Edvantage Science AP Chemistry 2 WorkbookPLUS Chapter 7.7

7.5

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7.5 The Electrolytic Cell Warm Up Complete the following table showing pairs of devices that perform opposite energy transformations. Each pair of devices transforms a different form of energy to and from electrical energy.

Energy Conversion

Device

light bulb

speaker

From

To

electrical energy

light energy

light energy

electrical energy

electrical energy electrical energy

heating element

electrical energy electrical energy

motor

electrical energy electrical energy

electrolytic cell

electrical energy

chemical energy electrical energy

The Structure and Function of the Electrolytic Cell

Current electricity is a flow of electrical charge. Although electrical charge is always carried by a particle, that particle is not always an electron. In solutions, ions are the particles responsible for carrying the charge. A chemical is an electrolyte if its aqueous solution conducts electricity. Whenever electricity is conducted through a molten electrolyte or an electrolyte solution, the process is called electrolysis and the apparatus is called an electrolytic cell.

For an electric current to exist, there must be a continuous path from the source’s anode to its cathode. This continuous path is called a circuit. It would be simple if there were a chemical species in the electrolytic cell solution that picked up the electrons from one electrode and ferried them over to the other electrode but this is not the case. Instead, an oxidizing agent picks up the electrons from one electrode while an entirely different chemical species, a reducing agent, donates electrons at the other electrode. The circuit is complete. It doesn’t matter that the electrons being removed from one electrode are not the same ones being released at the other electrode. “If you’ve seen one electron you’ve seen them all.” It’s like a con game where the electrons are being switched and the source never suspects.

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In an electrolytic cell, oxidation occurs at the anode and reduction occurs at the cathode just like in an electrochemical cell. Figure 7.5.1 shows an electrolytic cell. cathode

anode e–

cathode (graphite) Ni2+ + 2e–

Ni

– +

I– Ni2+

e–

anode (graphite) 2 I–

I2 + 2e–

Figure 7.5.1 An electrolytic cell

The schematic symbol at the top of Figure 7.5.1 represents the electrochemical cell that serves as the source of direct current (DC) driving the electrolytic cell. Note that the electrochemical cell’s anode is connected to the electrolytic cell’s cathode. This makes sense if you think it through.

Oxidation (electron loss) occurs at the anode; therefore, this is where the voltaic cell emits electrons.

In the electrolytic cell, those electrons are accepted by a chemical species, which is thereby reduced.

The site of reduction is the cathode. Likewise, the electrochemical cell’s cathode is connected to the electrolytic cell’s anode.

An electrolytic cell may superficially resemble an electrochemical cell but it performs the reverse energy transformation. In other words, an electrochemical cell generates electricity whereas an electrolytic cell uses electricity.

Electrolytic cells transform electrical energy into chemical energy.

The non-spontaneous reactions of electrolytic cells “read” in the reverse direction in the SRP table (Table A7) as the spontaneous reactions of electrochemical cells. In electrolytic cells, the oxidation half-reaction is above the reduction half-reaction. Try the following trick to remember which direction electrons are passed in the SRP table. Electrons are very “light” so they spontaneously float upward in the table but they must be pushed downward.

oxidation ½ reaction

I2 + 2e− ← H2SO3 + 4 H+ + 2e−  Cu2+ + 2e−  SO4 2− + 4 H+ + 2e− 

2 H+ + 2e−  Sn2+ + 2e−  Ni2+ + 2e− →

2 I− . . . . . . . . . . . . . . . . +0.54 V S + 3 H2O . . . . . . . . . . . +0.45 V Cu. . . . . . . . . . . . . . . . . . +0.34 V H2SO3 + H2O. . . . . . . +0.17 V H2 . . . . . . . . . . . . . . . . . . . 0.00 V Sn. . . . . . . . . . . . . . . . . . –0.14 V Ni. . . . . . . . . . . . . . . . . . . –0.26 V

2 I− → I2 + 2e− Ni2+ + 2e− → Ni Ni2+ + 2 I− → I2 + Ni

reduction ½ reaction

The net E° for the non-spontaneous reactions that occur in electrolytic cells is negative. The net E° gives us an estimation of how much voltage will be required to drive this electrolytic cell. In other words, the SRP table not only allows you to determine how much voltage an electrochemical cell will generate but also allows you to determine how much voltage an electrolytic cell will require to

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operate. This value will normally be less than the voltage actually required. The difference between the voltages the SRP table predicts will be necessary to drive electrolytic cells and the voltages actually required to drive those cells is referred to as overpotential.

The standard reduction potentials in the SRP table are derived solely from thermodynamic data. They represent the difference between the chemical potential energy of the reactants and the chemical potential energy of the products. However, the voltages required to drive electrolytic cells are also influenced by kinetic factors such as the reaction’s activation energy and the reactants’ localized concentrations at the electrodes. A cell’s overpotential thus depends on the specific reactions involved and the cell’s design.

Sample Problem 7.5.1 — Predicting the Voltage Required to Operate an Electrolytic Cell Predict the voltage required to operate the electrolytic cell in Figure 7.5.1.

What to Think About

How to Do It

1. Write each half-reaction and its standard potential. The oxidation potential of I− is – 0.54 V because it is the reverse of the reduction potential of I2 provided in the SRP table.

2I− → I2 + 2e− E° = – 0.54 V Ni2+ + 2e− → Ni E° = – 0.26 V Ni2+ + 2I− → I2 + Ni E° = – 0.80 V

2. Add the reduction potential to the oxidation potential.

A voltage of at least 0.80 V would be required to operate this cell.

Practice Problems 7.5.1 — Predicting the Voltage Required to Operate an Electrolytic Cell Predict the voltage required to operate each of the following electrolytic cells given the pair of half-reactions occurring within each cell.

1. Ag → Ag+ + e− and Ni2+ + 2e− → Ni

2. 2 F− → F2 + 2e− and Cu2+ + 2e− → Cu

3. Sn → Sn2+ + 2e− and Al3+ + 3e− → Al

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Comparing

Electrochemical and Electrolytic Cells Electrochemical Cell Oxidizing Agents

Reducing Agents

strong

weak

A+ + e–

+

A

–

Z +e weak

In electrochemical cells, the strongest available reducing agent (lowest on the right in the SRP table) passes electrons up to the strongest available oxidizing agent (highest on the left in the SRP table).

Thus, the half-reactions that occur in electrochemical cells are the ones that are farthest apart in the SRP table and generate the greatest possible voltage.

In electrolytic cells, the strongest available reducing agent (lowest on the right in the SRP table) passes electrons down to the Electrolytic Cell strongest available oxidizing agent (highest on the left in the SRP table). Thus, the half-reactions that occur in electrolytic cells are Oxidizing Reducing Agents Agents the ones closest together in the SRP table and require the least voltage to drive them (Figure 7.5.2). Table 7.5.1 summarizes the strong weak differences between an electrochemical cell and an electrolytic cell.

G+ + e–

G

M+ + e–

M

Z strong

weak

strong

Figure 7.5.2 Using the SRP table for electrochemical and electrolytic cells

Table 7.5.1 Contrasting the Electrochemical Cell and the Electrolytic Cell

Electrochemical Cell

• • • • • • • •

exothermic “makes” electricity transforms chemical energy into electrical energy is a DC voltage source 2 half-cells spontaneous redox reaction E° is positive salt bridge or equivalent V

Electrolytic Cell

• • • • • • • •

endothermic “takes” electricity transforms electrical energy into chemical energy requires a DC voltage source 1 cell non-spontaneous redox reaction E° is negative no salt bridge DC

The oxidation half-reaction is below the reduction half-reaction in the SRP table.

Think of the electrons floating upward.

The oxidation half-reaction is above the reduction half-reaction in the SRP table.

Think of the electrons being pushed downward.

The half-reactions that are farthest apart in the SRP table will occur, generating the greatest possible voltage.

The half-reactions that are closest together in the SRP table will occur, requiring the least possible voltage to drive the cell.

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Quick Check What am I (an electrochemical cell, an electrolytic cell, or both)?

1. I have two half cells. _________________________ 2. My oxidation half-reaction is Ni → Ni2+ + 2e−.

My reduction half-reaction is Fe2+ + 2e− → Fe. _________________________ 3. Oxidation occurs at my anode. _________________________ 4. I transform chemical energy into electricity. _________________________ 5. In order to flow, the electrical charge requires a complete path or circuit. _________________________ 6. You can use the SRP table to calculate how much voltage it “takes” to operate me. _________________________ 7. My E° is + 0.94 V. _________________________

Electrolytic Cell Types

Type 1 Cells

There are three types of electrolytic cells: • In type 1 cells, inert electrodes are immersed in a molten ionic compound.

Only the molten ions (1 type of thing) can be oxidized and reduced.

• In type 2 cells, inert electrodes are immersed in an aqueous ionic solution.

Only the ions or H2O (2 types of things) can be oxidized and reduced.

• In type 3 cells, non-inert electrodes are immersed in an aqueous ionic solution. The ions, H2O, or the anode itself (3 types of things) can be oxidized and the ions or H2O can be reduced. Metals never form anions so the cathode itself is never reduced.

Type 1 cells consist of inert electrodes (typically carbon or platinum) immersed in a molten ionic fluid.

Electrolysis (electro-lysis) literally means separation by electricity. In a type 1 cell, the cations pick up the electrons and are reduced at the cathode, while the anions drop off electrons and are oxidized at the anode. In that way, the cations and anions are separated.

Consider an electrolytic cell having carbon electrodes immersed in molten magnesium chloride.

The half-reactions occurring at the electrodes are: Cathode Anode

Mg2+(l) + 2e− → Mg(l) 2 Cl−(l) → Cl2(g) + 2e−

Magnesium chloride has a higher melting point than magnesium so the magnesium formed at the cathode is also liquid.

Type 2 Cells

Type 2 cells consist of inert electrodes immersed in an aqueous ionic solution. The first question generally asked about an electrolytic cell is, “What half-reactions occur within this cell?” A type 1 cell has only one type of chemical species that can react, the molten ions. In type 2 cells, either the dissolved ions or the water can react. A chemist uses the SRP table to determine whether a dissolved ion or water is reduced and whether a dissolved ion or water is oxidized.

The oxidation and the reduction of water have particularly high overpotentials. Under electrolytic conditions, these half-reactions must be moved relative to the others as shown on the SRP table. This is significant as type 2 and type 3 electrolytic cells conduct currents through aqueous solution. Because of the overpotential effect, water is a weaker reducing agent in electrolytic cells than Br− and Cl− ions, and a weaker oxidizing agent in electrolytic cells than Zn2+, Cr3+, and Fe2+ ions.

Water doesn’t usually react in the spontaneous reactions of electrochemical cells but frequently reacts in the non-spontaneous reactions of electrolytic cells. Water can act as both a weak reducing agent (low oxidation potential) and as a weak oxidizing agent (low reduction potential). Since it is high on the right side of the SRP table, there are not many species above it on the left that it will reduce spontaneously. However, there are many below it that it will reduce non-spontaneously. Likewise, since it is low on the left side of the table, there are not many species below it on the right that it will oxidize spontaneously, but there are many species above it on the right that it will oxidize nonspontaneously.

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Sample Problem 7.5.2(a) — Predicting the Half-Reactions That Will Occur in a Type 2 Electrolytic Cell Identify the half-reactions that occur in the electrolysis of an aqueous solution of manganese(II) bromide.

How to Do It

1. Identify the oxidizing agent(s) and the reducing agent(s).

Mn2+ and H2O are oxidizing agents.

Br− and H2O are reducing agents.

2. Use the SRP table to determine which species is the strongest oxidizing agent and which species is the strongest reducing agent.

4. Write the two half-reactions balancing the transfer of electrons if necessary.

Overpotential Effect

Increasing strength

ClO4− + 8 H+ + 8e−  Cl2 + 2e−  Cr2O7 2− + 14 H+ + 6e− 

½ O2 + 2 H+ + 2e−  MnO2 + 4 H+ + 2e−  IO3− + 6 H+ + 5e−  Br2 + 2e−  AuCl4 + 3e−  NO3− + 4 H+ + 3e− 

½ O2(g) + 2 H+ (10−7 M) + 2e−  Ag+ + e−  Cu+ + e−  Cu2+ + 2e−  SO42− + 4 H+ + 2e− 

2 H+ + 2e−  Pb2+ + 2e−  Ni2+ + 2e− 

2 H2O + 2e−  Fe2+ + 2e−  Ag2S(s) + 2e−  Cr3+ + 3e−  Zn2+ + 2e−  Te(s) + 2 H+ + 2e− 

2 H2O + 2e−  Mn2+ + 2e− 

Cl− + 4 H2O 2 Cl− 2 Cr3+ + 7 H2O H2O Mn2+ + 2 H2O ½ I2 + 3 H2O 2 Br− Au + 4 Cl− NO + 2 H2O H2O Ag Cu Cu H2SO3 + H2O H2 Pb Ni H2 + 2 OH− (10−7 M) Fe(s) 2 Ag(s) + S2− Cr(s) Zn(s) H2Te H2(g) + 2 OH− Mn(s)

Increasing strength

3. The lowest available species on the right will pass electron(s) down to the highest available species on the left. Be mindful of the overpotential effect!

Oxidizing Reducing Agents Agents

Overpotential Effect

What to Think About

Br− is a stronger reducing agent than H2O (overpotential effect).

H2O is a stronger oxidizing agent than Mn2+.

2 Br− → Br2 + 2e− 2 H2O + 2e− → H2 + 2 OH−

2 Br− + 2 H2O → Br2 + 2 OH− + H2

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Type 3 Cells

Type 3 cells consist of non-inert electrodes immersed in an aqueous ionic solution. A type 3 cell adds the possibility that a metal anode itself could be oxidized.

Sample Problem 7.5.2(b) — Predicting the Half-Reactions That Will Occur in a Type 3 Electrolytic Cell Identify the half-reactions that occur in an electrolytic cell consisting of copper electrodes in a solution of CrBr3.

How to Do It

1. Identify the oxidizing agent(s) and the reducing agent(s).

Cr3+ and H2O are oxidizing agents.

Br−, H2O and Cu are reducing agents.

2. Use the SRP table to determine which species is the strongest oxidizing agent and which species is the strongest reducing agent.

4. Write the two half-reactions balancing the transfer of electrons if necessary.

Overpotential Effect

Increasing strength

ClO4− + 8 H+ + 8e−  Cl2 + 2e−  Cr2O7 2− + 14 H+ + 6e− 

½ O2 + 2 H+ + 2e−  MnO2 + 4 H+ + 2e−  IO3− + 6 H+ + 5e−  Br2 + 2e−  AuCl4 + 3e−  NO3− + 4 H+ + 3e− 

½ O2(g) + 2 H+ (10−7 M) + 2e−  Ag+ + e−  Cu+ + e−  Cu2+ + 2e−  SO42− + 4 H+ + 2e− 

2 H+ + 2e−  Pb2+ + 2e−  Ni2+ + 2e− 

2 H2O + 2e−  Fe2+ + 2e−  Ag2S(s) + 2e−  Cr3+ + 3e−  Zn2+ + 2e−  Te(s) + 2 H+ + 2e− 

2 H2O + 2e−  Mn2+ + 2e− 

Cl− + 4 H2O 2 Cl− 2 Cr3+ + 7 H2O H2O Mn2+ + 2 H2O ½ I2 + 3 H2O 2 Br− Au + 4 Cl− NO + 2 H2O H2O Ag Cu Cu H2SO3 + H2O H2 Pb Ni H2 + 2 OH− (10−7 M) Fe(s) 2 Ag(s) + S2− Cr(s) Zn(s) H2Te H2(g) + 2 OH− Mn(s)

Increasing strength

3. The lowest available species on the right will pass electron(s) down to the highest available species on the left.

Oxidizing Reducing Agents Agents

Overpotential Effect

What to Think About

Cu is a stronger reducing agent than Br− or H2O. Cr3+ is a stronger oxidizing agent than H2O (overpotential effect).

3 Cu → 3 Cu2+ + 6e− 3+

2 Cr + 6e− → 2 Cr

3 Cu + 2 Cr3+ → 3 Cu2+ + 2 Cr

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Practice Problems 7.5.2 — Predicting the Half-Reactions That Will Occur in an Electrolytic Cell 1. For each of the following, identify the type of electrolytic cell and the half-reactions occurring within it: (a) carbon electrodes in MgI2(l)

(b) platinum electrodes in CaSO4(aq)

(c) iron electrodes in NaCl(aq)

2. Draw an electrolytic cell having carbon electrodes in NiBr2(aq). Label the DC source, its terminals, and the anode and cathode of the electrolytic cell. Show each ion in the solution migrating toward the appropriate electrode. Write the half-reaction that occurs at each electrode and predict the voltage required to operate this cell.

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Applications: Electrowinning, Electroplating, Electrorefining

–+ e– cathode Me Cu

Electrolytic cells are used extensively in mining and other metallurgy-related industries.

Electrowinning is a metallurgical term for the electrolytic recovery of a metal from a solution

containing its ions. The metal ions are reduced at the cathode where they deposit as metal. The world’s largest zinc and lead smelter is the Cominco plant at Trail, B.C., Canada, Cominco electrowins zinc from zinc sulfate at this smelter.

Electroplating is a form of electrowinning in which a conductive material, usually a metal, is coated with a thin layer of a different metal. Electroplating is usually performed to provide a surface property such as wear resistance, corrosion protection, or lustre to a surface that lacks these properties. The technique is used widely in the manufacture of electronic and optic components and e– sensors, and to chrome plate bathroom fixtures and automobile parts.

Consider an electrolytic cell having a metal object as its cathode, a copper anode, and a solution anode of CuSO4 (Figure 7.5.3). Cu is a stronger reducing agent than SO42− or H2O. Cu2+ is a stronger oxidizing Cu agent than H2O (or SO42− under acidic conditions). The half-reactions that occur in this cell are therefore:

Cu2+ 2–

SO4

Figure 7.5.3 Electroplating a

metal (Me) with copper

Cu → Cu2+ + 2e− Cu2+ + 2e− → Cu Note that no net reaction occurs in this electrolytic cell. Copper ions from the solution are reduced at the cathode and plate out as metallic copper. Those copper ions are replaced by the oxidation of the copper anode. In effect, copper is simply being transferred from the anode to the cathode. This process can continue indefinitely as long as the anode is replaced after it has been consumed.

Electrorefining is another application of electrowinning. Electrorefining is the electrolytic purification of a metal. The metal of an impure anode is oxidized to ions that then migrate to the pure cathode where they are reduced back to the metal. The impurities remain behind. Sample Problem 7.5.2(b) would be an example of electrorefining if the anode were composed of impure copper and the cathode composed of pure copper. An impurity is anything that makes something impure. In chemistry, impurities are minority substances within a majority substance. Just because a substance is present in a material as an impurity doesn’t mean that it isn’t valuable. Many rare and valuable metals are recovered from the impurities left behind as anode mud during electrolysis. The electrorefining of lead by the Betts process, pioneered by Cominco’s Trail Operations in 1902, is still the last step of lead production at the Trail complex. Significant quantities of silver and gold are recovered from the anode mud in this process. The cell voltage used is insufficient to oxidize silver or gold, which are much weaker reducing agents than lead.

In 1885, aluminum was more valuable than gold. The most common mineral of aluminum is bauxite (Al2O3 ∙ 3 H2O). Aluminum cannot be produced from bauxite by electrowinning because water is a Applications: The

stronger oxidizing agent than Al3+. Producing aluminum from the electrolysis of molten Al2O3 (type Héroult-Hall Process 1 cell) is very expensive because Al2O3’s melting point is over 2050°C. In 1886, two 23-year-olds, for Producing Aluminum Paul Héroult of France and Charles Hall of the USA independently discovered that aluminum oxide dissolves in molten cryolite carbon anodes + (Na3AlF6(l)). This aluminum oxide-cryolite mixture melts at about CO2 – 1000°C, making it much more economical to electrolyze than aluminum oxide alone. The Héroult-Hall process still consumes a tremendous amount of energy, both to melt the aluminum oxidealuminum cryolite mixture and to electrolyze it. The aluminum produced in the carbon oxide/cryolite electrolytic cell is also molten at 1000°C. Molten aluminum is denser cathode than the aluminum oxide-cryolite mixture so it runs off through Al openings at the bottom of the cell where it collects.

Figure 7.5.4 The Héroult-Hall process

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Applications: The

Chloralkali Industry

Sodium hydroxide and chlorine are both produced by the electrolysis of aqueous sodium chloride. The chemical industry based on this process is known as the chloralkali industry. Both sodium hydroxide and chlorine are among the world’s top 10 chemicals (by mass) produced annually. Between 12 and 14 million tonnes of each are produced annually in the United States alone, for sales of approximately $4 billion. Sodium hydroxide and chlorine are used in the manufacture of a tremendous variety of everyday products from pharmaceuticals to plastics.

In the electrolysis of an aqueous sodium chloride solution, chloride ions are oxidized: 2 Cl− → Cl2 + 2e− and water molecules are reduced: 2 H2O + 2e− → H2 + 2 OH− The net reaction is: 2 NaCl + 2 H2O → Cl2 + H2 + 2 NaOH The sodium ions are uninvolved spectators in this process. They are dissolved with chloride ions in the reactants and with hydroxide ions in the products. Note that hydrogen gas is also produced in this process but it is generated more easily by other means.

Applications:

Impressed Current Cathodic Protection

A form of cathodic protection is called “galvanic” cathodic protection because the reactions involved are spontaneous (positive E°) like those of a galvanic cell. Another form of cathodic protection called impressed current cathodic protection involves non-spontaneous (negative E°) reactions like those of an electrolytic cell. In this form of cathodic protection, the structure to be protected is connected to the negative terminal of a direct current source such as a rectifier or a battery. The positive terminal is usually connected to an inert electrode that must also be immersed in the solution or buried in the material containing the oxidizing agent(s). In this arrangement, the protected structure becomes the cathode in an electrolytic cell. Impressed current systems are generally less expensive and more effective than galvanic systems (particularly when used in combination with protective coatings).

Despite the advantages of impressed current systems, galvanic systems continue to be more commonly used.

Quick Check 1. What is electrowinning? ________________________________________________________________________________________ 2. What discovery was the key to designing an economical process for refining aluminum? ________________________________________________________________________________________ 3. What two chemicals are produced in the chloralkali industry? ________________________________________________________________________________________ 4. In what form of cathodic protection does the metal being protected act as the cathode in an electrolytic cell? ________________________________________________________________________________________

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