Skip to content

Electrolysis And Faraday's Laws

Hardchemistry

During electrolysis of an aqueous CuSO₄ solution using platinum electrodes, 0.635 g of copper is deposited at the cathode. The volume of O₂ liberated at STP at the anode is (Atomic mass of Cu = 63.5, F = 96500 C mol⁻¹):

Select the correct option:

🔒 Solution Hidden from View

Submit your answer to unlock the detailed step-by-step solution.

About This Question

Subject
chemistry
Chapter
redox reactions and electrochemistry
Topic
electrolysis and faraday's laws
Difficulty
Hard
Year
2025
Tags
ElectrolysisCuSO₄Gas Volume at STP

Solution

Correct Answer:

112 mL

At the cathode: Cu²⁺ + 2e⁻ → Cu. Moles of Cu deposited = 0.635/63.5 = 0.01 mol. Electrons transferred = 0.01 × 2 = 0.02 mol. At the anode (with Pt electrodes, water is oxidised): 2H₂O → O₂ + 4H⁺ + 4e⁻. So moles of O₂ = moles of electrons / 4 = 0.02/4 = 0.005 mol. Volume at STP = 0.005 × 22400 mL = 112 mL. The cathode and anode processes are linked by the same total charge passed through the cell. Since 4 electrons are needed to produce one molecule of O₂, the moles of O₂ are always one-quarter of the total moles of electrons, which in turn are determined by the copper deposition stoichiometry.

This hard difficulty chemistry question is from the chapter redox reactions and electrochemistry, covering the topic of electrolysis and faraday's laws. It appeared in the 2025 exam.

Looking for more practice? Explore all chemistry questions or browse redox reactions and electrochemistry questions on RankGuru.