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Question

# In the electrolysis of an aqueous solution of NaOH, 2.8 litres of oxygen was liberated at the anode at NTP. How much hydrogen in litres, was liberated at the cathode?

A
5.6
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B
5.60
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Solution

## Electrolysis of NaOHsolution : At cathode : 2H++2e−→H2 At anode : 4HO−→2H2O+O2+4e− Number of equivalents of O2(g) liberated at anode = Number of equivalents of H2(g) liberated at cathode. O2(g) liberated =volume of O2(g) liberated at NTPvolume occupied by 1 equivalent of O2(g) at NTP=2.8L5.6L=12 The same number of equivalents of hydrogen liberated at the cathode =12 ∴volume of hydrogen = number of equivalents × volume occupied by 1 equivalent of hydrogen at NTP =12×11.2 =5.6 litres.

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