Question

What happens during electrolysis of Brine?

Answer 1

1. Brine is a concentrated solution of common salt $\mathrm{NaCl}$(Sodium chloride) in water.
2. Products of electrolysis depend on the nature of material being electrolysed and the type of electrodes being used.
3. The products of electrolysis depend on the different oxidising and reducing species present in the electrolytic cell and their standard potentials.
4. Electrolysis of brine is a large scale process to manufacture salt.
5. The products of electrolysis of brine solution gives Sodium hydroxide($\mathrm{NaOH}$), Chlorine(${\mathrm{Cl}}_2$) and Hydrogen(${\mathrm{H}}_2$).
6. Hence, these products need to be separated before they react to another compound.
7. The chemical reaction at cathode is given as,
   $$\begin{gathered} {\mathrm{Na}}^{+}\left(\mathrm{aq}\right)+{\mathrm{e}}^{-}\to \mathrm{Na}\left(\mathrm{s}\right){{\mathrm{E}}^{°}}_{\mathrm{cell}}=-2.71\mathrm{V} \\ \mathrm{Sodium} \\ {\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{e}}^{-}\to \frac{1}{2}{\mathrm{H}}_2\left(\mathrm{g}\right){{\mathrm{E}}^{°}}_{\mathrm{cell}=}0.00\mathrm{V} \\ (\mathrm{Hydrogen}\mathrm{gas}) \end{gathered}$$
8. The reaction with higher value of ${\mathrm{E}}^{°}$is preferred and ${\mathrm{H}}^{+}$(aq) is produced by the dissociation of water(${\mathrm{H}}_2\mathrm{O}$).
   ${\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\to {\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)$
9. Therefore, the net reaction at cathode maybe written as
   ${\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)+{\mathrm{e}}^{-}\to \frac{1}{2}{\mathrm{H}}_2\left(\mathrm{g}\right)+{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)$
10. Reactions at anode is given as,
    $$\begin{gathered} 2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\to {\mathrm{O}}_2+4{\mathrm{H}}^{+}+4{\mathrm{e}}^{-}{{\mathrm{E}}^{°}}_{\mathrm{cell}}=1.23\mathrm{V} \\ \left(\mathrm{water}\right) \end{gathered}$$
    ${\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)\to \frac{1}{2}{\mathrm{Cl}}_2\left(\mathrm{g}\right)+{\mathrm{e}}^{-}{{\mathrm{E}}^{°}}_{\mathrm{cell}}=1.36\mathrm{V}$
11. The reaction at anode with lower value of ${\mathrm{E}}^{°}$is preferred.
12. Thus the net reactions is summarised as
    $$\begin{gathered} \mathrm{NaCl}\left(\mathrm{aq}\right)\stackrel{{\mathrm{H}}_2\mathrm{O}}{\to }{\mathrm{Na}}^{+}\left(\mathrm{aq}\right)+{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right) \\ \mathrm{Cathode}:{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)+{\mathrm{e}}^{-}\to \frac{1}{2}{\mathrm{H}}_2\left(\mathrm{g}\right)+{\mathrm{OH}}^{-}\left(\mathrm{aq}\right) \\ \mathrm{Anode}:{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)\to \frac{1}{2}{\mathrm{Cl}}_2\left(\mathrm{g}\right)+{\mathrm{e}}^{-} \\ \mathrm{Net}\mathrm{reaction}: \\ \mathrm{NaCl}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\to {\mathrm{Na}}^{+}\left(\mathrm{aq}\right)+{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)+\frac{1}{2}{\mathrm{H}}_2\left(\mathrm{g}\right)+\frac{1}{2}{\mathrm{Cl}}_2\left(\mathrm{g}\right) \end{gathered}$$