxide (2) manganese dioxide
(3) potassium hydroxide (4) sodium phosphate
6. Which of the following reactions occurs at the cathode of a common dry cell?
(1) Mn → Mn2+ + 2e–
(2) () 24 3 MnONHeMnOOHNH +− ++→+
(3) 2 3 23 Zn2NH2ClZnCl2NH +−++→⋅
(4) 2 ZnZn2e+−→+
Answer Key
(1) 3 (2) 3 (3) 2 (4) 4
(5) 2 (6) 2
2.9 CORROSION
Natural tendency of conversion of a metal into its undesirable compound on interaction with the environment is known as corrosion. It is an oxidative deterioration of metals.
Rusting of iron (Fe 2 O 3 .xH 2 O), tarnishing of silver (Ag 2S), and development of green coating on copper (CuCO3(OH)2) are some important examples of corrosion. Metallic corrosion causes enormous damage to bridges, buildings, ships, etc.
2.9.1 Mechanism of Corrosion
Process of corrosion may be chemical or electrochemical. Electrochemical corrosion is considered as the anodic dissolution of metal undergoing corrosion. Anodic dissolution involves oxidation of metal. M →Mn+ + ne–
Electrochemical corrosion occurs if the environmental conditions of metal favour the formation of a galvanic cell with the metal acting as anode.
If oxygen is not uniformly distributed on the surface of the metal, metal corrodes at those points when concentration of oxygen is less.
The portion of metal with access to high concentration of oxygen functions as cathode and with low oxygen as anode.
Metal with differential oxygenation functions as a galvanic cell. The portion of metal access to lower oxygen concentration acts as anode and undergoes anodic dissolution, as shown in Fig.2.15.
Fig. 2.15 Corrosion of iron
Rusting of iron is a common example of differential of oxygenation type corrosion. The mechanism corrosion is quite complex. At a particular spot of object of iron, it behaves as anode, at which oxidation takes place and metal dissolves.
At anode (oxidation),
Fe(s)→Fe2+ + 2e–; E° = – 0.44 V
Electrons released at anodic spot move through the metal and go to another spot and reduce oxygen in the presence of proton. The spot behaves as cathode, where oxygen is reduced.
At cathode (reduction)
O2(g) + 4H+(aq) + 4e– →2H2O(ℓ); E° = 1.23V. The overall reaction is
Since E cell is positive, the free energy change is negative, suggesting the corrosion is spontaneous. Further, ferrous iron is oxidised to ferric iron in the atmospheric oxygen and forms rust, which is chemically hydrated ferric oxide, Fe2O3 . xH2O.
Hence, for the rusting process to occur, the iron must be exposed to both water and oxygen.
2.9.2 Preventive Methods of Corrosion
The main principle underlying the method of prevention of corrosion is to separate the metal from the environment. Some of the methods are:
1. Painting the metal object protects the surface of the metal from the impact of the environment.
Metal surface can also be covered by using chemicals like bisphenol.
2. Prevention of metal surface from electrical conducting media.
3. Alloying the metal with more anodic metal, e.g., coating of iron with zinc (galvanisation of iron).
Iron pole in sea water undergoes corrosion faster than iron on beach, because iron in sea water suffers high salt concentration because dissolved ions make water a better conductor of electric current.
CHAPTER REVIEW
Electrical Conductivity
■ The substance that allows the flow of electricity through it is called electrical conductor.
■ Electrical conductors are of two types electronic and electrolytic.
TEST YOURSELF
1. Metal can be prevented from rusting by (1) Connecting iron to more electropositive metal – a case of cathodic protection
(2) Connecting iron to more electropositive metal –a case of anodic protection
(3) Connecting iron to less electropositive metal –a case of anodic protection
(4) Connecting iron to less electropositive metal –a case of cathodic protection
2. The composition of rust is (1) Fe2O3xH2O (2) Fe2O3.2H2O (3) Fe2O36H2O (4) Fe2O3
3. Galvanization is applying a coating of (1) Cr (2) Cu (3) Zn (4) Pb
4. Rusting of iron is catalysed by which of the following?
(1) Fe (2) Zn (3) N2 (4) H+
5. The Zn acts as sacrificial or cathodic protection to prevent rusting of iron because (1) OP OP E of ZnE of Fe οο <
(2) OP OP E of ZnE of Fe οο > (3) OP OP E of ZnE of Fe οο =
(4) Zn is cheaper than Fe Answer Key
(1) 1 (2) 1 (3) 3 (4) 4 (5) 2
■ Electronic conductors conduct electricity due to mobility of free electrons. Examples: Graphite, gas carbon, petroleum coke, metals, alloys and solid salts like CdS, CuS, etc.,
■ Electrolytic conductors conduct electricity due to mobility of free ions. Examples: Aqueous solutions of salts, acids and bases and fused salts like NaCl, KCl, etc.
■ Solutions of alkali metals in liquid ammonia are mixed conductors. They contain solvated cations and solvated electrons.
■ Substances that do not conduct electric current in the molten state or in solution are called non-electrolytes. Examples: sugar, glucose, urea, fructose, etc.
■ Non-electrolytes are non-polar covalent substances. They do not undergo ionisation.
■ All electrolytes spontaneously dissociate into charged particles when dissolved in water.
MA M+ + A–
■ The degree of ionisation (a) is the fraction of the total number of ionised molecules in a solution.
■ The extent of ionisation is different for different electrolytes, and it depends on the nature of the electrolyte, concentration of electrolyte, and temperature.
■ The degree of ionisation increases with increase in dilution, and at infinite dilution, it approaches unity.
■ Strong electrolytes ionise to a greater extent. Examples: Strong acids, strong bases, and soluble salts
■ Weak electrolytes ionise to a lesser extent. Examples: Weak acids, weak bases and sparingly soluble salts
■ Degree of ionisation of electrolytes increases with increase in temperature.
■ Arrhenius theory is applicable to only weak electrolytes.
Electrolytic Conductance
■ Resistance to the flow of electricity in a solution is directly proportional to the length and inversely proportional to the area of the tube in which the solution is taken.
■ Resistivity is the resistance of a conductor of 1 m length and 1 m2 area of cross section.
■ Units of specific resistance is ohm. metre (Ω.m)
■ For a given electrolytic cell, the quantity a is constant, which is called cell constant
■ Unit of cell constant is taken as m–1 or cm–1
■ Conductance is the ease of flow of electric current through the conductor. Conductance is the reciprocal of resistance.
■ Unit of conductance is ohm –1 or Ω–1 (or) mho. In SI system, it is seimen (S).
■ Specific conductance or conductivity (K, kappa) is the conductance of a conductor of 1 m length and 1m2 area of cross section.
■ Conductance taken as the reciprocal of resistivity.
■ Units of specific conductance is ohm–1.m–1 (or) S.m–1.
■ Conductance of all the ions produced by one gram equivalent weight of an electrolyte in V mL of solution is called equivalent conductance,
Λ eq = κ .V
■ Λ eq = 1000 N κ× , where N is normality .
■ Units of equivalent conductivity are Ω–1 . cm2. equiv–1 (or) S.cm2. eq–1.
■ Conductance of all the ions produced by one gram mole of an electrolyte in Vml of solution is called molar conductivity, Λ m = κ .V
■ Λ m = 1000 M κ× , where M is molarity.
■ Units of Λ m are Ω–1.cm2.mol–1 or S.m2.mol–1
■ The conductance of increase in all electrolytes increases with temperature.
■ With decrease in concentration of solution or increase in dilution, Λ m increases, Λ m increases and κ decreases.
■ At infinite dilution, concentration approaches zero. Molar conductance at this dilution is called limiting molar conductivity m 0.
■ Equivalent conductance at this dilution is called limiting equivalent conductivity ^0.
■ In case of strong electrolytes, Λ m or m increases to less extent with dilution.
■ Strong electrolytes are almost completely ionised at all concentrations. Increase in Λ m or m , with dilution is only due to decrease in the interionic forces.
■ For strong electrolytes, variation of conductance with dilution is given by Huckel Onsagar equation.
C0 KC ∧=∧−
Kohlrausch Law
■ Kohlrausch law states that, the limiting equivalent conductance of an electrolyte is the sum of equivalent conductivities of cations and anions.
() 0 00mxyABxy+−Λ=λ+λ
Electrolysis
■ The chemical decomposition of an electrolyte by the influence of electric current is called electrolysis.
■ In an electrolytic cell, electrical energy is converted into chemical energy.
■ During electrolysis, reduction (electronation) takes place at cathode and oxidation (de-electronation) takes place at anode.
■ During electrolysis, electrons flow from anode to cathode in the external circuit.
■ When a solution contains different kinds of cations and anions, during electrolysis, the cation of the least el ectropositive element
and the anion of the least electronegative element are discharged first.
■ Preferential discharge of some cations: K+ < Ca2+ < Na+ < Mg2+ < Al3+ < Zn2+ < Fe2+ < H+ < Cu2+ < Ag+< Au+
■ Preferential discharge of some anions: F– < PO43– < SO42– < NO3– < OH– < HSO4– < Cl– < Br– < I–
■ Electrodes that do not involve in chemical reaction during electrolysis are called inert electrodes. Examples: graphite, platinum, gold, etc.
■ Active electrodes involve in chemical reaction during electrolysis. In general, anodic metal dissolves in solution. Example: Cu anode in aqueous CuSO 4.
■ Quantitative relationship of electrolysis was given by Faraday. There are two Faraday’s laws.
■ The mass of the substance evolved or deposited or dissolved at an electrode is directly proportional to the quantity of electricity passed through the electrolyte.
■ Mass, m = eQ, where e is electrochemical equivalent
■ Quantity of electricity ( Q) is taken as the product of current strength in amperes (A) and time in seconds ( t).
■ Q = it. C is A s.
■ If same quantity of electricity is passed into different electrolytes (connected in series), the ratio of the amount of substances deposited at the respective electrode is equal to the ratio of their equivalent weights.
■ 11 22 mE mE = Here, m1, m2 are masses of elements and E1, E2 are chemical equivalents
Electrochemical Cells
■ The cell in which chemical energy is converted to electrical energy is called as galvanic or voltaic cell.
■ In galvanic cell, oxidation takes place at anode and electrons are released, and reduction takes place at cathode and electrons are used up.
■ Electrons travel along external circuit from anode to cathode. Cathode is +ve electrode and anode is –ve electrode. These are connected by salt bridge.
■ Daniel cell is an example of galvanic cell. It is constructed using Zn and Cu electrodes and a porous partition.
■ The representation of the Daniel cell is Zn(s) / Zn2+(aq) // Cu2+(aq) / Cu(s)
■ The standard emf of a Daniel cell is given as 1.1 volt.
■ Single electrode potential is the potential developed at metal and metal ion interface or non-metal and non-metal ion interface.
■ Non-metallic element is generally gas. Single electrode of non-metal / nonmetal ion needs an inert metal rod like Pt. Example: Pt, H2(g) / H+(aq)
■ The magnitude of potential developed by a single electrode depends on the nature of metal or non-metal, number of electrons involved in half-cell reaction, concentration of ions in solution and temperature.
Electrochemical Series
■ Normal hydrogen electrode, NHE or SHE is represented as: Pt, H2(g)(1atm) /H+(1M).
■ The potential of SHE is assumed to be ‘zero’ volts.
■ A saturated calomel electrode is now used as secondary reference electrode. It is represented as Hg/ Hg 2Cl 2 (s), KCl (saturated).
■ Potential of saturated calomel electrode is taken as –0.2422 volts.
■ Standard electrode potential (E o ) is the potential developed by single electrode if concentration of ions is 1 M at 298 K and 1 atmosphere pressure.
■ The potentials are standard reduction potentials only (SRP).
■ Standard oxidation potential (SOP) is equal to SRP in magnitude, but with opposite sign.
■ Electrochemical series is the arrangement of various electrode systems in the ascending order of their SRP values. It is also called activity series.
■ All the metals that are present above hydrogen in the electrochemical series are called active metals. They have –ve SRP values.
■ Active metals liberate hydrogen from dilute mineral acids.
■ All the elements that are present below hydrogen in the activity series have +ve SRP values. They do not liberate H2 from acid.
■ A metal with less SRP has high tendency to lose electrons, undergoes oxidation and acts as reducing agent. It is represented on LHS in galvanic cell and replaces other metal ions that have more SRP from their salt solution.
■ Metal with more SRP has high tendency to gain electrons. It undergoes reduction and acts as an oxidant. It is represented on RHS in galvanic cell and is replaced by other metals that have less SRP from its salt solution.
■ An element with a lower reduction potential is a more powerful reducing agent.
■ Reduction power of some metals is in the order
Li > K > Ca > Zn > Cu
■ An element with a higher reduction potential is a more powerful oxidising agent.
■ The order of oxdising ability of halogens is F2 > Cl2 > Br2 > I2
■ The potential difference between the two electrodes when no current is drawn from the cell is known as the emf of the cell.
■ If the emf of the cell is negative, the cell reaction is non-spontaneous and that cell cannot be constructed.
Nernst Equation
■ If the salt bridge is removed, the cell reaction stops, the emf of the cell becomes zero and ions move randomly.
■ Nernst equation gives the dependence of the electrode potential on the concentration of ion.
■ Nernst equation holds good for reversible electrodes.
■ For metal electrodes, the half-cell reaction is Mn++ne– M.
The Nernst equation is:
0 n M 2.303RT EElog nF M +
■ Simplified form of Nernst equation for metal electrodes at 298 K is
0 0.059
EElogC n =+
■ For non-metal electrodes, the half cell reaction is A + ne – A n–. The Nernst equation is: n 0 EElog2.303RT[A] nF[A] =−
■ Simplified form of Nernst equation for non-metal electrodes at 298 K is:
0 0.059 EElogC n =−
■ For hydrogen electrode, E = – 0.059 pH
■ Concentration cell is a galvanic cell in which both the electrodes are of same type but the electrolytes have different concentrations.
Cu|CuSO4(C1)|| CuSO4(C2)|Cu
■ The cell potential of the concentration cell can be calculated using the equations:
Ecell= 2.303RT nF log 2 1 C C = 0.0591 n = log 2 1 C C
■ When the cell reaction is in equilibrium, the cell emf is zero. The Nernst equation is E0 = 2.303RT nF logKc (or) E0 = 0.0591 n logKc
■ The change in free energy is represented by ΔG. For spontaneous reaction, ΔG will have a negative value (ΔG = –ve)
■ Electrical work done in one second is the product of electrical potential and total charge passed.
■ To obtain maximum work from a galvanic cell, the charge has to be passed reversibly.
■ The reversible work done by a galvanic cell is the decrease in its free energy.
W max = – ΔG.
■ When no current is drawn from the cell, the emf of the cell is E, the amount of charge passed is nF (n is the number of electrons and F is faraday).
ΔG = – nFE
■ From standard free energy, the equilibrium constant can be calculated
ΔG° = –RT lnkc; 0 c G2.303RTlogK∆=−
Batteries
■ Battery is a single galvanic cell or more than one cell connected in series. Commercial batteries are of two types.
■ Primary cell becomes dead after some time and cannot be used again. Examples: Dry cell, mercury cell
■ Secondary battery can be recharged and can be used again. Examples: Lead storage battery, nickel–cadmium cell, alkali storage cell (Edison’s battery).
■ Dry cell is also called Leclanche cell. It is a Zn container. Cathode is a graphite rod, surrounded by powdered MnO 2 and carbon. Gap between cathode and anode is filled with moist paste of NH4Cl and ZnCl2, which acts as electrolyte. Cell potential is 1.5 volt.
■ Mercury cell is used for low current devices. Anode is ZnHg. Cathode is a paste of HgO and carbon. Electrolyte is a paste of KOH and ZnO. Cell potential is 1.35 volts.
■ Lead storage battery is used in automobiles and inverters. Anode is lead grids filled with spongy lead. Cathode is lead grids filled with PbO2.
■ Cathode and anode in lead storage battery arranged alternately and separated by thin fiber glass sheets. Electrolyte is 38% (by weight) solution of H 2SO4.
■ Overall cell reaction in lead storage battery
Pb(s) + PbO2(s) + 2H2SO4(aq) discharging charging
2PbSO4(s) + 2H2O(l)
■ During the working of storage cell, H2SO4 is used up (during discharge). The cell reactions are reversed. Each set of anode and cathode produces a potential of 2 volts.
■ Nickel–Cadmium cell has longer life than lead storage battery. Anode is Cd(s), cathode is Ni(OH) 3(s) and electrolyte is moist NaOH or KOH. Cd acts as reductant and Ni(OH)3 as oxidant.
Corrosion
■ Corrosion is the process of gradual destruction of metal by the environment
■ Rusting of iron, tarnishing of silver and green coating on copper are examples of corrosion.
■ Corrosion is due to the formation of oxide or other salts on the metal. Corrosion is an electrochemical phenomenon.
■ During rusting of iron, at a particular place, iron undergoes oxidation, forming hydrated ferric oxide (rust).
Fe2O3 + xH2O → Fe2O3.xH2O
■ Prevention of corrosion of iron can be prevented by painting iron surface, by electroplating the iron with corrosion resistant metals like chromium and connecting with sacrificial anodes.
■ Coating of iron surface with zinc metal is called galvanization.
■ Metals used as sacrificial anode are more electropositive elements like Zn, Mg, Al and their alloys.
Exercises
Level - I
Electrical Conductivity
1. Which one of the following materials conducts electricity ?
(1) diamond
(2) barium sulphate
(3) crystalline sodium chloride
(4) fused potassium chloride
2. Which of the following is conductor of electricity
(1) diamond
(2) graphite
(3) carborundum
(4) silica
3. In metallic conductor the current is conducted by flow of (1) electrons (2) atoms
(3) ions (4) molecules
4. Choose the wrong statement
(1) Electrical conductance of an electrolytic conductor increases with increase in temperature.
(2) Electrical conductance of a metallic conductor increases with increase in temperature.
(3) Electrical conductance of a metallic conductor decreases with increase in temperature.
(4) Degree of dissociation of an electrolyte increases with dilution.
5. Which of the following is not a conductor of electricity?
(1) Solid NaCl
(2) Cu
(3) Fused NaCl
(4) Brine solution
6. Which of the following is an example of weak electrolyte
(1) H3BO3 (2) H2SO4
(3) HNO3 (4) HCIO3
Electrolytic Conductance
7. Conductivity (units Siemen’S) is directly proportional to the area of the vessel and the concentration of the solution in it and is inversely proportional to the length of the vessel, then the unit of constant of proportionality is
(1) S m mol–1
(2) Sm2 mol–1
(3) S–2 m2 mol
(4) S2 m2 mol-2
8. If the specific conductance and conductance of a solution is same, then its cell constant is equal to
(1) 1 (2) 2
(3) 10 (4) 100
9. The distance between two electrodes of a cell is 2.5 cm and area of each electrode is 5 cm2 the cell constant (in cm–1) is (1) 2 (2) 12.5
(3) 7.5 (4) 0.5
10. Which of the following solutions of NaCl has the higher specific conductance ?
(1) 0.001N (2) 0.01N
(3) 0.1 N (4) 1 N
11. Molar conductivity of a solution is 1.26 × 102 Ω–1cm2mol–1 Its molarity is 0.01M. Its specific conductivity will be
(1) 1.26 × 10–5
(2) 1.26 × 10–3
(3) 1.26 × 10–4
(4) 0.0063
12. The unit of specific conductivity is
(1) Ohms cm–1
(2) Ohms cm–2
(3) Ohm–1 cm
(4) Ohm–1 cm–1
13. Effect of dilution on conduction is as follows
(1) Specific conductance increases, molar conductance decreases
(2) Specific conductance decreases, molar conductance increases
(3) Both increase with dilution
(4) Both decrease with dilution
14. The value of molar conductivity of HCl is greater than that of NaCl at a particular temperature because
(1) Molecular mass of HCl is greater than that of NaCl
(2) Mobility of H+ ions is more than that of Na+ ions
(3) HCl is strongly acidic
(4) Ionisation of HCl is larger than that of NaCl
15. The unit of equivalent conductivity is
(1) Ohm cm
(2) Ohm–1cm–2(g equivalent)–1
(3) S cm–2
(4) Ohm cm2 (g equivalent)
Kholrausch Law
16. According to Kohlrausch law, the limiting value of molar conductivity of an electrolyte A2B is (1)
(1) 100 m m K C λ=
(2) 000 m vv +− +−λ=λ+λ
(3) 1000 eq eq K C λ=
(4) 0 mca λ=λ+λ
18. If the molar conductance values of Ca2+ and Cl– at infinite dilution are respectively 118.88 × 10–4 m2 mho mol–1 and 77.33×10–4 m2 mho mol–1 then that of CaCl2 is (in m2 mho mol–1)
(1) 118.88 × 10-4
(2) 154.66 × 10-4
(3) 273.54 × 10-4
(4) 196.21 × 10-4
19. The molar conductivities 00 NaOAcHCl and ΛΛ at infinite dilution in water at 25°C and 91.0 and 426.2 S cm2 / mol respectively. To caculate the additional value required is (1) 0 NaCl Λ (2) 2 0 HO Λ (3) 0 KCl Λ (4) 0 NaOH Λ
20. The equivalent conductances of two strong electrolytes at infinite dilution in H2O (where ions move freely through a solution) at 25°C are given below 3 2 CHCOONa 91.0Scm/ equiv Λ= o 2 HCl 426.2Scm/ equiv Λ= o
What additional information / quantity one need to calculate 0 Λ of an aqueous solution of acetic acid ?
(1) 3 of CHCOOK Λ o
17. The equation representing Kohlrausch law from the following is(v+ are number of cations and v- are number of anions )
(2) + Ë of H o
(3) 0 2 of ClCHCOOH Λ
(4) 0 of NaCl Λ
21. Calculate HOAc ∞ Λ using appropriate molar conductances of the electrolytes listed above at infinite dilution in H 2 O at 25°C
Electrolyte
() *21 Scmmol Λ 150 145 426 91 127 (1) 217 (2) 390 (3) 552 (4) 517
22. The difference in molar conductances between KCl and KNO 3 at infinite dilution at 298K is 34.9. What is the difference in molar conductances between KNO3 and LiNO3 at infinite dilution in the same units at 298K
(1) 34.9 (2) 68.8 (3) 3.49 (4) 6.88
23. Which of the following equation gives correct relation between molar conductance of strong electrolyte, and it’s concentration
(1) MKC ∞ λ=λ+
(2) ()MKC ∞ λ=λ−
(3) ()MKC ∞ λ=λ−
(4) MKC ∞ λ=λ−
24. The ionic equivalent conductivity of -2 24 , COK + and Na + ions are X, Y and Z Scm 2 Eq –1 respectively. The 0 eq Λ of (NaOOC–COOK) based on the data given is
y xz
++
(1) Scmeq2-1 2
z xy
++
(2) Scmeq2-1 2
(3) () Scmeq2-1 ++ xyz
25. Kohlrausch law is useful in the calculation of
(A) pH of weak electrolyte (B) degree of dissociation of weak electrolyte
(C) K w of water
(D) K sp of sparingly soluble salt
(1) A, C and D only
(2) A and C only (3) A, B and C only (4) A, B, C and D
Electrolysis
26. Three Faradays of electricity was passed through on aqueous solution of iron (II) bromide. The mass of Iron metal (at mass 56) deposited at the cathode is (1) 56 g (2) 84 g (3) 112 g (4) 168 g
27. What is the amount of Al deposited on the electrolysis of moltenAl 2 O 3 when a current of 9.65 A is passed for 10.0 seconds
(1) 0.9 g (2) 0.009 g (3) 9 g (4) 0.09 g
28. Product at anode in electrolysis of dilute sulphuric acid is (1) H2S (2) H2 3) SO2 (4) O2
29. Process which occur at anode in the electrolysis of fused sodium chloride is (1) Na+ is oxidised (2) Cl– is oxidised (3) Cl is reduced (4) Na is reduced
xyz
(4) Scmeq2-1 22
++
30. Electro chemical equivalent is highest for (1) Sodium (2) Hydrogen (3) Aluminium (4) Magnesium
31. How many coulombs of electricity are required for the reduction of 1 mol of 2 4 MnOtoMn−+ ?
(1) 96500 C
(2) 1.93 ×105C
(3) 4.83×105C
(4) 9.65×106C
32. 1 faraday can deposit/liberate one gram atom of metal at cathode during the electrolysis of (inert electrodes are used)
(1) aq. AgNO3
(2) aq. CuSO4
(3) aq. NaCl
(4) aq. ZnCl2
33. If three faraday of electricity are passed through the solutions of AgNO3,CuSO4, and AuCl3 the molar ratio of the cations deposited at the cathodes will be
(1) 1 : 1: 1 (2) 1 : 2 : 3
(3) 3 : 2 : 1 (4) 6 : 3 : 2
34. If a current of 1.5 ampere flows through a metallic wire for 3 hours, then how many electrons would flow through the wire?
(1) 2.25×1022 electrons
(2) 1.13×1023 electrons
(3) 1.01×1023 electrons
(4) 4.5×1023 electrons
35. 810 g of acidulated water is subjected to electrolysis using 9.65 × 104 amperes. Time required for electrolysis is (1) 45 seconds (2) 90 seconds
(3) 180 seconds (4) 360 seconds
Electro Chemical Cells
36. For cell reaction,Zn+Cu 2+→ Zn 2++Cu, cell representation is:
(1) Zn/ Zn2+||Cu2+/Cu
(2) Cu/Cu2+||Zn2+/Zn
(3) Cu|Zn2+||Zn|Cu2+ (4) Cu2+|Zn||Zn2+|Cu
37. Which of the following is the cell reaction that occurs when the following half-cells are combined?
I2+2e–→2I–(1M); E°=+0.54V
Br2+2e–→2Br–(1M); E°=+1.09V
(1) 2Br–+I2→Br2+2I–
(2) I2+Br–→ 2I– +2Br–
(3) 2I–+Br2→I2+2Br–
(4) 2I–+2Br2→I2+Br2
38. For the galvanic cell,Cu|Cu2+||Ag+|Ag.
Which of the following observations is not correct?
(1) Cu acts as anode and Ag acts as cathode
(2) Ag electrode loses mass and Cu electrode gains mass.
(3) Reaction at anode Cu→Cu2++2e–
(4) Copper is more reactive than silver
39. In the cell, Zn| Zn2+||Cu2+|Cu, the negative terminal is
(1) Cu (2) Cu2+
(3) Zn (4) Zn2+
40. The cell reaction of the galvanic cell : ()()()() 2+2+ saqaql CuCuHgHg is
(1) Hg+Cu2+→Hg2++Cu
(2) Hg+Cu2+→Cu++Hg+
(3) Cu+Hg→CuHg
(4) Cu+Hg2+→Cu2++Hg
41. Which of the following is the correct cell representation for the given cell reaction?
Zn+H2SO4→ ZnSO4+H2
(1) Zn| Zn2+||H+|H2
(2) Zn| Zn2+||H+,H2|Pt
(3) Zn|ZnSO4|H2SO4|Zn
(4) Zn|H2SO4| ZnSO4|H2
42. Following reactions are taking place in a Galvanic cell, Zn→ Zn2++2e–;Ag++e–→Ag
Which of the given representations is the correct method of depicting the cell?
(1) Zn(s)|Zn2+(aq)||Ag+(aq)|Ag(s)
(2) Zn2+|Zn||Ag|Ag+
(3) Zn(aq)|Zn2+(s)||Ag+(s)|Ag(aq)
(4) Zn(s)|Ag+(aq)||Zn2+(aq)|Ag(s)
43. Regarding the standard hydrogen electrode, the incorrect statement is
(1) H2 gas is bubbled at 1 bar
(2) 2 M HCl is electrolyte
(3) SRP is arbitrarily taken as Zero. (4) It is a primary reference electrode.
44. Which cell will measure the standard electrode potential of copper electrode?
(1) Pt(s)|H 2(g 0.01 bar) |H + (aq 1M) | Cu2+(aq 1M) Cu
(2) Pt(s)|H 2 (g 0.1 bar) |H + (aq 1M)| Cu2+(aq 2M)Cu
(3) Pt(s)|H2(g 1bar) |H+(aq 1M)| Cu2+(aq 1M)Cu
(4) Pt(s)|H 2(g 0.1 bar) |H +(aq 0.1M) | Cu2+(aq 1M) Cu
45. Reduction potential of a hydrogen electrode at pH = 1 is:
(1) 0.059 volt (2) 0 volt
(3) –0.059 volt (4) 0.59 volt
Electrochemical Series
46. Given that 22 O/HO E=+1.23V
2-2284 SO/SO E=2.05V2 Br/Br E=+1.09V
3+ Au/Au E=+1.4V
The strongest oxidizing agent is:
(1) 2-SO28 (2) Au3+ (3) Br2 (4) O2
47. Which of the following is not an application of electro chemical series?
(1) To compare the relative oxidizing and reducing power of substances.
(2) To predict evolution of hydrogen gas on reaction of metal with acid.
(3) To predict spontaneity of a redox reaction.
(4) To calculate the amount of metal deposited on cathode.
48. Zn gives hydrogen with H2SO4 and HCl but not with HNO3 because
(1) Zn acts as oxidizing agent when reacts with HNO3
(2) HNO 3 is weaker acid than H 2SO 4 and HCl
(3) Zn is above the hydrogen in electrochemical series
(4)NO3 is reduced in preference toH+ ion.
49. Which of the following is the correct order in which metals displace each other from the salt solution of their salts?
(1) Zn,Al,Mg,Fe,Cu
(2) Cu,Fe,Mg,Al, Zn
(3) Mg,Al, Zn,Fe,Cu
(4) Al,Mg,Fe,Cu,Zn
50. The metal that cannot displace hydrogen from dilute HCl is:
(1) Al (2) Fe
(3) Cu (4) Zn
51. E° of Mg2+/Mg,Zn2+/Zn and Fe2+/Fe are are –2.37V, –0.76V and –0.44V respectively. Which of the following statements is correct
(1) Zn will reduce Fe2+
(2) Zn will reduce Mg2+
(3) Mg oxidises Fe
(4) Zn oxidises Fe
52. Four metals A, B, C & D are having respectively standard electrode potential are –3.05, –1.66, –0.40 & 0.80 V. Which one will be the strong reducing agent
(1) A (2) B
(3) C (4) D
53. E° for Fe 2+ +2e→Fe is–0.44 V; E° Zn2++2e→Zn is –0.76V Then
(1) Zn is more electropositive than Fe
(2) Fe is more electropositive than Zn
(3) Zn is more electronegative
(4) No reaction takes place
54. The correct statement among the following, about electrochemical series (S.R.P. increasing order) is
(1) the metals occupying top positions in the series do not liberate hydrogen with dilute acids
(2) the substances which are stronger reducing agents and stronger oxidising agents are placed below & top respectively
(3) a metal higher in the series will displace the metal from its solution which is lower in the series
(4) various electrodes are arranged in a series in the descending order of their potentials
55. The E0 values for
Al+|Al=+0.55 V and Tl+|Tl =–0.34V
Al+3|Al = -1.66V and Tl+3|Tl=+1.26V
identify the incorrect statement.
(1) Al is more electropositive than Al
(2) Tl+3 is a good reducing agent than Tl+
(3) Al+ is unstable in solution
(4) Tl can be easily oxidized to Tl+ than Al+3
Nernst Equation
56. Emf of the cell Zn(s)/Zn+2(C1)//Fe2+(C2)/ Fe(s) can be increased by
(1) increasing the size of Iron electrode
(2) increasing [Fe2+]
(3) increasing [Zn2+]
(4) All of these
57. The emf of the cell in which of the following reaction, Zn (s)+Ni 2+(0.1M)→ Zn 2+ (1.0 M)+Ni (s) occurs is found to 0.5105 V at 298 K. The standard emf of the cell is
(1) 0.4810 V (2) 0.5696 V
(3) –0.5105 V (4) 0.5400 V
58. The Nernst equation giving dependence of electrode potential on concentration is
(1) 0 n+ M 2.303RT E=E+log nF M
(2)
0 M 2.303RT E=E+lognFM
(3) n+ 0 M 2.303RT E=E-lognFM
(4) 0 n+ 2.303RT E=E-logM nF
59. A copper electrode is in contact with copper sulphate solution. The reduction potential of copper electrode when copper sulphate solution is diluted by 10 times will be
(1) decreases by 0.0295 V
(2) increases by 0.0295 V
(3) decreases by 0.059 V
(4) increases by 0.059 V
60. The standard EMF of Daniel cell is 1.10 volt. The maximum electrical work obtained from the Daniel cell
(1) 212.3 kJ
(2) 175.4 kJ
(3) 106.15 kJ
(4) 53.07 kJ
61. Reduction potential of Chlorine electrode in contact with 0.001 M HCl solution is2 1 Pt,Cl/Cl 2 E=+1.36V
(1) 1.54 V (2) 1.24 V (3) 1.39 V (4) 1.45 V
62. The relationship between free energy and electrode potential is
(1) D G=–nFE (2) D G=nFE
(3) nFE G R ∆= (4) H G nFE ∆ ∆=
63. The potential of hydrogen electrode depends on
(A) Concentration of acid in which the platinum foil is immersed
(B) Temperature
(C) Pressure at which hydrogen gas is bubbled through the solution
(1) A, B & C
(2) A & B only
(3) A only
(4) A & C only
64. Reduction potential of Pt,Cl 2 (1atm)/ HCl(0.1M) is () 2 0 / 1.36 ClCl EV =+ (1) + 1.36V (2) + 1.39V (3) + 1.45V (4) + 1.42V
65. The correct Nernst equation for the given cell
Pt(s)|Br2(l)|Br–(M)||H+(M)|H2(g)(1bar)|Pt(s) is (1)
66. The pressure of H2 required to make the potential of H 2 electrode zero in pure water at 298 K is:
(1) 10–10 atm (2) 10–4 atm (3) 10–14 atm (4) 10–12 atm
67. The Nernst equation for following galvanic cell ()()() +32g XMYM AlAlBrBr;Pt is (1) 0 36 0.0591 E=E-log[X][Y] 3
(2) 0 23 0.0591 E=E-log[X] [Y] 6
(3) 0 26 0.0591 E=E-log[X][Y] 3 (4) 03 0.0591 E=E-log[X][Y] 3
68. What will be value of E cell for reaction ()()() -3 + +2 s 0.01M 10M Mg+2AgMg+2Ag given 0 cellE is 3.17 V (1) 2.67 (2) 2.13 (3) 3.27 (4) 3.051
69. Cu electrode is dipped in 0.1M of its solution at 298K. Reduction potential of electrode is +2 0 Cu/Cu E=+0.34V (1) +0.301V (2) +0.34V (3) +0.31V (4) +0.37V
70. SRP values of Ni, Cl2 electrodes are -0.25 & 1.36 V respectively. Then the EMF of the following cell is Ni/Ni+2(0.01M)||Cl–(0.01M)/Cl2,Pt. (1) 1.79 V (2) –1.79 V (3) 1.17 V (4) 1.11 V
Batteries
71. Which of the following reactions occurs at the cathode of a common dry cell
(1) Mn→Mn2++2e–
(2) MnO2+NH+4+e–→MnO(OH)+NH3
(3) Zn+2+2NH3+2Cl–→ZnCl2.2NH3
(4) Zn→Zn2++2e–
72. Dry cell is commonly used in transistors and clocks. Identify correct statement about dry cell.
(1) Cathode is zinc
(2) Anode is carbon
(3) zinc is oxidised
(4) carbon is oxidised
73. The reaction taking place at cathode in the lead storage battery during its discharging is
(1) Pb+SO4–2→ PbSO4 + 2e–
(2) PbO2+4H++SO4–2+2e–→PbSO4+2H2O
(3) PbSO 4 +2H 2 O → PbO 2 + 4H + +SO 4 –2 +2e–
(4) PbSO4+2e–→ Pb+SO4–2
74. A secondary cell is one
(1) can be recharged by itself
(2) can be recharged by passing current through it in the same direction
(3) can be recharged by passing current through it in the opposite direction (4) cannot recharged.
75. The overall reaction of a hydrogen –oxygen fuel cell is:
(1) 2H2(g)+O2(g)→2H2O(l) (2) 2H2(g)+4OH–(aq)→4H2O (l) +4e–(3) O2(g)+2H2O (l) +4e→4OH–(aq) (4) 4OH–(aq)+4e–→2H2O (l)
76. With respect to fuel cell prepared from H2 and O2 gases, the false statement is (1) It is free from pollution
(2) This is more efficient than conventional method of generating electricity
(3) The reaction occuring at anode is O2(g)+2H2O+4e–→4OH–
(4) These take little time to go into operation.
Level - II
Electrical Conductivity
1. Which of the following is not a semiconductor
(1) CuO
(2) Silicon
(3) Ge
(4) Mixed metal oxides at 150 K
2 Which of the following is not true ?
(1) Metallic conduction is due to the movement of electrons in the metal
(2) Electrolytic conduction is due to mobility of ions in solution
(3) The current carrying ions are not necessarily discharged at the electrodes
(4) The metallic conduction increases with increase in temperature, whereas electrolytic conductions decreases with increase of temperature
3. Amongst the following, the form of water with the lowest ionic conductance at 298 K is:
(1) sea water
(2) water from well
(3) saline water used for intravenous injection
(4) distilled water
4. Aqueous solution of which of the following compounds is the best conductor of electric current?
(1) Ammonia NH3
(2) Fructose C6H12O6
(3) Acetic acid, C2H4O2
(4) Hydrochloric acid, HCl
5. The correct trend of conductivity (in S m–1) at 298.15 K is represented in
(1) Ag > Cu > Au > 0.1M HCl
(2) Ag >Au > Cu > 0.1M HCl
(3) 0.1 M HC l> Ag> Au > Cu
(4) 0.1 M HCl> Ag > Cu > Au
6. The decrease in electrical conductivity of metals with increase in temperature is due to increase in
(1) the velocity of electrons
(2) the resistance of the metal (3) the number of electrons
(4) the number of metal atoms
Electrolytic Conductance
7. Which of the following solutions of KCl has higher specific conductance than others?
(1) 1N (2) 0.1 N
(3) 0.01 N (4) 0.001 N
8. Molar conductivity of a solution is 1.26×10 2 Ω –1 cm 2 mol –1 . Its molarity is 0.01M. Its specific conductivity will be
(1) 1.26×10–5 (2) 1.26×10–3
(3) 1.26×10–4 (4) 0.0063
9. The conductivity of 0.001 M acetic acid is 5 × 10–5 S cm–1 and Λ° is 390.5 S cm2 mol–1 then the calculated value of dissociation constant of acetic acid would be
(1) 81.78 × 10–4
(2) 81.78 × 10–5
(3) 18.78 × 10–6
(4) 18.78 × 10–7
10. The specific conductance of salt of 0.01M concentration is 1.06×10 –4 mho.cm –1 . Molar conductance of same solution in mho.cm2.mol–1 is
(1) 106.1 (2) 10.61 (3) 1.061 (4) 1.061×10–4
11. Molar conductance of an electrolyte increase with dilution according to the
equation: O mm Ac Λ=Λ−
Which of the following statements are true?
(A) This equation applies to both strong and weak electrolytes.
(B) Value of the constant A depends upon the nature of the solvent.
(C) Value of constant A is same for both BaCl2 and MgSO4
(D) Value of constant A is same for both BaCl2 and Mg(OH)2
Choose the most appropriate answer from the options given below:
(1) (A) and (B) only
(2) (A), (B) and (C) only (3) (B) and (C) only
(4) (B) and (D) only
12. The conductivity of a solution containing 2.08 g of anhydrous barium chloride in 200 mL solution is 6×10–3 Ohm.cm–1. The molar conductivity of the solution (in Ohm cm–1 mol–1)is x × 102. The value of x is (Atomic mass of Ba = 137, Cl = 35.5)
(1) 1.2 (2) 2.4 (3) 3.6 (4) 3.0
13. specific conductivity of 0.04 M “AB” is ______S m–1. (Λ in mho cm2 mol–1 and c in mole/litre)
(1) 100 (2) 1 (3) 10–2 (4) 10–1
14. Molar conductance of 0.2 M weak mono protic acid HA is 4.95 mho cm 2 mol −1 . Molar conductance of 0.002 M HA at the same temperature would be nearly (‘α‘ is too small compared to 1)
(1) 495 mho cm2 mol-1
(2) 4.95 mho cm2 mol-1
(3) 49.5 mho cm2 mol-1
(4) 0.0495 moh cm2 mol-1
Kohlrausch Law
15. m ° Λ for NaCl, HCl and NaA are 126.4, 425.9 and 100.5 S cm2 mol–1 respectively. If the conductivity of 0.001M HA is 5×10–5 S cm–1, degree of dissociation of HA is
(1) 0.25 (2) 0.50
(3) 0.75 (4) 0.125
16. At 25°C, the molar conductances at infinite dilution for the strong electrolytes NaOH, NaCl and BaCl 2 are 248 × 10 –4 , 126 × 10 –4 and 280 × 10 –4 S m 2 mol –1 respectively. m λ Ba(OH)2 in Sm2 mol–1 is.
(1) 52.4 × 10–4
(2) 524 × 10–4
(3) 402 × 10–4
(4) 262 × 10–4
17. The limiting molar conductivities (Λ°) for NaCl, KBr an KCl are 126, 152 and 150 S cm2 mol–1 respectively. Then Λ° for NaBr is
(1) 128 S cm2 mol–1
(2) 302 S cm2 mol–1
(3) 278 S cm2 mol–1
(4) 288 S cm2 mol–1
18. The equivalent conductances of two strong electrolytes at infinite dilution in H 2 (where ions move freely through a solution) at 25°C are given below:
3 2 91.0S cmeq Λ= CHCOONa HCl =426.2Scm/ eq Ë
What additional information/quantity one needs to calculate Λ°of an aqueous solution of acetic acid?
(1) Λ° of NaCl
(2) Λ° of CH3COOK
(3) The limiting equivalent conductance of () H Λ+
(4) Λ° of chloroacetic acid (ClCH2COOH)
19. Molar conductivity of BaCl 2 ,KCl and KOH is x,y,z mho cm2 mol–1 respectively. Equivalent conductance of Ba(OH)2 is
(1) x+2z–2y
(2) x + z – y
(3) x2z2y 2 +−
(4) 2(x+2z–2y)
20. Molar ionic conductance of Ca+2 is x Sm2 mol –1 and that of PO 4 –3 is y S m 2 mol –1 equivalent conductance of calcium phosphate is....S m2 eq–1
(1) x + y (2) (3x + 2y)
(3) 6(3x + 2y) (4) 3x2y 6 +
21. Kohlrausch’s law is applicable
(1) for concentrated solution (2) at infinite dilution
(3) for concentrated as well as dilute solutions
(4) none
22. The conductivity of saturated solution of AgCl is found to 1.86 × 10 –6 ohm –1 cm –1 and that of water is 6×10 –8 ohm –1 cm–1.If λ°AgCl is 138 ohm–1 cm2 eq–1 the solubility product of AgCl is
(1) 1.3 × 10–5 (2) 1.69 × 10–10
(3) 2 × 10–10 (4) 2.7 × 10–10
23. Molar ionic conductivities of a bivalent electrolyte are 57 and 73. The molar conductivity of the solution will be (1) 130 S cm2 mol–1
(2) 65 S cm2 mol–1
(3) 260 S cm2 mol–1
(4) 187 S cm2 mol–1
24. At 25°C, the ionic mobility of CH3COO–,H+ respectively 4.1×10–4, 3.63×10–3 cm/s. The conductivity of 0.001M CH3COOH is 5×10–5 S cm1.Dissociation constant of CH3COOH is cm.
(1) 1.53 × 10–5 (2) 3 × 10-4
(3) 3 × 10-5 (4) 3 × 10-6
25. At 298 K, molar conductance of 0.1 M weak monoacidic base, BOH is 39.6 mho. cm2.mol–1. Molar conductance of BOH at infinite dilution is 396 mho.cm 2.mol –1 . pH of centimolar solution of BOH is
(1) 11 (2) 10.699 (3) 12 (4) 9.301
Electrolysis
26. During the electrolysis of acidulated water, the mass of hydrogen obtained is ‘ x’ times that of O2 and the volume of H2 is ‘y’ times that of O2. The ratio of ‘y’ and ‘ x’ is a
(1) 16 (2) 8
(3) 0.125 (4) 0.25
27. One coulomb of charge passes through solutions of AgNO 3 and CuSO 4 . The ratio of the amounts of silver and copper deposited on platinum electrodes used for electrolysis is
(1) 108 : 63.5 (2) 54 : 31.75
(3) 108 : 31.75 (4) 215.8 : 31.75
28. How many coulombs of electricity are required for the oxidation of one mole of water to dioxygen?
(1) 9.65 × 104 C (2) 1.93 × 104 C
(3) 1.93 × 105 C (4) 19.3 × 105 C
29. What is the time (in sec) required for depositing all the silver present in 125 mL of 1M AgNO3 solution by passing a current of 241.25 amperes ? (1F = 96500 coulombs)
(1) 10 (2) 50
(3) 1000 (4) 100
30. A const ant current was flown for one minute through a solution of KI. At the end of experiment, liberated I2 consumed 150 mL of 0.01Msolution of Na 2 S 2 O 3 . Average rate of current flow is
(1) 9.65A (2) 4.825A
(3) 2.4125A (4) 1.206A
31. When 6 ×1022 electrons are used in the electrolysis of a metallic salt, 1.9 gm of the metal is deposited at the cathode. The atomic weight of that metal is 57. So, oxidation state of the metal in the salt is
(1) +2 (2) +3
(3) +1 (4) +4
32. The density of copper is 8 g/cc. Number of coulombs required to plate an area of 10 cm × 10 cm on both sides to a thickness of 10 –2 cm using CuSO 4 solution as electrolyte is
(1) 48,250 (2) 24,314 (3) 96,500 (4) 10,000
33. When 3.86 amperes current are passed through an electrolyte for 50 minutes, 2.4 g of a divalent metal is deposited. The gram atomic weight of the metal (in grams) is (1) 24 (2) 12 (3) 64 (4) 40
34. When 965 amp current is passed through aqueous solution of salt X using platinum electrodes for 10 sec, the volume of gases liberated at the respective electrodes is in 1:1 ratio. Then X is (1) MgSO4 (2) AgCl (3) MgCl2 (4) KNO3
35. When electricity is passed through molten AlCl3, 13.5 g of Al is deposited. The number of Faradays must be (1) 0.5 (2) 1.0 (3) 1.5 (4) 2.0
36. 965 amp current is passed through molten metal chloride for one minute and 40 seconds during electrolysis. The mass of metal deposited is 9 gm at the cathode. The valency of metal atom (at.wt = 27) is
(1) 4 (2) 3
(3) 2 (4) 1
37. On passing current through molten KCl, 19.5 g of K is deposited. The amount of Al deposited by the same quantity of electricity when passed through molten AlCl3 is
(1) 4.5 g (2) 9 g
(3) 13.5 g (4) 2.7 g
Electrochemical Cells
38. The cell reaction of the galvanic cell, ()()()() 22 sl |||| ++ aqaq CuCuHgHg is
(1) Hg+Cu2+→Hg2++Cu
(2) Hg+Cu2+→Hg2++Cu+
(3) Hg+Cu2+→Cu+Hg
(4) Cu+Hg2+→Cu2++Hg
39. The cell for which the cell reaction is H2+Cu2+→2H++Cu is represented as
(1) Cu||Cu2+||H+|H2(g)
(2) H2(g)||H+||Cu2+|Cu
(3) Pt,H2(1 atm)|H+||Cu2+(aq)|Cu
(4) Pt,H2|H2+(aq) (1atm)|Cu2+|Cu
40. The chemical reaction
2AgCl ( s ) +H2 ( g )→2HCl(aq)+2Ag(s) taking place in a galvanic cell is represented by the notation
(1) Pt(s)|H2(g),1 bar|1M KCl(aq)|AgCl(s)|Ag(s)
(2) Pt (s) |H 2(g) ,1 bar|1M KCl (aq) | |1MAg+(s)|Ag(s)
(3) Pt(s)|H2(g),1 bar|1M HCl(aq)|AgCl(s)|Ag(s)
(4) Pt (s) |H 2(g) ,1 bar|1M HCl (aq) | Ag+(s)|AgCl(s)
41. Which is correct for a spontaneous cell reaction?
(1) Zn+2Ag+→ Zn2++2Ag
(2) 2Ag+ Zn2+ → 2Ag++ Zn
(3) Cu + Zn2+→ Cu2++ Zn
(4) Fe + Zn2+→ Fe2++ Zn
42. Saturated solution of KNO 3 is used to construct salt bridge because
(1) velocity of K+ is greater than that of NO–3
(2) velocity of NO–3 is greater than that of K+
(3) velocities of both K+ and NO–3 are nearly the same
(4) KNO3 is sparingly soluble in water
43. The difference of potential of two electrodes in a galvanic cell is known as (1) EMF
(2) Potential difference
(3) Electrode difference
(4) Ionic difference
44. The potential of single electrode depends upon
A) the nature of the electrode
B) temperature
C) concentration of the ion with respect to which it is reversible
(1) A only (2) B only
(3) C only (4) A, B and C
45. Standard electrode potential for equation 1 is 2.8 V, determine the standard electrode potential for equation (2). () 2 22...1FeF +→ () 2 1 FeF..2 2 +→
(1) 2.8 V (2) 1.4 V
(3) – 2.8 V (4) – 1.4 V
46. E0 for Cl2(g)+2e–1→2Cl–1(aq) is +1.36 V; then E0 for Cl–1 (aq)→½ Cl2(g)+e–1 is:
(1) +1.36 V (2) – 1.36V
(3) – 0.68 V (4) +0.68V
47. The following statements is correct with respect to both electrolytic cell and Galvanic cell
(1) In both cells, the anode is a negative electrode.
(2) In both cells, the cathode is a negative electrode.
(3) in both cells, reduction reaction takes place at the cathode
(4) in both cells, oxidation reaction takes place at the cathode
48. The cell reaction of the galvanic cell, ()()()() 2+ 2+ saq aql ||| CuCuH | gHg is
(1) Hg+Cu2+→Hg2+ +Cu
(2) Hg+Cu2+→Hg++Cu+
(3) Hg+Cu2+→+Cu+Hg2+
(4) Cu+Hg2+→Cu2++Hg
49 Incorrect statement about Daniel cell among the following is
(1) When it acts like galvanic cell, copper electrode is the negative electrode
(2) When it acts like galvanic cell, zinc electrode is the anode
(3) When connected to external emf of lesser than 1.1 V, Zn(s) reduces () 2 Cuaq + ion
(4) When connected to external emf of greater than 1.1 V, it acts like electrolytic cell
Electrochemical Series
50. Zn gives H2 gas with H2SO4 and HCl but not with HNO3 because (1) Zn acts as oxidant when reacts with HNO3
(2) HNO3 is weaker acid than H2SO4 and HCl
(3) In electrochemical series Zinc is above hydrogen (4) NO –3 is reduced in preference to H3O+
51. Which reaction is not feasible?
(1) 2KI+Br2→2KBr+I2 (2) 2KBr+I2→2KI+Br2 (3) 2KBr+Cl2→2KCI+Br2 (4) 2H2O+2F2→4HF+O2
52. At 298 K the standard reduction potentials for the following half reactions are given as
Zn2+(aq)+2e–→Zn(s);–0.762 V
Cr3+(aq)+3e–→Cr(s);–0.740 V 2H+(aq)+2e–→H2(s);–0.00 V Fe3+(aq)+e–→Fe2+(aq);+0.770 V
The strongest reducing agent is (1) Zn(S) (2) H2(g) (3) Cr(s) (4) Fe2+(aq)
53. A gas X at 1 atm is bubbled through a solution containing a mixture of 1M Y– and 1M Z – at 25°C. if the reduction potential of Z > Y > X, then (1) Y will oxidize X but not Z (2) Y will oxidize both X and Z (3) Y will oxidize Z but not X (4) Y will reduce both X and Z
54. E° of Fe|Fe2+ is 0.44 V, E° of Cu|Cu2+ is –0.32 V. Then in the cell
(1) Cu oxidises Fe2+ ion
(2) Cu2+ oxidises iron
(3) Cu reduces Fe2+ ion
(4) Cu2+ ion reduces Fe
55. The standard potentials at 25°C for the half reactions are given against them below:
Zn2++2e–→Zn; E°=–0.762V
Mg2++2e–→Mg; E°=–2.37V
When Zn dust is added to a solution of MgCl2
(1) Magnesium is precipitated
(2) Zinc dissolves in the solution
(3) Zinc chloride is formed
(4) No reaction takes place
56. When
2 00 || E0.80V and E0.76V++ ==− AgAgZnZn , which of the following is correct?
(1) Ag+ can be reduced by H2
(2) Ag can oxidise H2 into H+ ion
(3) Zn+2 can be reduced by H2 (4) Ag can reduce Zn+2 ion
57. Given 32 4 00 /| E0.74V;E1.51V +−+ =−= CrCrMnOMn
23 27 2 00 O/ / E1.33V;E1.36V −+ == CrCrClCl
Based on the data given above, strongest oxidizing agent will be
(1) Cl– (2) Cr3+ (3) Mn2+ (4) MnO–4
58. Adding powdered lead and iron to a solution that is 1M in both Pb2+ and Fe2+ ions would results to a reaction in which
60. Given:
Co3++e→Co2+ ; E°=+1.81 V
Pb4+2e–→Pb2+ ; E°=+1.67 V
Ce4++e–→Ce3+ ; E°=+1.61 V
Bi3++3e–→Bi ; E°=+0.20 V
Oxidizing power of the species will increase in the order:
(1) Ce4+<Pb4+<Bi3+<Co3+
(2) Co3+<Pb4+<Co4+<Bi3+
(3) Bi3+<Ce4+<Pb4+<Co3+ (4) Co3+<Ce4+<Bi3+<Pb4+
Nernst Equation
61. For 2 27O14H6eCr −+−++→
30 2 27HO E1.33V.++=+ Cr
[Cr3+]=15 millimole, E = 1.067V. The pH of the solution is nearly equal to (1) 2 (2) 3 (3) 5 (4) 4
62. The reduction potential at pH=14 for the Cu2+/Cu couple is
2+ 2 0 -19 s ) p Cu ( /C O u CuH ] [givenE=0.34V and K1×1 = 0
] [givenE=0.34V and K1×1 = 0 (1) 0.34V (2) – 0.34V (3) 0.22V (4) – 0.22V
22 00 // E0.44V,E0.13V ++ = −= FeFePbPb
()
(1) more amount of Pb and Fe is formed
(2) more amount of Pb2+ and Fe2+ formed
(3) more amount of Pb and Fe 2+ is formed
(4) more amount of Pb 2+ and Fe is formed.
59. If a spoon of copper metal is placed in a solution of ferrous sulphate
(1) Cu will precipitate
(2) iron will precipitate
(3) Cu and Fe will precipitate
(4) no reaction will take place
63. A hydrogen electrode placed in a buffer solution of sodium cyanide and hydrogen cyanide in the ratio x:y and y:x has the electrode potential values “ a ” and “ b ” volts respectively at 25°C. If a – b = 35.52 mV. Then y:x is (1) 1 (2) 2 (3) 3 (4) 6
64. The EMF of the cell Ni/Ni 2+(0.01 M)// Cl–(0.01 M) Cl2 Pt is___V. If the SRP of nickel and chlorine electrodes are – 0.25 V and +1.36 V respectively (1) + 1.61 (2) – 1.61 (3) + 1.79 (4) – 1.79
65. The solution of nickel sulphate in which nickel rod is dipped is diluted 10 times.
The potential of nickel
(1) decreases by 60 mV
(2) increases by 30 V
(3) decreases by 30 m V
(4) decreases by 60 V
66. Ag + I– →AgI + e–; E° = 0.152 V
Ag → Ag+e–; E° = 0.800V
What is the approximate value of log Ksp for AgI? (2.303 RT/F = 0.059 V)
(1) – 37.83 (2) – 16.13
(3) – 8.12 (4) + 8.612
67. K sp of AgCl is 10–10 M2 Electrode potential of Ag/AgCl electrode in contact with 1M KCl at 25°C is ( 0 / E0.8V AgAg+=+ )
(1) 1.08 V (2) 0.592 V
(3) 0.8 V (4) 0.20 V
68. For a cell reaction involving a two electron change, the standard emf of the cell is found to be 0.295 V at 25°C. The equilibrium constant of the reaction at 25°C will be
(1) 1 ×10–10
(2) 29.5 ×10–2
(3) 10
(4) 1 ×1010
69. For the following cell Zn 2+ ||Cd 2+ |Cd E cell=0.30 V and E ° cell=0.36 V. Then the valueof [Cd2+]/[Zn2+] is (1) 10 (2) 1 (3) 0.1 (4) 0.01
70. The standard emf of a galvanic cell involving cell reaction with n = 2 is found to be 0.295 V at 25°C. The equilibrium constant of the reaction would be (F=96500 C mol–1; R=8.314 JK–1 mol–1) (1) 1.0 × 1010 (2) 2.0 × 1011 (3) 4.0 × 1012 (4) 1.0 × 102
71. The Gibbs energy for the decomposition of Al2O3 at 500° C is as follows 23 2 24 AlOAl+O 33 → ; D r G =+960 kJ. The voltage needed for the electrolytic reduction of aluminium oxide Al2O3 at 500° C at least.
(1) – 4.5 V (2) – 3.0 V (3) – 2.5 V (4) – 5.0 V
72. What is reaction quotient,Q, for the cell Ni(s) / Ni2+(0.190 M) //Cl–(0.40 M) /Cl2(g), Pt(s) (1atm)
(1) 3.1×10-1 (2) 1.3×10-1
(3) 8.0×10-2 (4) 3.0×10-2
73. During the discharge of lead storage battery, weight of PbSO 4 obtained at cathode involving of 1F of electricity is(At.Wt. of Pb = 208)
(1) 304 g (2) 152 g (3) 456 g (4) 608 g
Batteries
74. In a fuel cell, methanol is used as fuel and oxygen gas is used as an oxidiser. The reaction is
At 298K, standard Gibbs energies of formation for CH 3 OH (I) ,H 2 O (l) and CO 2(g) are –166.2,–237.2 and –394.4 kJ respectively. If standard enthalpy of combustion of methanol is –726 kJ, efficiency of the fuel cell will be (1) 80% (2) 87% (3) 90% (4) 97%
75. For lead storage battery pick the correct statements
A) During charging of battery, PbSO4 on anode is converted into PbO2
B) During charging of battery, PbSO4on cathode is converte d into PbO2
C) Lead storage battery, consists of grid of lead packed with PbO2 as anode D) Lead storage battery has ~38% solution of sulphuric acid as an electrolyte
Choose the correct answer from the options given below:
(1) B, D only (2) B, C, D only (3) A, B, D only (4) B, C only
76. Equivalent mass of H2SO4 in lead storage battery is (M = Molar mass) (1) 2M (2) M (3) M/2 (4) M/3
FURTHER EXPLORATION
1. A layer of chromium metal 0.25 mm thick is to be plated on a auto bumper with a total area of 0.32 M2 from a solution containing CrO4–2? What current flow is required for this electro plating if the bumper is to be plated in 60 sec the density of chromium metal is 7.20 g/cc.
(1) 4.9 × 103 A (2) 1.78 × 103 A
(3) 5.3 × 104 A (4) 10.69 × 104 A
2. In the electrochemical conversion (Kolbe’s eletrolysis) of R–COONa to R–R, 1 A current was passed for 965 seconds. Calculate the amount of R–R formed in this process (Faraday constant = 96,500 C mol –1) (1) 10 m mol (2) 5 m mol
MATCHING TYPE QUESTIONS
1. Match the terms given in Column I with the items given in Column II.
Column - I Column - II
(a) Λm(Molar conductance) (I) intensive property
(b) E°cell (II) depends on number of ions per unit volume
77. The anodic half – cell of lead battery is recharged using electricity 0.05 Faraday. The amount of PbSO 4 electrolysed in grams during the process is (molar mass of PbSO4=303 g mol–1)
(1) 22.8 (2) 15.2 (3) 7.6 (4) 11.4
78. A lead storage battery has been used for one month (30 days) at the rate of one hour per day by drawing a constant current of 2 amperes. H2SO4 consumed by the battery is (1) 1.12 mol (2) 2.24 mol (3) 3.36 mol (4) 4.48 mol
(3) 100 m mol (4) 50 m mol
3. In H2–O2 fuel cell the reaction occurring at cathode is
(1) 2H2 + O2 → 2H2O(l)
(2) H+ + OH– → H2O
(3) O2 + 2H2O + 4e– → 4OH–
(4) H+ + e– → 0.5 H2
4. The time period to coat a metal surface of 80 cm2 with 5 × 10–3 cm thick layer of silver (density =1.05 g cm–3) with the passage of 3A current through a silver nitrate solution is
(1) 125 seconds (2) 250 seconds
(3) 175 seconds (4) 200 seconds
(c) k (Conductivity) (III) extensive property
(d) Δ r G cell (IV) increases with dilution
Choose the correct answer from the options given below:
(A) (B) (C) (D)
(1) II I III IV
(2) I II IV III
(3) IV I II III
(4) II IV I III
2. Match the terms given in Column I(conductance) with the items given in Column II(units).
Column-I
Column–II
(A) Conductance (p) cm–1
(B) Specific conductance (q) ohm–1cm2mol–1
(C) Cell constant (r) ohm–1
(D) Molar conductance (s) ohm–1cm–1
Choose the correct answer from the options given below:
(A) (B) (C) (D)
(1) r s p q
(2) s r p q
(3) r s q p
(4) p s q r
3. Match material in column I with corresponding conductivity (S m –1) in column II
Column-I
Column-II
(A) Silver (I) 6.2×103
(b) Teflon (II) 1×10–18
(c) Germanium (III) 3.5×10–5
(d) Pure water (IV) 2
Choose the correct answer from the options given below:
(A) (B) (C) (D)
(1) I II IV III
(2) II I IV III
(3) IV II I III
(4) I III II IV
4. Match the terms given in Column I with the items given in Column II.
Column- I
Column - II
(A) Dilute solution of HCl (P) O2 evolved at anode
(B) Dilute solution of HCl (Q) H2 evolved at cathode
(C) Concentrate solution of NaCl (R) Cl2 evolved at anode
(D) Fairly concentrate solution of AgNO3 (S) Ag deposition at cathode
Choose the correct answer from the options given below:
(A) (B) (C) (D)
(1) P,Q P,Q R,Q P,S
(2) S P R Q
(3) P,R P S Q,S
(4) Q P R S
5. Match the terms given in Column I with the items given in Column II.
Column–I Column- II
(A) Electrolyte is a paste of KOH and ZnO (p) H2–O2 fuel cell
(B) Electrolyte is 38% H2SO4 solution (q) Mercury cell
(C) Water is the final product of cell reaction (r) Lead storage battery
(D) Cell in which metal changes from +3 state to +2 state in reaction at cathode (s) Nickel–Cadmium cell
Choose the correct answer from the options given below: (A) (B) (C) (D)
(1) r q p s
(2) q p r s
(3) q r p s
(4) q r s p
6. Match column I(batteries) with column II(change in oxidation state at cathode)
Column - I Column - II
(A) Fuel cell (I) +4 to +2
(B) Dry cell (II) +3 to +4
(C) Lead storage battery (III) 0 to - 2
(D) Ni-Cd cell (IV) +4 to +3 (V) +3 to +2
Choose the correct answer from the options given below:
(A) (B) (C) (D)
(1) III IV I V
(2) III I IV V
(3) III IV V I
(4) V IV I III
7. Match column I (electrolyte) with column II (product of electrolysis using inert electrodes)
Column- I
Column- II
(A) aq. CuCl2 (I) Ag and O2
(B) aq. Na2SO4 (II) H2 and Cl2
(C) aq. AgNO3 (III) H2 and O2
(D) aq. NaCl (IV) Cu and Cl2
Choose the correct answer from the options given below:
STATEMENT TYPE QUESTIONS
Each question has two statements. Statement I and statement II. Mark the correct answer as
(1) if both statement I and statement II are correct,
(2) if both statement I and statement II are incorrect,
(3) if statement I correct but statement II is incorrect,
(4) if statement I incorrect but statement II is correct.
(A) (B) (C) (D)
(1) III IV I II
(2) IV III II I
(3) III IV II I
(4) IV III I II
8. Match the terms given in Column I with the items given in Column II.
Column - I Column - II
(A) Electronic conductor (I) Aqueous urea solution
(B) Nonelectrolyte (II) Solid sodium
(C) Electrolytic dissociation (III) Electrolytic conductor
(D) Arrhenius (IV) Radioactivity increases
(V) Conductivity raises with temperature
Choose the correct answer from the options given below:
(A) (B) (C) (D)
(1) V I II III
(2) V II I IV
(3) II I V III
(4) II V I IV
1. S-I : In electrolysis the quantity of electricity needed for depositing 1 mole of silver (from AgNO3) is different from that required for 1 mole of copper (from CuSO4).
S-II : The atomic weights of silver and copper are different.
2. S-I : Conductivity always increases with decrease in the concentration of electrolyte
S-II : Molar conductivity always increases with decrease in the concentration of electrolyte.
3. S-I : The increase in molar conductance with dilution is more in HCl than in HCOOH
S-II : On dilution the degree of dissociation of strong electrolyte increases while that of weak electrolyte decreases
4. S-I : aq.CuSO4 on electrolysis using Pt electrodes gives Cu at cathode and O2 at anode
S-II : aq.CuSO4 on electrolysis using Cu
ASSERTION AND REASON QUESTIONS
In each of the following questions, a statement of Assertion (A) is given, followed by a corresponding statement of Reason (R). Mark the correct answer as
(1) if both (A) and (R) are true and (R) is the correct explanation of (A),
(2) if both (A) and (R) are true but (R) is not the correct explanation of (A),
(3) if (A) is true but (R) is false,
(4) if both (A) and (R) are false.
1. (A) : molar conductivity of weak electrolyte at infinite dilution cannot be determined experimentally
(R) : Kohlrausch law helps to find the molar conductivity of a weak electrolyte at infinite dilution
2. (A) : If an aqueous solution of NaCl is electrolysed, the product obtained at the cathode is H2 gas and not Na.
(R) : Gases are liberated faster than the metals during the electrolysis of an electrolyte
electrodes gives Cu at cathode and Cu+2 ions at anode
5. S-I : Copper cannot reduce dilute HNO3
S-II : Copper cannot reduce dilute HCl.
6. S-I : ° cellE should have a positive value for galvanic cell to function.
S-II : cathodeanodeEE < for a feasible reaction, with reference to SRP
7. S-I : E lectrolysis of Brine solution using inert electrodes liberates H 2 at cathode
S-II : SRP of value of NHE is greater than SRP value of Na+/Na
3. (A) : The cell potential of mercury cell remains constant during its life.
(R) : The overall reaction does not involve any ion in solution whose concentration can change during its life time.
4. (A) : In a galvanic cell, the cathode is a positive electrode.
(R) : The tendency of metal ions from solution to deposit on a metal electrode makes the cathode positively charged.
5. (A) : In equation D r G=–nFEcell value of D r G Depends on n
(R) : Ecell is intensive property and D r G is an extensive property.
6. (A) : Galvanised iron does not rust
(R) : Zinc has a more negative electrode potential than iron
7. (A) : Copper is dissolved at anode and deposited at cathode when Cu
electrodes are used and electrolyte is 1 M CuSO4(aq) solution.
(R) : SOP of Cu is less than SOP of water and SRP of Cu is greater than SRP of water.
8. (A) : Molar conductivity of CH3COOH increases with dilution
(R) : The graph between λmv/s (C)1/2 is a straight line
9. (A) : Molar conductivity of 0.1 NH4OH solution is less than that of 0.001 M NH4OH solution
(R) : Dilution increases the degree of
BRAIN TEASERS
1. The time period to coat a metal surface of 80 cm2 with 5×10–3 cm thick layer of silver (density 1.05 g cm –3) with the passage of 3 A current through a silver nitrate solution is
(1) 115 seconds (2) 125 seconds
(3) 135 seconds (4) 145 seconds
2. A factory produces 40 kg of calcium in two hours by electrolysis. How much aluminum can be produced by same current in 2 hours if current efficiency is 50%?
(1) 22 kg (2) 18 kg
(3) 9 kg (4) 27 kg
3. Consider the galvanic cell,Pt(s)|H 2 (1 bar)|HCl (aq) (1M)|Cl 2(g) (1 bar)|Pt(s) After running cell for some time, the concentration of the electrlolyte is automatically raised to 3 M HCl. Molar conductivity of the 3 M HCl is about 240 S cm2mol–1 and limiting molar conductivity of HCl is about 420 S cm2mol–1. If Kb of water is 0.52 K kg/mol, what is the boiling point of the electrolyte at the end of the experiment?
(1) 375.6 K (2) 377.8 K
ionisation of NH4OH
10. (A) : A blue colour is obtained when a copper wire is immersed in AgNO3 solution
(R) : Silver reduces Cu2+ to copper
11. (A) : The molar conductance of weak electrolytes is low as compared to that of strong electrolytes at moderate concentrations
(R) : Weak electrolytes at moderate concentrations dissociate to a much greater extent when compared to strong electrolytes
K
4. Conductivity of a saturated solution of a sparingly soluble salt AB at 298 K is 1.85×10–6 S–1. Solubility product of the salt AB at 298 K is [given λ°M(AB)=1.4×10–4
S–2 mol–1]
(1) 5.7×10–12 (2) 1.32×10–12 (3) 7.5×10–12 (4) 1.74×10–12
5. In a Cu voltameter, mass deposited in 30 seconds is 100 g. Carefully analyse the current-time graph shown below and identify the incorrect statement.
(1) Electrochemical equivalent for Cu is 50
(2) A constant current of 66.66 mA would also discharge the same amount in the same time
(3) 33.33 g got discharged in 10 seconds
(4) 50 g got discharged in 15 seconds
6. Cell notation of a galvanic cell is represented as Ni(s)|Ni +2 ( M 1 ) || Cl –
( M 2)|Cl 2(1atm),Pt(s).Emf of the cell is maximum when the value of M1 and M2 are
(1) M1 = M2= 0.01 M
(2) M1 = 1M; M2 = 1M
(3) M1 = 0.1 M; M2 = 0.01 M
(4) M1 = 0.01 M; M2 = 0.01 M
7. The potential of the cell containing two hydrogen electrodes as represented below Pt,H2(g)|H+(10–6 M)||H+(10–4 M)|H2(g)Pt at 298 K is
(1) –0.118 V (2) –0.0591 V
FLASH BACK (Previous NEET Questions)
1. Find the emf of the cell in which the following reaction takes place at 298 K
Ni(s)+2Ag+(0.001M)→Ni2+(0.001M)+2 Ag(s)
(Given
0 cell 2.303RT
E=10.5V,=0.059 at 298K F )
(1) 1.385 V (2) 0.9615 V
(3) 1.05 V (4) 1.0385 V
2. Given below are half cell reactions: 2 42 854 MnOHeMnHO −+−+ ++→+
4 2 2 / 1.510 + =− MnMnO EV 22 1 22 2 OHeHO +− ++→ 22 / 1.223 OHO EV =+ Will the permanganate ion, 4MnO liberate O 2 from water in the presence of an acid?
(1) No, because 0 0.287 cell EV =−
(2) Yes, because 0 cell E2.733V =−
(3) No, because 0 cell E2.733V =+
(3) 0.118 V (4) 0.0591 V
8. Iron powder is added to 1.0 M solution of CdCl2 at 298 K. The reaction occurring is Cd 2+ (aq) +Fe (s) → Cd (s) +Fe 2+ (aq) . If the standard potential of a cell producing this reaction is 0.037 V, the concentrations of Cd2+ and Fe2+ ions in the above reaction at equilibrium respectively will be (Hint : 101.3=20)
(1) 0.05 M, 0.95 M
(2) 0.95 M, 0.05 M
(3) 0.40 M, 0.60 M
(4) 0.60 M, 0.40 M
(4) Yes, because 0 cell E0.287V =+
3. At 298K, the standard electrode potentials of Cu2+/Cu,Zn2+/Zn,Fe2+/Fe and Ag+/Ag are 0.34 V,–0.76V,–0.44V and 0.80V respectively
On the basis of standard electrode potential predict Which of the following reaction can not occur?
(1) CuSO4(aq)Fe (s) →FeSO4(aq)+Cu (s)
(2) FeSO4(aq)Zn (s) →ZnSO4(aq)+Fe (s)
(3) 2CuSO4 (aq) +2Ag (s) →2Cu (s) +Ag2SO4 (aq)
(4) CuSO4 (aq) Zn (s) →ZnSO4 (aq) +Cu(s)
4. The molar conductance of NaCl, HCl and CH 3 COONa at infinite dilution are 126.45, 426.16 and 91.0 S cm 2 mol–1 respectively. The molar conductance of CH3COOH at infinite dilution is. Choose the right option for your answer
(1) 540.48 S cm2 mol-1
(2) 201.28 S cm2 mol-1
(3) 390.71 S cm2 mol-1
(4) 698.28 S cm2 mol-1
5. The molar conductivity of 0.007 M acetic acid is 20 S cm 2 mol -1 . What is
the dissociation constant of acetic acid? Choose the correct option.
(1) 2.50 x 10-5 mol L-1
(2) 1.75 x 10-4 mol L-1
(3) 2.50 x 10-4 mol L-1
(4) 1.75 x 10-5 mol L-1
6. On electrolysis of dil. sulphuric acid using Platinum (Pt) electrode, the product obtained at anode will be :
(1) Oxygen gas
(2) H2S gas
(3) SO2 gas
(4) Hydrogen gas
7. The number of Faradays (F) required to produce 20 g of calcium from molten CaCl2 (Atomic mass of Ca =40 g mol–1) is
(1) 2 (2) 3
(3) 4 (4) 1
8. For a cell involving one electron, E°cell=0.59V at 298 K, the equilibrium constant for the cell reaction is [Given that 2.303 0.059 RT F = V at T=298 K]
CHAPTER TEST
1. Correct statement(s) is/are:
A. Electrical conductance through metals is called metallic or electronic conductance.
B. The electronic conductance do not depend on temperature.
C. The conductance of electricity by ions present in the solutions is called electrolytic or ionic conductance.
D. The conductivity of electrolytic (ionic) solutions depends on the nature of the electrolyte added, the nature of the solvent and its viscosity, temperature etc.
(1) 1.0 ×1030
(2) 1.0 ×105
(3) 1.0 ×102
(4) 1.0 ×1010
9. In the electrochemical cell :
Zn|ZnSO 4 (0.1 M) ||CuSO 4 (1.0 M)|Cu the emf of this Daniel cell is E 1. When the concentration of ZnSO4 is changed to 10 M and that of CuSO 4 changed to 0.01 M,the emf changes to E2.From the following, which one is the relationship between E1 and E2 (Given, RT/F=0.059)
(1) E1<E2
(2) E1>E2
(3) E2=0≠E1
(4) E1 =E2
10. The pressure of H2 required to make the potential of H 2 electrode zero in pure water at 298 K
(1) 10–10 atm
(2) 10–4 atm
(3) 10–14 atm
(4) 10–12 atm
(1) A and B
(2) A, C and D
(3) Only A and D
(4) Only B
2. Given below are two statements. one is labelled as Assertion A and the other is labelled as Reason R.
(A) : Mercury cell does not give steady potential.
(R) : In the cell reaction, ions are not involved in solution
In the lig ht of the above statements, choose the most appropriate answer from the options given below.
(1) if both (A) and (R) are true and (R) is the correct explanation of (A),
(2) if both (A) and (R) are true but (R) is not the correct explanation of (A),
(3) if (A) is true but (R) is false,
(4) if both (A) and (R) are false.
3. Which of the following pair(s) is/are incorrectly matched?
(i) R(resistance) – Ohm
(ii) ρ (resistivity) – Ohm metre
(iii) C (conductance) - seimen or mho
(iv) k (conductivity) - Ohm metre-1
(1) i, ii and iii (2) ii and iii
(3) i, ii and iv (4) iv only
4. Given below are two statements.
S–I : Conductivity of weak electrolytes decreases with dilution whereas molar conductance increases.
S–II : On dilution, number of ions per milliliter decreases, but total number of ions increases considerably.
In the light of the above statements, choose the most appropriate answer from the options given below.
(1) if both statement I and statement II are correct,
(2) if both statement I and statement II are incorrect,
(3) if statement I correct but statement II is incorrect,
(4) if statement I incorrect but statement II is correct.
5. The correct match is List I List II
(A) Faraday’s first law (I) e × 96500
(b) Chemical equivalent (II) 12 12 mm EE =
(c) Electrolytic dissociation (III) S.H.E
(d) Pt,H2(atm)/ H+(1M) (IV) m=eQ
(V) Salt bridge
Choose the correct answer from the options given below: (A) (B) (C) (D)
(1) IV I III II
(2) IV V II III
(3) I I II III
(4) IV I II III
6. If the resistance and specific resistance of a solution have same magnitude, then its cell constant is equal to
(1) 0 (2) 10
(3) 100 (4) 1
7. Calculate the EMF of the cell, Na/Na+=(1 M)/Fe2+(1M)/Fe
Given, / + o NaNa E =–2.71V and E° Fe+2/Fe =–
0.44V
(1) +1.98 V
(2) –2.47 V
(3) –3.15 V
(4) +2.27 V
8. The pH of an acid solution in contact with hydrogen electrode is one. Reduction potential of Hydrogen electrode is
(1) –3.9 mV
(2) –118 mV
(3) –59 mV
(4) –11.8 mV
9. Which among the following is incorrect statement?
(1) When Lead storage battery is discharged ,H2SO4 is consumed
(2) When Lead storage battery is discharged H2SO4 is formed
(3) Zinc is used to protect corrosion of iron
(4) Efficiency of fuel cell is about 70%
10. The efficiency of a cell is 60%. Its cell reaction is A (s)+B 2+ (aq)→A 2+ (aq)+B; D H = –240 kJ The standard electrode potential of cell is
(1) 0.746 V (2) 0.37 V
(3) 1.48 V (4) 2.96 V
11. A dilute aqueous solution of CuSO 4 is electrolysed using platinum electrodes. The products at the anode and cathode are:
(1) O2, H2 (2) H2, O2
(3) O2, Cu (4) S2O82-, H2
12. Given,
Which of the following reactions under standard condition will not take place in the specified direction?
(1) Ni2+(aq)+Cu (s) →Ni (s) +Cu2+ (aq)
(2) Cu(s)+2Ag+ (aq) → Cu2+ (aq) +2Ag(s)
(3) Cu+2 (aq) +H2 (g) → Cu (s) +2H+(aq)
(4) Zn (s) +2H+ (aq) → Zn2+ (aq) +H2(g)
13. Molar conductance of 0.1 M weak mono basic acid(HA) is 36 mho cm 2 mol–1. At
the same temperature, molar conductance of HA at infinite dilution is 360 mho cm2 mol–1 pH of HA will be
(1) 2 (2) 3
(3) 2.7 (4) 3.3
14. For a cell involving two electron transfer,E°=0.59V at 298 K. Equilibrium constant for the cell reaction will be
(1) 105 (2) 1010
(3) 1020 (4) 108
15. Resistance of 0.2 M solution of an electrolyte is 50Ω. The specific conductance of the solution is 1.4 Sm–1 . The resistance of 0.5 M solution of the same electrolyte is 280 Ω The molar conductivity of 0.5 M solution of the electrolyte in Sm2 mol–1 is
(1) 5 ×10–4
(2) 5 ×10–3
(3) 5 ×103
(4) 5 ×102
16. Given below are two statements. one is labelled as Assertion A and the other is labelled as Reason R.
(A) : Cu is less reactive than hydrogen (R) : 2 0 / CuCu E + is negative.
In light of the above statements, choose the most appropriate answer from the options given below.
(1) Both A and R are true and R is the correct explanation of A
(2) Both A and R are true and R is not correct explanation of A
(3) A is true but R is false
(4) A is false but R is true
17. Match the laws given in Column I with expressions given in Column II
Column I
Column II
(A) Lead storage battery (p) During discharging density of H2SO4 decreases
(b) Mercury cell (q) Reaction at cathode 4e–+O2+2H2O → 4OH–
(c) Fuel cell (r) Reaction at cathode MnO2+NH4++e–→MnO(OH)+NH3
(d) Leclanche cell (s) Suitable for low current devices like hearing aids
(A) (B) (C) (D)
(1) s p q r
(2) p r q s
(3) q r p s
(4) p s q r
18. The weight of silver (at. wt=108) displaced by a quantity of electricity which displaces 5600 mL of O2 at STP will be
(1) 5.4 g (2) 10.8 g
(3) 54.0 g (4) 108.0 g
19. A graph is plotted between Λ c (y-axis) versus C (x-axis) for KCl solution. The nature of the graph will be ( C = concentration of KCl)
(1) A straight line with positive slope and negative y–intercept
(2) A straight line with negative slope and positive y–intercept
(3) A straight line passing through origin
(4) A straight line with positive slope and positive y–intercept
20. Given below are two statements.
S–I : Increase in molar conductance with dilution is more in HBr than in HCOOH.
S–II : HBr (aq) is a stronger electrolyte whereas HCOOH (aq) is a weaker electrolyte.
In the light of the above statements, choose the most appropriate answer from the options given below.
(1) Statements I and II are correct
(2) Statement I is correct and statement II is incorrect
(3) Statement I is incorrect and statement II is correct
(4) Statements I and II are incorrect
21. Given below are two statements.
S–I : In a galvanic cell cathode has a positive potential with respect to solution.
S–II : Potential of a individual half cell cannot be measured.
In the light of the above statements, choose the most appropriate answer from the options given below.
(1) Both statement I and statement II are correct
(2) Both statement I and statement II are incorrect
(3) Statement I is correct and statement II is incorrect
(4) Statement I is incorrect and statement II is correct
22. Given below are two statements. one is labelled as Assertion A and the other is labelled as Reason R.
(A) : The molar conductance of weak electrolytes is less when compared
to that of strong electrolytes at moderate concentration
(R) : The degree of dissociation of weak electrolytes is less than that of strong electrolytes at moderate concentration
In the light of the above statements, choose the most appropriate answer from the options given below.
(1) Both ‘A’ & ‘R’ are correct and ‘R’ is the correct explanation of ‘A’
(2) Both ‘A’ & ‘R’ are correct and ‘R’ is not the correct explanation of ‘A’
(3) ‘A’ is true & ‘R’ is false
(4) Both ‘A’ & ‘R’ are false
23. E 1 , E 2 and E 3 are the emf values of the three galvanic cells respectively
(i) Zn| Zn2+(1M)||Cu2+(0.1M)|Cu
(ii) Zn| Zn2+(1M)||Cu2+(M)|Cu
(iii) Zn| Zn2+(0.1M)||Cu2+(1M)|Cu
Which one of the following is true?
(1) E2>E 3 >E1
(2) E 3 >E2>E1
(3) E1>E2>E 3
(4) E1>E 3 >E2
24. The S.R.Ps of Cu2+/Cu,Hg2+/Hg and Zn2+/ Zn are respectively 0.34 V, 0.85 V and -0.76 V. The wrong statement is
(1) Cu reduces Hg2+
(2) Zn reduces Cu2+
(3) Hg reduces Zn2+
(4) Zn reduces both Cu2+ and Hg2+
25. Which of the following is a good conductor of electricity?
(1) Diamond
(2) Graphite
(3) Solid NaCl
(4) Wood
26. Given below are two statements. one is labelled as Assertion A and the other is labelled as Reason R.
(A) : Λ m = kV , on dilution k decreases but Λ m increases.
(R) : On dilution, decrease in k is more than compensated by increase in its V
In the light of the above statements, choose the most appropriate answer from the options given below .
(1) (A) is true but (R) is false
(2) Both (A) and (R) are true but (R) is not the correct explanation of (A)
(3) Both (A) and (R) are true and (R) is the correct explanation of (A)
(4) Both (A) and (R) are false
27. The products of electrolysis of aqueous NaCl solution using inert electrodes are
(1) Na at cathode and Cl2 at anode
(2) H2 at cathode and Cl2 at anode
(3) H2 at cathode and O2 at anode
(4) Na at cathode and O2 at anode
28. Match the items of Column I and Column II
Column I
Column II
(A) k (specific conductance) (I) current × time
(b) Λ m (molar conductance) (II) Λ c / Λ o
(c) α(degree of dissociation) (III) 1000 M × k
(d) Q (quantity of electricity) (IV) G R ∗
(V) Salt bridge
Choose the correct answer from the options given below:
(A) (B) (C) (D)
(1) IV III I IV
(2) I II III IV
(3) IV III II I
(4) III I II IV
29. Cost of Electricity to liberate 24 g of Oxygen is Rs 3000. Cost of Electricity to deposit 24 g of magnesium will be
(1) Rs 2000
(2) Rs 1500
(3) Rs 4300
(4) Rs 2500
30. S.R.P values of some electrodes are given below.
2 0 / E2.37V + =− MgMg
E1.36V; =+ ClCl
2 1 / 2
0 / E0.8V; +=+AgAg
2 0 / E0.76V + =− ZnZn
From the above data, identify the best reducing agent
(1) Zn (2) Mg
(3) Cl2 (4) Ag
31. Molar conductance at infinite dilution of Ammonium chloride, Sodium hydroxide and Sodium chloride are X , Y and Z respectively. Molar conductance at infinite dilution of Ammonium hydroxide at the same temperature would be (all measurements expressed in SI units)
(1) X–Y + Z
(2) X+ Y–Z
(3) Y + Z –X
(4) X + Z –Y
32. Brine solution on electrolysis using inert electrodes gives ----- and ----- at anode and cathode respectively.
(1) Cl2 , H2 (2) Cl2,Na (3) O2, H2 (4) O2 ,Na
33. The following statements is correct with respect to both electrolytic cell and Galvanic cell
(1) in both cells, anode is shown by +ve sign
(2) in both cells, cathode is shown by –ve sign
(3) in both cells, reduction reaction takes place at the cathode
(4) in both cells, oxidation reaction takes place at the cathode
34. Aqueous CuSO4 is subjected to electrolysis using copper electrodes by passing 0.2 F of electricity. Correct statement about the process is
(1) 0.2 mol of oxygen is liberated at anode
(2) 2.24 litres of hydrogen (at STP) is liberated at cathode
(3) 0.2 moles of copper decomposes at anode
(4) 0.1 mol of copper is deposited at cathode
35. 2 0 -12-1 224 ohmcmg eq Λ= ClCHCOONa
Λ= HCl
0 -12-1 203 ohmcmgeq
0 -12-1 203 ohmcmgeq Λ= HCl
What is the value of 2 0 ClCHCOOH Λ ?
(1) 288.5 Ohm–1 cm2 g eq–1
(2) 289.5 Ohm–1 cm2 g eq–1
(3) 388.8 Ohm–1 cm2 g eq–1
(4) 59.5 Ohm–1 cm2 g eq–1
36. During the electrolysis of acidulated water, the mass of hydrogen obtained is ‘ x’ times that of O2 and the volume of is ‘ y’ times that of O2. The ratio of ‘y’ and ‘ x’ is
(1) 16 (2) 8
(3) 0.125 (4) 0.25
37. S.R.P values of some electrodes are given below.
2 2 1 / / 2 2.37 V;1.36 V ; + =−=+oo MgMgClCl EE
2 // 0.8 V ;0.76 V ; ++=+=oo AgAgZnZn EE
From the above data, identify the best reducing agent
(1) Zn (2) Mg
(3) Cl2 (4) Ag
38. For the given cell representation ()()()() 23 |||| ++ saqaqs MgMgAlAl total number of electrons transferred is (1) 2 (2) 3 (3) 6 (4) 5
39. For a particular galvanic cell, the cell notation is Cu | Cu2+||Ag+|Ag. The metal which dissolves is (1) Cu (2) Ag (3) Both Cu and Ag (4) Neither Cu nor Ag
40. Reason for increase in electrical conduction of electrolyte with increase in temperature is A) increase in the number of ions B) increase in the speed of ions C) increase in the degree of dissociation of electrolyte
(1) A and B only (2) B and C only (3) A and C only (4) A, B and C
41. When an electric current is passed through acidified water, 112mL of hydrogen gas at N.T.P was collected at
the cathode in 965 seconds. The current passed in ampere is (1) 1.0 (2) 0.5 (3) 0.1 (4) 2.0
42. Given 32 00//0.72,0.42 ++ =−= CrCrFeFe EVEV
The potential for the cell ()() 32 /0.1||0.01| CrCrMFeMFe ++ is (1) 0.339 V (2) – 0.339 V (3) – 0.26 V (4) 0.26 V
43. How long (approximate) should water be electrolyzed by passing through 100 amperes current so that the oxygen released can completely burn 27.66 g of diborane (Atomic weight of B = 10.8 u)
(1) 6.4 hours
(2) 0.8 hours
(3) 3.2 hours
(4) 1.6 hours
44. Eight grams of aq.CuSO4 is subjected to electrolysis by passing 1.5×1022 electrons using inert electrodes. Weight of cupric ions left over after electrolysis is (Atomic wt. of copper = 64)
(1) 6.4 g (2) 7.2 g
(3) 1.6 g (4) 2.4 g
45. Electrolysis using inert electrodes neither changes the composition of the electrolyte nor changes the pH of the solution in the case of (1) aq NaCl
(2) aq CuSO4
(3) aq K2SO4
(4) aq H2SO4
ANSWER KEY
Neet Drill Level - I
(51) 2 (52) 1 (53)
Further Exploration (1) 4 (2) 2 (3) 3 (4) 1
Matching Type Questions
Statement Type Questions (1) 1 (2) 2 (3) 2 (4) 1 (5) 4 (6) 3 (7) 1
Assertion and Reason Questions (1) 2 (2) 3 (3) 1 (4) 1 (5) 2 (6) 2 (7) 3 (8) 3 (9) 1 (10) 3 (11) 3
Brain Teasers
Flashback
Chapter Test