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

We know,
`becauseE_(Ag^(+)//Ag)=E_(Ag^(+)//Ag)+(0.059)/(1)log_(10)[Ag^(+)]`...(1)
Also `K_(SP_(Agl))=[Ag^(+)][I^(-)]`
`because[Ag^(+)]=[I^(-)]` (for a saturated solution)
`therefore[Ag^(+)]=sqrt(K_(SP_(Agl)))sqrt(8.7xx10^(-17))=9.32xx10^(-9)`...(2)
`therefore` By Eq. (1)
`E_(Ag^(+)//Ag)=0.799+(0.059)/(1)log_(10)(9.32xx10^(-9))`
`=0.799-0.474=0.32V`
Also `AgrarrAg^(+)+e,E_(OP)^(0)=-0.799V`
`(AgI_((S))+erarrAg+I^(-))/(AgIrarrAg^(+)+I^(-))`
`thereforeE_(cell)=E_(OP_(Ag//Ag^(+)))^(0)-(0.059)/(1)log[Ag^(+)]+E_(RP_(I^(-)//AgI//Ag))^(0)+(0.059)/(1)log.(1)/([I^(-)])`...(3)
`becauseE_(cell)=0` at equilibrium, thus, from Eq. (3)
`E_(O_(Ag//Ag^(+)))^(0)+E_(RP_(I^(-)//AgI//Ag))^(0)=(0.059)/(1)log[Ag^(+)][I^(-)]`
`=(0.059)/(1)logK_(SP_(AgI))`
`-0.799+E_(RP_(I^(-)//AgI//Ag))^(0)=(0.059)/(1)log8.7xx10^(-17)`
`orE_(RP_(I^(-)//AgI//Ag))^(0)=-0.948+0.799=-0.149V`