Question

In electrolysis of dilute H\(_2\)SO\(_4\), what is liberated at anode?

• A. H\(_2\)
• B. SO\(_4^{2-}\)
• C. SO\(_2\)
• D. O\(_2\)

Hint:

In electrolysis of aqueous solutions, water often gets oxidized at the anode, especially when the anions like SO\(_4^{2-}\) are difficult to oxidize.

Answer 1

The Correct Option is D

When dilute sulphuric acid (H\(_2\)SO\(_4\)) undergoes electrolysis, the solution includes these ions:
\[\n \text{H\(_2\)O} \rightleftharpoons \text{H}^+ + \text{OH}^-, \quad \text{H\(_2\)SO\(_4\)} \rightarrow 2\text{H}^+ + \text{SO}_4^{2-}\n \]\ At the cathode (reduction): - H\(^+\) ions are reduced, forming H\(_2\) gas: \[\n 2\text{H}^+ + 2e^- \rightarrow \text{H}_2(g)\n \]\ At the anode (oxidation): - OH\(^-\) ions (from water) are oxidized instead of SO\(_4^{2-}\): \[\n 4\text{OH}^- \rightarrow 2\text{H}_2\text{O} + \text{O}_2(g) + 4e^-\n \]\ Therefore, oxygen gas (O\(_2\)) is produced at the anode.