The equilibrium constant of cell reaction: `Ag_((s))+Fe^(3+) iff Fe^(2+)+Ag^(+)` is 0.335 at `25^(@)C`. Calculate the standard emf of the cell `Ag//Ag^(+), Fe^(3+), Fe^(2+)//Pt`. Calculate `E^(@)` of the half cell `Fe^(3+), Fe^(2+)//Pt` is 0.7791 V Calculate `E^(@)` of `Fe^(3+), Fe^(2+)//Pt` half cell.

Explanation

Given : K = 0.335,
`E^(@)=(Ag^(+).Ag)=0.7991`
n = 1, F = 96495 C
Formula :
`E^(@)=(-2.303RT)/(nF)logK`
Solution :
`" "=(-0.591)/1log0.335`
Standard emf of the cell
`" "=-0.0280 V`
`E^(@)""_(cell)" "=E^(@)""_(R)-E^(@)""_(L),E^(@)""_(R)=?`
`therefore E^(@)""_(R)" "=E^(@)""_(cell)-E^(@)""_(L)`
`" "=-0.0280+0.7991=0.771V`
`E^(@)""_((Fe^(3+), Fe^(2+)))=0.771V`.