Electrolysis: Faraday’s Laws

How much copper does a current plate out, how long does it take, or how much current do you need? Every mole of electrons that flows through a cell deposits or releases a fixed amount of substance. Enter the product, the number of electrons per formula unit and two of mass, current and time for the third, with the charge, the amount of substance, the gas volume for gases and, with the cell voltage, the energy used.

ElectrolysisElectroplatingFaraday

Inputs

Input examples

Type the substance that is deposited or released, such as Cu, Al or H_2 (type _ for a subscript). z is the number of electrons needed per formula unit: 2 for Cu from Cu²⁺, 3 for Al from Al³⁺, 1 for Ag, 2 for H₂ (2H⁺ + 2e⁻), 2 for Cl₂ (2Cl⁻ → Cl₂ + 2e⁻) and 4 for O₂ from water. The current efficiency is the share of the charge that makes the product (side reactions waste the rest). The gas volume uses the ideal gas law; the energy needs the cell voltage (optional).

Product

Electrons per formula unit z

Find

Current I

Time t

Current efficiency η — %

Cell voltage — V, optional

Counting Electrons

In 1834 Michael Faraday found that the amount of substance an electric current releases is proportional to the charge that flows. We now know why: each copper ion needs exactly two electrons to become a copper atom. A current of 1 ampere is 1 coulomb per second, and one mole of electrons carries 96 485 coulombs, the Faraday constant.

So the charge tells you the moles of electrons, z tells you how many moles of product that makes, and the molar mass turns it into grams. This is how electroplating (chromium, nickel, silver, gold), the production of aluminium, chlorine and sodium hydroxide, the refining of copper and the making of hydrogen by electrolysis are dimensioned.

In practice not every electron does useful work: some go into side reactions, such as hydrogen forming at a cathode where nickel is plated. The current efficiency accounts for that. The voltage times the charge is the electrical energy used; aluminium smelters need about 13–15 kWh for every kilogram.

F96 485 C per mole of electrons

1 A for 1 h3600 C = 0.0373 mol e⁻

1 Fdeposits 1/z mole of product

η < 100%side reactions waste charge

Product Over Time

mass

Result

Enter the values above

Worked Techniques

Charge, Then Moles

Convert the time to seconds first. The charge divided by F gives the moles of electrons.

Example: 2 A for 30 minutes moves 0.0373 mol of electrons.

Electrons to Product

Divide the moles of electrons by z for the moles of product, then multiply by the molar mass.

Example: the same charge gives 4.02 g of silver (z = 1) but only 0.335 g of aluminium (z = 3).

Gas Volume and Energy

For a gas, the ideal gas law turns moles into litres. The voltage across the cell times the charge is the energy, in joules (3.6 MJ = 1 kWh).

Example: 10 A for an hour makes 0.187 mol of H₂: 4.62 L at 25 °C and 1 bar.

Sources & Verification

References

P. Atkins and J. de Paula, Atkins’ Physical Chemistry, Oxford University Press — electrolysis and Faraday’s laws.
CODATA 2018 recommended values, NIST — F = 96 485.33212 C/mol, R = 8.314462618 J/(mol·K).
IUPAC standard atomic weights — the molar masses, as on the Molar Mass page.