Reaction of Halogens with water
Why hypochlorous acid is formed in second reaction why doesn't it proceed like the first one?
  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:

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

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:
$$\ce{Cl2 &lt;=&gt; Cl- + Cl+}\tag C$$
Here the chlorine is oxidize by the other chloride molecule and reacts with the water to form $\ce{HCl}$ and $\ce{HOCl}$ 

$$\ce{Cl+ + OH- &lt;=&gt; HClO} \qquad\text{and}\qquad \ce{Cl- + H+ &lt;=&gt; HCl}\tag D$$


  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 

$$\ce{Cl+ + Cl- + OH- +H+ \implies Cl2 + H2O &lt;=&gt; HCl + HOCl}\tag2$$

Now for the why. The answer comes from the electronegativities of the elements. $(X_\ce{F} = 4.0 &gt; X_\ce{O} = 3.5 &gt; X_\ce{Cl} = 3.2)$ 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:

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