Question

On electrolysis of aqueous $N{a}_{2}S{O}_4$ solution using Platinum $\left(Pt\right)$ electrodes, the product obtained at anode will be:

• A. $O_2$ gas
• B. $H_{2}S$ gas
• C. $S{O}_2$ gas
• D. $H_2$ gas

Answer 1

The correct option is A $O_2$ gas

Aqueous $N{a}_{2}S{O}_4$ solution on electrolysis produces $O_2$ gas at anode and $H_2$ gas at cathode.
At anode:

$2H_{2}O\left(l\right)\to O_2\left(g\right)+4H^{+}\left(aq\right)+4e^{-}$

At cathode:
$H_{2}O\left(l\right)+e^{-}\to \frac{1}{2}{H}_2\left(g\right)+O{H}^{-}\left(aq\right)$