Faraday's Law (Electrolysis) Equation, Derivation & Rearrangements

Core idea

Overview

Faraday's First Law of Electrolysis quantifies the relationship between the mass of a substance produced at an electrode and the total electric charge passed through the electrolytic cell. It established the electrochemical equivalent, proving that chemical changes are directly proportional to the quantity of electricity used in a circuit.

When to use: Apply this equation when calculating the yield of a metal during electroplating or determining the time/current required to produce a specific mass of product. It is applicable for DC electrolysis where the reaction stoichiometry and electron transfer per ion are known. It assumes ideal conditions where the current efficiency is 100%.

Why it matters: This principle is the foundation of industrial metallurgy, particularly in the production of aluminum via the Hall-Héroult process and the refining of copper. It allows engineers to predict the efficiency and production rates of chemical manufacturing plants. It also plays a vital role in the design and charging protocols of secondary batteries.

Symbols

Variables

m = Mass Deposited, Q = Charge Passed, M = Molar Mass, F = Faraday Constant, z = Valency

m

Mass Deposited

g

Q

Charge Passed

C

M

Molar Mass

g / m o l

F

Faraday Constant

C / m o l

z

Valency

V a r iab l e

Walkthrough

Derivation

Understanding Faraday's Laws of Electrolysis

Links the amount of substance produced to charge passed by converting charge to moles of electrons.

Current is constant.
100% current efficiency (no side reactions).

1

Calculate Charge:

Charge (C) equals current (A) times time (s).

Q = I \cdot t

2

Convert to Moles of Electrons:

Divide by Faraday constant F ( $\approx$ 96485\; $C mol$ ^{-1}) to get moles of electrons transferred.

n (e^{-}) = \frac{Q}{F}

Result

n (e^{-}) = \frac{Q}{F}

Source: Edexcel A-Level Chemistry — Redox and Electrochemistry

Free formulas

Rearrangements

Solve for $m$

Make m the subject

m = \frac{QM}{F z}

m is already the subject of the formula.

Difficulty: 1/5

Solve for $Q$

Faraday's Law (Electrolysis): Make Q the subject

Q = \frac{m F z}{M}

Start with Faraday's Law (Electrolysis) and rearrange to make Q (Charge Passed) the subject. This involves clearing the denominator and then isolating Q.

Difficulty: 2/5

Solve for $M$

Make M the subject

M = \frac{m F z}{Q}

To make M the subject of Faraday's Law (Electrolysis), first clear the denominator Fz, then divide by Q.

Difficulty: 2/5

Solve for $F$

Make F the subject

F = \frac{QM}{m z}

Start from Faraday's Law (Electrolysis). To make F the subject, multiply both sides by Fz to clear the denominator, then divide by mz to isolate F.

Difficulty: 2/5

Solve for $z$

Make z the subject

z = \frac{QM}{F m}

Start from Faraday's Law (Electrolysis). To make z the subject, first multiply by Fz to clear the denominator, then divide by Fm.

Difficulty: 2/5

The static page shows the finished rearrangements. The app keeps the full worked algebra walkthrough.

Visual intuition

Graph

The graph is a straight line passing through the origin, showing that mass deposited is directly proportional to the charge passed. For a chemistry student, this means that a small charge results in a tiny amount of substance collected, while a large charge leads to a proportionally greater mass of product. The most important feature is the constant slope, which confirms that doubling the charge passed will always result in exactly doubling the mass deposited.

Graph type: linear

Why it behaves this way

Intuition

Imagine electrons flowing through a solution like tiny carriers, each delivering a specific 'packet' of charge that enables a fixed number of ions to transform into neutral atoms and accumulate as a solid layer on an

m

The total mass of a substance deposited or consumed at an electrode.

Directly measures the tangible outcome of the electrolysis process; more current and time means more mass.

Q

The total electric charge (in coulombs) that passes through the electrolytic cell.

Represents the 'amount of electricity' that drives the chemical reaction; it's the product of current and time (Q=It).

M

The molar mass of the substance being deposited or reacted, expressed in g/mol.

Determines how much mass one mole of the substance contributes; heavier atoms/molecules result in more mass for the same number of moles.

F

The Faraday constant, representing the magnitude of electric charge per mole of electrons (approximately 96485 C/mol).

A fundamental conversion factor that links the macroscopic electrical charge to the microscopic number of moles of electrons involved in the reaction.

z

The charge number (or valence) of the ion, representing the number of electrons transferred per mole of the substance during the electrode reaction.

Indicates the stoichiometry of the electron transfer; a higher 'z' means fewer moles of substance are produced per unit of charge, as each ion requires more electrons.

Free study cues

Insight

Canonical usage

The equation is typically used with SI-derived units where mass (m) is in grams (g), molar mass (M) in grams per mole (g mol^-1), charge (Q) in Coulombs (C), and the Faraday constant (F) in Coulombs per mole (C mol^-1).

Common confusion

Students often mix units for molar mass (g/mol vs kg/mol) and mass (g vs kg) without ensuring consistency, leading to incorrect magnitudes.

Unit systems

$m$ g · Commonly calculated in grams (g) for practical chemistry problems, though the SI base unit for mass is kilogram (kg).

$Q$ C · Total electric charge, often calculated as current (A) multiplied by time (s), i.e., Q = It.

$M$ g mol^-1 · Molar mass. Must be consistent with the unit chosen for mass (m). If m is in g, M should be in g mol^-1.

$F$ C mol^-1 · The Faraday constant, representing the charge per mole of electrons.

$z$ dimensionless · The charge number, representing the number of moles of electrons transferred per mole of substance in the half-reaction. It is an integer.

One free problem

Practice Problem

Calculate the mass of copper deposited from a CuSO₄ solution when a charge of 5000 Coulombs is passed through the cell. (Copper molar mass = 63.55 g/mol, Valency z = 2).

Charge Passed5000 C

Molar Mass63.55 g/mol

Faraday Constant96485 C/mol

Valency2

Solve for: $m$

Hint: Copper in CuSO₄ exists as Cu²⁺ ions, so the valency (z) is 2.

The full worked solution stays in the interactive walkthrough.

Where it shows up

Real-World Context

Calculating mass of copper deposited during electroplating.

Study smarter

Tips

Ensure the valency z represents the actual number of electrons transferred for one unit of the substance produced.
Charge Q is often derived from Current (I) × Time (t); always convert time to seconds.
The Faraday constant F is approximately 96485 Coulombs per mole of electrons.
Match the units of mass (m) and molar mass (M), typically both in grams.

Avoid these traps

Common Mistakes

Forgetting to include valency z.
Using wrong units for Q.

Keep going

Related Formulas

Common questions

Frequently Asked Questions

Links the amount of substance produced to charge passed by converting charge to moles of electrons.

Apply this equation when calculating the yield of a metal during electroplating or determining the time/current required to produce a specific mass of product. It is applicable for DC electrolysis where the reaction stoichiometry and electron transfer per ion are known. It assumes ideal conditions where the current efficiency is 100%.

This principle is the foundation of industrial metallurgy, particularly in the production of aluminum via the Hall-Héroult process and the refining of copper. It allows engineers to predict the efficiency and production rates of chemical manufacturing plants. It also plays a vital role in the design and charging protocols of secondary batteries.

Forgetting to include valency z. Using wrong units for Q.

Calculating mass of copper deposited during electroplating.

Ensure the valency z represents the actual number of electrons transferred for one unit of the substance produced. Charge Q is often derived from Current (I) × Time (t); always convert time to seconds. The Faraday constant F is approximately 96485 Coulombs per mole of electrons. Match the units of mass (m) and molar mass (M), typically both in grams.

References

Sources

Atkins' Physical Chemistry
IUPAC Gold Book: Faraday constant
Wikipedia: Faraday's laws of electrolysis
NIST CODATA
IUPAC Gold Book
Edexcel A-Level Chemistry — Redox and Electrochemistry

Faraday's Law (Electrolysis)

Overview

Variables

Derivation

Calculate Charge:

Convert to Moles of Electrons:

Rearrangements

Graph

Intuition

Insight

Practice Problem

Real-World Context

Tips

Common Mistakes

Related Formulas

Standard Cell Potential

Frequently Asked Questions

Sources