Question

In electrolysis of dilute $H_{2}S{O}_4$, what is liberated at anode?

• A. $H_2$
• B. $S{O}_4^{2-}$
• C. $S{O}_2$
• D. $O_2$

Answer 1

Answer: (D)

$O_2$

When a dilute solution of sulfuric acid is electrolysed, gases are produced at both the anode and the cathode.

The gas produced at the cathode burns with a 'pop' when a sample is lit with a lighted splint. This shows that the gas is hydrogen.

The gas produced at the anode relights a glowing splint dipped into a sample of the gas. This shows that the gas is oxygen.

At cathode,

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

At anode,

$4O{H}^{-}\to 2H_{2}O+O_2+4e^{-}$

Hence option (D) is correct.