Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy, compare the oxidising power of F_2 and Cl_2.

Considering the parameters such as bond dissociation enthalpy, electro

1.	Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy, compare the oxidising power of F_2 and Cl_2.
Answer» Solution :The electrode potential of `F_2(+ 2.87 V)` is much higher than that of `Cl_2 (+ 1.36 V)`, therefore, `F_2` is a much stronger OXIDISING agent than `Cl_2`. This can be explained as under: Electrode potential depends upon three factors : (i) bond dissociation energy, (ii) ELECTRON gain enthalpy and (ii) hydration energy. Although electron gain enthalpy of fluorine is less NEGATIVE (-333 kJ `"mol"^(-1)`) than that of CHLORINE (-349 kJ `"mol"^(-1)`), the bond dissociation energy of F – F bond is much lower (158.8 kJ `"mol"^(-1)`) than that of CI - Cl bond (242.6 kJ `"mol"^(-1)`) and hydration energy of `F^(-)` ion (515 kJ `"mol"^(-1)`) is much higher than that of `Cl^(-)` ion (381 kJ `"mol"^(-1)`). The last two factors more than compensate the less negative electron gain enthalpy of `F_2`. As a result, electrode potential of `F_2` is higher than that of `Cl_2` and hence `F_2` is a much stronger oxidising agent than `Cl_2`.