Electrolysis And Faraday's Laws
During the electrolysis of molten NaCl, 23 g of sodium is deposited at the cathode. The quantity of electricity passed through the electrolyte is (Atomic mass of Na = 23 g mol⁻¹, 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
- Medium
- Year
- 2025
Solution
Correct Answer:
96500 C
Using Faraday's first law: m = (M × Q)/(n × F), where m = mass deposited, M = molar mass, n = number of electrons involved, Q = charge, and F = Faraday constant. Na⁺ + e⁻ → Na, so n = 1. Rearranging: Q = (m × n × F)/M = (23 × 1 × 96500)/23 = 96500 C. This means exactly 1 Faraday of electricity (96500 C) deposits one mole (23 g) of sodium. This is a direct consequence of the stoichiometry of the reduction half-reaction, where each sodium ion requires only one electron for reduction to metallic sodium.
This medium 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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