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(A) The oxidizing strength of a halogen in aqueous solution is determined by the standard electrode potential $(E^{\circ})$,which depends on the energ

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Bond dissociation enthalpy,electron gain enthalpy,hydration enthalpy

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(A) The oxidizing strength of a halogen in aqueous solution is determined by the standard electrode potential $(E^{\circ})$,which depends on the energy changes involved in the process: $\frac{1}{2} X_2(g)$ $\rightarrow X(g)$ $\rightarrow X^-(g)$ $\rightarrow X^-(aq)$.This process involves three main energy terms:$1$. Bond dissociation enthalpy $(\frac{1}{2} \Delta_{diss} H^{\circ})$: Energy required to break the $X-X$ bond.$2$. Electron gain enthalpy $(\Delta_{eg} H^{\circ})$: Energy released when an electron is added to the atom.$3$. Hydration enthalpy $(\Delta_{hyd} H^{\circ})$: Energy released when the ion is hydrated in water.Fluorine $(F_2)$ has a low bond dissociation enthalpy,high negative electron gain enthalpy,and very high negative hydration enthalpy compared to chlorine $(Cl_2)$. The combination of these three factors makes $F_2$ a much stronger oxidizing agent in water.

Combination of which parameters explains the higher oxidizing strength of $F_2$ as compared to $Cl_2$ in water?

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