Question

Which of the following is not correct regarding the electrolytic preparation of
H2O2 ?
(1) Lead is used as chathode,
(2) 50% H2SO, is used.
(3) Hydrogen is liberated at anode.
(4) Sulphuric acid undergoes oxidation,​

Answer 1

Answer:

H

2

O

2

can be prepared by electrolysis of 50% H

2

SO

4

.

Here hydrogen is liberated at cathode

H

2

SO

4

⇌2H

+

+SO

4

2−

At anode:

2HSO

4

−

⇌H

2

S

2

O

8

+2e

−

H

2

S

2

O

8

+H

2

O⟶2H

2

SO

4

+H

2

O

2

At cathode:

2H

+

+2e

−

⟶H

2

Answer 2

AnSwer :-

$\Large\star$H$_2$O$_2$ can be prepared by electrolysis 50% of 50 H$_2$SO$_4$. In this method, Hydrogen is liberated at cathode.

$\sf H_2SO_4 ⇌ 2H^+ + 2HSO_4$

At anode $\sf 2HSO_4 \longrightarrow H_2S_2O_8 + 2e^-$

$\sf H_2S_2O_8 + 2H_2O \longrightarrow 2H_2SI_4 + H_2O_2$

At cathode $\sf 2H^+ + 2e^- \longrightarrow H_2 \uparrow$

Thus, option (3) Hydrogen is liberated at anode is ur answer.