    Question

# Find the potential of a half - cell that has the reaction, Ag2S+2e−→2Ag+S2− in a solution buffered at pH = 3 and which is also saturated with 0.1 M H2S. For H2S:K1=10−8 and K2=2×10−13, Ksp(Ag2S)=2×10−48, 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 VSince the solution is buffered at pH = 3, so H2SKa=K1×K2⇌2H++S2− ∴Ka=10−8×2×10−13=[H+]2[S2−][H2S] ⇒2×10−21=(10−3)2[S2−]0.1 ⇒[S2−]=2×10−16 M Again Ag2S(s)Ksp⇌2Ag++S2− ∴Ksp=[Ag+]2[S2−] ⇒2×10−48=[Ag+]2×2×10−16 ⇒[Ag+]=10−16 M So, for cell Ag2S+2e−→2Ag+S2− when [S2−]=2×10−16 M Which is equivalent to emf of the cell Ag++e−→Ag when [Ag+]=10−16 M Hence E=E0−0.0591log1[Ag+]⇒E=0.8−0.0591log110−16=0.8−0.944=−0.144 V  Suggest Corrections  0      Similar questions  Related Videos   Redox Reactions
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