Sodium metal is prepared by Down’s process. This process involvesthe electrolysis of a fused mixture of NaCl and $\mathrm{CaCl}_2$ at 873 K. Duringelectrolysis, sodium is discharged at the cathode and $\mathrm{Cl}_2$ is obtained atthe anode as a by-product. Cathode: $\mathrm{Na}^{+}(\text{melt}) + e^{-} \rightarrow \mathrm{Na}_{(s)}$Anode : $\mathrm{Cl}^{-}(\text{melt}) \rightarrow \mathrm{Cl}_{(g)} + e^{-}$ $2 \mathrm{Cl}_{(g)} \rightarrow \mathrm{Cl}_{2(g)}$If an aqueous solution of NaCl is electrolysed, $\mathrm{H}_2$ is evolved at thecathode and $\mathrm{Cl}_2$ is obtained at the anode. The reason being that the E°of Na $^{+}$ / Na redox couple is much lower (E° = –2.71V) than that of $\mathrm{H}_2\mathrm{O} \left( E^0_{\mathrm{H}_2\mathrm{O}/\mathrm{H}_2} \right) = -0.83\,\text{V} )$ and hence water is reduced to $\mathrm{H}_2$ in preference toNa$^{+}$ions. However, NaOH is obtained in the solution. Anode : $\mathrm{Cl}^{-}_{(aq)} \rightarrow \mathrm{Cl}_{(g)} + e^{-}$ $2 \mathrm{Cl}_{(g)} \rightarrow \mathrm{Cl}_{2(g)}$Cathode