GCSE Chemistry - How to Solve: Electrolysis Calculations & Half-Equations

How to Solve: Electrolysis Calculations & Half-Equations

Complete Guide (GCSE / A-Level Chemistry & Physics – Ready to Ace Your Exam!)

Introduction

"Mastering electrolysis calculations and half-equations unlocks 4-6 marks in GCSE Chemistry (Paper 1) and 8-12 marks in A-Level (Paper 1 & 3). That’s the difference between a Grade 5 and a Grade 7—or a B and an A—and it’s how you predict how much copper you’ll get from a battery charger, or why your phone’s lithium-ion battery degrades over time!"

WHAT YOU NEED TO KNOW FIRST

Before you start, you must already understand:
1. Atomic structure & ions – What cations (+) and anions (–) are, and how they move in a circuit.
2. Balancing chemical equations – How to write and balance full equations (e.g., 2H₂ + O₂ → 2H₂O).
3. Moles & Avogadro’s number – How to convert between mass, moles, and number of particles (6.02 × 10²³).

If you’re shaky on any of these, pause here and review them first—electrolysis builds on these!

KEY TERMS & FORMULAS

Key Terms

Formulas

1. Charge (Q) = Current (I) × Time (t)
2. Q = charge (C)
3. I = current (A)
4. t = time (s)

5. MEMORISE THIS – You’ll use it in every electrolysis calculation.

6. Moles of electrons (n) = Charge (Q) ÷ Faraday constant (F)

7. n = moles of electrons (mol)
8. Q = charge (C)
9. F = 96,500 C/mol (given on exam sheet)

10. MEMORISE F = 96,500 C/mol – Examiners love testing this.

11. Moles of product = Moles of electrons ÷ Electrons per ion (from half-equation)

12. Example: If Cu²⁺ + 2e⁻ → Cu, then 1 mole of Cu needs 2 moles of electrons.

13. MEMORISE – The number of electrons in the half-equation directly tells you the ratio.

14. Mass = Moles × Molar mass (Mr)

15. MEMORISE – Standard moles formula, but critical for electrolysis.

STEP-BY-STEP METHOD

Follow these exact steps for every electrolysis calculation. No shortcuts!

Step 1: Identify the half-equations

• Write the oxidation (anode) and reduction (cathode) half-equations.
• Check charges balance (electrons lost = electrons gained).
• Example:
• Cathode (reduction): Cu²⁺ + 2e⁻ → Cu
• Anode (oxidation): 2Cl⁻ → Cl₂ + 2e⁻

Step 2: Calculate total charge (Q)

• Use Q = I × t.
• Convert time to seconds (if given in minutes/hours).
• Example: If I = 2 A and t = 30 minutes (1800 s), then Q = 2 × 1800 = 3600 C.

Step 3: Calculate moles of electrons (n)

• Use n = Q ÷ F.
• F = 96,500 C/mol (given on exam sheet).
• Example: n = 3600 ÷ 96,500 = 0.0373 mol e⁻.

Step 4: Use the half-equation to find moles of product

• Look at the number of electrons in the half-equation.
• Example: Cu²⁺ + 2e⁻ → Cu → 2 moles of e⁻ make 1 mole of Cu.
• So, moles of Cu = moles of e⁻ ÷ 2 = 0.0373 ÷ 2 = 0.01865 mol.

Step 5: Calculate mass of product

• Use mass = moles × Mr.
• Example: Mr of Cu = 63.5, so mass = 0.01865 × 63.5 = 1.18 g.

Step 6: Check units & significant figures

• Mass should be in grams (g).
• Charge in coulombs (C).
• Current in amps (A).
• Time in seconds (s).
• Round to 2 or 3 significant figures (unless the question specifies).

WORKED EXAMPLES

Example 1 – Basic: Copper Electrolysis

Question: A current of 1.5 A is passed through copper(II) sulfate solution for 20 minutes. Calculate the mass of copper deposited at the cathode. (Mr of Cu = 63.5, F = 96,500 C/mol)

Solution:
1. Half-equation (cathode): Cu²⁺ + 2e⁻ → Cu (reduction)
2. Charge (Q): - t = 20 × 60 = 1200 s - Q = I × t = 1.5 × 1200 = 1800 C
3. Moles of electrons (n): - n = Q ÷ F = 1800 ÷ 96,500 = 0.01865 mol e⁻
4. Moles of Cu: - From half-equation, 2 mol e⁻ → 1 mol Cu - So, moles of Cu = 0.01865 ÷ 2 = 0.009325 mol
5. Mass of Cu: - mass = moles × Mr = 0.009325 × 63.5 = 0.592 g
6. Final answer: 0.592 g of copper is deposited.

What we did and why: - We started with the half-equation to know how many electrons make 1 mole of Cu. - Q = I × t gave us total charge. - n = Q ÷ F converted charge to moles of electrons. - The half-equation ratio told us moles of Cu. - mass = moles × Mr gave the final answer.

Example 2 – Medium: Aluminium Electrolysis

Question: Aluminium is extracted by electrolysing molten aluminium oxide (Al₂O₃). A current of 50,000 A is used for 24 hours. Calculate the mass of aluminium produced. (Mr of Al = 27, F = 96,500 C/mol)

Solution:
1. Half-equation (cathode): Al³⁺ + 3e⁻ → Al (reduction)
2. Charge (Q): - t = 24 × 60 × 60 = 86,400 s - Q = I × t = 50,000 × 86,400 = 4,320,000,000 C
3. Moles of electrons (n): - n = Q ÷ F = 4,320,000,000 ÷ 96,500 = 44,766.84 mol e⁻
4. Moles of Al: - From half-equation, 3 mol e⁻ → 1 mol Al - So, moles of Al = 44,766.84 ÷ 3 = 14,922.28 mol
5. Mass of Al: - mass = moles × Mr = 14,922.28 × 27 = 402,901.56 g
6. Convert to kg: - 402,901.56 g = 402.9 kg
7. Final answer: 403 kg of aluminium is produced.

What we did and why: - Large numbers? No problem—just keep units consistent (seconds, not hours!). - 3 electrons per Al³⁺ meant we divided moles of e⁻ by 3. - Industrial-scale electrolysis—this is how real aluminium smelters work!

Example 3 – Exam-Style: Disguised Question

Question: A student sets up an electrolysis experiment with silver nitrate solution (AgNO₃). A current of 0.8 A is passed for 15 minutes. The student measures 0.80 g of silver deposited. Calculate the percentage yield of silver. (Mr of Ag = 108, F = 96,500 C/mol)

Solution:
1. Half-equation (cathode): Ag⁺ + e⁻ → Ag (reduction)
2. Charge (Q): - t = 15 × 60 = 900 s - Q = I × t = 0.8 × 900 = 720 C
3. Moles of electrons (n): - n = Q ÷ F = 720 ÷ 96,500 = 0.00746 mol e⁻
4. Theoretical moles of Ag: - From half-equation, 1 mol e⁻ → 1 mol Ag - So, moles of Ag = 0.00746 mol
5. Theoretical mass of Ag: - mass = moles × Mr = 0.00746 × 108 = 0.806 g
6. Percentage yield: - % yield = (actual mass ÷ theoretical mass) × 100 - % yield = (0.80 ÷ 0.806) × 100 = 99.3%
7. Final answer: The percentage yield is 99.3%.

What we did and why: - Percentage yield is a common exam trap—always calculate theoretical mass first! - 1 electron per Ag⁺ made the ratio simple. - Rounding errors? We kept 3 sig figs until the final step.

COMMON MISTAKES

EXAM TRAPS

1-MINUTE RECAP (Night Before the Exam)

"Right, listen up—this is your electrolysis cheat sheet for tomorrow. Here’s what you must remember:

1. Half-equations first – Write oxidation (anode) and reduction (cathode). Electrons lost = electrons gained.
2. Q = I × t – Charge = current × time. Convert time to seconds—no excuses!
3. n = Q ÷ F – Moles of electrons = charge ÷ 96,500. F is given, but memorise it anyway.
4. Ratio from half-equation – If it’s 2e⁻ → 1 Cu, divide moles of e⁻ by 2. If it’s 1e⁻ → 1 Ag, keep it the same.
5. Mass = moles × Mr – Standard moles formula, but only use the Mr of the element being made.
6. Watch for traps – Solutions vs. molten? Check for competing ions. Gas volume? Use 24 dm³ per mole.

That’s it. Follow the steps, don’t skip, and you’ll smash those 6-12 marks. Now go get some sleep—you’ve got this!"