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

(N/A) The stronger oxidizing power of fluorine compared to chlorine is due to the following factors:
$(i)$ Low enthalpy of dissociation of the $F-F$ bond.
$(ii)$ High hydration enthalpy of the $F^-$ ion.
These factors collectively make the standard electrode potential $(E^{\Theta})$ of fluorine more positive, thereby making it a stronger oxidizing agent.

Property	$F$
Atomic number	$9$
Atomic mass $/g \ mol^{-1}$	$19.00$
Electronic configuration	$[He] 2s^2 2p^5$
Covalent radius $/pm$	$64$
Ionic radius $X^- / pm$	$133$
Ionisation enthalpy $/kJ \ mol^{-1}$	$1680$
Electron gain enthalpy $/kJ \ mol^{-1}$	$-333$
Electronegativity	$4.0$
$\Delta_{Hyd} H(X^-) / kJ \ mol^{-1}$	$515$
Bond dissociation enthalpy $/kJ \ mol^{-1}$	$158.8$
$E^{\Theta} / V$	$2.87$