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29/09/2020 8:27 pm
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During electrolysis of water the volume of O2 liberated is 2.24 dm3. The volume of hydrogen liberated, under same conditions will be
(a) 2.24 dm3
(b) 1.12 dm3
(c) 4.48 dm3
(d) 0.56 dm3
1 Answer
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29/09/2020 8:30 pm
(c) 4.48 dm3
Explanation:
2H2O(2 vol.) + Electrolysis → 2H2(2 vol.) + O2(1 vol.)
Thus, the volume of hydrogen liberated is twice that of the volume of oxygen liberated. When 2.24 dm3 of oxygen is liberated the volume of hydrogen liberated will be 2 × 2.24 dm3 or 4.48 dm3.