Question

A solution of sodium sulphate in water is electrolysed using inert electrodes. The products at the cathode and anode are respectively
(IIT-JEE, 1987)

• A. $H_2,O_2$
• B. $O_2,H_2$
• C. $O_2,Na$
• D. $O_2,S{O}_3$

Answer 1

The correct option is A $H_2,O_2$

In aqueous solution: reduction potential of $H_{2}O>$ reduction potential $N{a}^{\oplus }.$
Hence, $H_{2}O$ undergoes reduction at cathode to give $H_{2\left(g\right)}$
$H_{2}O+e^{-}\to \frac{1}{2}{H}_{2\left(g\right)}+\stackrel{\ominus }{O}{H}_{\left(aq\right)}$
Similarly, oxidation potential of $H_{2}O>$ oxidation potential of $S{O}_4^{2-}$ ions.
Hence, oxidation of $H_{2}O$ occurs at anode to give $O_{2\left(g\right)}.$
$H_{2}O\to 2H^{\oplus }+\frac{1}{2}{O}_{2\left(g\right)}+2e^{-}$