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`.

Explanation

`Ag|0.1N AgNO_(3)||1N KBr, AgBr(s)|Ag`
The above cell is concentration cell
`:. E^(c-)._(cell)=0`
`Ag^(o+)._((anode))rarr Ag^(o+)(0.1N)+e^(-)`
`Ag^(o+)._((cathode))+e^(-)rarr Ag_((cathode))`
`ulbar(Ag^(o+)._((cathode))rarr Ag^(o+)(0.1N))`
Since `AgNO_(3)` is `81.3%` ionized
`:. [Ag^(o+)]_(a)=0.1xx0.813=0.0813N`
`-0.64=0-0.059log .(0.0813)/([Ag^(o+)._((cathode))])`
`:. [Ag^(o+)]_((cathode))=3.2xx10^(-12)M`
Since `1 N KBr` is `75.5%` dissociated
`:. [Br^(c-)]=1xx0.755=0.755 M`
`K_(sp)[Ag^(o+)][Br^(c-)]=3.2xx10^(-12)xx0.755`
`2.416xx10^(-12)`
`S=sqrt(2.416xx10^(-12))`
`=1.544xx10^(-6)M`