A galvanic cell is constructed with `Ag//Ag^(+)` as one elecrode and `Fe^(2+)//Fe^(3+)` as the second electrode. Calculate the concentration of `Ag^(+)` ions at which the E.M.F. of the cell will be zero at equimolar concentrations of `Fe^(2+) and Fe^(3+)` ions. Given `E_(Ag^(+)//Ag)^(@)=0.80V,E_(Fe^(3+)//Fe^(2+))^(@)=0.77V`

Explanation

Given electrode potential values show that EMF of the cell will be positive only if reduction occurs at silver electrode. Therefore, the cell reaction will be `Fe^(2+)+Ag^(+)toFe^(3+)+Ag`
Hence, the cell may be represented as: `Fe^(2+)|Fe^^(3+)||Ag^(+)|Ag`
`thereforeE_(cell)^(@)=E_(Ag^(+)//Ag)^(@)-E_(Fe^(3+)//Fe^(@+))^(@)=0.80-0.77=0.03` V
Applying nernst equation to the above reaction, `E_(Cell)=E_(Cell)^(@)-(0.0591)/(1)"log"([Fe^(3+)][Ag])/([Fe^(2+)][Ag^(+)])`
But `[Fe^(2+)]=[Fe^(3+)]` (given) and [Ag]=1.
`thereforeE_(cell)=E_(Cell)^(@)-0.0591"log"(1)/([Ag^(+)]) therefore0=0.03+0.0591log[Ag^(+)]`
or `log[Ag^(+)]=-(0.03)/(0.0591)=-0.5076=1.4924` or `[Ag^(+)]`=antilog `overline(1).4924=3.1xx10^(-1)=0.31M`