The emf of the cell `Ag|AgI|CI(0.05M)||AgNO_(3)(0.05M)|Ag` is 0.79 V. Calculate the solubility product of AgI.

Answer

`K_(SP)=1xx10^(-18)`

Explanation

`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)`