The reaction of fluorine with water is very different from that of chlorine with water. $$\ce{2F2 + 2(H2O)->O2 + 2HF}\tag1$$

$$\ce{Cl2 + H2O ->HCl + HOCl}\tag2$$

I want to understand two main things:

  1. Why hypochlorous acid is formed in second reaction why doesn't it proceed like the first one?
  2. Has it got to anything with reduction potentials? chlorine and oxygen are very close and is that reason?

I found this where a related question was discussed here I expect the answer to describe what happens in reaction of any halogen with water and how does its reduction potential play a role in it also are $\ce{HOI}$ and $\ce{HOBr}$ also formed?


2 Answers 2

  1. Why hypochlorous acid is formed in second reaction why doesn't it proceed like the first one?
  2. Has it got to anything with reduction potentials? Cl and O are very close and is that reason?

Your intuition in 2. is a good guess and could answer the question mathematically but would be an indirect explanation. If you look closer at reaction 1 you will see that oxygen is oxidized by fluoride where as in reaction 2 oxygen is not oxidized. From a re-dox perspective in the first reaction your equation 1 looks something like:

$$\ce{H2O -> 2H+ + 1/2O2 + 2e-}\tag A\\ \text{and}$$ $$\ce{F2 + 2e- -> 2F-}\tag B$$

Here the fluorine molecule takes 2 electrons from the oxygen atom in an irreversible process. But for equation 2 the redox reaction is more like: $$\ce{Cl2 <=> Cl- + Cl+}\tag C$$ Here the chlorine is oxidize by the other chloride molecule and reacts with the water to form $\ce{HCl}$ and $\ce{HOCl}$

$$\ce{Cl+ + OH- <=> HClO} \qquad\text{and}\qquad \ce{Cl- + H+ <=> HCl}\tag D$$

side note: $\ce{HOCl}$ would more properly be presented as $\ce{HClO}$ since chlorine has a $+1$ formal charge but not $\ce{ClHO}$ since $\ce{ClO-}$ is the conjugate base.

Note that the reactions for clorine are revresable now whereas fluorine was not. Combining equations C and D give your original equation 2

$$\ce{Cl+ + Cl- + OH- +H+ \implies Cl2 + H2O <=> HCl + HOCl}\tag2$$

Now for the why. The answer comes from the electronegativities of the elements. $(X_\ce{F} = 4.0 > X_\ce{O} = 3.5 > X_\ce{Cl} = 3.2)$ as you can see fluorine's electronegativity is greater than oxygen's which allows flourine to oxidize oxygen and oxygen's electronegativity is greater than chlorine's which mean chlorine cannot oxidize it. Further Hydrogen is already oxidized and so the only thin for chlorine to oxidize is itself.

Since the other halogens also cannot oxidize oxygen, this equation can be generalized for all of the halogens that are not fluorine as well:

$$\ce{X2 + H2O -> HX^{(I-)} + HX^{(I+)}O}\tag{X = Cl, Br, I}$$


In addition to A.K's answer there is a more subtle effect, having to do with the fact that $\ce{HF}$ unlike the other hydrogen halides is only a weak acid. Thereby, hydrogen can readily bond to fluorine forming $\ce{HF}$ molecules, facilitating displacement of oxygen as the element. We cannot form $\ce{HCl}$ and heavier congener molecules so easily in water, so we don't break all the original bonds between hydrogen and oxygen and thus we get products with hydroxyl groups like $\ce{HOCl}$.


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