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`.

Explanation

`K_(SP)` of `AgI = [Ag^(+)][I^(-)] = [Ag^(+)][0.05]`
For given cell `E_(Cell) = E_(OP_(Ag)+E_(RP_(Ag))` …(1)
`= E_(OP_(Ag//Ag^(+)))-(0.059)/(1)"log"[Ag^(+)]_(L.H.S.)`
`+E_(RP_(Ag^(+)//Ag))-(0.059)/(1)"log"[Ag^(+)]_(R.H.S.)`
`E_(Cell) = (0.059)/(1)"log"[Ag^(+)]_(R.H.S.)/[Ag^(+)]_(L.H.S.)`
`[because E_(OP_(Ag//Ag^(+)))^(@)= E_(RP_(Ag^(+)//Ag))^(@)]`
`= 0.788 = "log"(0.05)/([Ag^(+)]_(L.H.S.))`
`:. [Ag^(+)]_(L.H.S.) = 2.203 xx 10^(-15)`
By eq. (1) `K_(SP) = [2.203 xx 10^(-15)][0.05]`
`K_(SP_(AgI))= 1.10 xx 10^(-16)`