Which of the following represents correct order of oxidising power of halogens with water?

Whenever one of the halogens is involved in oxidising a species in solution, the halogen is reduced to a halide ion associated with water molecules. The following reactions illustrate this process ${ }^{\circ}$ $\mathrm{F}_2(g)+2 e^{-} \longrightarrow 2 \mathrm{~F}^{-}(a q)$ $\begin{aligned} & \mathrm{Cl}_2(g)+2 e^{-} \longrightarrow 2 \mathrm{Cl}^{-}(a q) \\ & \mathrm{Br}_2(l)+2 e^{-} \longrightarrow 2 \mathrm{Br}^{-}(a q) \\ & \mathrm{I}_2(g)+2 e^{-} \longrightarrow 2 \mathrm{I}^{-}(a q)\end{aligned}$ Smaller the size of ion, higher is the energy released when the ion is hydrated. Since, the size of ions increases in the order $\mathrm{F}^{-} < \mathrm{Cl}^{-} < \mathrm{Br}^{-} < \mathrm{I}^{-}$. The order of hydration energy is $$ \mathrm{F}^{-}(a q)>\mathrm{Cl}^{-}(a q)>\mathrm{Br}^{-}(a q)>\mathrm{I}^{-}(a q) $$ Because of very high hydration enthalpy of the fluoride ion, $\mathrm{F}_2$ gets very easily reduced followed by $\mathrm{Cl}_2$, then $\mathrm{Br}_2$ and $\mathrm{I}_2$. Therefore, the correct order of oxidising power of halogens is $\mathrm{F}_2>\mathrm{Cl}_2>\mathrm{Br}_2>\mathrm{I}_2$.