A galvanic cell at a temperature of 25.0°C is powered by the following redox reaction: 2MnO4^-(aq)+16H+(aq)+5Pb(s)-->2M n^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

particle   p is an electron and particle q is a neutron