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_(RPAg)`
`=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))`
`[E_(OP_(Ag//Ag^+))^0=E_(RP_(Ag^+//Ag))]0.788=log(0.05)/([Ag^+]_(L.H.S))`
`:. [Ag^+]_(L.H.S)=2.203xx10^(-15)`
by equation(i)
`K_(sp)=[2.303xx10^(-15)][0.05] K_(sp Agl)=1.10xx10^(-16)`