Question

On passing electricity through dil. $H_{2}S{O}_4$ solution the substance liberated at the cathode is:

• A. $H_2\left(g\right)$
• B. $O_2\left(g\right)$
• C. $H^{+}$ ion
• D. $S{O}_4^{2-}$ ion

Answer 1

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

Solution:

On passing electric current through dilute $H_{2}S{O}_4,$ $H_2$ liberated at cathode whereas $O_2$ liberated at anode.

Redox reaction is as follows:

At cathode:

$2H^{+}+2e^{-}\to H_2$ [Reduction]

At anode:
$4O{H}^{-}\to 2H_{2}O+O_2+4e^{-}$ [oxidation]

Overall reaction is as follows:

$2H_{2}O\to 2H_2+O_2$

Option A is correct.