Explain why `E^(@)` for `Mn^(3+)//Mn^(2+)` couple is more positive than that for `Fe^(3+)//Fe^(2+)`(At. Nos. Mn=25,Fe=26)?

Explanation

`._(25)Mn^(2+) = [Ar] 3d^(5), . _(25) Mn^(3+) =[Ar] 3d^(4), ._(26) Fe^(2+) = [ Ar] 3d^(6) , . _(26) Fe^(3+)= [ Ar] 3d^(5)`
Thus, `Mn^(2+)` has more stable configuration than `Mn^(3+)` while `Fe^(3+)` has more stable then `Fe^92+)`. Consequently, large third ionisation enthalpy is required to change `Mn^(2+)` to `Mn^(3+)`. As `E^(@)` is the sum of enthalpy of atomisation, ionization enthalpy and hydration enthalpy, therefore, `E^(@)` for `Mn^(3+)//Mn^(2+)` couple is more positive than `Fe^(3+) //Fe^(2+)` .
Note `,` The large positive `E^(@)` for `Mn^(3+)//Mn^(2+)` means that `Mn^(3+)` can be easily reduced to `Mn^(2+)` , i.e. `Mn^(3+)` is less stable . `E^(@)` value for `Fe^(3+) //Fe^2+)` is positive but small, i.e., `Fe^(3+)` can also be reduced to `Fe^(2+)` but less easily. Thus,`Fe^(3+)` is more stable than `Mn^(3+)` . It also explains why `+3` state of Mn is of little importance.