# Electrolysis and Faraday’s Law

> AP Chemistry · Unit 9: Applications of Thermodynamics
> Source: https://www.owlsprep.com/study/ap-chemistry-u9-electrolysis-and-faraday-s-law/

This subtopic covers electrolytic cell operation, Faraday’s quantitative laws of electrolysis, product prediction for molten and aqueous electrolytes, and step-by-step stoichiometric calculations for AP Chemistry exam questions.

**Prerequisites:** Balancing redox half-reactions; Standard reduction potential ranking; Basic mass-mole stoichiometry

## Learning objectives

- Distinguish between electrolytic and galvanic cell structure and polarity
- Apply Faraday’s laws to calculate product mass, current, or time for electrolytic processes
- Predict products of electrolysis for molten salts and aqueous solutions
- Solve quantitative AP-style problems involving electrolysis stoichiometry

## 1. 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: an external battery pulls electrons from the anode (making it positive) and pushes electrons onto the cathode (making it negative).

> **tip**
>
> Polarity reversal is one of the most commonly tested distinctions between cell types.

For molten salts, only the salt's ions are available to react: cations reduce at the cathode, anions oxidize at the anode. The core relationship between current, time, and total charge is:

$$Q = I \times t$$

Faraday’s first law states the mass of product formed is proportional to the total charge passed through the electrolyte.

**Worked example:** 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 hour = 3600 s. Half-reaction:
2. $$\text{Mg}^{2+}(l) + 2e^- \rightarrow \text{Mg}(s), \text{ so 2 moles of electrons are required per 1 mole of Mg}$$
3. Calculate total charge passed:
4. $$Q = I \times t = 2.5 \text{ A} \times 3600 \text{ s} = 9000 \text{ C}$$
5. Calculate moles of electrons using Faraday’s constant:
6. $$n_{e^-} = \frac{Q}{F} = \frac{9000}{96500} \approx 0.0933 \text{ mol }e^-$$
7. Convert moles of electrons to mass of Mg:
8. $$\text{Moles Mg} = 0.0933 \text{ mol }e^- \times \frac{1 \text{ mol Mg}}{2 \text{ mol }e^-} = 0.0466 \text{ mol Mg} \\ \text{Mass Mg} = 0.0466 \text{ mol} \times 24.31 \text{ g/mol} \approx 1.1 \text{ g}$$

> **Exam tip:** Always convert time to seconds as your first step in any electrolysis calculation. Current is defined as coulombs per second, so using time in other units will give an answer off by orders of magnitude.

## 2. Faraday’s Laws and Quantitative Calculations

Faraday’s second law of electrolysis states the mass of product formed by a given charge is proportional to the molar mass of the product divided by the number of electrons transferred per mole of product ($z$). Combining the two laws gives the combined formula:

$$m = \frac{I \times t \times M}{z \times F}$$

Where $m$ is mass of product (g), $M$ is molar mass (g/mol), and $z$ is moles of electrons per mole of product. You do not need to use the combined formula: solving step-by-step reduces algebraic error and is equally acceptable on the AP exam. Faraday’s constant is provided on the AP Chemistry equation sheet.

**Worked example:** 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, identify $z$ and $M$:
2. $$\text{Ag}^+(aq) + e^- \rightarrow \text{Ag}(s), \text{ so } z = 1 \text{ mol }e^- \text{ per mole of Ag}, M_{\text{Ag}} = 107.87 \text{ g/mol}$$
3. Calculate moles of Ag and moles of electrons:
4. $$\text{Moles Ag} = \frac{5.0 \text{ g}}{107.87 \text{ g/mol}} \approx 0.0464 \text{ mol Ag}, n_{e^-} = 0.0464 \times 1 = 0.0464 \text{ mol }e^-$$
5. Calculate total charge required:
6. $$Q = n_{e^-} \times F = 0.0464 \text{ mol} \times 96500 \text{ C/mol} \approx 4480 \text{ C}$$
7. Solve for time:
8. $$t = \frac{Q}{I} = \frac{4480 \text{ C}}{1.5 \text{ A}} \approx 3.0 \times 10^3 \text{ s} (\approx 50 \text{ minutes})$$

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

## 3. Electrolysis of Aqueous Solutions

When electrolyzing aqueous solutions, water can act as both an oxidizing and reducing agent, so you must always compare the reduction potentials of all possible species (including water) to predict products. The relevant half-reactions for water are:

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

The general rule for product prediction is:

- The species with the highest (most positive) reduction potential will be reduced at the cathode
- The species with the highest (most positive) oxidation potential will be oxidized at the anode

Overpotential (extra voltage for gas formation) is rarely tested on the AP exam, unless explicitly specified in the question.

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

1. List all possible species available to react: $\text{Na}^+$, $\text{I}^-$, $\text{H}_2\text{O}$
2. Compare possible cathode reduction reactions: $\text{Na}^+ + e^- \rightarrow \text{Na}$ has $E^\circ = -2.71 \text{ V}$, while water reduction has $E^\circ = -0.83 \text{ V}$. The higher reduction potential belongs to water, so water is reduced, and the cathode product is $\text{H}_2(g)$.
3. Compare possible anode oxidation reactions: $2\text{I}^- \rightarrow \text{I}_2 + 2e^-$ has $E^\circ_{oxidation} = +0.54 \text{ V}$, while water oxidation has $E^\circ_{oxidation} = +1.23 \text{ V}$. The higher oxidation potential belongs to iodide, so $\text{I}^-$ is oxidized, and the anode product is $\text{I}_2(aq)$.
4. Final balanced half-reactions:
5. $$\text{Cathode: } 2\text{H}_2\text{O}(l) + 2e^- \rightarrow \text{H}_2(g) + 2\text{OH}^-(aq) \\ \text{Anode: } 2\text{I}^-(aq) \rightarrow \text{I}_2(aq) + 2e^-$$

> **Exam tip:** If asked for the overall reaction, balance electrons between the two half-reactions before combining, just like any other redox reaction.

## 4. AP-Style Additional Worked Examples

**Worked example:** 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

1. Since current and time are identical, total charge and total moles of electrons transferred are the same for both experiments. Write the half-reaction for Cu:
2. $$\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$$
3. Calculate moles of electrons from the copper data:
4. $$n_{e^-} = 2 \times \frac{0.80 \text{ g}}{63.5 \text{ g/mol}} \approx 0.0252 \text{ mol }e^-$$
5. Write the half-reaction for gold and calculate moles and mass of Au:
6. $$\text{Au}^{3+} + 3e^- \rightarrow \text{Au}, \text{ so } \text{moles Au} = \frac{0.0252}{3} = 0.0084 \text{ mol} \\ \text{Mass Au} = 0.0084 \text{ mol} \times 197 \text{ g/mol} \approx 1.7 \text{ g}$$
7. The correct answer is B.

**Worked example:** 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) 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: $2\text{H}_2\text{O}(l) \rightarrow \text{O}_2(g) + 4\text{H}^+(aq) + 4e^-$) using a 2.00 A current?

1. Part (a): Write the half-reaction for copper, convert time to seconds, calculate charge:
2. $$\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}(s), t = 40.0 \text{ min} \times 60 \text{ s/min} = 2400 \text{ s}, Q = 0.500 \text{ A} \times 2400 \text{ s} = 1200 \text{ C}$$
3. Calculate moles of copper and moles of electrons, solve for F:
4. $$\text{Moles Cu} = \frac{0.390 \text{ g}}{63.55 \text{ g/mol}} \approx 0.00614 \text{ mol}, n_{e^-} = 2 \times 0.00614 = 0.01228 \text{ mol} \\ F = \frac{Q}{n_{e^-}} = \frac{1200}{0.01228} \approx 97700 \text{ C/mol }e^-$$
5. Part (b): The calculated Faraday's constant 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 copper gives a larger calculated number of moles of electrons, and since $F = Q/n_{e^-}$, a larger $n_{e^-}$ gives a smaller calculated F.
6. Part (c): Calculate moles of O₂, moles of electrons, charge, then time:
7. $$\text{Moles O}_2 = \frac{10.0 \text{ g}}{32.00 \text{ g/mol}} = 0.3125 \text{ mol}, n_{e^-} = 4 \times 0.3125 = 1.25 \text{ mol} \\ Q = 1.25 \text{ mol} \times 96500 \text{ C/mol} = 120625 \text{ C}, t = \frac{120625 \text{ C}}{2.00 \text{ A}} = 60312.5 \text{ s} \approx 16.8 \text{ hours}$$

## Common pitfalls

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

## Cheatsheet

| Category | Formula / Rule | Notes |
| --- | --- | --- |
| Charge from current | $Q = I \times t$ | $Q$ (C), $I$ (A), $t$ (s). Always convert time to seconds. |
| Moles of electrons | $n_{e^-} = \frac{Q}{F}$ | $F \approx 96500 \text{ C/mol }e^-$, provided on AP equation sheet. |
| Combined mass formula | $m = \frac{I t M}{z F}$ | $z$ = moles of electrons per mole of product. |
| Electrolytic cell polarity | Anode = (+), Cathode = (-) | Oxidation at anode, reduction at cathode always holds; polarity reversed from galvanic cells. |
| Cathode product rule | Highest (most positive) reduction potential is reduced | Always include water's reduction half-reaction for aqueous solutions. |
| Anode product rule | Highest (most positive) oxidation potential is oxidized | Always include water's oxidation half-reaction for aqueous solutions. |
| Water reduction | $2\text{H}_2\text{O}(l) + 2e^- \rightarrow \text{H}_2(g) + 2\text{OH}^-(aq)$ | $E^\circ = -0.83 \text{ V}$ |
| Water oxidation | $2\text{H}_2\text{O}(l) \rightarrow \text{O}_2(g) + 4\text{H}^+(aq) + 4e^-$ | $E^\circ_{oxidation} = +1.23 \text{ V}$ |

## What's next

Electrolysis and Faraday’s law is a core foundation for the remaining topics in Unit 9 that connect electrochemistry to thermodynamics. Mastering the stoichiometry of electron transfer and Faraday’s quantitative relationships is critical for solving problems involving cell potential, Gibbs free energy, and the Nernst equation, where you will often need to relate measured current to reaction quantities. This topic connects back to earlier units on redox reactions and stoichiometry, and industrial electrochemistry applications are common as context for AP Chemistry FRQ questions. Building a solid understanding of the rules for product prediction and calculation will help you avoid common pitfalls on both exam sections.

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