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

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