JEE Main · 2021 · 27 Jul 2021 · Shift 1 · p-Block Elements

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

(A) $HO_{3} SOSO_{3} H$
(B) $HSO_{4}^{-}$
(C) $HO_{2} SOSO_{2} H$
(D) $HO_{3} SOOSO_{3} H$ ✓ Correct

Answer: (D)

Solution

Electrolysis of a concentrated acidified sulphate solution discharges hydrogensulphate ions at the anode, and two of the resulting radicals couple through oxygen:

2 HS O_{4}^{-} ⟶ H_{2} S_{2} O_{8} + 2 e^{-}

The product is peroxodisulphuric acid, Marshall's acid, in which two

- S O_{3} H

units are joined by a peroxide linkage and each sulphur stays in the

+ 6

state. The product is therefore

H O_{3} S - O - O - S O_{3} H

JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Single correct option

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