Question

Which of the following substance gets liberated at the anode during the electrolysis of aqueous NaCl?

• 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 B

$C{l}_2\left(g\right)$

Water plays a 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 so hydrogen gas is evolved at the cathode. Similarly, $C{l}^{-}$ has a greater tendency than water to get oxidized so it gets oxidized and produces chlorine gas. Oxidation takes place at anode so chlorine gas will be produced at the anode. The reaction taking place at the cathode and the anode during electrolysis of $NaCl$ solution is:

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

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