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Question

The total change in free energy during the cell reaction in Cu|Cu2+1M||A1Mg+|Ag would be:

A
80.1kJ
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B
40.05kJ
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C
31.36kJ
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D
62.72kJ
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Solution

The correct option is A 80.1kJ
Given,
Cu2++eCu+;ΔGo1=Eo1F×1

Cu++eCu;ΔGo2=Eo2F×1
----------------------------
by adding these equations

Cu2++2eCu;

ΔGo3=[ΔGo1+ΔGo2]=[0.15F+0.5F]

or 2×Eo3F=0.65F
(as ΔG=nFEo)
Eo3=0.325V
So for reverse reaction.
CuCu2++2e;Eo=0.325V

CuCu2++2e;Eo=0.325V
2Ag++2e2Ag;Eo=0.799V
Eocell=EoOPCu/Cu2++EoRPAg+/Ag
=0.325+0.799=0.474V
Now according to Nernst equation:

E=Eo0.0591(logQ)/n

Ecell=Eocell+0.0592log[Ag+]2[Cu2+]
=+0.474+0.0592log(0.1)21=0.415V

The decrease in Gibbs energy =ΔG=nEF
=2×0.415×96500
=80095J=80.1kJ


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