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Question

Find the potential of a half - cell that has the reaction, Ag2S+2e2Ag+S2 in a solution buffered at pH = 3 and which is also saturated with 0.1 M H2S.
For H2S:K1=108 and K2=2×1013, Ksp(Ag2S)=2×1048, E0Ag+|Ag=0.8 V

A
0.432 V
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B
1.658 V
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C
0.245 V
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D
0.144 V
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Solution

The correct option is D 0.144 V
Since the solution is buffered at pH = 3, so
H2SKa=K1×K22H++S2
Ka=108×2×1013=[H+]2[S2][H2S]
2×1021=(103)2[S2]0.1 [S2]=2×1016 M
Again
Ag2S(s)Ksp2Ag++S2
Ksp=[Ag+]2[S2] 2×1048=[Ag+]2×2×1016 [Ag+]=1016 M
So, for cell Ag2S+2e2Ag+S2 when [S2]=2×1016 M
Which is equivalent to emf of the cell Ag++eAg when [Ag+]=1016 M
Hence E=E00.0591log1[Ag+]E=0.80.0591log11016=0.80.944=0.144 V

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