The product obtained from the electrolytic oxidation of acidified sulphate solutions, is :

Solution

Chemistry ma’am · 3 steps

Sulphur is already +6 and cannot be oxidised further, so the oxygens join into a peroxide bridge.

1. What gets oxidised. In HSO₄⁻ sulphur is already at +6, its highest oxidation state. At the anode the electrons must come from oxygen, which goes from -2 to -1.

   2HSO₄⁻ → H₂S₂O₈ + 2e⁻

2. Peroxide link. Two HSO₄⁻ units each lose one electron and join through an O-O bond. The product is peroxodisulphuric acid (Marshall's acid), HO₃S-O-O-SO₃H, and sulphur stays at +6 in it.

   Option	Acid	Oxidation state of S	Bridge
   (a) HSO₄⁻	hydrogen sulphate ion	+6	none, this is the starting ion
   (b) HO₃SOOSO₃H	peroxodisulphuric acid, H₂S₂O₈	+6	O-O peroxide
   (c) HO₂SOSO₂H	pyrosulphurous acid, H₂S₂O₅	+4	single O
   (d) HO₃SOSO₃H	pyrosulphuric acid, H₂S₂O₇	+6	single O

3. So. Only (b) contains the peroxide O-O link that anodic oxidation produces. The product is H₂S₂O₈.

The idea to keep. When the central atom is already in its highest oxidation state, oxidation happens at oxygen. Look for a product with an O-O peroxide bond.

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