A galvanic cell at a temperature of 25.0°C is powered by the following redox reaction:

2MnO4^ - (aq) + 16H + (aq) + 5Pb (s) - ->2Mn^2 + (aq) + 8H2O (l) + 5Pb^2 + (aq)

Suppose the cell is prepared with 1.87 M MnO-4 and 1.37 M H + in one half-cell and 3.23 M Mn+2 and 6.62 M Pb+2 in the other. Calculate the cell voltage under these conditions. Round your answer to 3 significant digits.

1.63 V

Explanation:

Let us state the reaction equation again for the purpose of clarity;

2MnO4^ - (aq) + 16H + (aq) + 5Pb (s) - ->2Mn^2 + (aq) + 8H2O (l) + 5Pb^2 + (aq)

The reduction potentials for the two half reaction equations are;

MnO 4 - (aq) + 8H + (aq) + 5e - → Mn2 + (aq) + 4H2O (l) Eo=1.51 V

Pb2 + (aq) + 2e - → Pb (s) Eo = - 0.13 V

E°cell = E°red - E°Ox

E°cell = 1.51 - (-0.13)

E°cell = 1.51 + 0.13

E°cell = 1.64 V

But Q = [Mn^2+]^2 [Pb^2+]^5/[MnO4^-]^2 [H^+]^16

Q = [3.23]^2 [6.62]^5/[1.87]^2 [1.37]^16

Q = 10.43 * 12714.22/3.4969 * 154

Q = 132609.3/538.5226

Q = 246.25

From Nernst equation

E = E° - 0.0592/n log Q

Where n=10

E = 1.64 - 0.0592/10 log 246.25

E = 1.64-0.0142

E = 1.63 V