Apply Faraday’s law and stoichiometry to electrochemical cells

Alberta Grade 12 Chemistry

Electrochemical Changes

Alberta Chemistry 30 — Outcome 30-B2.8k

Connect a copper wire to a battery, dip two electrodes into a copper sulfate solution, and watch a shiny pink layer of copper slowly build up on one electrode. Nothing appears to be reacting — no bubbles, no colour change in the water around you — yet solid metal is forming out of a clear blue solution. The explanation lies in a flow of electrons that you cannot see but can measure precisely. In this lesson you will learn how to count those electrons using Faraday's law, and then use stoichiometry to predict exactly how much copper — or any other electroactive substance — is produced or consumed.

What you will learn

Explain how electric charge is related to the number of electrons transferred in an electrochemical cell.
Use Faraday's law to convert between charge (coulombs), current (amperes), and time (seconds).
Apply mole-ratio stoichiometry to find the mass of a substance deposited or dissolved at an electrode.
Identify and correct common errors in Faraday's law calculations.

Prerequisite Review: Charge, Current, and the Mole

Before connecting Faraday's law to chemistry, you need three quantities from previous courses. First, electric charge is measured in coulombs (C). Second, electric current is the rate of charge flow: one ampere (A) means one coulomb passes a point every second. Third, the mole is the chemist's counting unit — one mole of any particle contains

6.022 \times 10^{23}

particles.

The relationship between current, charge, and time is: charge equals current multiplied by time. In symbols,

Q = I \times t

, where

Q

is charge in coulombs,

I

is current in amperes, and

t

is time in seconds. This equation is your starting point for every Faraday's law problem. If a cell runs at 2.00 A for 500 s, the total charge that flows is

2.00 \times 500 = 1000

C.

You also need to recall half-reactions. In an electrolytic or galvanic cell, oxidation happens at the anode and reduction happens at the cathode. The half-reaction at the cathode in copper electroplating is

\text{Cu}^{2+}\,\text{(aq)} + 2\,e^- \rightarrow \text{Cu}\,\text{(s)}

. This tells you that two electrons are needed to deposit one atom of copper — or, scaling up, two moles of electrons are needed per mole of copper deposited.

Q = I \times t

$Q = I \times t$ links charge, current, and time.
Current is measured in amperes (A); charge in coulombs (C); time in seconds (s).
A half-reaction shows how many electrons are transferred per ion or atom.
Oxidation occurs at the anode; reduction occurs at the cathode.

Faraday's Constant and the Mole of Electrons

Michael Faraday discovered that the mass of substance produced at an electrode is directly proportional to the total charge that flows through the cell. The key bridge between electrical measurement and chemical amount is Faraday's constant, given the symbol

F

. Faraday's constant is the charge carried by exactly one mole of electrons:

F = 9.65 \times 10^4 \text{ C/mol}

(from the Alberta Chemistry 30 Data Booklet).

Think about it at the particle level. Each electron carries a charge of

1.60 \times 10^{-19}

C. Multiply that by Avogadro's number (

6.022 \times 10^{23}

electrons per mole) and you get approximately

9.65 \times 10^{4}

C per mole of electrons — exactly Faraday's constant. So

F

is not a new or mysterious quantity; it is just the charge of one mole of electrons.

To find the number of moles of electrons transferred during an electrolysis, divide the total charge

Q

by Faraday's constant:

n_{e^-} = \frac{Q}{F}

. Once you know the moles of electrons, you can use the mole ratio from the balanced half-reaction to find moles of the substance produced or consumed, and then convert to grams using molar mass — exactly the stoichiometry you practised in Chemistry 20.

n_{e^-} = \frac{Q}{F} = \frac{I \times t}{F}

Faraday's constant: $F = 9.65 \times 10^4 \text{ C/mol}_{e^-}$ .
$n_{e^-} = \frac{Q}{F}$ converts coulombs to moles of electrons.
The mole ratio from the half-reaction connects moles of electrons to moles of substance.
Moles of substance $\times$ molar mass = mass in grams.

The Full Calculation Pathway

Every Faraday's law stoichiometry problem follows the same four-step path. Step 1: calculate the total charge using

Q = I \times t

. Step 2: convert charge to moles of electrons using

n_{e^-} = Q / F

. Step 3: use the mole ratio from the balanced half-reaction to find moles of the target substance. Step 4: convert moles of substance to grams using the molar mass.

This four-step path works in any direction. If you know the mass deposited and need to find the current or time, simply reverse the steps: convert mass to moles, use the mole ratio to find moles of electrons, multiply by

F

to get charge, then divide by

I

to get time (or divide by

t

to get current). The algebra is straightforward as long as you write units at every step.

The mole ratio is the most important thing to read carefully from the half-reaction. For

\text{Al}^{3+}\,\text{(aq)} + 3\,e^- \rightarrow \text{Al}\,\text{(s)}

, three moles of electrons are needed per mole of aluminum. For

\text{Ag}^+\,\text{(aq)} + 1\,e^- \rightarrow \text{Ag}\,\text{(s)}

, only one mole of electrons is needed per mole of silver. Misreading this ratio is the single most common error in these problems.

m = \frac{I \times t}{F} \times \frac{1}{n} \times M

Step 1: $Q = I \times t$ . Step 2: $n_{e^-} = Q/F$ . Step 3: mole ratio from half-reaction. Step 4: multiply by molar mass.
The electron coefficient in the half-reaction is the mole ratio denominator.
The pathway can be run forwards (find mass) or backwards (find time or current).
Always carry units through every calculation step.

Applying the Pathway to Different Electrode Reactions

The same four-step pathway applies to any electrode reaction — deposition of a metal, dissolution of an anode, or production of a gas. For example, in the electrolysis of water, hydrogen gas forms at the cathode by the half-reaction

2\,\text{H}^+\,\text{(aq)} + 2\,e^- \rightarrow \text{H}_{2(g)}

. Here two moles of electrons produce one mole of hydrogen gas. If you need the volume of gas instead of a mass, you would convert moles of

\text{H}_2

to volume using molar volume at the stated conditions — a step identical to Chemistry 20 gas stoichiometry.

For anode dissolution, the direction reverses:

\text{Cu}\,\text{(s)} \rightarrow \text{Cu}^{2+}\,\text{(aq)} + 2\,e^-

. The stoichiometric ratio is the same (2 mol electrons per mol Cu), but now copper is being removed from the anode rather than deposited on the cathode. The mass calculation is identical in structure.

It is important to treat each half-cell separately. The charge that flows through the external circuit is the same for both electrodes, but the mass change at each electrode depends on that electrode's own half-reaction and the molar mass of its own substance. Do not mix the two half-reactions together when calculating mass.

n_{\text{substance}} = n_{e^-} \times \frac{1 \text{ mol substance}}{n \text{ mol } e^-}

The four-step pathway works for metal deposition, metal dissolution, and gas production.
Read the electron coefficient from the specific half-reaction for each electrode.
Both electrodes experience the same total charge $Q$ .
Mass or volume changes are calculated separately for each electrode using its own half-reaction.

Common Electrode Half-Reactions and Their Electron Ratios

Half-Reaction (cathode, reduction)	Electrons per Mole of Substance	Mole Ratio (mol substance / mol e⁻)
$\text{Ag}^+\,\text{(aq)} + e^- \rightarrow \text{Ag}\,\text{(s)}$	1	1 / 1
$\text{Cu}^{2+}\,\text{(aq)} + 2\,e^- \rightarrow \text{Cu}\,\text{(s)}$	2	1 / 2
$\text{Ni}^{2+}\,\text{(aq)} + 2\,e^- \rightarrow \text{Ni}\,\text{(s)}$	2	1 / 2
$\text{Al}^{3+}\,\text{(aq)} + 3\,e^- \rightarrow \text{Al}\,\text{(s)}$	3	1 / 3
$2\,\text{H}^+\,\text{(aq)} + 2\,e^- \rightarrow \text{H}_{2(g)}$	2	1 / 2

Worked example

Mass of Nickel Deposited by Electrolysis

An electrolytic cell deposits nickel metal from a solution of nickel(II) sulfate. A constant current of 3.20 A flows for 25.0 minutes. Calculate the mass of nickel deposited at the cathode. The molar mass of nickel is 58.69 g/mol. The relevant half-reaction is:

\text{Ni}^{2+}\,\text{(aq)} + 2\,e^- \rightarrow \text{Ni}\,\text{(s)}

Step 1 — Convert time to seconds
Faraday's law requires time in seconds because the ampere is defined as coulombs per second. Multiply minutes by 60 s/min.
$t = 25.0 \text{ min} \times 60 \text{ s/min} = 1500 \text{ s}$
Step 2 — Calculate total charge
Use $Q = I \times t$ . Current is 3.20 A and time is 1500 s.
$Q = 3.20 \text{ A} \times 1500 \text{ s} = 4800 \text{ C}$
Step 3 — Convert charge to moles of electrons
Divide the total charge by Faraday's constant to find moles of electrons that flowed through the circuit.
$n_{e^-} = \frac{4800 \text{ C}}{9.65 \times 10^4 \text{ C/mol}} = 0.04974 \text{ mol } e^-$
Step 4 — Apply the mole ratio from the half-reaction
The half-reaction shows 2 mol of electrons deposit 1 mol of nickel. Divide moles of electrons by 2 to get moles of nickel.
$n_{\text{Ni}} = 0.04974 \text{ mol } e^- \times \frac{1 \text{ mol Ni}}{2 \text{ mol } e^-} = 0.02487 \text{ mol Ni}$
Step 5 — Convert moles of nickel to grams
Multiply moles of nickel by the molar mass of nickel to get the mass deposited.
$m_{\text{Ni}} = 0.02487 \text{ mol} \times 58.69 \text{ g/mol} = 1.46 \text{ g}$

Answer: The mass of nickel deposited is 1.46 g (3 significant digits).

Check: Reasonableness check: roughly 0.025 mol of Ni at about 59 g/mol gives approximately 1.5 g, which is consistent with 1.46 g. Units cancel correctly at every step: C ÷ (C/mol) = mol; mol × (g/mol) = g.

Common mistakes and how to avoid them

Using time in minutes instead of seconds in

Q = I \times t

, which gives a charge value that is 60 times too small.

Correction: Always convert time to seconds before multiplying by current. One minute = 60 s.

Reading the electron coefficient from the wrong half-reaction — for example, using the anode's coefficient when calculating mass deposited at the cathode.

Correction: Identify which electrode you are calculating for, write its specific half-reaction, and read the electron coefficient from that reaction only.

Forgetting to divide moles of electrons by the electron coefficient — for example, treating moles of electrons as equal to moles of Cu when the half-reaction shows 2 electrons per Cu atom.

Correction: Always apply the mole ratio:

n_{\text{substance}} = n_{e^-} \div n

where

n

is the electron coefficient in the balanced half-reaction.

Using an incorrect or unbalanced half-reaction, leading to a wrong electron coefficient and therefore a wrong mass.

Correction: Check that charges and atoms are balanced on both sides of the half-reaction before reading the electron coefficient.

Rounding intermediate results too early, which causes the final answer to be inaccurate.

Correction: Carry at least one extra significant digit through intermediate steps and round only the final answer to the correct number of significant digits.

Lesson summary

Faraday's law states that the mass of substance produced or consumed at an electrode is directly proportional to the total electric charge that flows through the cell.
The total charge is found from $Q = I \times t$ , where $I$ is current in amperes and $t$ is time in seconds.
Faraday's constant $F = 9.65 \times 10^4$ C/mol is the charge carried by one mole of electrons; it converts coulombs to moles of electrons using $n_{e^-} = Q / F$ .
The balanced half-reaction provides the mole ratio between electrons and the substance of interest.
The four-step calculation pathway is: calculate $Q$ , find $n_{e^-}$ , apply mole ratio, then convert to grams using molar mass.
The same pathway works in reverse to find the current or time needed to produce a given mass of substance.

Check your understanding

Question 1

A current of 2.00 A flows through an electrolytic cell for 965 s. What is the total charge delivered to the cell?

482.5 C
963 C
1930 C
9.65 × 10⁴ C

Show answer and explanation

1930 C

Q = I \times t = 2.00 \text{ A} \times 965 \text{ s} = 1930 \text{ C}

. Current times time gives charge in coulombs.

Question 2

A total charge of

9.65 \times 10^4

C passes through an electrolytic cell. How many moles of electrons does this represent?

0.500 mol
1.00 mol
2.00 mol
$9.65 \times 10^4$ mol

Show answer and explanation

1.00 mol

n_{e^-} = Q / F = (9.65 \times 10^4 \text{ C}) / (9.65 \times 10^4 \text{ C/mol}) = 1.00 \text{ mol}

. One Faraday of charge equals exactly one mole of electrons.

Question 3

Silver is deposited by the half-reaction

\text{Ag}^+\,\text{(aq)} + e^- \rightarrow \text{Ag}\,\text{(s)}

. If 0.200 mol of electrons flows through the cell, what mass of silver (molar mass 107.9 g/mol) is deposited?

5.40 g
10.8 g
21.6 g
53.9 g

Show answer and explanation

21.6 g

The mole ratio is 1 mol Ag per 1 mol

e^-

, so

n_{\text{Ag}} = 0.200

mol. Mass

= 0.200 \text{ mol} \times 107.9 \text{ g/mol} = 21.6 \text{ g}

.

Question 4

Aluminum is produced by the half-reaction

\text{Al}^{3+}\,\text{(aq)} + 3\,e^- \rightarrow \text{Al}\,\text{(s)}

. Compared with depositing the same number of moles of silver (1 electron per ion), producing the same number of moles of aluminum requires:

One-third as many moles of electrons.
The same number of moles of electrons.
Three times as many moles of electrons.
Nine times as many moles of electrons.

Show answer and explanation

Three times as many moles of electrons.

Each mole of Al requires 3 mol of electrons, while each mole of Ag requires only 1 mol of electrons. So producing the same number of moles of Al needs three times as many moles of electrons.

Key terms

Electrolytic cell: An electrochemical cell that uses electrical energy from an external source to drive a non-spontaneous chemical reaction.
Faraday's constant (F): The total electric charge carried by one mole of electrons; $F = 9.65 \times 10^4$ C/mol, as given in the Alberta Chemistry 30 Data Booklet.
Coulomb (C): The SI unit of electric charge. One coulomb is the charge delivered when one ampere of current flows for one second.
Ampere (A): The SI unit of electric current. One ampere means one coulomb of charge passes a point in one second.
Half-reaction: A balanced equation that shows either the oxidation or the reduction process separately, including the electrons gained or lost.
Mole ratio: The ratio of moles of one substance to moles of another, taken directly from the coefficients of a balanced equation or half-reaction.
Cathode: The electrode where reduction (gain of electrons) occurs in an electrochemical cell.
Anode: The electrode where oxidation (loss of electrons) occurs in an electrochemical cell.

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About this lesson and its review

Published by DoAssignment. This reviewed lesson follows Alberta Grade 12 Chemistry (CHEM30), expectation 30-B2.8k. It is a study resource, not an official curriculum publication.

Before publication, content is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability. Errors can still occur, so corrections are welcomed.

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30-B2.8k · Apply Faraday’s law and stoichiometry to electrochemical cells

Alberta Grade 12 Chemistry

Alberta Chemistry 30 — Outcome 30-B2.8k

What you will learn

Prerequisite Review: Charge, Current, and the Mole

Faraday's Constant and the Mole of Electrons

The Full Calculation Pathway

Applying the Pathway to Different Electrode Reactions

Common Electrode Half-Reactions and Their Electron Ratios

Mass of Nickel Deposited by Electrolysis

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Question 4

Key terms

Continue through CHEM30

About this lesson and its review