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

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`.

Text Solution

Verified by Experts

`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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