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Chemistry: Atoms First

# 16.7Electrolysis

Chemistry: Atoms First16.7 Electrolysis

### Learning Objectives

By the end of this section, you will be able to:
• Describe electrolytic cells and their relationship to galvanic cells
• Perform various calculations related to electrolysis

In galvanic cells, chemical energy is converted into electrical energy. The opposite is true for electrolytic cells. In electrolytic cells, electrical energy causes nonspontaneous reactions to occur in a process known as electrolysis. The charging electric car pictured in Figure 16.1 at the beginning of this chapter shows one such process. Electrical energy is converted into the chemical energy in the battery as it is charged. Once charged, the battery can be used to power the automobile.

The same principles are involved in electrolytic cells as in galvanic cells. We will look at three electrolytic cells and the quantitative aspects of electrolysis.

### The Electrolysis of Molten Sodium Chloride

In molten sodium chloride, the ions are free to migrate to the electrodes of an electrolytic cell. A simplified diagram of the cell commercially used to produce sodium metal and chlorine gas is shown in Figure 16.19. Sodium is a strong reducing agent and chlorine is used to purify water, and is used in antiseptics and in paper production. The reactions are

$anode:2Cl−(l)⟶Cl2(g)+2e−ECl2/Cl−°=+1.3 Vcathode:Na+(l)+e−⟶Na(l)ENa+/Na°=−2.7 V¯overall:2Na+(l)+2Cl−(l)⟶2Na(l)+Cl2(g)Ecell°=−4.0 Vanode:2Cl−(l)⟶Cl2(g)+2e−ECl2/Cl−°=+1.3 Vcathode:Na+(l)+e−⟶Na(l)ENa+/Na°=−2.7 V¯overall:2Na+(l)+2Cl−(l)⟶2Na(l)+Cl2(g)Ecell°=−4.0 V$

The power supply (battery) must supply a minimum of 4 V, but, in practice, the applied voltages are typically higher because of inefficiencies in the process itself.

Figure 16.19 Passing an electric current through molten sodium chloride decomposes the material into sodium metal and chlorine gas. Care must be taken to keep the products separated to prevent the spontaneous formation of sodium chloride.

### The Electrolysis of Water

It is possible to split water into hydrogen and oxygen gas by electrolysis. Acids are typically added to increase the concentration of hydrogen ion in solution (Figure 16.20). The reactions are

$anode:2H2O(l)⟶O2(g)+4H+(aq)+4e−Eanode°=+1.229 Vcathode:2H+(aq)+2e−⟶H2(g)Ecathode°=0 V¯overall:2H2O(l)⟶2H2(g)+O2(g)Ecell°=−1.229 Vanode:2H2O(l)⟶O2(g)+4H+(aq)+4e−Eanode°=+1.229 Vcathode:2H+(aq)+2e−⟶H2(g)Ecathode°=0 V¯overall:2H2O(l)⟶2H2(g)+O2(g)Ecell°=−1.229 V$

Note that the sulfuric acid is not consumed and that the volume of hydrogen gas produced is twice the volume of oxygen gas produced. The minimum applied voltage is 1.229 V.

Figure 16.20 Water decomposes into oxygen and hydrogen gas during electrolysis. Sulfuric acid was added to increase the concentration of hydrogen ions and the total number of ions in solution, but does not take part in the reaction. The volume of hydrogen gas collected is twice the volume of oxygen gas collected, due to the stoichiometry of the reaction.

### The Electrolysis of Aqueous Sodium Chloride

The electrolysis of aqueous sodium chloride is the more common example of electrolysis because more than one species can be oxidized and reduced. Considering the anode first, the possible reactions are

$(i)2Cl−(aq)⟶Cl2(g)+2e−Eanode°=+1.35827 V(ii)2H2O(l)⟶O2(g)+4H+(aq)+4e−Eanode°=+1.229 V(i)2Cl−(aq)⟶Cl2(g)+2e−Eanode°=+1.35827 V(ii)2H2O(l)⟶O2(g)+4H+(aq)+4e−Eanode°=+1.229 V$

These values suggest that water should be oxidized at the anode because a smaller potential would be needed—using reaction (ii) for the oxidation would give a less-negative cell potential. When the experiment is run, it turns out chlorine, not oxygen, is produced at the anode. The unexpected process is so common in electrochemistry that it has been given the name overpotential. The overpotential is the difference between the theoretical cell voltage and the actual voltage that is necessary to cause electrolysis. It turns out that the overpotential for oxygen is rather high and effectively makes the reduction potential more positive. As a result, under normal conditions, chlorine gas is what actually forms at the anode.

Now consider the cathode. Three reductions could occur:

$(iii)2H+(aq)+2e−⟶H2(g)Ecathode°=0 V(iv)2H2O(l)+2e−⟶H2(g)+2OH−(aq)Ecathode°=−0.8277 V(v)Na+(aq)+e−⟶Na(s)Ecathode°=−2.71 V(iii)2H+(aq)+2e−⟶H2(g)Ecathode°=0 V(iv)2H2O(l)+2e−⟶H2(g)+2OH−(aq)Ecathode°=−0.8277 V(v)Na+(aq)+e−⟶Na(s)Ecathode°=−2.71 V$

Reaction (v) is ruled out because it has such a negative reduction potential. Under standard state conditions, reaction (iii) would be preferred to reaction (iv). However, the pH of a sodium chloride solution is 7, so the concentration of hydrogen ions is only 1$××$ 10−7 M. At such low concentrations, reaction (iii) is unlikely and reaction (iv) occurs. The overall reaction is then

$overall: 2H2O(l)+2Cl−(aq)⟶H2(g)+Cl2(g)+2OH−(aq)Ecell°=−2.186 Voverall: 2H2O(l)+2Cl−(aq)⟶H2(g)+Cl2(g)+2OH−(aq)Ecell°=−2.186 V$

As the reaction proceeds, hydroxide ions replace chloride ions in solution. Thus, sodium hydroxide can be obtained by evaporating the water after the electrolysis is complete. Sodium hydroxide is valuable in its own right and is used for things like oven cleaner, drain opener, and in the production of paper, fabrics, and soap.

### Chemistry in Everyday Life

#### Electroplating

An important use for electrolytic cells is in electroplating. Electroplating results in a thin coating of one metal on top of a conducting surface. Reasons for electroplating include making the object more corrosion resistant, strengthening the surface, producing a more attractive finish, or for purifying metal. The metals commonly used in electroplating include cadmium, chromium, copper, gold, nickel, silver, and tin. Common consumer products include silver-plated or gold-plated tableware, chrome-plated automobile parts, and jewelry. We can get an idea of how this works by investigating how silver-plated tableware is produced (Figure 16.21).

Figure 16.21 The spoon, which is made of an inexpensive metal, is connected to the negative terminal of the voltage source and acts as the cathode. The anode is a silver electrode. Both electrodes are immersed in a silver nitrate solution. When a steady current is passed through the solution, the net result is that silver metal is removed from the anode and deposited on the cathode.

In the figure, the anode consists of a silver electrode, shown on the left. The cathode is located on the right and is the spoon, which is made from inexpensive metal. Both electrodes are immersed in a solution of silver nitrate. As the potential is increased, current flows. Silver metal is lost at the anode as it goes into solution.

$anode: Ag(s)⟶Ag+(aq)+e−anode: Ag(s)⟶Ag+(aq)+e−$

The mass of the cathode increases as silver ions from the solution are deposited onto the spoon

$cathode: Ag+(aq)+e−⟶Ag(s)cathode: Ag+(aq)+e−⟶Ag(s)$

The net result is the transfer of silver metal from the anode to the cathode. The quality of the object is usually determined by the thickness of the deposited silver and the rate of deposition.

### Quantitative Aspects of Electrolysis

The amount of current that is allowed to flow in an electrolytic cell is related to the number of moles of electrons. The number of moles of electrons can be related to the reactants and products using stoichiometry. Recall that the SI unit for current (I) is the ampere (A), which is the equivalent of 1 coulomb per second (1 A = 1 $CsCs$). The total charge (Q, in coulombs) is given by

$Q=I×t=n×FQ=I×t=n×F$

Where t is the time in seconds, n the number of moles of electrons, and F is the Faraday constant.

Moles of electrons can be used in stoichiometry problems. The time required to deposit a specified amount of metal might also be requested, as in the second of the following examples.

### Example 16.8

#### Converting Current to Moles of Electrons

In one process used for electroplating silver, a current of 10.23 A was passed through an electrolytic cell for exactly 1 hour. How many moles of electrons passed through the cell? What mass of silver was deposited at the cathode from the silver nitrate solution?

#### Solution

Faraday’s constant can be used to convert the charge (Q) into moles of electrons (n). The charge is the current (I) multiplied by the time
$n=QF=10.23 Cs×1 hr×60 minhr×60 smin96,485 C/mol e−=36,830 C96,485 C/mole−=0.3817 mol e−n=QF=10.23 Cs×1 hr×60 minhr×60 smin96,485 C/mol e−=36,830 C96,485 C/mole−=0.3817 mol e−$

From the problem, the solution contains AgNO3, so the reaction at the cathode involves 1 mole of electrons for each mole of silver

$cathode: Ag+(aq)+e−⟶Ag(s)cathode: Ag+(aq)+e−⟶Ag(s)$

The atomic mass of silver is 107.9 g/mol, so

$mass Ag=0.3817 mol e−×1 mol Ag1 mol e−×107.9 g Ag1 mol Ag=41.19 g Agmass Ag=0.3817 mol e−×1 mol Ag1 mol e−×107.9 g Ag1 mol Ag=41.19 g Ag$

Check your answer: From the stoichiometry, 1 mole of electrons would produce 1 mole of silver. Less than one-half a mole of electrons was involved and less than one-half a mole of silver was produced.

#### Check Your Learning

Aluminum metal can be made from aluminum ions by electrolysis. What is the half-reaction at the cathode? What mass of aluminum metal would be recovered if a current of 2.50 $××$ 103 A passed through the solution for 15.0 minutes? Assume the yield is 100%.

### Answer:

$Al3+(aq)+3e−⟶Al(s);Al3+(aq)+3e−⟶Al(s);$ 7.77 mol Al = 210.0 g Al.

### Example 16.9

#### Time Required for Deposition

In one application, a 0.010-mm layer of chromium must be deposited on a part with a total surface area of 3.3 m2 from a solution of containing chromium(III) ions. How long would it take to deposit the layer of chromium if the current was 33.46 A? The density of chromium (metal) is 7.19 g/cm3.

#### Solution

This problem brings in a number of topics covered earlier. An outline of what needs to be done is:
• If the total charge can be determined, the time required is just the charge divided by the current
• The total charge can be obtained from the amount of Cr needed and the stoichiometry
• The amount of Cr can be obtained using the density and the volume Cr required
• The volume Cr required is the thickness times the area

Solving in steps, and taking care with the units, the volume of Cr required is

$volume=(0.010 mm×1 cm10 mm)×(3.3m2×(10,000cm21m2))=33 cm3volume=(0.010 mm×1 cm10 mm)×(3.3m2×(10,000cm21m2))=33 cm3$

Cubic centimeters were used because they match the volume unit used for the density. The amount of Cr is then

$mass=volume×density=33cm3×7.19 gcm3=237 g Crmass=volume×density=33cm3×7.19 gcm3=237 g Cr$
$mol Cr=237 g Cr×1 mol Cr52.00 g Cr=4.56 mol Crmol Cr=237 g Cr×1 mol Cr52.00 g Cr=4.56 mol Cr$

Since the solution contains chromium(III) ions, 3 moles of electrons are required per mole of Cr. The total charge is then

$Q=4.56 mol Cr×3mol e−1 mol Cr×96485 Cmol e−=1.32×106CQ=4.56 mol Cr×3mol e−1 mol Cr×96485 Cmol e−=1.32×106C$

The time required is then

$t=QI=1.32×106C33.46 C/s=3.95×104s=11.0 hrt=QI=1.32×106C33.46 C/s=3.95×104s=11.0 hr$

Check your answer: In a long problem like this, a single check is probably not enough. Each of the steps gives a reasonable number, so things are probably correct. Pay careful attention to unit conversions and the stoichiometry.

#### Check Your Learning

What mass of zinc is required to galvanize the top of a 3.00 m $××$ 5.50 m sheet of iron to a thickness of 0.100 mm of zinc? If the zinc comes from a solution of Zn(NO3)2 and the current is 25.5 A, how long will it take to galvanize the top of the iron? The density of zinc is 7.140 g/cm3.

### Answer:

11.8 kg Zn requires 382 hours.

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