Question

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

• A. 22.4 L
• B. 11.2 L
• C. 2.24 L
• D. 1.12 L

Hint:

A quick way to solve this is to remember that during the electrolysis of water, the volume of hydrogen produced is always double the volume of oxygen produced (V\(_H_2\) = 2V\(_O_2\)). Moles of O\(_2\) = 16/32 = 0.5 mol. Volume of O\(_2\) at STP = 0.5 \(\times\) 22.4 = 11.2 L. Volume of H\(_2\) = 2 \(\times\) Volume of O\(_2\) = 2 \(\times\) 11.2 L = 22.4 L.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The overall reaction for the electrolysis of water dictates the stoichiometric ratio of Hydrogen gas produced to Oxygen gas produced.
Step 2: Key Formula or Approach:
Write the balanced chemical equation for the electrolysis of water ($2H_2O \rightarrow 2H_2 + O_2$). Convert the given mass of $O_2$ to moles, use the stoichiometric ratio to find moles of $H_2$, and then convert it to volume at STP using $V = n \times 22.4$ L.
Step 3: Detailed Explanation:
1. Balanced Equation:
The electrolysis of water is given by: $2H_{2}O(l) \rightarrow 2H_{2}(g) + O_{2}(g)$.
This shows that $2$ moles of $H_{2}$ are produced for every $1$ mole of $O_{2}$. The molar ratio is $n(H_2) : n(O_2) = 2 : 1$.

2. Calculate moles of O$_{2$:}
Mass of $O_{2}$ = 16 g. Molar mass of $O_{2}$ = 32 g/mol.
Moles of $O_{2} = \frac{16}{32} = 0.5$ mol.

3. Calculate moles of H$_{2$:}
From the $2:1$ ratio, moles of $H_{2} = 2 \times 0.5 = 1.0$ mol.

4. Calculate volume at STP:
Volume of 1 mole of an ideal gas at standard temperature and pressure (STP) is 22.4 L.
Volume of $H_{2} = 1.0 \text{ mol} \times 22.4 \text{ L/mol} = 22.4$ L.
Step 4: Final Answer:
The volume of hydrogen gas liberated is 22.4 L.