Question

The substance liberated at the cathode during the electrolysis of $NaCl$ solution is:

• A. $H_2\left(g\right)$
• B. $C{l}_2\left(g\right)$
• C. $Na\left(s\right)$
• D. $H_{2}O\left(l\right)$

Answer 1

The correct option is A $H_2\left(g\right)$

Water plays a vital role during the electrolysis of $NaCl$ solution. Since water can be both oxidized and reduced, it competes with the dissolved $N{a}^{+}$ and $C{l}^{-}$ ions. Having a greater tendency to get reduced than sodium ions, water gets reduced and produces hydrogen. Reduction takes place at cathode and hydrogen gas is evolved at the cathode.

Similarly, $C{l}^{-}$ has a greater tendency than water to get oxidised so it gets oxidised and produces chlorine gas. Oxidation takes place at anode and chlorine gas will be produced at the anode. The reactions taking place during the electrolysis of $NaCl$ solution are:

At Cathode: $H_{2}O\left(l\right)+2e^{-}\to H_2\left(g\right)+2O{H}^{-}$

At Anode: $2C{l}^{-}\to C{l}_2\left(g\right)+2e^{-}$