In the electrolysis of dilute tetraoxosulphate (VI) acid using platinum electrodes, the products obtained at the anode are cathode respectively are

Step 1: Analyze the question We need the products at the anode and the cathode when dilute H₂SO₄ is electrolysed with inert platinum electrodes. Note that the order is anode first, then cathode.

Step 2: Concept

• The anode is the positive electrode, where oxidation (loss of electrons) occurs. Anions move there.
• The cathode is the negative electrode, where reduction (gain of electrons) occurs. Cations move there.
• Platinum is inert, so it does not take part in the reaction.

Step 3: Ions present Dilute H₂SO₄ contains:

• H⁺ and SO₄²⁻ from the acid
• H⁺ and OH⁻ from water

Step 4: Reactions

At the cathode (H⁺ is the only cation): 2H⁺ + 2e⁻ → H₂(g) Hydrogen gas is liberated.

At the anode (OH⁻ and SO₄²⁻ compete): OH⁻ is lower in the electrochemical series than SO₄²⁻, so it is preferentially discharged. In dilute solution, OH⁻ is discharged: 4OH⁻ → 2H₂O + O₂ + 4e⁻ Oxygen gas is liberated.

The sulphate ion stays in solution, so no sulphur is ever produced. This rules out D and E. Overall, water is decomposed: 2H₂O → 2H₂ + O₂, with H₂ and O₂ in a 2:1 volume ratio.

Step 5: Answer Anode: oxygen. Cathode: hydrogen.

Exam tip: Remember "dilute acid with inert electrodes is just electrolysis of water". Hydrogen goes to the cathode (negative), oxygen to the anode (positive). Also watch the order of the question: it asks for the anode first. Option A is the trap with the reverse order.