Question

Considering the parameters such as bond dissociation enthalphy, electron gain enthalpy and hydration entalpy, compare the oxidising power of `F_2` and `Cl_2`.

Answer 1

Elctrode potenential depends on:
`E=(1)/(2)Delta_(diss)H^(ɵ)(X_(2))+Delta_(eg)H^(ɵ)(X)+Delta_(hyd)H^(ɵ)(X^(ɵ))`
Although, electron gain enthalpy of fluorine is less negative than that of chlorine the bond dissociation enthalpy of `F-F` bond is much lower than `Cl-Cl` bond and hydration enthalpy of `F^(ɵ)` ion is much higher than that of `Cl^(ɵ)` ion. The later two factors more than compensate the less negative electron gain enthalpy of `F_2`. Consequently electrode potential of `F_2` is higher than `Cl_2` and `F_2` is a much stronger oxidising agent than `Cl_2`.