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,
`because E_(Ag^(+)//Ag)=E_(Ag^(+)//Ag)+(0.059)/(1)log_(10)[Ag^(+)]` ...1
Also `K_(SP_(AgI))=[Ag^(+)][I^(-)]`
`because [Ag^(+)] = [I^(-)]` (for a saturated solution)
`:. [Ag^(+)] = sqrt(K_(SP_(AgI))) = sqrt(8.7 xx 10^(-17)) = 9.32 xx 10^(-9)` ...(2)
`:.` By Eq. `(1)`
`E_(Ag^(+)//Ag) = 0.799 + (0.059)/(1)log_(10)(9.32 xx 10^(-9))`
`= 0.799 - 0.474 = 0.32 V`
Also `{:(AgrarrAg^(+)+e, E_(OP)^(@)=-0.799 V),(AgI_(s)+erarrAg+I^(-)):}/(AgIrarrAg^(+)+I^(-))`
`:. E_(cell) = E_(OP_(Ag//Ag^(+)))^(@) - (0.059)/(1)log[Ag^(+)]`
`E_(RP_(I^(-)//AgI//Ag))^(@) - (0.059)/(1)log.(1)/([I^(-)])` ...(3)
`because E_(cell) = 0` at equilibrium, thus, from Eq. `(3)`
`E_(OP_(Ag//Ag^(+)))^(@) + E_(RP_(I^(-)//AgI//Ag))^(@) = (0.059)/(1)log[Ag^(+)][I^(-)]`
`= (0.059)/(1)logK_(SP_(AgI))`
`-0.799 + E_(RP_(I^(-)//AgI//Ag))^(@) = (0.059)/(1)log8.7 xx 10^(-17)`
or `E_(RP_(I^(-)//AgI//Ag))^(@) = -0.948 + 0.799
`= -0.419 V`