Question

The electrolysis of aqueous sodium chloride (brine) is an important industrial process for the production of chlorine and sodium hydroxide. In fact, this process is the second largest consumer of electricity in the United States, after the production of aluminium. Write a balanced equation for the electrolysis of aqueous sodium chloride (hydrogen gas is also produced).

Short answer

Upon electrolysis of aqueous sodium chloride, the balanced equation obtained is ${\text{2Na}}^{\text{+}}{\text{(aq) + 2H}}_{\text{2}}{\text{O(l) + 2Cl}}^{\text{-}}\text{(aq)}\to {\text{2NaOH(aq) +H}}_{\text{2}}{\text{(g) + Cl}}_{\text{2}}\text{(g)}$.

Step-by-step solution

Electrolysis Definition

Electrolysis is the process of converting ionic compounds into their component parts by passing a direct electric current through the complex in a fluid form. Cations are reduced while anions are oxidised at the cathode.

Reduction Potentials of Sodium and Water

Electrolysis of aqueous sodium chloride is a process for the production of chlorine and sodium hydroxide. Standard reduction potentials for the reduction of${\text{Na}}^{\text{+}}$and${\text{H}}_{\text{2}}\text{O}$are shown below.

$$\begin{gathered} {\text{Na}}^{\text{+}}{\text{+e}}^{\text{-}}\to {\text{NaE}}^0\text{=-2.7V} \\ {\text{2H}}_{\text{2}}{\text{O + 2e}}^{\text{-}}\to {\text{H}}_{\text{2}}{\text{+ 2OH}}^{\text{-}}{\text{E}}^0\text{=-0.83V} \end{gathered}$$

We can clearly see that the water is more easily reduced than ${\text{Na}}^{\text{+}}$ which is why sodium is not produced in this process under normal circumstances.

Electrolysis equation of aqueous sodium chloride

Therefore, hydrogen is produced at the cathode and not sodium. At the anode is produced chlorine gas.

After the electrolysis, a solution contains dissolved NaoH and NaCl .

$$\begin{gathered} \text{NaCl(aq)}\to {\text{Na}}^{\text{+}}{\text{(aq) + Cl}}^{\text{-}}\text{(aq)} \\ {\text{cathode: 2H}}_{\text{2}}{\text{O(l) + 2e}}^{\text{-}}\to {\text{H}}_{\text{2}}{\text{(g) + 2OH}}^{\text{-}}\text{(aq)} \\ {\text{anode: 2Cl}}^{\text{-}}\text{(aq)}\to {\text{Cl}}_{\text{2}}{\text{(g) + 2e}}^{\text{-}} \\ {\text{2Na}}^{\text{+}}{\text{(aq) + 2H}}_{\text{2}}{\text{O(l) + 2Cl}}^{\text{-}}\text{(aq)}\to {\text{2NaOH(aq) +H}}_{\text{2}}{\text{(g) + Cl}}_{\text{2}}\text{(g)} \end{gathered}$$

Therefore, the balanced equation is obtained as –${\text{2Na}}^{\text{+}}{\text{(aq) + 2H}}_{\text{2}}{\text{O(l) + 2Cl}}^{\text{-}}\text{(aq)}\to {\text{2NaOH(aq) +H}}_{\text{2}}{\text{(g) + Cl}}_{\text{2}}\text{(g)}$