Page 164 - Modern Analytical Chemistry

1400-CH06 9/9/99 7:40 AM Page 147

Chapter 6 Equilibrium Chemistry 147

RT
o
E= E – ln Q
nF
Substituting appropriate values for R and F, assuming a temperature of 25 °C
(298 K), and switching from ln to log* gives the potential in volts as

. 0 05916
o
E= E – log Q 6.24
n
The standard-state electrochemical potential, E°, provides an alternative way of
expressing the equilibrium constant for a redox reaction. Since a reaction at equilib-
rium has a ∆G of zero, the electrochemical potential, E, also must be zero. Substi-
tuting into equation 6.24 and rearranging shows that
RT
o
E = log K 6.25
nF
Standard-state potentials are generally not tabulated for chemical reactions, but are
calculated using the standard-state potentials for the oxidation, E° ox , and reduction
half-reactions, E° red . By convention, standard-state potentials are only listed for re-
duction half-reactions, and E° for a reaction is calculated as
E° reac = E° red – E° ox
where both E° red and E° ox are standard-state reduction potentials.
Since the potential for a single half-reaction cannot be measured, a reference half-
reaction is arbitrarily assigned a standard-state potential of zero. All other reduction
potentials are reported relative to this reference. The standard half-reaction is
–
+
2H 3 O (aq)+2e t 2H 2 O(l)+H 2 (g)
Appendix 3D contains a listing of the standard-state reduction potentials for se-
lected species. The more positive the standard-state reduction potential, the more
favorable the reduction reaction will be under standard-state conditions. Thus,
under standard-state conditions, the reduction of Cu 2+ to Cu (E° = +0.3419) is
2+
more favorable than the reduction of Zn to Zn (E° = –0.7618).

EXAMPLE 6. 5
Calculate (a) the standard-state potential, (b) the equilibrium constant, and
+
2+
(c) the potential when [Ag ] = 0.020 M and [Cd ] = 0.050 M, for the
following reaction taking place at 25 °C.
+
2+
Cd(s) + 2Ag (aq) t Cd (aq) + 2Ag(s)
SOLUTION
+
(a) In this reaction Cd is undergoing oxidation, and Ag is undergoing
reduction. The standard-state cell potential, therefore, is
o
0
1
E = E o Ag / Ag – E o Cd 2 + / Cd = .7996 V – (– .4030 V) = .2026 V
0
+
(b) To calculate the equilibrium constant, we substitute the values for the
standard-state potential and number of electrons into equation 6.25.
0 05916
.
.
1 2026 = log K
2
*ln(x) = 2.303 log(x)