Question

A voltaic cell consists of an Mn/Mn2+ half-cell and a Pb/Pb2+ half-cell. Calculate [Pb2+] when [Mn2+] is 1.5 M and Ecell is 0.17 V.

Answer 1

Pb²⁺ + 2e- --------------> Pb Eo = - 0.13 V

Mn²⁺ + 2e- ---------------> Mn Eo = - 1.18 V

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Mn + Pb²⁺ ----------------> Mn²⁺ + Pb

E^ocell = -0.13 - (-1.18 ) = + 1.05 V

no of electrons transferred n = 2

Given that Ecell = + 0.17 V

[Mn2+] = 1.5 M

E_cell = E^o_cell - (0.059/n) Iog {[Mn2+]/[Pb2+]} where n = no of electrons transferred

+ 0.17 = + 1.05 - (0.059/2) Iog {[1.5 ]/[Pb2+]}

(0.059/2) Iog {[1.5 ]/[Pb2+]} = 0.88

[Pb2+] = 2.2 x 10^-30 M