Although `Delta_(eg)H` of fluorine is less negative than that of chlorine, but fluorine is a stronger oxidising agent than chlorine. Why?

Explanation

The oxidising capability of a substance can be determined by its electrode potential. Higher the electrode potential, stronger is the oxidising agent. The electrode potential, in turn, depends upon the following three factors : (i) Bond dissociation enthalpy (ii) Electron gain enethalpy and, (iii) Hydration energy.
Although electron gain enthalpy of flrorine is less negative than that of chlorine but
(a) enthalpy of disociation of F-F bond (158.8 kJ `mol^(-1)`) is much lower than that of CI-CI bond (242.6 kJ `mol^(-1)`), and
(b) hydration enthalpy of `F^(-)` ion (515 kJ `mol^(-1)`) is much higher than that of `CI^(-)` ion (381 kJ `mol^(-1)`).
Because of these two reasons, electrode potential of `F_(2)`(+ 2.87 V) is much higher than that of `CI_(2)` (+ 1.36 V) and hence `F_(2)` is a stronger oxidising agent than `Cl_(2)`