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Question

The emf of the following cell is 0.7995 V.

Pt|H2(1atm)|HNO3(1M)||AgNO3(1M)|Ag
If we add enough KCl to the Ag half cell so that the final Cl is 1M, then the measured emf of the cell is 0.222V. The Ksp of AgCl would be :

A
1×109.8
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B
1×1019.6
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C
2×1010
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D
2.64×1014
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Solution

The correct option is B 1×109.8
2Ag++H22H++2Ag
E=Eo0.05912log(H+)2PH2×(Ag+)2 ---- 1

Substitute E=0.222V,Eo=0.7995V,PH2=1 atm and [H+]=1M to get,
[Ag+]=109.8 from equation 1.
Now,

Ksp=[Ag+][Cl]=(109.8)×(1)=1×109.8

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