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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
    Commented Apr 8, 2021 at 14:52
  • $\begingroup$ Related $\endgroup$
    – user60158
    Commented Apr 15, 2022 at 4:15
  • $\begingroup$ The hydration enthalpy is the amount of energy released on dilution of one mole of gaseous ions or we can say the heat energy released when new bonds are made between the ions and the water molecules is called hydration enthalpy. The magnitude of hydration enthalpy depends on the charge density of the ions. The charge density is more in case of small ions and that's why the smaller ions have high hydration enthalpy. The higher is the charge density the higher will be the force of attraction between ion and the water polar end. So the value of hydration enthalpy is higher for small ions (...) $\endgroup$
    – user119245
    Commented Apr 15, 2022 at 4:28
  • $\begingroup$ (...) Both the oxidising power of an element and the hydration enthalpy are related with its charge density. The higher the charge density greater will be the oxidizing power ( or strength of oxidising agent) and also the higher will be hydration enthalpy. So, the hydration enthalpy and strength of oxidizing agent are directly related if one is high then other will also be high (...) $\endgroup$
    – user119245
    Commented Apr 15, 2022 at 4:28
  • $\begingroup$ (...)As Fluorine has high charge density due to its very small size therefore, it has high oxidising power and also have high hydration enthalpy. As explained above because of small size and higher charge density the attraction between fluorine ion and water will more so the fluorine generates a large amount of heat when it forms its hydrated ion and hence, the hydration enthalpy of fluorine will also be high. $\endgroup$
    – user119245
    Commented Apr 15, 2022 at 4:29

2 Answers 2

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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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In aqueous solutions, halogens are powerful oxidising agents (especially fluorine). Let us confine our discussion to fluorine and chlorine. Oxidising ability can be demonstrated using the enthalpy change for the following reaction.

$$\ce{1/2 X2(g) + e- -> X-(aq)}$$

Let $\Delta H$ be the enthalpy change for the above mentioned reaction. This reaction can be achieved in three steps as follows:

  1. Dissociation of halogen into gaseous atoms. $$\ce{ 1/2X2(g) -> X(g)}$$ Let $D_0$ be bond dissociation enthalpy.

  2. Converting gaseous atoms into gaseous ions. $$\ce{ X(g) + e- -> X-(g)}$$ Let $\Delta_\mathrm{eg}H$ be electron gain enthalpy or electron affinity value.

  3. Hydration of gaseous ions into hydrated ions. $$\ce{X-(g) -> X-(aq)}$$ Let $\Delta_\mathrm{hyd}H$ be the hydration enthalpy.

Based on Hess law,

$$\Delta H = D_0 + \Delta_\mathrm{eg}H + \Delta_\mathrm{hyd}H$$

What we need now is this net enthalpy $\Delta H$. The more negative the enthalpy, the more the oxidising power. As fluorine has low bond dissociation enthalpy and high hydration enthalpy it act as powerful oxidising agent than chlorine. So, now it can easily be understood why we should consider enthalpy of hydration in comparing oxidising power.

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  • $\begingroup$ Why are we considering only aqueous solutions? Is fluorine not a stronger oxidizing agent than chlorine outside aqueous solutions? $\endgroup$
    – user60158
    Commented Apr 21, 2022 at 4:36
  • $\begingroup$ Fluorine is a stronger oxidising agent than chlorine even outside the aqueous medium because of its high electronegativity. In general, we use water as a medium for many redox reactions that you perform in the lab. Is your doubt how hydration energy comes into picture if we perform reactions outside water ? $\endgroup$
    – Infinite
    Commented Apr 21, 2022 at 13:48
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    $\begingroup$ You may think that hydration energy is indirectly linked to oxidising power. Because we can't define hydration energy in mediums other than water. So, hydration energy is a consequence for mediums other than water, but it is a factor which decides oxidising power in aqueous solutions. $\endgroup$
    – Infinite
    Commented Apr 21, 2022 at 13:54
  • $\begingroup$ Mostly, this kind of books we usually use generalises the statements rather than discussing particular situations. $\endgroup$
    – Infinite
    Commented Apr 21, 2022 at 14:00
  • $\begingroup$ I have awarded the bounty to you, but I have a request. Please include the comments that you just made about hydration energy and oxidizing power in your answer (and possibly expand on those comments!) $\endgroup$
    – user60158
    Commented Apr 21, 2022 at 14:12

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