How to Solve: Electrolysis Problems

How to Solve: Electrolysis Problems

Introduction

"Imagine your phone battery dies—how does charging it actually work? Electrolysis isn’t just a textbook topic; it’s how we extract metals, purify water, and even power electric cars. Master this, and you’ll crush exam questions on Faraday’s laws, charge calculations, and real-world applications—guaranteed."

What You Need To Know First

Before tackling electrolysis problems, you must understand: 1. Basic circuit concepts – Current (I), charge (Q), and time (t) relationships. 2. Faraday’s laws of electrolysis – How charge relates to mass deposited at electrodes. 3. Mole concept & Avogadro’s number – How atoms, moles, and mass connect.

If any of these are unclear, pause and review them first.

Key Vocabulary

Formulas To Know

Step-by-Step Method

Follow these steps for every electrolysis problem:

1. Identify the ion and its charge (e.g., Cu²⁺ → n = 2).
2. Write the half-reaction (e.g., Cu²⁺ + 2e⁻ → Cu).
3. Extract given values (current, time, molar mass, etc.).
4. Calculate total charge (Q = I × t).
5. Use Faraday’s law to find mass (m = (I × t × M) / (n × F)).
6. Check units (A, s, g/mol, C/mol).
7. Round to significant figures (usually 2 or 3).

WORKED EXAMPLE (Step-by-Step)

Problem: A current of 3 A is passed through CuSO₄ solution for 2 hours. Calculate the mass of copper deposited at the cathode.

Step 1: Ion = Cu²⁺ → n = 2. Step 2: Half-reaction: Cu²⁺ + 2e⁻ → Cu. Step 3: Given: I = 3 A, t = 2 hours = 7200 s, M(Cu) = 63.5 g/mol, F = 96,500 C/mol. Step 4: Q = I × t = 3 × 7200 = 21,600 C. Step 5: m = (I × t × M) / (n × F) = (3 × 7200 × 63.5) / (2 × 96,500). Step 6: m = (1,371,840) / (193,000) ≈ 7.11 g. Step 7: Round to 3 s.f. → 7.11 g.

What we did and why: - Converted time to seconds (SI units). - Used Faraday’s law to link charge, moles, and mass. - Confirmed Cu²⁺ needs 2 electrons per atom.

Worked Examples

Example 1 – Basic

Problem: How much silver (Ag) is deposited if 0.5 A flows for 30 minutes in AgNO₃ solution? Solution: 1. Ion: Ag⁺ → n = 1. 2. Half-reaction: Ag⁺ + e⁻ → Ag. 3. Given: I = 0.5 A, t = 30 × 60 = 1800 s, M(Ag) = 108 g/mol. 4. Q = 0.5 × 1800 = 900 C. 5. m = (0.5 × 1800 × 108) / (1 × 96,500) = 1.01 g.

What we did and why: - Simple 1-electron transfer (n = 1). - Time converted to seconds. - Direct application of Faraday’s law.

Example 2 – Medium

Problem: A student electrolyses Al₂O₃ with 5 A for 1 hour. Calculate the mass of aluminum produced. (Al³⁺ + 3e⁻ → Al) Solution: 1. Ion: Al³⁺ → n = 3. 2. Given: I = 5 A, t = 3600 s, M(Al) = 27 g/mol. 3. Q = 5 × 3600 = 18,000 C. 4. m = (5 × 3600 × 27) / (3 × 96,500) = 1.68 g.

What we did and why: - 3-electron transfer (n = 3). - Molar mass of Al used. - Confirmed units (A, s, g/mol).

Example 3 – Exam Style

Problem: A factory uses electrolysis to purify copper. If 2.5 kg of copper is needed daily, what current (in kA) is required if the process runs for 8 hours? (Cu²⁺ + 2e⁻ → Cu) Solution: 1. Ion: Cu²⁺ → n = 2. 2. Given: m = 2500 g, t = 8 × 3600 = 28,800 s, M(Cu) = 63.5 g/mol. 3. Rearrange m = (I × t × M) / (n × F) → I = (m × n × F) / (t × M). 4. I = (2500 × 2 × 96,500) / (28,800 × 63.5) = 265.5 A ≈ 0.266 kA.

What we did and why: - Rearranged formula for current. - Converted kg to g and hours to seconds. - Rounded to 3 s.f. for final answer.

Common Mistakes

Exam Traps

1-Minute Recap

"Alright, last-minute review! Electrolysis problems always follow the same steps: 1. Find the ion charge (n) – Cu²⁺? n = 2. Al³⁺? n = 3. 2. Calculate charge (Q = I × t) – Convert time to seconds! 3. Use Faraday’s law (m = (I × t × M) / (n × F)) – Plug in numbers carefully. 4. Check units – Grams, seconds, coulombs. No shortcuts!

Common traps? Time in minutes, wrong n, or forgetting F = 96,500 C/mol. Double-check your half-reaction—anode = oxidation, cathode = reduction. You’ve got this!