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Question

The solubility product of AgBr(s) is 5×1013 at 298 K. If the standard reduction potential of the half cell E0Br|AgBr|Ag is 0.07 V.
Find the value of the E0Ag+/Ag

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Solution

AgBr(s)Ag+(aq)+Br(aq)......(1)
ΔG1=ΔG01+RTln Q
At equilibrium,
ΔG1=0
Q=Ksp
ΔG01=RT ln Ksp

Ag+(aq)+eAg(s)........(2)
ΔG02=nFE0Ag+/Ag

Adding equation (1) and (2)
AgBr(s)+eAg(s)+Br.......(3)
ΔG03=nFE0Br/AgBr/Ag

Adding equation (1) and (2) gives (3)

ΔG03=ΔG01+ΔG02
nFE0Br/AgBr/Ag=RT ln KspnFE0Ag+/Ag

nFE0Br/AgBr/Ag=RT ln Ksp+nFE0Ag+/Ag

E0Br/AgBr/Ag=RTnF ln Ksp+E0Ag+/Ag

n=1 for given electrode reaction,

E0Br/AgBr/Ag=RTF ln Ksp+E0Ag+/Ag

E0Br/AgBr/Ag=2.303RTF log Ksp+E0Ag+/Ag

E0Br/AgBr/Ag=0.0591 log Ksp+E0Ag+/Ag

E0Br/AgBr/Ag=0.0591 log 50.0591×13 log 10+E0Ag+/Ag

0.07=0.0410.77+E0Ag+/Ag

E0Ag+/Ag=0.799 V

E0Ag+/Ag0.8 V

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