The `e.m.f` of cell `Ag|AgI_((s)),0.05MKI|| 0.05 M AgNO_(3)|Ag` is `0.788 V`. Calculate solubility product of `AgI`.

The `e.m.f` of cell `Ag|AgI_((s)),0.05MKI|| 0.05 M AgNO_(3)|Ag` is `0.

1.	The `e.m.f` of cell `Ag\|AgI_((s)),0.05MKI\|\| 0.05 M AgNO_(3)\|Ag` is `0.788 V`. Calculate solubility product of `AgI`.
Answer» `E_(cell)=E^(@)_(cell)=(0.591)/(1)log[Al^(+)]_(LHS)/[(Ag^(+))]_(RHS)` `0.8=0-(0.0591)/(1)log(C_(1))/(0.2)` `log(0.2)/(C_(1))=(0.8)/(0.0591)=(13.54)` `(0.2)/(C_(1))="Antilog"(13.54)` `(0.2)/(C_(1))=3.467xx10^(13)` `C_(1)=(0.2)/(3.467xx10^(13))` `C_(1)=5.77xx10^(-15)` `K_(sp)=[Ag^(+)][I^(-)]=5.77xx10^(-15)xx0.19` `k_(sp)=1.1xx10^(-16)`

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The `e.m.f` of cell `Ag|AgI_((s)),0.05MKI|| 0.05 M AgNO_(3)|Ag` is `0.788 V`. Calculate solubility product of `AgI`.

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