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
1.5A
2.3A
3.4A
4.2A.
Answers
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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
ANSWER== 4.2A
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