Question

In the electrolysis of acidulate water 11.2L of hydrogen gas is liberated at cathode. The volume of oxygen gas liberated at the anode is

Answer 1

1. The amount of O₂ released would be $\frac{11.2}{2}=5.6L$.
2. This is because,one molecule of water has 2H and one 1O.
3. So the volume of O₂ gas would be half of that of volume of H₂.