Question

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

Answer 1

Step I. Calculation of `[Ag^(+)]` in solution.
The ionisation of AgI may be represented as :
`AgI(s) underset(S)hArr Ag^(+)(aq)+underset(S)I^(-)(aq)`
`K_(sp)=[Ag^(+)(aq)][I^(-)(aq)]=SxxS`
or `" "` S i.e., `[Ag^(+)]=(K_(sp))^(1//2)=(8.7xx10^(-17))^(1//2)=9.327xx10^(-9) M`
Step II. To calculate the reduction potential of `Ag^(+)//Ag`
`Ag^(+)(aq)+e^(-) to Ag(s)`
Here `" " [Ag^(+)(aq)]=9.327xx10^(-9)M,E_(Ag^(+)//Ag)^(@)=0.799" V"`, n=1
According to Nernst equation,
`E_((Ag^(+)//Ag))^(@)=E_((Ag^(+)//Ag))^(@)-(0.0591)/(n)"log"(1)/([Ag^(+)(aq)])=0.799-(0.0591)/(1)"log"(1)/((9.327xx10^(-9)))`
`=0.799+0.0591 log (9.327xx10^(-9)=0.799+0.0591(log 9.327-9 log 10)`
`=0.799+0.0591(0.9697-9)=0.799+0.0591(-8.030)`
`=0.799-0.474=0.325" V"`.