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Question

Consider the following cell
Pb(s)|Pb2+(0.5M)||Cu2+(0.001M)|Cu
EoCu=0.34V,EoPb=0.123V at 25oC
ΔG for the cell reaction is ______ kJ

A
50.92
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B
115.45
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C
73.92
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D
89.4
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Solution

The correct option is D 89.4
Gibbs free energy is additive,
ΔGfullreaction=nFEcell
ΔGfullreaction=2×96500×(.34(.123), here n= Valancy factor
ΔGfullreaction=89.4KJ

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