A half cell reaction: Ag2S(s)+2e−→3Ag(s)+S2− is carried out in a half cell PtAg2S|Ag,H2S(0.1M)at[H⊕]=10−3, The EMF of the half cell is: [E⊖Ag⊕|Ag=0.80V,Ka(H2S)=10−21,andKspofAg2S=10−49]
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 H2S⇌2H⊕+S2− Ka=[H⊕]2[S2−][H2S]=(10−3)2×[S2−]0.1 ∴[S2−]=10−21×0.110−16 Since [Ag⊕]2[S2−]=Ksp ∴[Ag⊕]=√Ksp[S2−]