Fundamentals of Electrochemistry - W.H. Freeman

1504T_ch14_270-297 2/14/06 8:42 Page 282
Box 14-3
Latimer Diagrams: How to Find E for a New Half-Reaction
A Latimer diagram displays standard reduction potentials E°,
connecting various oxidation states of an element. 11 For example,
in acid solution, the following standard reduction potentials are
observed:
The standard free-energy change, G°,
given by
G° nFE°
for a reaction is
?
When two reactions are added to give a third reaction, the sum of
the individual G° values must equal the overall value of G°.
1.318
IO 1.589
4
IO 1.154 1.430 0.535
To apply free energy to our problem, we write two reactions
3
HOI I 2
(s) I
Oxidation whose sum is the desired reaction:
(7) (5) (1) (0) (1) state
of iodine
1.210
IO 3 6H 5e E 11.210 ° ! 1 2 I 2 (s) 3H 2 O
As an example to understand what the Latimer diagram means, the
notation IO 3 ¡ 1.154 HOI stands for the balanced equation
IO 3
5H 4e T HOI 2H 2 O
We can derive reduction potentials for arrows that are not
shown in the diagram by using G°. For example, the reaction
shown by the dashed line in the Latimer diagram is
IO 3
6H 6e T I 3H 2 O
E° 1.154 V
To find E° for this reaction, express the reaction as a sum of reactions
whose potentials are known.
G° 1 5F(1.210)
1
2 I 2 (s) e E° 20.535
! I G° 2 1F(0.535)
IO 3
6H 6e E° 3?
! I 3H 2 O
But, because G° 1 G° 2 G° 3 , we can solve for E° 3 :
G° 3 G° 1 G° 2
6FE° 3 5F(1.210) 1F(0.535)
E° 3
5(1.210) 1(0.535)
6
G° 3 6FE° 3
1.098 V
Electrons flow toward more positive potential.
Figure 14-8 and note that electrons always flow toward a more positive potential. Therefore,
electrons in the circuit flow from cadmium to silver. The separation of the two halfcells
is 1.242 V. This diagram works the same way even if both half-cell potentials are
positive or both are negative. Electrons always flow toward more positive potential.
Different Descriptions of the Same Reaction
In Figure 14-4, the right half-reaction can be written
AgCl(s) e T Ag(s) Cl E° 0.222 V
E E° 0.059 16 log [Cl ] 0.222 0.059 16 log (0.033 4) 0.309 3 V
(14-17)
(14-18)
The Cl in the silver half-reaction was derived from 0.016 7 M CdCl 2 (aq).
Suppose that a different, less handsome, author had written this book and had chosen to
describe the half-reaction differently:
Ag e T Ag(s) E° 0.799 V
(14-19)
This description is just as valid as the previous one. In both cases, Ag(I) is reduced to Ag(0).
If the two descriptions are equally valid, then they should predict the same voltage. The
Nernst equation for Reaction 14-19 is
1
E 0.799 0.059 16 log
[Ag ]
Electrons flow
from Cd to Ag
Cd 2+ Cd
E = –0.461 V
Ag + Ag
E = 0.781 V
Figure 14-8 An intuitive view of cell
potentials. 2 Electrons always flow to the right in
this diagram.
–1.0 – 0.5 0 0.5 1.0
E(cell) = 0.781 – (–0.461)
= 1.242 V
282 CHAPTER 14 Fundamentals of Electrochemistry