EXAMPLE 21-6 The Nernst Equation
A cell is constructed at 25°C as follows. One half-cell consists of the Fe^3 /Fe^2 couple in which
[Fe^3 ]1.00 Mand [Fe^2 ]0.100 M; the other involves the MnO 4 /Mn^2 couple in acidic
solution in which [MnO 4 ]1.00 10 ^2 M,[Mn^2 ]1.00 10 ^4 M,and [H]1.00
10 ^3 M.(a) Find the electrode potential for each half-cell with these concentrations, and
(b) calculate the overall cell potential.
Plan
(a) We can apply the Nernst equation to find the reduction potential of each half-cell with the
stated concentrations. (b) As in Section 21-15, we write the half-reaction with the more posi-
tive potential (aftercorrection) along with its potential. We reverse the other half-reaction and
change the sign of its Evalue. We balance the electron transfer and then add the half-reactions
and their potentials to find the overall cell potential.
Solution
(a) For the MnO 4 /Mn^2 half-cell as a reduction,
MnO 4 8H 5 e88nMn^2 H 2 O E^0 1.507 VEE^0 0.0
n592
log
[Mn[
OM
4n
2][H]
] 81.507 V
0.0
5592
log1.507 V
0.0
5592
log (1.00^1022 )1.507 V0.0
5592
(22.0)1.246 V(b) For the Fe^3 /Fe^2 half-cell as a reduction,
Fe^3 e88nFe^2 E^0 0.771 VEE^0 0.0
n592
log[
[F
Fe
e2
3
]
]0.771 V
0.0
1592
log0
1.1
.00
00
0.771 V
0.0
1592
log (0.100)0.771 V0.0
1592
(1.00)0.830 VThe corrected potential for the MnO 4 /Mn^2 half-cell is greater than that for the Fe^3 /Fe^2
half-cell, so we reverse the latter, balance the electron transfer, and add.
E(corrected)
MnO 4 8H 5 e88nMn^2 4H 2 O 1.246 Vrr
5(Fe^2 88nFe^3 e) 0.830 VrrMnO 4 8H5Fe^2 88nMn^2 4H 2 O5Fe^3 Ecell0.416 VThe reaction in Example 21-6 is product-favored (spontaneous) under the stated condi-
tions, with a potential of 0.416 volt when the cell starts operation.As the cell discharges
and current flows, the product concentrations, [Mn^2 ] and [Fe^3 ], increase. At the same
time, reactant concentrations, [MnO 4 ], [H], and [Fe^2 ], decrease. This increases
1.00 10 ^4
(1.00 10 ^2 )(1.00 10 ^3 )^821-19 The Nernst Equation 879