Faraday's Law of Electrolysis

mass = (I × t × M) ÷ (n × F): the mass deposited or liberated in electrolysis is directly proportional to the total electric charge passed.

m = \frac{I \cdot t \cdot M}{n \cdot F}

mass = (current x time x molar mass) / (n x Faraday constant); moles of electrons = I x t / F

What it means

Current is charge per second, so multiplying current by time gives the total charge that flowed. Dividing that charge by Faraday's constant converts it into moles of electrons, the half-reaction tells you how many electrons produce one mole of product, and the molar mass turns that into grams. The chain is beautifully direct: double the current or double the time, and you double the mass.

Variables

Symbol	Name	Unit	Meaning
$m$	Mass of product	g	The mass of substance deposited or liberated at the electrode.
$I$	Current	A (C/s)	The steady electric current flowing through the cell.
$t$	Time	s	The duration of electrolysis, in seconds.
$M$	Molar mass	g/mol	The molar mass of the deposited or liberated substance.
$n$	Electrons per product	mol e⁻/mol	Moles of electrons required per mole of product, from the balanced half-reaction.
$F$	Faraday constant	C/mol	The charge of one mole of electrons, 96485 C/mol.

Assumptions

Current efficiency is 100% — no side reactions steal charge from the desired half-reaction.
Time t is in seconds; convert minutes or hours before multiplying.
n comes from the balanced half-reaction for the product of interest.
M in g/mol gives the mass m in grams.

Worked example

Problem

Copper is electroplated by Cu²⁺ + 2e⁻ → Cu(s) at a current of 2.00 A for 30.0 min. What mass of copper deposits?

Given:

Half-reaction: Cu²⁺ + 2e⁻ → Cu(s), so n = 2
Current I = 2.00 A
Time t = 30.0 min
M(Cu) = 63.546 g/mol
F = 96485 C/mol

Convert time to seconds: t = 30.0 min × 60 = 1800 s.
Total charge: Q = I × t = 2.00 A × 1800 s = 3600 C.
Moles of electrons: 3600 C / 96485 C/mol = 0.0373 mol.
Moles of copper: 0.0373 mol / 2 = 0.0187 mol.
Mass: 0.0187 mol × 63.546 g/mol = 1.19 g.

1.19 g of Cu deposited

About 1.2 g of copper plates out in half an hour at 2 amps — real electroplating plants run exactly this calculation at industrial scale.