Faraday's Law of Electrolysis
Relates electric charge to mass deposited
Formula
m = (Q × M) / (n × F)
m: mass deposited (g), Q: charge (C), M: molar mass (g/mol), n: electrons per ion, F: Faraday constant (96485 C/mol).
Example
Given: Q = 10,000 C, ion Cu²⁺ (n = 2), M = 63.546 g/mol.
m = (10,000 × 63.546) / (2 × 96485) g
m ≈ (635,460) / 192,970 ≈ 3.29 g
Answer: ≈ 3.29 g of Cu deposited
Notes & Pitfalls
- Use correct valence n; for Ag⁺, n = 1; for Al³⁺, n = 3.
- Ensure molar mass M corresponds to the deposited species.
- Account for efficiency if the process is not 100% current efficient.