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My textbook says that the greater oxidising behavior of fluorine than chlorine can be attributed to the high hydration enthalpy of $\ce{F-}$ ions and the low dissociation enthalpy of $\ce{F-F}$ bond.

While the second reason makes perfect sense to me, I can't reason why hydration enthalpy comes into play. I know that these elements act as oxidising agents in aqueous media, but I still can't understand why we need to consider hydration enthalpies.

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    $\begingroup$ @NisargBhavsar Oxford dictionary says "oxidize/-zing, UK English also oxidise/-sing". So "s" is not exclusive even in UK. $\endgroup$
    – Poutnik
    Apr 8 at 14:52
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Fluorine is a stronger oxidizing agent than chlorine. So fluorine oxidizes chloride ion as per the following reaction:

$$\ce{F2 + 2Cl- -> 2F- + Cl2}$$

It can be observed that low enthalpy of dissociation of $\ce{F-F}$ bond lowers the energy of activation of the reaction mentioned above and hence favours the formation of products. Due to the small size and high electronegativity of fluoride ion, it has a higher hydration enthalpy or in other words it's easily covered by a lot of solvent molecules. This factor greatly reduces the rate of the reverse reaction. This process also shifts the equilibrium towards the products. This is how hydration enthalpy plays an important role in the oxidizing capability of fluorine.

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