Electrolysis And Faraday's Laws
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):
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About This Question
- Subject
- chemistry
- Chapter
- redox reactions and electrochemistry
- Topic
- electrolysis and faraday's laws
- Difficulty
- Hard
- Year
- 2025
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.
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