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`.

Answer

`0.325V,-0.149V`

Explanation

`K_(sp)` of `AgI=[Ag^(o+)][I^(c-)]=8.7xx10^(-17)`
In a saturated solution of `AgI`
`[Ag^(o+)]=[I^(c-)]`
So, `[Ag^(o+)]=sqrt(K_(sp))=sqrt(8.7xx10^(-17))=9.33xx10^(-9)M`
By Nerst equation of `Ag^(o+)+e^(-) rarr Ag(s)`
`E_(cell)=E^(c-)._(cell)-(0.0591)/(n)log.(1)/(Ag^(o+))`
`=0.799-(0.0591)/(1)log.(1)/(9.33xx10^(-9))`
`=0.325V`
For standard `I^(o+)|Ag|Ag` electrode , `[I^(o+)]=1.0M`
`K_(sp)` of `AgI=[Ag^(o+)][I^(c-)]`
`8.7xx10^(-17)=[Ag^(o+)][1.0]`
`:. [Ag^(o+)]=(8.7xx10^(-17))/(1.0)=8.7xx10^(-17)`
`:.` Electrode potential
`E_(cell)=E^(c-)._(cell)-(0.0591)/(n)log.(1)/([Ag^(o+)])`
`E_(I^(c-)|Ag|Ag)=0.799-(0.0591)/(1)log.(1)/(8.7xx10^(-17))`
`=-0.149V`