Question

the passage of electricity through dilute H2SO4 for 16 minutes liberates a total of 224 ml of H2 the strength of current in ampere will be

Answer 1

Answer:

Electrolysis of dilute sulphuric acid:-

At cathode-

2H

+

+2e

−

⟶H

2

At anode-

4OH

−

+⟶2H

2

O+2O

2

+4e

−

Overall reaction-

H

2

O⟶2H

2

+O

2

∴ Amount of substance liberated at cathode = 2 moles of H

2

=4g

Also, amount of substance liberated at anode = 1 mole of O

2

=32g

∴ Ratio of the amount of substance liberated at the cathode and anode = 4:32=1:8

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