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

H S O_{4}^{-}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

H O_{2} S O S O_{2} H

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $H O_{3} S O O S O_{3} H$
Anode:
$2 H_{2} S O_{4} ⟶ H_{2} S_{2} O_{8} + 2 H^{+} + 2 e^{-}$

$2 H_{2} O ⟶ O_{2} + 4 H^{+} + 4 e^{-}$

Cathode :

$e^{-} + H^{+} ⟶ \frac{1}{2} H_{2}$

Main product of electrolysis of conc. $H_{2} S O_{4}$ is
$H O_{3} S O O S O_{3} H or (H_{2} S_{2} O_{8})$

Suggest Corrections

Similar questions

When SO₂ is passed through an acidified solution of K2Cr2O7, then chromium sulphate is formed. Change in the oxidation state of Cr is from.

S O_{2}

25

H_{2} S O_{4}

34.25

0.1 N

K_{2} C r_{2} O_{7}

25

22.45

K_{2} C r_{2} O_{7}

F e_{2} (S O_{4})_{3}

Join BYJU'S Learning Program