Mass Deposited in Electrolysis (Faraday's Laws)

Core idea

Overview

Faraday's Laws of Electrolysis provide a quantitative relationship between the amount of substance produced or consumed at an electrode during electrolysis and the quantity of electricity passed through the electrolyte. This equation, derived from Faraday's second law, allows for the calculation of the mass of a substance deposited or liberated, given the current, time, molar mass, number of electrons involved, and Faraday's constant. It is fundamental to understanding and predicting the outcomes of electrochemical processes.

When to use: This equation is applied when you need to calculate the mass of a substance produced or consumed during an electrolysis reaction. It can also be rearranged to find the current, time, molar mass, or the number of electrons transferred per mole of substance. Ensure all units are consistent, especially time in seconds and current in Amperes.

Why it matters: Faraday's Laws are crucial for industrial applications such as electroplating (e.g., chrome plating, silver plating), the refining of metals (e.g., copper, aluminum), and the production of elements like chlorine and sodium. It's also essential in battery technology, corrosion prevention, and in analytical techniques like coulometry, providing a quantitative basis for electrochemical engineering and research.

Symbols

Variables

I = Current, t = Time, M = Molar Mass, n = Number of Electrons (n), F = Faraday Constant

I

Current

A

t

Time

s

M

Molar Mass

g/mol

n

Number of Electrons (n)

Variable

F

Faraday Constant

C/mol

Walkthrough

Derivation

Formula: Mass Deposited in Electrolysis (Faraday's Laws)

This formula combines the relationship between charge, current, time, and the stoichiometric relationship between charge and moles of substance.

The current efficiency is 100%, meaning all the charge passed contributes to the desired electrochemical reaction.
The current (I) is constant throughout the electrolysis process.
The molar mass (M) and the number of electrons (n) involved in the half-reaction are accurately known.

1

Relate Charge to Current and Time:

The total electric charge (Q) passed through the electrolyte is the product of the current (I) and the time (t) for which the current flows. (Units: Coulombs = Amperes ×seconds).

Q = I t

2

Relate Charge to Moles of Electrons:

The total charge (Q) is also equal to the number of moles of electrons ( $n_{e}$ ) multiplied by the Faraday constant (F), which is the charge carried by one mole of electrons (approximately 96485 C/mol).

Q = n_{e} F

3

Equate Charge Expressions:

By equating the two expressions for charge, we link the electrical parameters (I, t) to the chemical quantity (moles of electrons).

I t = n_{e} F

4

Relate Moles of Electrons to Moles of Substance:

From the balanced half-reaction, 'n' represents the number of moles of electrons required to deposit or produce one mole of the substance. So, the total moles of electrons ( $n_{e}$ ) is 'n' times the moles of the substance ( $n_{s u b s t an ce}$ ). Thus, $I t = n \times n_{s u b s t an ce} \times F$ .

n_{e} = n \times n_{s u b s t an ce}

5

Relate Moles of Substance to Mass:

The number of moles of a substance ( $n_{s u b s t an ce}$ ) is equal to its mass (m) divided by its molar mass (M). Substituting this into the previous equation gives $I t = n \times \frac{m}{M} \times F$ .

n_{s u b s t an ce} = \frac{m}{M}

6

Rearrange for Mass (m):

Rearranging the equation to make 'm' the subject yields the final formula for the mass deposited or produced during electrolysis.

m = \frac{I tM}{n F}

Note: This is a direct application of Faraday's Second Law of Electrolysis.

Result

m = \frac{I tM}{n F}

Source: AQA A-level Chemistry — Physical Chemistry (3.1.10.3 Electrolysis)

Free formulas

Rearrangements

Solve for $I$

Mass Deposited in Electrolysis: Make I the subject

I = \frac{mn F}{tM}

To make $I$ (Current) the subject of Faraday's Law, multiply mass by $n F$ and divide by $tM$ .

Difficulty: 2/5

Solve for $t$

Mass Deposited in Electrolysis: Make t the subject

t = \frac{mn F}{I M}

To make $t$ (Time) the subject of Faraday's Law, multiply mass by $n F$ and divide by $I M$ .

Difficulty: 2/5

Solve for $n$

Mass Deposited in Electrolysis: Make n the subject

n = \frac{I tM}{m F}

To make $n$ (Number of Electrons) the subject of Faraday's Law, multiply $I tM$ and divide by $m F$ .

Difficulty: 2/5

Solve for $F$

Mass Deposited in Electrolysis: Make F the subject

F = \frac{I tM}{mn}

To make $F$ (Faraday Constant) the subject of Faraday's Law, multiply $I tM$ and divide by $mn$ .

Difficulty: 2/5

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

Visual intuition

Graph

Graph unavailable for this formula.

The graph displays a straight line starting at the origin, showing that the mass deposited increases proportionally as the current increases. For a chemistry student, this means that applying a larger current to the system results in a greater amount of substance being produced, while a smaller current yields less mass. The most important feature of this linear relationship is that doubling the current will exactly double the mass of the substance deposited over the same period of time.

Graph type: linear

Why it behaves this way

Intuition

Imagine a steady stream of charge (current over time) flowing through a chemical bath, where each 'packet' of charge (moles of electrons)

m

The total mass of the substance deposited or produced at an electrode.

This is the direct outcome of the electrolysis, quantifying how much material is formed or consumed. More current, longer time, or a heavier substance (higher molar mass) will generally result in a greater mass.

I

The electric current flowing through the electrolytic cell, representing the rate of charge flow.

A higher current means more electrons are moving through the circuit per second, leading to a faster rate of electrochemical reaction and thus more substance produced in a given time.

t

The duration for which the electric current is applied to the electrolytic cell.

A longer time allows more total charge to pass through the cell, resulting in a greater extent of the electrochemical reaction and thus more substance produced or consumed.

M

The molar mass of the substance being deposited or produced.

For a given number of moles of substance produced, a higher molar mass means a greater total mass will be deposited. It's the 'weight per unit' of the material.

n

The number of moles of electrons required to deposit or produce one mole of the substance, determined by the stoichiometry of the half-reaction.

This factor represents how 'electron-intensive' the reaction is. If more electrons are needed per mole of substance (higher 'n'), then for a fixed amount of charge, fewer moles of the substance can be produced, leading

F

Faraday's constant, the magnitude of electric charge per mole of electrons.

This is a fundamental constant that converts the total charge (It) into moles of electrons. It acts as a scaling factor, linking the electrical quantity (charge) to the chemical quantity (moles of electrons).

Free study cues

Insight

Canonical usage

This equation is typically used with SI units for current, time, and Faraday's constant, while molar mass is often in g/mol, leading to mass in grams.

Common confusion

A common mistake is using time in minutes or hours instead of seconds, or using molar mass in kg/mol while expecting mass in grams, or vice-versa without proper conversion.

Dimension note

The variable 'n' represents the stoichiometric number of electrons transferred per mole of substance, making it a dimensionless quantity.

Unit systems

$m$ g - The mass of substance deposited or produced. Often expressed in grams when molar mass is in g/mol.

$I$ A - Electric current, measured in Amperes.

$t$ s - Time duration for which the current flows, measured in seconds.

$M$ g mol^-1 - Molar mass of the substance. Commonly used in g/mol in chemistry, though kg/mol is the strict SI unit.

$n$ dimensionless - The number of moles of electrons transferred per mole of substance in the balanced half-reaction.

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

Ballpark figures

Quantity:

One free problem

Practice Problem

A current of $0.50 A$ is passed through a solution of $C u S O_{4}$ for $30 minutes$ . Calculate the mass of copper deposited at the cathode. (Molar mass of Cu = $63.5 g/mol$ ).

Current0.5 A

Time1800 s

Molar Mass63.5 g/mol

Number of Electrons (n)2

Faraday Constant96485 C/mol

Solve for: mass

Hint: Remember to convert time from minutes to seconds and determine the number of electrons (n) for copper deposition.

The full worked solution stays in the interactive walkthrough.

Where it shows up

Real-World Context

In Electroplating a thin layer of gold onto jewelry for aesthetic and protective purposes, Mass Deposited in Electrolysis (Faraday's Laws) is used to calculate Mass Deposited from Current, Time, and Molar Mass. The result matters because it helps connect measured amounts to reaction yield, concentration, energy change, rate, or equilibrium.

Study smarter

Tips

Always convert time to seconds (s) and current to Amperes (A) before calculation.
Correctly determine 'n', the number of moles of electrons transferred per mole of the substance being deposited/produced, from the balanced half-reaction.
Use the correct value for the Faraday constant (F = 96485 C/mol).
Ensure the molar mass (M) is for the specific substance being deposited or produced (e.g., for $C l_{2}$ , use the molar mass of $C l_{2}$ , not Cl).
Remember that this formula assumes 100% current efficiency.

Avoid these traps

Common Mistakes

Using time in minutes or hours instead of seconds.
Incorrectly determining the value of 'n' (number of electrons) from the half-reaction.
Using the atomic mass instead of the molar mass for diatomic elements (e.g., using 35.5 for Cl instead of 71.0 for $C l_{2}$ ).
Errors in unit conversions, especially for current or time.

Keep going

Related Formulas

Common questions

Frequently Asked Questions

This formula combines the relationship between charge, current, time, and the stoichiometric relationship between charge and moles of substance.

This equation is applied when you need to calculate the mass of a substance produced or consumed during an electrolysis reaction. It can also be rearranged to find the current, time, molar mass, or the number of electrons transferred per mole of substance. Ensure all units are consistent, especially time in seconds and current in Amperes.

Faraday's Laws are crucial for industrial applications such as electroplating (e.g., chrome plating, silver plating), the refining of metals (e.g., copper, aluminum), and the production of elements like chlorine and sodium. It's also essential in battery technology, corrosion prevention, and in analytical techniques like coulometry, providing a quantitative basis for electrochemical engineering and research.

Using time in minutes or hours instead of seconds. Incorrectly determining the value of 'n' (number of electrons) from the half-reaction. Using the atomic mass instead of the molar mass for diatomic elements (e.g., using 35.5 for Cl instead of 71.0 for $Cl_2$). Errors in unit conversions, especially for current or time.

In Electroplating a thin layer of gold onto jewelry for aesthetic and protective purposes, Mass Deposited in Electrolysis (Faraday's Laws) is used to calculate Mass Deposited from Current, Time, and Molar Mass. The result matters because it helps connect measured amounts to reaction yield, concentration, energy change, rate, or equilibrium.

Always convert time to seconds (s) and current to Amperes (A) before calculation. Correctly determine 'n', the number of moles of electrons transferred per mole of the substance being deposited/produced, from the balanced half-reaction. Use the correct value for the Faraday constant (F = 96485 C/mol). Ensure the molar mass (M) is for the specific substance being deposited or produced (e.g., for $Cl_2$, use the molar mass of $Cl_2$, not Cl). Remember that this formula assumes 100% current efficiency.

References

Sources

Atkins' Physical Chemistry
Bird, Stewart, and Lightfoot, Transport Phenomena
Wikipedia: Faraday's laws of electrolysis
IUPAC Gold Book: Faraday constant
NIST CODATA
IUPAC Gold Book
Britannica, The Editors of Encyclopaedia. 'Faraday's laws of electrolysis'. Encyclopedia Britannica, 22 Sep. 2023.
AQA A-level Chemistry — Physical Chemistry (3.1.10.3 Electrolysis)

Overview

Variables

Derivation

Relate Charge to Current and Time:

Relate Charge to Moles of Electrons:

Equate Charge Expressions:

Relate Moles of Electrons to Moles of Substance:

Relate Moles of Substance to Mass:

Rearrange for Mass (m):

Rearrangements

Graph

Intuition

Insight

Practice Problem

Real-World Context

Tips

Common Mistakes

Related Formulas

Charge (Q=It)

Moles (n=m/M)

Standard Electrode Potential

Frequently Asked Questions

Sources