Electrolysis and Faraday’s law — AP Chemistry Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: Electrolytic cell structure and polarity, Faraday’s first and second laws of electrolysis, charge-mole-product stoichiometry, electrolysis of aqueous solutions, and quantitative calculations for electrolytic processes.

You should already know: Balancing redox half-reactions. Standard reduction potential ranking. Basic mass-mole stoichiometry.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Chemistry style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Electrolysis and Faraday’s law?

Electrolysis is the process of using an external electric current to drive a non-spontaneous redox reaction, the reverse of the spontaneous reaction that occurs in a galvanic (voltaic) cell. It has widespread industrial use, including metal purification, electroplating, chlorine and sodium hydroxide production from brine, and copper refining. Faraday’s laws of electrolysis are the quantitative relationships that describe how much product is formed from a given amount of electric charge passed through an electrolyte.

According to the AP Chemistry CED, this topic accounts for ~7-9% of the exam weight within Unit 9 (Applications of Thermodynamics), and it appears in both multiple-choice (MCQ) and free-response (FRQ) sections. It is often combined with concepts of cell potential, Gibbs free energy, and redox stoichiometry for cross-unit testing. Notation conventions used on the AP exam: charge $Q$ is measured in coulombs (C), current $I$ in amperes (A, where 1 A = 1 C/s), Faraday’s constant $F\approx 96500\text{ C/mol }{e}^{-}$, and $n_{e^{-}}$ is moles of electrons transferred.

2. Electrolytic Cell Fundamentals

In all electrochemical cells, oxidation always occurs at the anode and reduction always occurs at the cathode. The key difference between electrolytic and galvanic cells is electrode polarity: because electrolysis uses an external battery to force electrons to flow against the spontaneous potential gradient, the battery pulls electrons away from the anode (making it positive) and pushes electrons onto the cathode (making it negative). This polarity reversal is one of the most commonly tested distinctions between cell types.

When electrolyzing a molten salt, only the salt’s ions are available to react, so the cation is reduced at the cathode and the anion is oxidized at the anode. For aqueous solutions, water can also be oxidized or reduced, so you must compare standard reduction potentials to determine which species reacts. In general, the species with the highest reduction potential will be reduced at the cathode, and the species with the lowest reduction potential (highest oxidation potential) will be oxidized at the anode.

The core relationship between current, time, and total charge is: $$Q=I\times t$$ Faraday’s first law of electrolysis states that the mass of product formed at an electrode is proportional to the total charge passed through the electrolyte: more charge equals more electrons transferred equals more product.

Worked Example

Problem: A current of 2.5 A is passed through molten magnesium chloride for 1.0 hour. What mass of magnesium metal is produced at the cathode?

1. Convert time to seconds and write the balanced reduction half-reaction: $1.0\text{ hour}=3600\text{ s}$. Half-reaction: ${\text{Mg}}^{2+}(l)+2e^{-}\to \text{Mg}(s)$, so 2 moles of electrons are required per 1 mole of Mg.
2. Calculate total charge passed: $Q=I\times t=2.5\text{ A}\times 3600\text{ s}=9000\text{ C}$.
3. Calculate moles of electrons using Faraday’s constant: $n_{e^{-}}=\frac{Q}{F}=\frac{9000}{96500}\approx 0.0933\text{ mol }{e}^{-}$.
4. Convert moles of electrons to mass of Mg: $\text{Moles Mg}=0.0933\text{ mol }{e}^{-}\times \frac{1\text{ mol Mg}}{2\text{ mol }{e}^{-}}=0.0466\text{ mol Mg}$. $\text{Mass Mg}=0.0466\text{ mol}\times 24.31\text{ g/mol}\approx 1.1\text{ g}$.

Exam tip: Always start any electrolysis calculation by converting time to seconds. Current is defined as coulombs per second, so using time in minutes or hours will give an answer that is off by orders of magnitude.

3. Faraday’s Laws and Quantitative Calculations

Faraday’s second law of electrolysis states that the mass of product formed by a given amount of charge is proportional to the molar mass of the product divided by the number of moles of electrons transferred per mole of product ($z$, the charge number of the ion in the half-reaction). Combining Faraday’s two laws gives a single combined formula for the mass of product formed: $$m=\frac{I\times t\times M}{z\times F}$$ where $m$ is mass of product (g), $M$ is molar mass of product (g/mol), and $z$ is moles of electrons per mole of product. This formula is just a combination of the individual stoichiometric steps: calculate charge, convert to moles of electrons, convert to moles of product, convert to mass. You can solve step-by-step or use the combined formula; both are acceptable on the AP exam, but step-by-step solving reduces algebraic error.

Faraday’s constant is provided on the AP Chemistry equation sheet, so you do not need to memorize its exact value, but remembering $F\approx 96500\text{ C/mol }{e}^{-}$ is useful for quick calculations.

Worked Example

Problem: How long will it take to plate out 5.0 g of silver metal from a solution of AgNO₃ using a constant current of 1.5 A?

1. Write the balanced reduction half-reaction: ${\text{Ag}}^{+}(aq)+e^{-}\to \text{Ag}(s)$, so $z=1\text{ mol }{e}^{-}$ per mole of Ag, and $M_{\text{Ag}}=107.87\text{ g/mol}$.
2. Calculate moles of Ag and moles of electrons: $\text{Moles Ag}=\frac{5.0\text{ g}}{107.87\text{ g/mol}}\approx 0.0464\text{ mol Ag}$, so $n_{e^{-}}=0.0464\times 1=0.0464\text{ mol }{e}^{-}$.
3. Calculate total charge required: $Q=n_{e^{-}}\times F=0.0464\text{ mol}\times 96500\text{ C/mol}\approx 4480\text{ C}$.
4. Solve for time: $t=\frac{Q}{I}=\frac{4480\text{ C}}{1.5\text{ A}}\approx 3.0\times 10^3\text{ s}$ (or ~50 minutes).

Exam tip: Always confirm the value of $z$ (moles of electrons per product) by writing the full balanced half-reaction. For example, 1 mole of O₂ produced from water requires 4 moles of electrons, not 1.

4. Electrolysis of Aqueous Solutions

When electrolyzing aqueous solutions, water can act as both an oxidizing and reducing agent, so you must always include water when comparing possible half-reactions. The relevant half-reactions for water are:

• Reduction at cathode: $2{\text{H}}_2\text{O}(l)+2e^{-}\to {\text{H}}_2(g)+2{\text{OH}}^{-}(aq)E^{\circ }=-0.83\text{ V}$
• Oxidation at anode: $2{\text{H}}_2\text{O}(l)\to {\text{O}}_2(g)+4{\text{H}}^{+}(aq)+4e^{-}{E}_{oxidation}^{\circ }=+1.23\text{ V}$

Cations with reduction potentials higher than that of water will be reduced instead of water; for example, Cu²⁺ ($E^{\circ }=+0.34$ V) will plate out as Cu metal, while Na⁺ ($E^{\circ }=-2.71$ V) is not reduced, and water is reduced to H₂ instead. Overpotential (extra voltage required for gas formation at electrodes) can occasionally change products, but the AP exam will specify if you need to account for this, otherwise use standard reduction potentials.

Worked Example

Problem: Predict the anode and cathode products for electrolysis of aqueous sodium iodide (NaI) at standard conditions, write the balanced half-reactions.

1. List all possible species: Na⁺, I⁻, H₂O.
2. Compare possible cathode reduction reactions: ${\text{Na}}^{+}+e^{-}\to \text{Na}{E}^{\circ }=-2.71\text{ V}$; $2{\text{H}}_2\text{O}+2e^{-}\to {\text{H}}_2+2{\text{OH}}^{-}{E}^{\circ }=-0.83\text{ V}$. The higher (more positive) reduction potential is -0.83 V, so water is reduced, cathode product is H₂(g).
3. Compare possible anode oxidation reactions: $2{\text{I}}^{-}\to {\text{I}}_2+2e^{-}{E}_{oxidation}^{\circ }=+0.54\text{ V}$; $2{\text{H}}_2\text{O}\to {\text{O}}_2+4{\text{H}}^{+}+4e^{-}{E}_{oxidation}^{\circ }=+1.23\text{ V}$. The higher oxidation potential is +0.54 V, so I⁻ is oxidized, anode product is I₂(aq).
4. Final half-reactions: Cathode: $2{\text{H}}_2\text{O}(l)+2e^{-}\to {\text{H}}_2(g)+2{\text{OH}}^{-}(aq)$; Anode: $2{\text{I}}^{-}(aq)\to {\text{I}}_2(aq)+2e^{-}$.

Exam tip: If you are asked for the overall reaction, make sure to balance electrons between the two half-reactions before combining them, just like any other redox reaction.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Assigning anode as negative and cathode as positive in an electrolytic cell. Why: Students memorize polarity from galvanic cells and incorrectly apply it to electrolytic cells without adjusting. Correct move: Remember "oxidation at anode, reduction at cathode" always holds, then check cell type: electrolytic = anode positive, cathode negative; galvanic = anode negative, cathode positive.
• Wrong move: Forgetting to convert time from hours/minutes to seconds when calculating charge. Why: Current is defined as coulombs per second, so mismatched units lead to answers off by a factor of 60 or 3600. Correct move: Convert time to seconds as the first step in any calculation, and write the unit conversion explicitly.
• Wrong move: Using the charge of the ion instead of the moles of electrons per mole of product. Why: Students assume 1 electron per product regardless of reaction stoichiometry, e.g. using $z=1$ for 1 mole of O₂, which requires 4 electrons. Correct move: Always write the balanced half-reaction before starting calculations to get the correct electron to product ratio.
• Wrong move: Forgetting to include water as a possible reactant for aqueous electrolysis. Why: Students only focus on the dissolved salt ions, leading to incorrect predictions like sodium metal from aqueous NaCl. Correct move: Always add water's two half-reactions to your list of possible reactions before comparing potentials.
• Wrong move: Rounding Faraday's constant to 100000 C/mol e⁻ for simplicity, leading to errors outside the accepted range. Why: Students try to simplify calculations and round too early, leading to 3-5% errors that can change MCQ answers. Correct move: Use 96500 C/mol e⁻, which matches the value on the AP equation sheet.

6. Practice Questions (AP Chemistry Style)

Question 1 (Multiple Choice)

A constant current is passed through a CuCl₂ solution for 20 minutes, depositing 0.80 g of Cu metal at the cathode. The same current is passed through an AuCl₃ solution for the same amount of time. What mass of Au metal is deposited? (Molar masses: Cu = 63.5 g/mol, Au = 197 g/mol) A) 0.83 g B) 1.7 g C) 2.5 g D) 3.3 g

Worked Solution: Since current and time are identical, total charge and total moles of electrons transferred are the same for both experiments. For Cu, the half-reaction is ${\text{Cu}}^{2+}+2e^{-}\to \text{Cu}$, so moles of electrons = $2\times (0.80\text{ g}/63.5\text{ g/mol})\approx 0.0252\text{ mol }{e}^{-}$. For Au, the half-reaction is ${\text{Au}}^{3+}+3e^{-}\to \text{Au}$, so moles of Au = $0.0252/3=0.0084\text{ mol}$. Mass of Au = $0.0084\text{ mol}\times 197\text{ g/mol}\approx 1.7\text{ g}$. The correct answer is B.

Question 2 (Free Response)

A student measures Faraday's constant by electrolyzing aqueous copper sulfate, plating copper onto a cathode. The student uses a 0.500 A current for 40.0 minutes, and measures a 0.390 g increase in cathode mass. (a) Calculate the experimental value of Faraday's constant from this data. (b) The student finds that copper(II) hydroxide impurity plated onto the cathode along with copper. Will the calculated Faraday's constant be higher, lower, or equal to the true value? Justify your answer. (c) How many hours will it take to produce 10.0 g of O₂ gas at the anode (half-reaction: $2{\text{H}}_2\text{O}(l)\to {\text{O}}_2(g)+4{\text{H}}^{+}(aq)+4e^{-}$) using a 2.00 A current?

Worked Solution: (a) Half-reaction: ${\text{Cu}}^{2+}+2e^{-}\to \text{Cu}(s)$. Time = $40.0\text{ min}\times 60\text{ s/min}=2400\text{ s}$. Charge $Q=0.500\text{ A}\times 2400\text{ s}=1200\text{ C}$. Moles Cu = $0.390\text{ g}/63.55\text{ g/mol}\approx 0.00614\text{ mol}$. Moles electrons = $2\times 0.00614=0.01228\text{ mol}$. $F=Q/n_{e^{-}}=1200/0.01228\approx 97700\text{ C/mol }{e}^{-}$.

(b) The calculated F will be lower than the true value. The impurity adds extra mass to the cathode, so the measured mass increase is larger than the actual mass of copper plated. A larger measured mass of Cu gives a larger calculated number of moles of electrons, and since $F=Q/n_{e^{-}}$, a larger $n_{e^{-}}$ gives a smaller calculated F.

(c) Moles O₂ = $10.0\text{ g}/32.00\text{ g/mol}=0.3125\text{ mol}$. Moles electrons = $4\times 0.3125=1.25\text{ mol}$. $Q=1.25\text{ mol}\times 96500\text{ C/mol}=120625\text{ C}$. $t=120625\text{ C}/2.00\text{ A}=60312.5\text{ s}=60312.5/3600\approx 16.8\text{ hours}$.

Question 3 (Application / Real-World Style)

Chromium electroplating is used to protect steel parts from corrosion. A factory wants to plate a 1.5 m² sheet of steel with a 0.020 mm thick layer of chromium on one side. The density of chromium is 7.19 g/cm³, and the plating solution contains Cr³⁺ ions. If the factory uses a constant current of 45 A, how many hours does the plating process take?

Worked Solution: Convert all units to cm to match density: Area = $1.5{\text{ m}}^2=15000{\text{ cm}}^2$, Thickness = $0.020\text{ mm}=0.0020\text{ cm}$. Volume of Cr = $15000{\text{ cm}}^2\times 0.0020\text{ cm}=30{\text{ cm}}^3$. Mass of Cr = $7.19{\text{ g/cm}}^3\times 30{\text{ cm}}^3=215.7\text{ g}$. Moles of Cr = $215.7\text{ g}/52.00\text{ g/mol}\approx 4.15\text{ mol}$. Half-reaction: ${\text{Cr}}^{3+}+3e^{-}\to \text{Cr}$, so moles electrons = $3\times 4.15=12.45\text{ mol}$. Charge $Q=12.45\text{ mol}\times 96500\text{ C/mol}\approx 1.20\times 10^6\text{ C}$. Time $t=1.20\times 10^6\text{ C}/45\text{ A}\approx 26700\text{ s}\approx 7.4\text{ hours}$. This means the process will take approximately 7.4 hours to plate the required thickness of chromium onto the steel sheet, which is a reasonable timeline for industrial electroplating.

7. Quick Reference Cheatsheet

8. What's Next

This chapter on electrolysis and Faraday’s law is the foundation for the remaining topics in Unit 9 that connect electrochemistry to thermodynamics: cell potential and Gibbs free energy, and non-standard cell potentials with the Nernst equation. Without mastering the stoichiometry of electron transfer and Faraday’s quantitative relationships, you will struggle to calculate reaction quotients for the Nernst equation, and to relate measured current to entropy changes for electrolytic reactions. This topic connects back to earlier units on redox reactions and stoichiometry, and feeds into industrial chemistry applications that are common as context for FRQ questions on the AP exam. Links to related topics: Cell Potential and Gibbs Free Energy, The Nernst Equation, Galvanic vs Electrolytic Cells, Industrial Electrochemistry