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: Write the half-reactions for the electrolysis of water.
At Anode (Oxidation): \[ 2\text{H}_2\text{O}(l) \rightarrow \text{O}_2(g) + 4\text{H}^+(aq) + 4e^- \] At Cathode (Reduction): \[ 4\text{H}^+(aq) + 4e^- \rightarrow 2\text{H}_2(g) \] (Alternatively, using water: \(4\text{H}_2\text{O} + 4e^- \rightarrow 2\text{H}_2 + 4\text{OH}^-\)) From these balanced half-reactions, we see that for every 1 mole of O\(_2\) produced, 2 moles of H\(_2\) are produced. The molar ratio is H\(_2\) : O\(_2\) = 2 : 1.
Step 2: Calculate the moles of O\(_2\) produced.

• Mass of O\(_2\) = 16 g
• Molar mass of O\(_2\) = 32 g/mol

\[ \text{Moles of O}_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{16 \text{ g}}{32 \text{ g/mol}} = 0.5 \text{ mol} \] Step 3: Calculate the moles of H\(_2\) produced.
Using the molar ratio from Step 1: \[ \text{Moles of H}_2 = 2 \times \text{Moles of O}_2 = 2 \times 0.5 \text{ mol} = 1 \text{ mol} \] Step 4: Calculate the volume of H\(_2\) at STP.
At Standard Temperature and Pressure (STP), 1 mole of any ideal gas occupies a volume of 22.4 liters. \[ \text{Volume of H}_2 = \text{moles} \times \text{molar volume at STP} = 1 \text{ mol} \times 22.4 \text{ L/mol} = 22.4 \text{ L} \] Step 5: Final Answer.
The volume of H\(_2\) gas liberated at the cathode is 22.4 L.