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.
$2Cl^{-} + 2H_2O \rightarrow Cl_2 + 2OH^{-} + H_2$
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.
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$Ag_{(s)} \, | \, Ag_2C_2O_{4(s)} \, | \, C_2O_4^{2-} (0.02 \ M) \, || \, Ag^{+}(0.5 \ M) \, | \, Ag_{(s)}$
$EMF$ of the cell is $0.264 \ V$ at $25 \ ^oC$. The $K_{sp}$ of $Ag_2C_2O_{4(s)}$ is:
$[Given: \frac{2.303RT}{F} = 0.06, \ \log 2 = 0.3]$
$Ag_{(s)} \, | \, Ag_2C_2O_{4(s)} \, | \, C_2O_4^{2-} (0.02 \ M) \, || \, Ag^{+}(0.5 \ M) \, | \, Ag_{(s)}$
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