Physical Chemistry Third Edition

364 8 The Thermodynamics of Electrochemical Systems

double cell. The procedure that led to Eq. (8.3-5) for the cell of Figure 8.4 can be

applied to any pair of electrodes that can be combined to make a cell without liquid

junction:To obtain the standard-state potential difference of any cell, take the standard

reduction potential of the right half-cell and subtract the standard reduction potential

of the left half-cell from it.This procedure allows us to use a table of standard electrode

potentials for half-cells, which is a much shorter table than a table containing entries

for all possible cells.

Exercise 8.6

a.Show that from a table ofNhalf-cell potentials, the potential differences forN(N−1)/ 2

cells can be calculated if each half-cell can be combined with every other half-cell to make

a cell.

b.Find the number of cells that can be made from 100 electrodes, assuming that each can be

paired with any one of the others.

Table A.13 in the appendix gives values for standard reduction potentials for a num-

ber of half-cells. Longer versions of such tables are available in handbooks. Unfor-

tunately, some older works use values that are the negative of the standard reduction

potentials (they are calledoxidation potentials). If you are not certain whether an

old table gives reduction potentials or oxidation potentials, look for an active metal

electrode such as sodium or potassium. If the table contains reduction potentials, the

half-cell potential of such a metal electrode will be negative.

There have been a number of theoretical approaches to the determination of

“absolute” electrode potentials(relative to the electric potential at a location infinitely

distant from all charges). All of them require the use of nonthermodynamic theories.

One work cites a value of− 4 .43 V (absolute) for the standard hydrogen electrode.^2

Other workers have come up with values ranging from this value to− 4 .73 V. We will

use only half-cell potentials relative to the standard hydrogen electrode.

EXAMPLE 8.3

Write the cell reaction equation and find the standard-state potential difference of the cell

Pt(s)|Cl 2 (g)|FeCl 2 (aq)|Fe(s)|Pt(s)

Solution

The half-cell reaction equations are

anode: 2Cl−−→Cl 2 (g)+ 2 e−

cathode: Fe^2 ++ 2 e−−→Fe(s)

The standard-state potential difference of the cell is

E◦− 0 .409 V−(+ 1 .3583 V)− 1 .767 V

SinceE◦is negative the cell reaction would spontaneously proceed in the reverse direction

under standard conditions.

(^2) H. Reiss and A. Heller,J. Phys. Chem., 89 , 4207 (1985).