NEET 2020 Chemistry Question 99 Solution

Dilute sulphuric acid dissociates into hydrogen ions and sulphate ions, while water also self-ionizes: $$\text{H}_2\text{SO}_4 \text{(aq)} \to 2\text{H}^+ + \text{SO}_4^{2-}$$ $$\text{H}_2\text{O (l)} \rightleftharpoons \text{H}^+ + \text{OH}^-$$

Hydrogen ions are the only reducible cations present. They migrate to the inert platinum cathode and undergo reduction: $$2\text{H}^+ \text{(aq)} + 2e^- \to \text{H}_2\text{ (g)}$$ Hydrogen gas is evolved at the cathode.

At the inert platinum anode, two competing oxidation half-reactions are possible: 1. Oxidation of water: $$2\text{H}_2\text{O (l)} \to \text{O}_2\text{ (g)} + 4\text{H}^+ + 4e^- \quad (E^\circ_{\text{cell}} = +1.23\text{ V})$$ 2. Oxidation of sulphate ions: $$2\text{SO}_4^{2-} \text{(aq)} \to \text{S}_2\text{O}_8^{2-} \text{(aq)} + 2e^- \quad (E^\circ_{\text{cell}} = +1.96\text{ V})$$

The reaction with the lower standard potential ($$E^\circ = +1.23\text{ V}$$) is thermodynamically favored. In dilute sulphuric acid, oxidation of water takes preference over sulphate oxidation, so oxygen gas ($$\text{O}_2$$) is evolved at the anode.