The standard reduction potential of the `Ag^(+)//Ag` electrode at 298 K is 0.799 V. Given that `K_(sp)` for Agl`=8.7xx10^(-17)`, evaluate the potential of `Ag^(+)//Ag` electrode in a saturated solution of AgI.

Explanation

`Agl_((s)) Leftrightarrow Ag_((aq))^(+)+I_((aq))^(-)`
`K_(sp)=[Ag^(+)] [I^(-)]=[Ag^(+)]^(2)`
`[Ag^(+)]=sqrt(K_(sp))=sqrt(8.7 xx 10^(-17)) =9.33 xx 10^(-9)M`
`Ag^(+) +theta to Ag`
`E_(Ag^(+)//Ag)=E_(Ag^(+)//Ag)^(0)-(0.0591)/(1) log""(1)/([Ag^(+)])`
`=0.799 -0.0591 log (1)/((9.33 xx 10^(-8))=0.799 -0.475=0.324V`
`Ag to Ag^(+) +e" "E_(Ag//Ag^(+))^(0)=0.799V`
`AgI+e to Ag_((s))+I^(-)" "E_(I^(-)//AgI//Ag)^(0)=xV`
`="Net cell reaction AgI"_((s)) to [Ag^(+)] [I^(-)]" "E_("cell")^(0)=(0.799+x)`
`K=[Ag^(+)] [I^(-)]=K_(sp)`
`E_("cell") =E_("cell")^(0)-(0.0591)/(1) log K_(sp) (AgI)`
`=0.799V +x=0.0591 log 8.7 xx 10^(-17)`
`rArr 0.799V =-0.949 V`
`x=E_(I^(-)//AgI//Ag)^(0)=-0.150V`