ToolNestr

Faraday's Law of Electrolysis Calculator

Solve m = (I×t×M)/(n×F) for the mass of metal deposited or dissolved during electrolysis, from current, time, molar mass, and the number of electrons transferred per ion. Two 3D diagrams show an electroplating cell and compare deposited mass at different plating times, with charts showing mass vs current and mass vs time.

Reviewed by the ToolNestr Editorial Team — July 2026

Disclaimer: This tool is provided for educational purposes to support learning in chemistry. It is not a substitute for professional laboratory, safety, or dosage calculations.
Chemistry
Mass deposited (m)
Charge passed (Q)

Electroplating in action

1. An electroplating cell

A metal object (cathode) sits in a solution of metal ions — current drives those ions to deposit onto the object's surface as a thin plated layer.

2. Deposited mass over time

A short plating time (thin layer) versus a long plating time (thick layer) — mass builds up linearly with time at constant current.

Faraday's law graphs

Mass deposited vs current (fixed t=3600s, Cu)
Mass deposited vs time (fixed I=2A, Cu)

How it works

The core idea in one line: every mole of electrons that flows through an electrolytic cell drives a fixed, predictable amount of chemical change at each electrode — so the total mass deposited or dissolved is directly proportional to the total electric charge passed.

Q = I × t

total electric charge passed — I in amps, t in seconds

m = (Q × M) / (n × F)

mass deposited — M = molar mass, n = electrons per ion, F = 96,485 C/mol

Charge and current are related simply by Q = I×t. Each mole of a metal ion needs exactly n moles of electrons to be reduced to metal (n depends on the ion's charge, like 2 for Cu²⁺ or 1 for Ag⁺), and each mole of electrons carries a charge of F = 96,485 coulombs (Faraday's constant). Combining these relationships, the moles of metal deposited equal Q/(nF), and multiplying by molar mass M converts moles into the mass m = (I×t×M)/(n×F) that actually plates out.

Worked example 1 — copper electroplating

Given: A current of I = 2 A flows for t = 3600 s (1 hour) through a copper sulfate solution. Copper's molar mass M = 63.55 g/mol, and Cu²⁺ requires n = 2 electrons.

Charge: Q = I × t = 2 × 3600 = 7200 C
Formula: m = (Q × M) / (n × F)
Substitute: m = (7200 × 63.55) / (2 × 96,485) = 457,560 / 192,970
Result: m ≈ 2.371 g of copper deposited

Just over 2 grams of copper plates out in one hour at 2 amps — real electroplating baths run for hours to build up a usable coating thickness.

Worked example 2 — silver electroplating

Given: A current of I = 0.5 A flows for t = 1800 s (30 minutes) through a silver nitrate solution. Silver's molar mass M = 107.87 g/mol, and Ag⁺ requires n = 1 electron.

Charge: Q = I × t = 0.5 × 1800 = 900 C
Formula: m = (Q × M) / (n × F)
Substitute: m = (900 × 107.87) / (1 × 96,485) = 97,083 / 96,485
Result: m ≈ 1.006 g of silver deposited

Silver deposits more efficiently per coulomb than copper here, partly because Ag⁺ needs only 1 electron per ion instead of 2.

Mass deposited by 1 coulomb of charge, by metal

For the same tiny amount of charge, different metals deposit different masses — driven by both molar mass and how many electrons each ion needs.

Metal ionnM (g/mol)mg per coulomb
Ag⁺ (silver)1107.871.118 mg
Cu²⁺ (copper) ★263.550.329 mg
Ni²⁺ (nickel)258.690.304 mg
Al³⁺ (aluminum)326.980.0932 mg

★ Computed as (M/n)/F × 1000 mg. Silver deposits the most mass per coulomb here, combining a large molar mass with needing only 1 electron per ion.

Where Faraday's law actually matters

✨ Electroplating and jewelry finishing

Jewelry, tableware, and electronics use electroplating to apply thin, controlled coatings of gold, silver, chromium, or nickel — Faraday's law lets manufacturers precisely time the process to hit a target coating mass.

🔋 Battery manufacturing and charge capacity

Battery charge capacity ratings are fundamentally tied to Faraday's law, since the amount of active material that can be oxidized or reduced sets the maximum charge (and therefore energy) the battery can store.

⚙️ Aluminum production (Hall-Héroult process)

Industrial aluminum smelting uses electrolysis on a massive scale, and Faraday's law is the basis for calculating the electrical energy and time needed to produce a target quantity of aluminum metal from its ore.

🧪 Electrorefining of metals

Faraday's law underlies electrorefining, where impure metal is dissolved from an anode and redeposited in pure form at a cathode, used industrially to purify copper and other metals to very high purity.

Common misconceptions

"The same current for the same time always deposits the same mass, regardless of the metal."

Mass deposited depends on the metal's molar mass and how many electrons its ion requires (n), not just on current and time — silver, copper, and aluminum all deposit different masses for the exact same charge passed.

"Faraday's constant is the charge of a single electron."

Faraday's constant (96,485 C/mol) is the charge carried by one full mole of electrons, not a single electron — it equals the elementary charge (1.602×10⁻¹⁹ C) multiplied by Avogadro's number.

"Doubling the current doubles the deposition rate, but doubling the time has no such effect."

Both current and time have exactly the same effect, since they only ever enter the formula multiplied together as charge (Q = I×t) — doubling either one alone doubles the mass deposited, just as doubling both would quadruple it.

"Electrolysis and a galvanic cell (battery) are the same process."

They are opposites: a galvanic cell generates electrical energy from a spontaneous chemical reaction (positive E°cell), while electrolysis uses external electrical energy to force a non-spontaneous reaction to occur — but Faraday's law of mass-charge proportionality applies to both.

Formula sources & further reading

The formulas here are standard, traceable to:

  • OpenStax, Chemistry 2e — Chapter 17, "Electrochemistry" (free, peer-reviewed). openstax.org
  • Brown, LeMay & Bursten, Chemistry: The Central Science — Chapter 20, Electrochemistry (Electrolysis).
  • Zumdahl & Zumdahl, Chemistry — Electrolysis and Faraday's laws.

m = (I×t×M)/(n×F) with F = 96,485 C/mol (Faraday's constant). Assumes 100% current efficiency (no side reactions). Results are rounded for display.

How to use this calculator

1

Enter current and time

Provide current in amps and duration in seconds (or use the minutes/hours toggle).

2

Enter molar mass and n

Provide the metal ion's molar mass and how many electrons it needs per ion.

3

Read the result

Mass deposited or dissolved solves live in grams.

Related tools

Frequently asked questions

What is Faraday's law of electrolysis?

Faraday's law states that the mass of a substance deposited or dissolved at an electrode is directly proportional to the total electric charge passed through the cell: m = (Q×M)/(n×F), where Q = I×t is the charge, M is molar mass, n is electrons per ion, and F is Faraday's constant (96,485 C/mol).

What is Faraday's constant?

Faraday's constant, F = 96,485 coulombs per mole, is the electric charge carried by one mole of electrons. It links the macroscopic quantity of charge passed through a cell to the microscopic number of electrons actually transferred.

Why does the number of electrons transferred (n) matter?

Different ions require different numbers of electrons to be reduced or oxidized — Cu²⁺ needs 2 electrons per copper atom deposited, while Ag⁺ needs only 1 per silver atom. For the same charge passed, an ion requiring fewer electrons deposits proportionally more mass.

How is charge related to current and time?

Charge Q = I × t, where I is current in amps and t is time in seconds. This is the same relationship used in basic circuit calculations — electrolysis just uses that charge to drive a chemical reaction at each electrode.

What is electroplating and how does Faraday's law apply?

Electroplating deposits a thin layer of metal onto a surface by passing a controlled current through a solution containing that metal's ions. Faraday's law lets engineers precisely calculate how long to run the current to achieve a target coating thickness or mass.

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