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

Explanation

In the saturated solution of AgI, the half-cell reaction are :
`{:(,AgI+e^(-) rarr Ag+I^(-),"Cathode (Reduction)"),(,Ag rarr Ag^(+)+e^(-),"Anode (Oxidation)"),("Cell reaction",bar(AgI rarr Ag^(+)+I^(-)),):}`
`E_(Ag^(+)//Ag)=E_(Ag^(+)//Ag)^(@)+0.0591 log [Ag^(+)]`
`[Ag^(+)][I^(-)]=K_(sp) (AgI)=[Ag^(+)]^(2)=[I^(-)]^(2)`
So, `[Ag^(+)]^(2)=8.7xx10^(-17)`
`[Ag^(+)]=sqrt(8.7xx10^(-17))=9.3xx10^(-9)`
Substituting the values of `E_(Ag^(+)//Ag)^(@)` and `[Ag^(+)]` in the above equation.
`E_(Ag^(+)//Ag)=0.799+0.0591 log (9.3xx10^(-9))`
`=0.324` volt
`E_(cell)^(@)=0.0591 log K_(sp) (AgI)`
`=0.0591 log (8.7xx10^(-17))`
`=-0.95` volt
`E_(cell)^(@)=` Oxid. pot. of anode + Red. pot. of cathode
Red. pot. of cathode `E_(I^(-)//AgI//Ag)^(@)=-0.95-(-0.799)`
`=-0.95+0.799`
`=-0.151` volt