Why is an external $emf$ of more than $2.2 \ V$ required for the extraction of $Cl_2$ from brine?

(N/A) The following reaction takes place during the extraction of $Cl_2$ from brine:
$2Cl^{-} + 2H_2O \rightarrow Cl_2 + 2OH^{-} + H_2$
The standard Gibbs free energy change for this reaction is $\Delta G^{\circ} = +422 \ kJ \ mol^{-1}$.
Using the relation $\Delta G^{\circ} = -nFE^{\circ}$,where $n = 2$ and $F = 96500 \ C \ mol^{-1}$,we calculate the standard cell potential:
$E^{\circ} = -\frac{\Delta G^{\circ}}{nF} = -\frac{422000 \ J \ mol^{-1}}{2 \times 96500 \ C \ mol^{-1}} \approx -2.2 \ V$.
Since the standard cell potential is negative,the reaction is non-spontaneous. Therefore,an external $emf$ greater than $2.2 \ V$ is required to drive the reaction.