Why is fluorine a oxidising agent? An oxidizing agent pulls the electron cloud of the substance being oxidized towards itself, for example:
$$\ce{F2 + 2 X- -&gt; 2F- + X2}\qquad (\ce{X} = \ce{Cl}, \ce{Br}, \ce{I})$$
We also know that fluorine has very high charge/mass ratio as it is very small in size and its electron gain enthalpy is know to be very high, i.e gain of electron is favourable thermodynamically.
Also, many sources state that:

Fluorine has low bond dissociation enthalpy and high hydration enthalpy due to which fluorine is a good oxidizing agent.
Fluorine is achieving octet, so electron gain is favorable.
I found another source stating that bond dissociation enthalpy is the major driving factor.

My questions(along with reasons why fluorine shouldn't be oxidising agent)

What has hydration enthalpy and bond dissociation enthalpy got to do with oxidizing power?
(As we know that lithium is the best reducing agent, and it also has high hydration enthalpy but we can’t say it is an oxidising agent.)
What am I missing here? What is the reason for such high standard electrode potential for fluorine?
Why wouldn't the electron cloud of fluorine atom repel the electron cloud of substance being oxidized? It seems paradoxical that the fluorine pulls electron cloud towards itself so easily.

Even in the post Why does chlorine have a higher electron affinity than fluorine? it is mentioned that electron cloud is repelled.
Is there any better explanation other than &quot;octet is being achieved&quot;, such that it resolves the repulsion issue, to explain why fluorine is a good oxidising agent?
If we consider

$\ce{F2(g) -&gt; 2 F(g)}$ with $\Delta H_\mathrm{F, bond}$

$\ce{2 F(g) + 2 e- -&gt; 2 F-(g)}$ with $\Delta H_\mathrm{F, aff}$

$\ce{2 F-(g)  -&gt;[H2O] 2 F-(aq)}$ with $\Delta H_\mathrm{F, hydr}$

and respectively

$\ce{Cl2(g) -&gt; 2 Cl(g)}$ with $\Delta H_\mathrm{Cl, bond}$

$\ce{2 Cl(g) + 2 e- -&gt; 2 Cl-(g)}$ with $\Delta H_\mathrm{Cl, aff}$

$\ce{2 Cl-(g)  -&gt;[H2O] Cl-(aq)}$ with $\Delta H_\mathrm{Cl, hydr}$

Then overall reaction enthalpy for $\ce{ F2(g) + 2 Cl-(aq) -&gt; Cl2(g) + 2 F-(aq)}$ is

$$\Delta H_\mathrm{r} = \left({\Delta H_\mathrm{F, bond} - \Delta H_\mathrm{Cl, bond}}\right) + \left({\Delta H_\mathrm{F, aff} - \Delta H_\mathrm{Cl, aff}}\right) + \left({\Delta H_\mathrm{F, hydr} - \Delta H_\mathrm{Cl, hydr}}\right)$$

The differences between fluorine and chlorine enthalpies of atomization(bond breaking ), electron affinity and ion hydration lead to the overall reaction enthalpy, that is pushing reaction toward chloride oxidation.

The truth is, that the equilibrium is driven by Gibbs energy, not enthalpy,

$$\Delta G^{\circ}_\mathrm{r}=\Delta H^{\circ}_\mathrm{r} - T \cdot \Delta S^{\circ}_\mathrm{r}=-RT\ln{K}$$ 

but for the rough reasoning you get the picture. Futhermore, the entropy change will not be major factor, due analogical structures.