Question

give reason: The ratio of hydrogen and oxygen formed at the cathode and anode is 2:1 by volume.

Answer 1

for cathode, 2OH- = 2OH + 2e = H2O2 + 2e = H2O + O2 +2e
for anode, H+ + e = 1/2 H2
therefore, H2: O2 = 1 : 0.5 = 2:1