Although electron gain enthalpy of fluorine is less negative as compared to chlorine, fluorine is a stronger oxidizing agent than chlorine. Why?

Explanation

The.oxdising capability of a substance can be determined by its electrode potential Higher the electrode potential, stronger is the oxidizing agent. The electrode potential, in turn, depends upon the following three factors:
(i) Bond dissociation enthalpy (ii) Electron gain enthalpy and (ii) Hydration energy. Although electron gain enthalpy of fluorine is less negative than that of chlorin e but (a) enthalpy of dissociation of F-F bond is lower than that of CI-Cl bond, and (6) Hydration enthalpy of Fion is higher than that of Crion. Because of these two reason, electrode potential of `F_(2)` (+ 2.87 V) is much higher than that of `Cl_(2)` (+ 1.36 V) and hence `F_(2)` is a stronger oxidizing agent than `Cl_(2)`.