The emf values of the cell reactions `Fe^(3+)+e^(-) rarr Fe^(2+) " and " Ce^(2+) rarr Ce^(3+) e^(-)` are 0.61 V and -0.85 V respectively. Construct the cell such that the free energy of the cell is negative. Calculate the emf of the cell.

Explanation

Given :
`E^(@)Fe^(3+)//Fe^(2+)=0.61V`
`E^(@)Ce^(3+)//Ce^(2+)=0.85V` (after reversing)
Solution :
`E^(@)=E_(R)-E_(L)`
`" "=0.85-0.61=0.24V`
Cell is,
`Fe^(3+)abs(Fe^(2+))abs(Ce^(3+))Ce^(2+)`
Left `" "` Right
`E^(@)Cell=0.24V`
`DeltaG^(@)=-nFE^(@)`
`DeltaG^(@)=-(1 times 96495 times 0.24)`
`" "` = -ve value