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

Explanation

Given, `E_(Ag^(+)|Ag)^(@)=0.799" (at 298 K) and "K_(sp)" of " Agl=8.7xx10^(-17)`
In a saturated solution of `Agl, [Ag^(+)]=[I^(-)]`
`therefore" "[Ag^(+)]=sqrt(K_(sp))=sqrt(8.7xx10^(-17))=9.33xx10^(-9)`
According to Nernst equation of `Ag^(+)+e rarr Ag(s)`
`therefore" "E_(Ag^(+)|Ag)=E_(Ag^(+)|Ag)^(@)-(0.059)/(n)log.(1)/([Ag^(+)])`
`"or, "E_(Ag^(+)|Ag)^(@)=0.799-(0.059)/(1)log.(1)/([Ag^(+)])`
`"or, "E_(Ag^(+)|Ag)^(@)=0.799-0.059log.(1)/(9.33xx10^(-9))`
`therefore" "E_(Ag^(+)|Ag)^(@)=0.325V`
The relation between `E_(I^(-)|Agl|Ag)^(@) and E_(Ag^(+)|Ag)^(@)` is
`E_(I^(-)|Agl|Ag)^(@)=E_(Ag^(+)|Ag)^(@)+0.059logK_(sp)`
`therefore" "E_(I^(-)|Agl|Ag)^(@)=0.799+0.059log(8.7xx10^(-17))=-0.148V`