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+)`

Explanation

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 and K_(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`