Correct Answer - The electrode potential of `F_(2) (+2.87 V)` is much higher than that of `Cl_(2) (+ 1.36 V)`, therefore, `F_(2)` a much stronger oxidizing agent than `Cl_(2)`.now, electrode potantial depends upon three factors : (i) bond dissociation energy (ii)electron gain enthalpy and (iii)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-CI` bond `(242.6 kJ mol^(-1))` and hydration energy of `F^(-)` ion `(515 kJ mol^(-1))` is much higher than that of `Cl-Cl` ion `(381 kJ mol^(-1))`.The later two factors more than compensate the less negative electron gain enthalpy of `F_(2)`.As a result, electrode potantial of `F_(2)` is higher than that of `Cl_(2)` and hence `F_(2)` is a much stronger oxidizing agent than `Cl_(2)`.
