Physical Chemistry Third Edition

8.3 Half-Cell Potentials and Cell Potentials 369

EXAMPLE 8.5

Assuming that∆H◦is constant, findE◦at 323.15 K for the Daniell cell. Neglect the liquid

junction potential.

Solution

E◦ 1 .1030 V at 298.15 K. We find from enthalpy changes of formation in Table A.8 of

Appendix A that∆H◦− 218 .66 kJ mol−^1. From Eq. (8.3-8)

E◦(323.15 K)

323 .15 K



1 .1030 V

298 .15 K

−

−218660 J mol−^1

(2)(96485 C mol−^1 )

[

1

323 .15 K

−

1

298 .15 K

]

 3. 699 × 10 −^3 VK−^1 − 2. 94 × 10 −^4 VK−^1

 3. 405 × 10 −^3 VK−^1

E◦(323.15 K)(3. 405 × 10 −^3 VK−^1 )(323.15 K) 1 .100 V

Exercise 8.9

FindE◦at 75◦C for the cell with a hydrogen electrode on the left, a silver–silver chloride

electrode on the right, and a solution of HCl. Assume that∆H◦is temperature-independent.

PROBLEMS

Section 8.3: Half-Cell Potentials and Cell Potentials

8.7Find the cell voltage of an electrochemical cell with a

Pb/PbSO 4 electrode on the left, a PbO 2 /PbSO 4 electrode

on the right, and a solution of H 2 SO 4 with a molality of

1.00 mol kg−^1 at 298.15 K. Assume that the mean ionic

activity coefficient of H 2 SO 4 at this molality is equal

to 0.36.

8.8 a.Using the extrapolation of Eq. (8.2-21), findE◦

for the cell of Figure 8.2. The following are

(contrived) data for the cell voltage at 298.15 K

withP(H 2 )P◦:

m/mol kg−^1 : 0.0100 0.0200 0.0300 0.0400 0.0500

E/volt: 0.4643 0.4305 0.4108 0.3970 0.3863

m/mol kg−^1 : 0.0600 0.0700 0.0800 0.0900 0.1000

E/volt: 0.3776 0.3703 0.3639 0.3583 0.3533

b.Using the extrapolation of Eq. (8.2-22), findE◦for the

cell of Figure 8.2.

8.9 a.Write the cell symbol for the cell with the half-reactions

Pb^2 ++ 2 e−−→Pb(s)

AgCl(s)+Cl−−→Ag(s)+e−

Why is a salt bridge needed? What substance would

you not use in the salt bridge?

b.Draw a sketch of the cell of part a.

c.Find the value ofEfor the cell if the activity of Pb^2 +is

0.100 and that of the Cl−is 0.200 on the molality scale.

State any assumptions.

8.10 Instead of the oxidation half-reaction used in Example 8.1,

use the reverse of the half-reaction

2H 2 O+ 2 e−−→H 2 +2OH− E◦− 0 .8277 V

to calculate the reversible cell potential. Since the OH−ion

is not in its standard state, the Nernst equation must be

applied.

8.11 In calculating theE◦value for a cell from half-reactionE◦

values, the half-reactionE◦values are added or subtracted

directly, without multiplying by the number of electrons in