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Question

We have cell
Cu/Cu2+//Ag+/Ag
Reactions are given as below
Cu2++eCu+E1=0.15V
Cu++eCuE2=0.5V
Ag++eAgE3=0.799V
Calculate Gibbs free Energy.

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Solution

we have cell
Cu/Cu2+//Ag+/Ag
oxidation half reaction
CuCu2++2e....(1)
Reduction half reaction
Ag++eAg2Ag+2e2Ag...(2)
So, cell reactions
Cu+2Ag+Cu2++2Ag...(3)
we have
Cu2++eCu+E1=0.15V...(4)
ΔG4=nEF=1EF=0.15FCu++eCuE2=0.5V....(5)
ΔG5=nEF=(1)EFΔG5=0.5F
Soaddingequations(4)&(5)wegetCu2++2eCu....(6)
ΔG6=ΔG4+ΔG5=0.15F+(0.5F)(2)EF=0.65FE=0.325V
We have E=0.325V
equation
Cu2++2eCuE=0.325V
which is reduction potention of electrode on left side
we have Ag++eAgE3=0.799VSo,2Ag++2e2Ag
E3 remains as it is E3=0.799V
which is reduction potential of right electrodes
so, we haveEcell=EREL=0.7990.325Ecell=0.474V
which is potential of cell that is for cell reaction equation..(3)
so ΔG3=(2)EF=2×0.474×96500ΔG3=91482J/mole

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