Which of the following electrolyte can be used to obtain H₂ S₂ O₈ by the process of electrolysis ?

Solution

Chemistry ma’am · 4 steps

At the anode, sulphate ions must beat water to oxidation, and they only win when the acid is concentrated.

1. Where it forms. H₂S₂O₈ (peroxodisulphuric acid) has an O–O link between two sulphur groups. It forms at the anode, when two sulphate-type ions each lose an electron and join up.

   2HSO₄⁻ → H₂S₂O₈ + 2e⁻

2. Competition with water. Water also wants to be oxidised at the anode, to O₂. Its oxidation needs much less potential (E° = 1.23 V for O₂/H₂O) than sulphate does (E° = 2.01 V for S₂O₈²⁻/SO₄²⁻). So in a dilute solution, water wins and you only get O₂.
3. Why concentrated. In concentrated H₂SO₄ there is a very large amount of HSO₄⁻ and very little free water. Then, at high current density, HSO₄⁻ is oxidised instead of water.
4. So. Electrolysing a concentrated solution of sulphuric acid gives H₂S₂O₈, option (a).

The idea to keep. In aqueous electrolysis, oxygen from water is the default at the anode. Another anion is oxidised only when it is present in large excess, so look for the word concentrated.

Answer from the NTA final answer key. Solution by PrepWiser.