Question

Statement 1: During electrolysis of acidified water, the volume of hydrogen given off is twice the volume of oxygen.
Statement 2: During the electrolysis of acidified water the product formed at the anode is oxygen gas and at the cathode is hydrogen gas.

• A. Statement 1 is incorrect, but statement 2 is correct.​​
• B. Only statement 1 is correct.​
• C. Statement 1 is correct, but statement 2 is incorrect.​
• D. Both statement 1 and statement 2 are correct.​

Answer 1

The correct option is D Both statement 1 and statement 2 are correct.​

• Cathode reaction during the electrolysis of water:
• $2H_{2}O\left(l\right)+4e^{-}\to 4{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)+2H_2\left(g\right)$
• Reduction at the cathode: $H_{2}O$ gains electrons, forms hydrogen gas.
• Anode reaction during the electrolysis of water:
• $2H_{2}O\left(l\right)\to O_2\left(g\right)+4H^{+}\left(\mathrm{aq}\right)+4e^{-}$
• Oxidation at the anode: $H_{2}O$ loses electrons, forms oxygen gas.
• ​The number of hydrogen molecules produced is twice the number of oxygen molecules. So, statement 1 is correct.
• Oxygen gas forms at anode and hydrogen gas forms at cathode. So, statement 2 is also correct.