The standard electrode potantials of the half cells Ag^(+)//Ag and Fe^(3+),Fe^(2+)//Pt are 0.7991V and 0.771V respectively. Calculate the equilibrium constant of the reaction : Ag_((s))+Fe^(3+)hArr Ag^(+)+Fe^(2+)

The standard electrode potantials of the half cells Ag^(+)//Ag and Fe^

1.	The standard electrode potantials of the half cells Ag^(+)//Ag and Fe^(3+),Fe^(2+)//Pt are 0.7991V and 0.771V respectively. Calculate the equilibrium constant of the reaction : Ag_((s))+Fe^(3+)hArr Ag^(+)+Fe^(2+)
Answer» Solution :The cell formed is `AG//Ag^(+),Fe^(3+),Fe^(2+)//Pt` `"At anode :"Ag(s)rarr Ag^(+)+E` `"At cathode :"Fe^(3+)+e rarr Fe^(2+)` `"Overall reaction :"Ag_((s))+Fe^(3+)rarr Fe^(2+)+Ag^(+)` emf of the cell is GIVEN by `(E_(R)-E_(L))` `E_("cell")^(@)=0.771-0.7991=-0.0281V` At equilibrium, `E_("cell")=E_("cell")^(@)-(RT)/(nF)ln""(a_(Ag)+a_(Fe^(2+)))/(a_(Fe^(3+)))` Since activity of solid silver is 1.0. `thereforen=1 andK_(eq)=(a_(Ag^(+))+a_(Fe^(2+)))/(a_(Fe^(3+)))` `therefore E_("cell")^(@)=(0.0591)/(n)logK_(eq)` `therefore logK_(eq)=(0.0281xx1)/(0.0591)=0.4751` `therefore K_(eq)=0.335`