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The e.m.f of cell Ag|AgI((s)),0.05M KI||...

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

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The correct Answer is:
`K_(SP)=1xx10^(-18)`
`AgtoAg_((c))^(+)+e^(-),Cxx(0.05)K_(SP)`
`Ag_((0.05M))^(+)+e^(-)toAg` if assume as concentration cell
`Ag_((0.05M))^(+)toAg_((c))^(+)`
`E=0.79=0-(0.0591)/(1)log((K_(sp))/(5xx10^(-2)xx5xx10^(-2)))`
`(K_(SP))/(25xx10^(-4))=4xx10^(-14)impliesK_(SP)=1xx10^(-16)`
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