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Question

A half cell reaction: Ag2S(s)+2e3Ag(s)+S2 is carried out in a half cell PtAg2S|Ag,H2S(0.1M)at[H]=103, The EMF of the half cell is:
[EAg|Ag=0.80V,Ka(H2S)=1021,andKspofAg2S=1049]

A
0.1735V
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B
0.19V
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C
+0.1735V
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D
+0.19V
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Solution

The correct option is A 0.1735V
H2S2H+S2
Ka=[H]2[S2][H2S]=(103)2×[S2]0.1
[S2]=1021×0.11016
Since [Ag]2[S2]=Ksp
[Ag]=Ksp[S2]
=10491016
=1033
2Ag+2e2Ag
ES2|Ag2S|Ag=EAg|Ag
ES2|Ag2S|Ag=EAg|Ag+0.0592log[Ag]2
=0.80+0.0592log1033
=0.1735V

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