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

A galvanic cell is constructed with Ag//Ag^(+) as one elecrode and Fe^

1.	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
Answer» Solution :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`