Question

Calculate the delta G of reaction for the following cell:

Pb (s) | Pb+2 (0.01393 M) || Cd+2 (0.0511 M) | Cd (s)

Answer 1

Cd²⁺ + 2e^- Cd           E^o = -0.40 V

Pb Pb²⁺ + 2e^-           E^o = 0.126 V

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Cd²⁺ + Pb Cd + Pb²⁺              E^o = -0.40 + 0.126 = -0.274 V

E(cell) = E°(cell) – [(0.0591)/n]•log Q

Q = [Pb²⁺]/[Cd²⁺] = 0.273

E(cell) = -0.274 - (0.0591/2) log (0.273) = -0.290 V

∆G = –nFE(cell) = -2 96500 (-0.290) = 55970 J/mol = 55.97 kJ/mol