Question

During the electrolysis of acidified water, 16 g of O$_2$ gas is formed. 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:

Always remember the stoichiometric ratio for the electrolysis of water ($2H_2 : 1O_2$) and the molar volume of a gas at STP (22.4 L/mol). These are fundamental constants in electrochemistry and gas laws.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks to determine the volume of hydrogen gas (H$_2$) liberated at the cathode during the electrolysis of acidified water, given the mass of oxygen gas (O$_2$) formed at the anode. The conditions are STP (Standard Temperature and Pressure).
Step 2: Key Formula or Approach:
1. Write the balanced chemical equation for the electrolysis of water to determine the stoichiometric ratio between O$_2$ and H$_2$.
2. Calculate the moles of O$_2$ from its given mass.
3. Use the stoichiometric ratio to find the moles of H$_2$.
4. At STP, 1 mole of any ideal gas occupies 22.4 L. Use this to find the volume of H$_2$.
Step 3: Detailed Explanation:
The overall reaction for the electrolysis of acidified water is:
\[ 2H_2O(l) \rightarrow 2H_2(g) + O_2(g) \]
From the balanced equation, 2 moles of H$_2$ gas are produced for every 1 mole of O$_2$ gas formed.
Given:
Mass of O$_2$ = 16 g
Molar mass of O$_2$ = $2 \times 16.0 \text{ g/mol} = 32.0 \text{ g/mol}$
Calculate moles of O$_2$:
\[ \text{Moles of O}_2 = \frac{\text{Mass of O}_2}{\text{Molar mass of O}_2} = \frac{16 \text{ g}}{32 \text{ g/mol}} = 0.5 \text{ mol} \]
Using the stoichiometric ratio (2 moles H$_2$ : 1 mole O$_2$):
\[ \text{Moles of H}_2 = 2 \times \text{Moles of O}_2 = 2 \times 0.5 \text{ mol} = 1.0 \text{ mol} \]
At STP, 1 mole of any ideal gas occupies 22.4 L.
\[ \text{Volume of H}_2 = \text{Moles of H}_2 \times 22.4 \text{ L/mol} = 1.0 \text{ mol} \times 22.4 \text{ L/mol} = 22.4 \text{ L} \]
Step 4: Final Answer:
The volume of H$_2$ gas liberated at the cathode under STP conditions is 22.4 L.