Calculate the solubility product constant of AgI from the following values of standard electrode potentials.
`E_(Ag^(+)//Ag)^(@)=0.80` volt and `E_(I//AgI//Ag)^(@)=-0.15` volt at `25^(@)C`

Explanation

Solubility product of `AgI =[Ag^(+)][I^(-)]`
Note : See chapter Ionic Equilibrium for solubility product.
two half reactions for the cell are :
`{:(,Ag rarr Ag^(+)+e^(-),"Anode (Oxidation)"),(,AgI+e^(-) rarr Ag+I^(-),"Cathode (Reduction)"),("Cell reaction",bar(AgI rarr Ag^(+)+I^(-)),):}`
Applying Nernst equation,
`E_(cell)=E_(cell)^(@)-0.0591/1"log"([Ag^(+)][I^(-)])/([AgI])`
At equilibrium, `E_(cell)=0` and `[AgI]=1`
So, `log [Ag^(+)][I^(-)]=E_(cell)^(@)/0.0591`
`E_(cell)^(@)=-0.80-0.15=-0.95` volt
`log [Ag^(+)][I^(-)]=- 0.95/0.0591=-16.0744`
Solubility product of `AgI=8.4xx10^(-17)`