Faraday’s Laws and Quantitative Electrolysis

Faraday’s laws of electrolysis state that the mass of a substance produced at an electrode is proportional to the electric charge passed and to its equivalent mass: m = (I × t × M) / (z × F), where I is the current, t the time, M the molar mass, z the number of electrons per ion and F = 96,485 C/mol. In brief: a current of 2.00 A passing for 30.0 minutes deposits 1.19 g of copper from a Cu²⁺ solution.

Faraday’s laws make electrolysis a quantitative tool. They allow chemists to calculate how much metal is deposited in electroplating, how long an electrolysis takes and how much gas is evolved. This article states the two laws, gives a systematic method for solving problems, provides a table of the mass deposited per unit of charge for common metals and works through four examples.

What are Faraday’s laws?

First law: the mass of a substance deposited or liberated at an electrode is directly proportional to the quantity of electric charge that passes through the cell: m ∝ Q.
Second law: for a given quantity of charge, the masses of different substances deposited are proportional to their equivalent masses (molar mass divided by the number of electrons transferred per ion).

In modern terms, one mole of electrons carries a charge equal to the Faraday constant, F = 96,485 C/mol (96,485.33 C/mol, an exact value since the 2019 redefinition of the SI units). To reduce 1 mol of a z-charged ion, z mol of electrons are needed. So the mass deposited is:

m = (Q / F) × (M / z) = (I × t × M) / (z × F)

How do you solve electrolysis problems?

Calculate the charge from the current and the time: Q = I × t (t in seconds).
Convert the charge to moles of electrons: n(e⁻) = Q / 96,485.
Use the half-reaction to find the moles of substance per mole of electrons (for example Cu²⁺ + 2e⁻ → Cu means 1 mol Cu per 2 mol e⁻).
Convert the moles to mass (with the molar mass) or to volume (for gases, with the ideal gas law).

To find the time needed, reverse the steps.

Mass deposited per mole of electrons

Equivalent masses and mass deposited per ampere-hour (3600 C = 0.03731 mol e⁻)
Metal	Ion	Molar mass (g/mol)	g per mol e⁻	g per ampere-hour
Silver	Ag⁺	107.87	107.87	4.025
Gold	Au³⁺	196.97	65.66	2.450
Copper	Cu²⁺	63.55	31.78	1.186
Zinc	Zn²⁺	65.38	32.69	1.220
Nickel	Ni²⁺	58.69	29.35	1.095
Aluminum	Al³⁺	26.98	8.99	0.336

Worked examples

Example 1: mass of copper plated

What mass of copper is deposited when a current of 2.00 A passes through a solution of CuSO₄ for 30.0 minutes?

Q = 2.00 × (30.0 × 60) = 3600 C.
n(e⁻) = 3600 / 96,485 = 0.03731 mol.
n(Cu) = 0.03731 / 2 = 0.01866 mol.
m = 0.01866 × 63.55 = 1.19 g.

Example 2: time needed to deposit silver

How long does it take to plate 5.00 g of silver from an AgNO₃ solution with a current of 1.50 A?

n(Ag) = 5.00 / 107.87 = 0.04635 mol, and one electron is needed per Ag⁺.
Q = 0.04635 × 96,485 = 4472 C.
t = Q / I = 4472 / 1.50 = 2981 s = 49.7 min.

Example 3: aluminum production

How much aluminum is produced per hour in a cell with a current of 100,000 A at 100 % efficiency?

Q = 100,000 × 3600 = 3.6 × 10⁸ C, so n(e⁻) = 3.6 × 10⁸ / 96,485 = 3731 mol.
Al³⁺ + 3e⁻ → Al, so n(Al) = 3731 / 3 = 1244 mol.
m = 1244 × 26.98 = 33,560 g = 33.6 kg per hour.

This corresponds to about 0.8 tonnes per day, and real cells operate at an efficiency of 90–95 %.

Example 4: volume of gases from water electrolysis

A current of 5.00 A passes through acidified water for 10.0 minutes. Find the volumes of H₂ and O₂ at 25 °C and 1 atm (molar volume 24.47 L/mol).

Q = 5.00 × 600 = 3000 C, so n(e⁻) = 0.03109 mol.
Cathode: 2H⁺ + 2e⁻ → H₂, so n(H₂) = 0.01555 mol and V = 0.01555 × 24.47 = 0.380 L.
Anode: 2H₂O → O₂ + 4H⁺ + 4e⁻, so n(O₂) = 0.03109 / 4 = 0.00777 mol and V = 0.190 L.

The volume ratio of hydrogen to oxygen is 2 : 1, as in the formula of water.

Example 5: finding the charge of an ion

A current of 3.00 A for 20.0 min deposits 1.15 g of a metal with a molar mass of 58.7 g/mol. n(metal) = 1.15 / 58.7 = 0.01959 mol; Q = 3.00 × 1200 = 3600 C, which gives 0.03731 mol of electrons. The ratio is 0.03731 / 0.01959 = 1.90, close to 2, so the ion has a charge of +2 (the metal is probably nickel, Ni²⁺). The small difference from 2 shows that the current efficiency was a little under 100 %. This kind of calculation was historically used to find the charge of metal ions.

Copper deposited from Cu²⁺ by a constant current of 2.00 A at 100% current efficiency, from m = I t M / (z F). The mass is directly proportional to the charge passed, so the graph is a straight line through the origin; 30.0 minutes gives 1.19 g, as in Example 1.

What is current efficiency?

In practice, not all of the current goes into the desired reaction: side reactions such as the evolution of hydrogen at a plating cathode use some of the charge. The current efficiency is the ratio of the actual mass deposited to the theoretical mass from Faraday’s law, times 100 %. Copper refining has an efficiency of about 95–98 %, chrome plating is typically only 10–20 %, and aluminum smelting about 90–95 %.

Who was Faraday?

Michael Faraday (1791–1867) was an English scientist who started as a bookbinder’s apprentice and became one of the greatest experimenters. In 1833 and 1834 he published the laws of electrolysis from experiments with a variety of electrolytes, and a “voltameter” in which the amount of gas from water decomposition was used to measure the quantity of electricity. His work gave the first quantitative evidence of the discrete nature of electric charge in matter, long before the electron was discovered in 1897. The unit of capacitance, the farad, and the Faraday constant are named after him.

Which mistakes are common?

Using minutes in Q = I × t: convert to seconds.
Forgetting z: divide the moles of electrons by the ion charge.
Mixing up the amount of electrons and the amount of substance: one mole of Cu needs two moles of electrons.
Ignoring current efficiency: real masses are lower.
Using the wrong half-reaction at the electrode: in aqueous solutions water may react instead of the metal ion.