The Foundations of Chemistry

Solution

The standard reduction potential for the Sn^4 /Sn^2 couple is 0.15 volt; that for the Cr^3 /Cr
couple is 0.74 volt. The equation for the reaction shows Cr being oxidized to Cr^3 , so the
sign of the E^0 value for the Cr^3 /Cr couple is reversed. The overall reaction, the sum of the
two half-reactions, has a cell potential equal to the sum of the two half-reaction potentials.

E^0

3(Sn^4  2 e88nSn^2 ) 0.15 V)

2(Cr88nCr^3  3 e) (0.74 V)

3Sn^4 2Cr88n3Sn^2 2Cr^3  E^0 cell0.89 V)

The positive value of E^0 cellindicates that the forward reaction is spontaneous.

G^0 nFE^0 cell


6

m

m

ol

o

r

l

x

e

n






9

V

.65

m



ol

10

e

(^4) J
(0.89 V)
5.2 105 J/mol rxn or 5.2 102 kJ/mol rxn
You should now work Exercise 95.
EXAMPLE 21-10 Calculation of K from Cell Potentials
Use the standard cell potential to calculate the value of the equilibrium constant, K,at 25°C
for the following reaction.
2CuPtCl 62 88n2CuPtCl 42 2Cl
Plan
We calculate E^0 cellfor the reaction as written. Then we use it to calculate K.
Solution
First we find the appropriate half-reactions. Cu is oxidized to Cu, so we write the Cu/Cu
couple as an oxidation and reverse the sign of its tabulated E^0 value. We balance the electron
transfer and then add the half-reactions. The resulting E^0 cellvalue can be used to calculate the
equilibrium constant, K,for the reaction as written.
2(Cu88nCue) (0.521 V)
PtCl 62  2 e88nPtCl 42 2Cl 0.68 V0)
2CuPtCl 62 88n2CuPtCl 42 2Cl E^0 cell0.16 V0)
Then we calculate K.
ln K
nF
R
E
T
0
cell12.5
Ke12.5 2.7 105
At equilibrium, K2.7 105.
The forward reaction is product-favored (spontaneous), and the equilibrium lies far to the right.
You should now work Exercises 96 and 98.
[Cu]^2 [PtCl 42 ][Cl]^2

[PtCl 62 ]
(2)(9.65 104 J/Vmol)(0.16 V)

(8.314 J/molK)(298 K)
21-21 The Relationship of E^0 cellto
G^0 and K 885
The quite negative value of
G^0 tells
us that the reaction is product-favored.
This tells us nothing about the speed
with which the reaction would occur.
As the problem is stated, we must keep
the equation as written. We must
therefore accept either a positive or a
negative value of E^0 cell. A negative
value of E^0 cellwould lead to K1.