Why is fluorine a 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.