During the electrolysis of acidified water, 16 g of O ₂ gas is formed at anode. The volume of H ₂ gas liberated at cathode under STP conditions is

Step-by-step solution

The overall reaction for the electrolysis of acidified water is: 2 H ₂ O 2 H ₂ + O ₂ Number of moles of O ₂ produced = 16 32 = 0.5 mol From the balanced chemical equation, 1 mole of O ₂ is produced for every 2 moles of H ₂ . Number of moles of H ₂ produced = 2 0.5 = 1 mol At STP, the volume occupied by 1 mole of an ideal gas is 22.4 L. Volume of H ₂ gas liberated = 1 22.4 = 22.4 L Answer: 22.4 L