JEE Main 2023 — S Block Elements Question with Solution

The electrolysis of brine, which is a solution of sodium chloride (NaCl) in water, involves the migration of ions towards electrodes. When an electric current is passed through the solution, positive sodium ions ($N{a}^{+}$) move towards the cathode (negative electrode), and negative chloride ions ($C{l}^{-}$) move towards the anode (positive electrode).

At the same time, water ($H_{2}O$), which also dissociates into $H^{+}$ and $O{H}^{-}$ ions, also contribute to the reactions at the electrodes.

At the anode, chloride ions are oxidized to chlorine gas:

$2C{l}^{-}\to C{l}_2(g)+2e^{-}$

At the cathode, water is reduced to hydrogen gas and hydroxide ions, instead of sodium ions being reduced to sodium metal. This is because water is more easily reduced than sodium ions:

$2H_{2}O+2e^{-}\to H_2(g)+2O{H}^{-}$

The sodium ions remain in the solution and combine with the hydroxide ions to form sodium hydroxide:

$N{a}^{+}+O{H}^{-}\to NaOH$

So the products of the electrolysis of brine are chlorine gas, hydrogen gas, and sodium hydroxide. Hydrochloric acid (HCl) is not produced during this process because the hydrogen and chloride ions do not combine in the electrolyte solution to form HCl. They are separately reduced and oxidized at the cathode and anode respectively.