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4 E°[sub cell] and ∆G°

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20.4 | E °cell and DG ° 937

(

)

joule

 coulombs 

maximum work = mole e − × 

= joule

×

 mole e 

coulombs

n

E cell

Combining Equations 20.3 and 20.4 gives us

DG = -n Ecell

(20.5)

At standard state we are dealing with the standard cell potential, so we can calculate the

standard free energy change.

DG ° = -n E °cell

DG related to Ecell

(20.6)

Referring back to Chapter 19, if DG has a negative value, a reaction will be spontaneous,

and this corresponds to a positive value of Ecell.

Up to now we have been careful to predict spontaneity for systems at standard state

where Ecell is equal to E °cell. Later in Section 20.5, we will see how to calculate Ecell under

nonstandard state conditions and precisely predict if a reaction is spontaneous.

DG ° related to E °cell

Example 20.7

Calculating the Standard Free Energy Change

Calculate DG ° for the following reaction, given that its standard cell potential is 0.320 V

at 25 °C.

NiO2(s) + 2Cl-(aq) + 4H+(aq) → Cl2( g ) + Ni2+(aq) + 2H2O

n Analysis:

We will need to use the relationship between E °cell and DG °. The given chemical equation allows us to determine n, which is used in the equation.

n Assembling

the Tools: Our tool for solving this problem is Equation 20.6.

DG ° = -n E °cell

We know  and E °cell, and the value of n, the total number of electrons transferred in the

balanced chemical equation, can be computed with the aid of principles developed in

Section 6.2. We will also use the Faraday constant, 1  = 9.65 × 104 C/mol e- (the

SI abbreviation for coulomb is C).

1 =

× 10 C

( 9.651 mol

)

e

4

n Solution:

To solve Equation 20.6 we need to determine n. By taking the coefficients

in the chemical equation to stand for moles, two moles of Cl- are oxidized to Cl2

and two moles of electrons are transferred to the NiO2. Therefore we conclude that

n = 2 mol e-. Using Equation 20.6, we have

(

0.320 J

 9.65 × 104 C 

∆G ° = −(2 mol e − ) × 

×

 1 mol e − 

C

)

= −6.18 × 104 J

= −61.8 kJ

n Is

the Answer Reasonable? Let’s do some approximate arithmetic. The Faraday constant equals approximately 100,000, or 105. The product 2 × 0.32 = 0.64, so DG °

should be about 0.64 × 105 or 6.4 × 104. The answer seems to be okay. Importantly, the

algebraic signs tell us the same thing: that is, the positive E °cell and the negative DG ° both

predict a spontaneous reaction.

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938 Chapter 20 | Electrochemistry

Practice Exercises

20.13 | A certain reaction has an E °cell of 0.107 volts and has a DG ° of -30.9 kJ. How

many electrons are transferred in the reaction? (Hint: See Equation 20.6.)

20.14 | Calculate DG ° for the reactions that take place in the galvanic cells described in

Practice Exercises 20.11 and 20.12.

E °cell and Equilibrium Constants

One of the useful applications of electrochemistry is the determination of equilibrium

constants. In Chapter 19 you saw that DG ° is related to the equilibrium constant by the

expression

DG ° = -RT ln Kc

where we used Kc for the equilibrium constant because electrochemical reactions occur in

solution. Earlier in this section we saw that DG ° is also related to E °cell

DG ° = -n E °cell

Therefore, E °cell and the equilibrium constant are also related. Equating the right sides of

the two equations, we have

Solving for E °cell gives4

-n E °cell = -RT ln Kc

E °cell =

Kc related to E °cell

RT

ln K c

n

(20.7)

For the units to work out correctly, the value of R must be 8.314 J mol-1 K-1, T must

be the temperature in kelvins,  equals 9.65 × 104 C per mole of e-, and n equals the

number of moles of electrons transferred in the reaction.

Example 20.8

Calculating Equilibrium Constants from E °cell

Calculate Kc for the reaction in Example 20.7.

n Analysis:

We now have an equation that relates the standard cell potential and the

equilibrium constant. We need to choose the appropriate values for each variable to

determine Kc.

n Assembling

the Tools: Equation 20.7 is our tool for solving this problem. We need to

collect the terms to insert in this equation to solve it. We need to find E °cell, n, R, T, and .

Remember that T is in kelvins and R must have the appropriate units of J mol-1 K-1.

Collecting the values needed, we find:

T = 25 °C = 298 K

 = 9.65 × 104 C per mole of e-

R = 8.314 J mol-1 K-1

E °cell = 0.320 V = 0.320 J C-1

n=2

4

For historical reasons, Equation 20.7 is sometimes expressed in terms of common logs (base 10 logarithms).

Natural and common logarithms are related by the equation

ln x = 2.303 log x

For reactions at 25 °C (298 K), all of the constants (R, T, and ) can be combined with the factor 2.303 to

give 0.0592 joules/coulomb. Because joules/coulomb equals volts, Equation 20.7 reduces to

E cell

° =

0.0592 V

log K c

n

where n is the number of moles of electrons transferred in the cell reaction as it is written.

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20.5 | Cell Potentials and Concentrations 939

n Solution:

tute values.

With all of the data, let’s rearrange Equation 20.7 for ln Kc and then substiln K c =

E °cell n 

RT

Substituting values,

ln K c =

0.320 J C −1 × 2 × 9.65 × 104 C mol −1

8.314 J mol −1 K −1 × 298 K

= 24.9

Taking the antilogarithm,

Kc = e 24.9 = 7 × 1010

n Is

the Answer Reasonable? As a rough check, we can look at the magnitude of E °cell and

apply some simple reasoning. When E °cell is positive, DG ° is negative, and in Chapter 19

you learned that when DG ° is negative, the position of equilibrium lies toward the product side of the reaction. Therefore, we expect that Kc will be large, and that agrees with

A more complete check would require evaluating the fraction used to compute ln Kc.

First, we should check to be sure we’ve substituted correctly into the equation for ln Kc.

Next, we could do some approximate arithmetic to check the value of ln Kc. Rounding all

numbers to one significant figure we get:

ln K c =

0.3 J C −1 × 2 × 10 × 104 C mol −1

10 J mol −1 K −1 × 300 K

=

6 × 104

= 2 × 101 = 20

3000

This is close to the 24.9 we calculated above, and we are confident the calculation was

done correctly.

20.15 | The calculated standard cell potential for the reaction

Practice Exercises

���

Cu2+(aq) + 2Ag(s) �

� Cu(s) + 2Ag+(aq)

is E °cell = -0.46 V. Calculate Kc for this reaction as written. Is this reaction spontaneous?

If not, what is Kc for the spontaneous reaction? (Hint: The tool described by Equation 20.7

is important here.)

20.16 | Use the following half-reactions and the data in Table 20.1 to write the spontaneous reaction. Write the equilibrium law for this reaction and use the standard cell voltage

to determine the value of the equilibrium constant.

Ag+(aq) + e- → Ag(s)

AgBr(s) + e- → Ag(s) + Br-(aq)

20.5 | Cell Potentials and Concentrations

At 25 °C when all of the ion concentrations in a cell are 1.00 M and when the partial

pressures of any gases involved in the cell reaction are 1.00 atm, the cell potential is equal

to the standard potential. When the concentrations or pressures change, however, so does

the potential. For example, in an operating cell or battery, the potential gradually drops

as the reactants are used up and as the cell reaction approaches its natural equilibrium

status. When it reaches equilibrium, the potential has dropped to zero—the battery is

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940 Chapter 20 | Electrochemistry

The Nernst Equation

The effect of concentration on the cell potential can be obtained from thermodynamics.

In Chapter 19, you learned that the free energy change is related to the reaction quotient

Q by the equation

DG = DG ° + RT ln Q

Walther Nernst (1864–1941) was

a professor of physical chemistry

who developed the equation

named after him. (Ullstein Bild/

The Granger Collection, New York)

Substituting for DG and DG ° from Equations 20.5 and 20.6 gives

-n Ecell = -n E °cell + RT ln Q

Dividing both sides by -n  gives

E cell = E °cell −

Nernst equation

n This is a heterogeneous reaction,

so we have not included the

concentration of the solid, Cu(s), in

the mass action expression.

RT

ln Q

n

(20.8)

This equation is commonly known as the Nernst equation,5 named after Walther Nernst, a

German chemist and physicist. Notice that if Q = 1, then ln Q = 0 and Ecell = E °cell.

In writing the Nernst equation for a galvanic cell, we will construct the mass action expression (from which we calculate the value of Q) using molar concentrations for ions and partial

pressures in atmospheres for gases.6 Thus, for the following cell using a hydrogen electrode

(with the partial pressure of H2 not necessarily equal to 1 atm) and having the reaction

Cu2+(aq) + H2( g ) → Cu(s) + 2H+(aq)

the Nernst equation would be written

E cell = E °cell −

RT

[H+ ]2

ln

n

[Cu 2+ ] PH2

Example 20.9

Calculating the Effect of Concentration on Ecell

Suppose a galvanic cell employs the following half-reactions.

���

Ni2+(aq) + 2e- �

E °Ni2+ = -0.25 V

� Ni(s)

3+

- ���

Cr (aq) + 3e � � Cr(s)

E °Cr3+ = -0.74 V

Calculate the cell potential when [Ni2+] = 4.87 × 10-4 M and [Cr3+] = 2.48 × 10-3 M.

n Analysis:

We will be calculating the cell potential for a system that does not have standard state concentrations by using the Nernst equation. As with previous problems, we

will need to write the appropriate chemical equations so that the variables can be identified and used properly.

n Assembling

the Tools: We start with the Nernst equation, Equation 20.8, to deter-

mine Ecell.

E cell = E °cell −

RT

ln Q

n

5

Using common logarithms instead of natural logarithms and calculating the constants for 25 °C gives

another form of the Nernst equation that is sometimes used:

0.0592 V

E cell = E cell

° −

log Q

n

6

Because of interionic attractions, ions do not always behave as though their concentrations are equal to their

molarities. Strictly speaking, therefore, we should use effective concentrations (called activities) in the mass

action expression. Effective concentrations are difficult to calculate, so for simplicity we will use molarities

and accept the fact that our calculations are not entirely accurate.

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20.5 | Cell Potentials and Concentrations 941

The balanced chemical equation is needed (Chapter 6) to determine the number of electrons transferred, n, and the correct form of the mass action expression (Chapter 15) from

which we calculate the value of Q.

n Solution: Nickel has the more positive (less negative) standard reduction potential, so

its half-reaction will occur as a reduction. This means that chromium will be oxidized.

Making electron gain (6e-) equal to electron loss (6e-), the cell reaction is found as follows.

3[Ni2+(aq) + 2e- → Ni(s)]

2[Cr(s) → Cr3+(aq) + 3e-]

3Ni2+(aq) + 2Cr(s) → 3Ni(s) + 2Cr3+(aq)

(reduction)

(oxidation)

(cell reaction)

The total number of electrons transferred is six, which means n = 6. Now we can write

the Nernst equation for the system.

E cell = E °cell −

RT

[Cr 3+ ]2

ln

n

[Ni 2+ ]3

Notice that we’ve constructed the mass action expression, from which we will calculate the

reaction quotient, using the concentrations of the ions raised to powers equal to their coefficients in the net cell reaction, and that we have not included concentration terms for the

two solids. This is the procedure we followed for heterogeneous equilibria in Section 15.4.

Next we need E °cell. Since Ni2+ is reduced,

E °cell = E °Ni2+ - E °Cr3+

= (-0.25 V) - (-0.74 V)

= 0.49 V

Now we can substitute this value for E °cell along with R = 8.314 J mol-1 K-1, T = 298 K,

n = 6,  = 9.65 × 104 C mol-1, [Ni2+] = 4.87 × 10-4 M, and [Cr3+] = 2.48 × 10-3 M

into the Nernst equation. This gives

E cell = 0.49 V −

8.314 J mol −1 K −1 × 298 K

(2.48 × 10 −3 )2

ln

(4.87 × 10 −4 )3

6 × (9.65 × 104 C mol −1 )

We notice that the volt, V, is 1 J/C so that we obtain

Ecell = 0.49 V - (0.00428 V) ln (5.32 × 104)

= 0.49 V - (0.00428 V) (10.882)

= 0.49 V - 0.0525 V

= 0.44 V

The potential of the cell is expected to be 0.44 V.

n Is

the Answer Reasonable? There’s no simple way to check the answer, but the small

numerical value of 0.44 V is reasonable. In addition, there are certain important points to

consider. First, check that you’ve combined the half-reactions correctly to (a) calculate E °cell

and (b) give the balanced cell reaction. Now check that we wrote the correct mass action

expression from the balanced equation and that we have the correct number of electrons transferred, n. Finally, check that the Kelvin temperature and R = 8.314 J mol-1 K-1 were used.

Example 20.10

The Spontaneous Reaction May Be Concentration Dependent

The reaction of tin metal with acid can be written as

Sn(s) + 2H+(aq) → Sn2+(aq) + H2( g )

Calculate the cell potential when (a) the system is at standard state, (b) the pH is 2.00, and

(c) the pH is 5.00. Assume that [Sn2+] = 1.0 M and the partial pressure of H2 is 1.00 atm.

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942 Chapter 20 | Electrochemistry

n Analysis:

We will be using the equations developed in this chapter for computing the

values for the standard cell potential and various cell potentials. For this we will need to

identify the substances that are oxidized and reduced in a chemical reaction and gather the

information needed to solve the appropriate equations. One way to approach this problem

is to write out the equations and then find values for the variables.

n Assembling

the Tools: Part (a) is at standard state, and we have used Equation 20.2

for combining standard reduction potentials to solve problems like that in Examples 20.5

and 20.6. In this reaction we identify the tin metal as the substance that is oxidized and

hydrogen ions are reduced.

For parts (b) and (c) we must use the Nernst equation as our tool, Equation 20.8. We

determine the number of electrons transferred, n, by noting that tin loses two electrons

and each hydrogen gains an electron. Therefore two electrons are transferred from tin to

the hydrogen ions. We can also set up Q, see Chapter 15, for the Nernst equation.

Substituting 1.00 for both [Sn2+] and PH2, the result is

Q =

[Sn 2+ ] PH2

1

= +2

+ 2

[H ]

[H ]

We can also calculate the hydrogen ion concentrations for the pH 2.00 and pH 5.00 solutions as 1.0 × 10-2 M and 1.0 × 10-5 M, respectively.

n Solution:

For part (a) we find the difference between standard reduction potentials as

E °cell = 0.00 - (-0.14) = +0.14 V

For parts (b) and (c) we substitute into the Nernst equation

RT

1

E cell = E °cell −

ln

n

[H+ ]2

8.314 J mol −1 K −1 × 298 K

1

ln

−1

4

(

.

1

0

10 −2 )2

×

2 × (9.65 × 10 C mol )

= 0.14 V − 0.12 = +0.02 V

(b) E cell = 0.14 V −

E cell

8.314 J mol −1 K −1 × 298 K

1

ln

(1.0 × 10 −5 )2

2 × (9.65 × 104 C mol −1 )

= 0.14 V − 0.30 = −0.16 V

(c) E cell = 0.14 V −

E cell

At standard state the reaction is spontaneous, and at pH = 2.00 it is spontaneous, but the

potential is barely positive. At pH 5.00 the reaction is not spontaneous, but the reverse

reaction will be.

n Is

the Answer Reasonable? The first question to answer is “does this make sense?” and

indeed it does. Looking at the natural logarithm part of the equation we see that as the

[H+] decreases, the ln term increases and makes a more negative adjustment to the E °cell.

Since a decrease in [H+] is an increase in pH, we expect Ecell will decrease as pH increases.

We cannot easily estimate natural logarithms, so a check of calculations may be easiest if the

values are entered into your calculator in a different sequence from your first calculation.

Practice Exercises

jespe_c20_918-973hr.indd 942

20.17 | A galvanic cell is constructed with a copper electrode dipping into a 0.015 M solution of Cu2+ ions and an electrode made of magnesium is immersed in a 2.2 × 10-6 M

solution of magnesium ions. What is the balanced chemical equation and the cell potential

at 25 °C? (Hint: Set up the Nernst equation for this reaction.)

20.18 | In Example 20.10, assume all conditions are the same except for the pH. At what

pH will the cell potential be zero?

20.19 | In a certain zinc–copper cell,

Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)

the ion concentrations are [Cu2+] = 0.0100 M and [Zn2+] = 1.0 M. What is the cell

potential at 25 °C?

11/18/10 4:59 PM

20.5 | Cell Potentials and Concentrations 943

Concentration from Ecell Measurements

One of the principal uses of the relationship between concentration and cell potential is

for the measurement of concentrations of redox reactants and products in a galvanic cell.

Experimental determination of cell potentials combined with modern developments in

electronics has provided a means of monitoring and analyzing the concentrations of all

sorts of substances in solution, even some that are not themselves ionic and that are not

involved directly in electrochemical changes. In fact, the operation of a pH meter relies on

the logarithmic relationship between hydrogen ion concentration and the potential of a

special kind of electrode (Figure 20.8).

Glass

electrode

n The ease of such operations and the

fact that they lend themselves well

to automation and computer analysis

make electrochemical analyses

especially attractive to scientists.

Reference

electrode

Hg2Cl2–Hg

KCl

solution

Dilute

HCl

Glass

membrane

Silver wire

coated with

AgCl

KCl

crystals

Fiber through

glass

Figure 20.8 | Electrodes used with a pH meter. The electrode on

the left is called a glass electrode. It contains a silver wire, coated with

AgCl, dipping into a dilute solution of HCl. This half-cell has a

potential that depends on the difference between the [H+] inside and

outside a thin glass membrane at the bottom of the electrode. On the

right is a reference electrode that forms the other half-cell. The

galvanic cell formed by the two electrodes produces a potential that is

proportional to the pH of the solution into which they are dipped.

Example 20.11

Using the Nernst Equation to Determine Concentrations

A laboratory was assigned the job of determining the copper(II) ion concentration in

thousands of water samples. To make these measurements an electrochemical cell was

assembled that consists of a silver electrode, dipping into a 0.225 M solution of AgNO3,

connected by a salt bridge to a second half-cell containing a copper electrode. The copper

half-cell was then filled with one water sample after another, with the cell potential measured for each sample. In the analysis of one sample, the cell potential at 25 °C was measured to be 0.62 V. The copper electrode was observed to carry a negative charge, so it

served as the anode. What was the concentration of copper ion in this sample?

n Analysis:

In this problem, we’ve been given the cell potential, Ecell, and we can calculate E c°ell. The unknown quantity is one of the concentration terms in Q that is part of

the Nernst equation. Once we assemble all of the variables for the Nernst equation, an

algebraic solution is possible.

n Assembling

the Tools: We will need Equation 20.8, the Nernst equation, and the

chemical reaction represented by the description of the galvanic cell. Because copper is

the anode, it is being oxidized. This also means that Ag+ is being reduced. Therefore, the

equation for the cell reaction is

Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s)

From the balanced equation we can determine the mass action expression, n, and E °cell.

Since R and Faraday’s constant are known, along with the concentration of silver ions, the

remaining variable is [Cu2+] that the question asks us to compute.

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944 Chapter 20 | Electrochemistry

n Solution:

Two electrons are transferred, so n = 2 and the Nernst equation is

E cell = E °cell −

RT

[Cu 2+ ]

ln

2

[ Ag + ]2

The value of E °cell can be obtained from the tabulated standard reduction potentials in

Table 20.1. Following our usual procedure and recognizing that silver ion is reduced,

E °cell = E °Ag+ - E °Cu2+

= (0.80 V) - (0.34 V)

= 0.46 V

Now we can substitute values into the Nernst equation and solve for the concentration

ratio in the mass action expression.

0.62 V = 0.46 V −

8.314 J mol−1 K −1 × 298 K

2 × (9.65 × 104 C mol−1 )

ln

[Cu 2+ ]

[ Ag+ ]2

Solving for ln ([Cu2+]/[Ag+]2) gives (retaining one extra significant figure)

ln

[Cu 2+ ]

= −12.5

[ Ag + ]2

Taking the antilog gives us the value of the mass action expression.

[Cu 2+ ]

= 4 × 10−6

[ Ag + ]2

Since we know that the concentration of Ag+ is 0.225 M, we can now solve for the Cu2+

concentration.

[Cu 2+ ]

= 4 × 10−6

(0 . 225)2

[Cu 2+ ] = 2 × 10−7 M (correctly rounded)

n Is

the Answer Reasonable? All we can do easily is check to be sure we’ve written the correct

chemical equation, on which all the rest of the solution to the problem rests. Be careful about

algebraic signs and that you select the proper value for R and the temperature in kelvins. Also,

notice that we first solved for the logarithm of the ratio of concentration terms. Then, after taking the (natural) antilogarithm, we substitute the known value for [Ag+] and solve for [Cu2+].

As a final point, notice that the Cu2+ concentration is indeed very small, and that it can

be obtained very easily by simply measuring the potential generated by the electrochemical cell. Determining the concentrations in many samples is also very simple—just change

the water sample and measure the potential again.

Practice Exercises

20.20 | A galvanic cell is constructed with a copper electrode dipping into a 0.015 M

solution of Cu2+ ions, and a magnesium electrode immersed in a solution with an

unknown concentration of magnesium ions. The cell potential is measured as 2.79 volts

at 25 °C. What is the concentration of magnesium ions?

Mg(s) + Cu2+(aq) → Mg2+(aq) + Cu(s)

(Hint: Use the Nernst equation to solve for [Mg2+].)

20.21 | In the analysis of two other water samples by the procedure described in

Example 20.11, cell potentials (Ecell) of 0.57 V and 0.82 V were obtained. Calculate the

Cu2+ ion concentration in each of these samples.

20.22 | A galvanic cell was constructed with a nickel electrode that was dipped into 1.20 M

NiSO4 solution and a chromium electrode that was immersed into a solution containing Cr3+

at an unknown concentration. The potential of the cell was measured to be 0.552 V, with the

chromium serving as the anode. The standard cell potential for this system was determined to

be 0.487 V. What was the concentration of Cr3+ in the solution of unknown concentration?

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20.6 | Electricity 945

Concentration Cells

Electron flow

The dependence of cell potential on concentration allows us to construct a

galvanic cell from two half-cells composed of the same substances, but having

different concentrations of the solute species. An example would be a pair of

copper electrodes dipping into solutions that have different concentrations of

Cu2+, say 0.10 M Cu2+ in one and 1.0 M in the other (Figure 20.9). When this

cell operates, reactions take place that tend to bring the two Cu2+ concentrations

toward the same value. Thus, in the half-cell containing 0.10 M Cu2+, copper is

oxidized, which adds Cu2+ to the more dilute solution. In the other cell, Cu2+

is reduced, removing Cu2+ from the more concentrated solution. This makes the

more dilute half-cell the anode and the more concentrated half-cell the

cathode.

Cu(s) | Cu2+(0.10 M) || Cu2+(1.0 M) | Cu(s)

anode

cathode

The half-reactions in the spontaneous cell reaction are

Cu(s) → Cu2+(0.10 M) + 2eCu2+(1.0 M) + 2e- → Cu(s)

Cu

1.0 M

Cu(NO3)2

Salt bridge

Cu

0.10 M

Cu(NO3)2

Figure 20.9 | A concentration cell.

When the circuit is completed, reactions

occur that tend to make the concentrations of Cu2+ the same in the two

half-cells. Oxidation occurs in the more

dilute half-cell, and reduction occurs in

the more concentrated one.

Cu2+(1.0 M) → Cu2+(0.10 M)

The Nernst equation for this cell is

E cell = E °cell −

[Cu 2+ ] dilute

RT

ln

n

[Cu 2+ ] conc

Because we’re dealing with the same substances in the cell, E °cell = 0 V. When the cell operates, n = 2, and we’ll take T = 298 K. Substituting values,

8.314 J mol −1 K −1 × 298 K

0.10

ln

−1

4

1.0

2 × 9.65 × 10 C mol

= 0.030 V

E cell = 0 V −

(

)

In this concentration cell, one solution is ten times more concentrated than the other,

yet the potential generated is only 0.03 V. In general, the potential generated by concentration differences are quite small. Yet they are significant in biological systems, where

electrical potentials are generated across biological membranes by differences in ion concentrations (e.g., K+). Membrane potentials are important in processes such as the transmission of nerve impulses.

These small differences in potential also illustrate that if Q, in a system that is not at

standard state, is between 0.10 and 10 we can conclude that Ecell ≈ E °cell and we can generalize the results in Sections 20.2 and 20.3 when predicting spontaneous reactions.

20.6 | Electricity

Large amounts of electricity are generated mechanically by rotating coils of wire in a

magnetic field in a turbine. To rotate the coils, superheated steam, flowing water, and

the wind are used in a variety of turbine designs. Superheated steam is produced reliably

by burning coal, oil, natural gas, and even municipal wastes. Such turbines are used

mainly in large central electrical generators near major population centers. Since the

energy of the fuel is not completely converted into electricity, the efficiency of these

generators ranges from 35 to 45% and they require a significant investment in pollution

control.

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946 Chapter 20 | Electrochemistry

The oldest known electric

battery in existence, discovered in

1938 in Baghdad, Iraq, consists of

a copper tube surrounding an iron

rod. If filled with an acidic liquid

such as vinegar, the cell could

produce a small electric current.

(Smith College Museum of Ancient

Inventions)

6V

+

2V

+

2V

+

2V

If three 2 volt cells are connected

in series, their voltages are additive

to provide a total of 6 volts. In

today’s autos, 12 volt batteries

containing six cells are the norm.

Hydroelectric power plants are claimed to have efficiencies around 90%. They are clean

and non-polluting. However, hydroelectric power often requires building large dams to

maintain a reservoir of water. Additionally they are often located large distances from their

customers, and transmission of electricity over long distances incurs significant losses.

Wind powered turbines and wind mills are also considered to be clean and non-polluting.

They are approximately 50% efficient and can be located very close to the end-user.

However, daily, even hourly, fluctuations in wind speed may introduce inconvenient variations in power production.

Chemical research is important in producing new turbine materials and combustion

processes to increase efficiencies of major electric generators. In this section we will concentrate on how chemical potential and intrinsic electrical properties of materials can be

used to generate smaller, but important, amounts of electricity.

Batteries

One of the most familiar uses of galvanic cells, popularly called batteries, is the generation

of portable electrical energy.7 These devices are classified as either primary cells (cells not

designed to be recharged; they are discarded after their energy is depleted) or secondary

cells (cells designed for repeated use; they are able to be recharged).

The common lead storage battery used to start an automobile is composed of a number of

secondary cells, each having a potential of about 2 V, that are connected in series so that

their voltages are additive. Most automobile batteries contain six such cells and give about

12 V, but 6, 24, and 32 V batteries are also available.

A typical lead storage battery is illustrated in Figure 20.10. The anode of each cell is

composed of a set of lead plates, the cathode consists of another set of plates that hold a

coating of PbO2, and the electrolyte is sulfuric acid. When the battery is discharging, the

electrode reactions are

PbO2(s) + 3H+(aq) + HSO4-(aq) + 2e- → PbSO4(s) + 2H2O

Pb(s) + HSO4-(aq) → PbSO4(s) + H+(aq) + 2e-

(cathode)

(anode)

The net reaction taking place in each cell is

PbO2(s) + Pb(s) + 2H+(aq) + 2HSO4−(aq)

2PbSO4(s) + 2H2O

2H2SO4

As the cell discharges, the sulfuric acid concentration decreases, which causes the density

of the electrolyte to drop. The state of charge of the battery can be monitored with a

hydrometer, which consists of a rubber bulb that is used to draw the battery fluid into a glass

tube containing a float (see Figure 20.11). The depth to which the float sinks is inversely

proportional to the density of the liquid—the deeper the float sinks, the lower is the density of the acid and the weaker is the charge on the battery. The narrow neck of the float is

usually marked to indicate the state of charge of the battery.

The principal advantage of the lead storage battery is that the cell reactions that occur

spontaneously during discharge can be reversed by the application of a voltage from an

external source. In other words, the battery can be recharged by electrolysis. The reaction

for battery recharge is

electrolysis

2PbSO 4 ( s ) + 2H 2O 

→ PbO2 ( s ) + Pb(s ) + 2H+ (aq ) + 2HSO4−(aq )

7

Strictly speaking, a cell is a single electrochemical unit consisting of a cathode and an anode. A battery is

a collection of cells connected in series.

jespe_c20_918-973hr.indd 946

11/18/10 4:59 PM

20.6 | Electricity 947

battery. A 12 volt lead storage

battery, such as those used in

most automobiles, consists of

six cells like the one shown

here. Notice that the anode and

cathode each consist of several

plates connected together. This

allows the cell to produce the

large currents necessary to start

a car.

(+)

(–)

storage battery

Alternating plates

of Pb and PbO2

H2SO4

electrolyte

PbO2

(cathode)

Pb

(anode)

Improper recharging of lead-acid batteries can produce potentially explosive H2 gas.

Most modern lead storage batteries use a lead–calcium alloy as the anode. That reduces the

need to have the individual cells vented, and the battery can be sealed, thereby preventing

spillage of the corrosive electrolyte.

Zinc–Manganese Dioxide Cells

The original, relatively inexpensive 1.5 V dry cell is the zinc–manganese dioxide cell, or

Leclanché cell (named after its inventor George Leclanché). It is a primary cell used to power

flashlights, toys, and the like, but it is not really dry (see Figure 20.12). Its outer shell is

made of zinc, which serves as the anode. The cathode—the positive terminal of the

battery—consists of a carbon (graphite) rod surrounded by a moist paste of graphite

powder, manganese dioxide, and ammonium chloride.

The anode reaction is simply the oxidation of zinc.

-

Zn(s) → Zn (aq) + 2e

2+

(anode)

Figure 20.11 | A battery

hydrometer. Battery acid is drawn

into the glass tube. The depth to

which the float sinks is inversely

proportional to the concentration

of the acid and, therefore, to the

state of charge of the battery.

(OPC, Inc.)

n This is “dry” compared to the

significant volume of liquid sulfuric

acid solution.

The cathode reaction is complex, and a mixture of products is formed. One of the major

reactions is

2MnO2(s) + 2NH4+(aq) + 2e- → Mn2O3(s) + 2NH3(aq) + H2O

(cathode)

Graphite

(cathode)

Paste of MnO2, NH4Cl,

and graphite powder

Porous spacer

Zinc shell

(anode)

Figure 20.12 | A cut-away view of a zinc–carbon

dry cell (Leclanché cell).

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