Faraday's Law Calculator
Calculate the mass deposited in electrolysis using Faraday’s law.
Mass Deposited
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Calculation Steps
Faraday's Laws of Electrolysis
Michael Faraday discovered the quantitative relationships governing electrolysis in the 1830s. His two laws describe how the amount of substance produced at an electrode depends on the electric charge passed and the chemical nature of the substance.
- First Law: The mass of a substance deposited or liberated at an electrode is directly proportional to the quantity of electric charge (Q) passed through the electrolyte.
- Second Law: For the same quantity of charge, the masses of different substances deposited are proportional to their equivalent weights (molar mass divided by the number of electrons transferred).
The Faraday constant, F = 96485 C/mol, represents the electric charge carried by one mole of electrons. It links the macroscopic world of measured charge to the number of moles of substance reacting at an electrode.
The Formula
Q = I × t
m = (Q × M) / (n × F) = (I × t × M) / (n × F)
- • m = Mass deposited (g)
- • Q = Electric charge (coulombs, C)
- • I = Current (amperes, A)
- • t = Time (seconds, s)
- • M = Molar mass of the substance (g/mol)
- • n = Number of electrons transferred per ion
- • F = Faraday constant = 96485 C/mol
The number of moles deposited equals Q / (n × F). Multiplying the moles by the molar mass M gives the mass deposited. Time must always be converted to seconds before computing the charge.
Electroplating Example
Problem: A copper(II) solution is electrolyzed with a current of 2 A for 1 hour. Copper has a molar mass of 63.55 g/mol and Cu²⁺ requires 2 electrons. How much copper is deposited on the cathode?
Step 1: Convert time to seconds
t = 1 hour × 3600 = 3600 s
Step 2: Calculate charge
Q = I × t = 2 A × 3600 s = 7200 C
Step 3: Calculate mass deposited
m = (Q × M) / (n × F) = (7200 × 63.55) / (2 × 96485) ≈ 2.37 g
Answer: ≈ 2.37 g of copper
Note: This calculator assumes 100% current efficiency, meaning all the charge passed contributes to depositing the substance of interest. In real cells, side reactions, gas evolution, and resistive losses can reduce the actual yield. Always verify molar masses and electron counts from reliable sources.