Question

In the electrolysis of water, if the mass of the gas collected at the anode is $m_a$ and the mass of the gas collected at the cathode is $m_c$, the value of $\frac{m_c}{m_a}$ is :

• A. 8
• B. 16
• C. $\frac{1}{16}$
• D. $\frac{1}{8}$

Hint:

Always remember: In the electrolysis of water, hydrogen (light gas) is collected at the cathode and oxygen (heavier gas) at the anode in a 2:1 volume ratio. Convert volume to mass using molar masses.

Answer 1

The Correct Option is A

Electrolysis of water: At the cathode, \( 2H_2O + 2e^- \rightarrow H_2 + 2OH^- \) (Hydrogen gas). At the anode, \( 2H_2O \rightarrow O_2 + 4H^+ + 4e^- \) (Oxygen gas). Hydrogen and oxygen are produced in a 2:1 volume ratio. Molar mass of hydrogen (\( H_2 \)) = 2 g/mol. Molar mass of oxygen (\( O_2 \)) = 32 g/mol. The mass ratio is calculated as: \[ m_c : m_a = \frac{2 \times 2}{1 \times 32} = \frac{4}{32} = \frac{1}{8}. \] Therefore, \[ \frac{m_c}{m_a} = 8. \] The correct answer is $(A)$.