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

`Agl(s)rarrAg^(+)(aq)+I^(-)(aq)`
`K_(sp)=[Ag^(++)]=sqrt(K_(sp))=9.33xx10^(-9)M`
`Ag^(+)+erarrAg`
`E_(Ag^(+)//Ag)=E^(@)_(Ag^(+)//Ag)-0.0591 log (1)/([Ag^(+)]`
`=0.799-0.0591 log (1)/([Ag^(+)])`
=0.799-0.0591 `log(1)/(9.33xx10^(-9))`
`=0.799-0.475=0.324 V`
`Ag rarrAg^(+)+e^(-) E^(@)=-0.799 V`
`Agl+e rarr Ag(s)+e^(-) E^(@)//Agl//Ag)=xV`
`Agl(s) rarr Ag^(+)I^(-) E_(cell)^(W)=(-0.799+X)`
`E_(cell)=e_(cell)^(@)-(0.059)/(1)log([Ag^(+)[I^(-)]`
At equilibrium `E_(cell)=0`
`rarr E_(cell)^(@)=0.0591log (K_(sp))`
`-0.799+xx=0.0591 log (8.7xx10^(-17))`
X=-0.150 V