Pearson Edexcel International GCSE Chemistry 4CH1: Electrolysis – Revision Notes

Condensed for the final weeks. For full explanations and worked examples, use the electrolysis study guide, then test yourself with the electrolysis practice questions.

These notes cover sub-topic 1(i) Electrolysis of the Pearson Edexcel International GCSE Chemistry (4CH1) specification, Issue 3 (September 2024): statements 1.55C–1.60C. The qualification is untiered. All six statements carry a C reference, so this sub-topic is Paper 2 only. Course links: 4CH1 Chemistry hub, printable checklist and the free all-topics diagnostic. The shorter electrolysis summary in the Principles of chemistry revision notes sits alongside this page.

Conduction (1.55C, 1.56C)

To conduct, a substance needs charged particles that are free to move.

Substance	State	Conducts?	Reason
Covalent compound	Any (solid, liquid, solution)	No	No ions; electrons held in bonds, none free to move
Ionic compound	Solid	No	Ions held in fixed positions in the lattice
Ionic compound	Molten	Yes	Lattice broken down; ions free to move
Ionic compound	Aqueous solution	Yes	Ions separated by water; free to move

In the electrolyte the ions move. In the wires, electrons move.

Key terms (1.57C)

Term	Meaning
Electrolysis	Breakdown of an ionic compound, molten or in solution, by a direct current
Electrolyte	The molten or dissolved ionic compound that conducts and is broken down
Cation	Positive ion; moves to the cathode
Anion	Negative ion; moves to the anode
Cathode	Negative electrode; reduction happens here
Anode	Positive electrode; oxidation happens here
Inert electrode	Does not react; graphite or platinum

Predicting products (1.58C)

Molten binary compound: metal at the cathode, non-metal at the anode.

Aqueous solution: water adds H⁺ and OH⁻ ions, so each electrode has a choice.

Method in steps – aqueous, inert electrodes

1. List all cations: the metal ion and H⁺.
2. Cathode: metal less reactive than hydrogen (Cu, Ag) → metal. Metal more reactive than hydrogen (K, Na, Ca, Mg…) → hydrogen.
3. List all anions: the salt’s anion and OH⁻.
4. Anode: halide (Cl⁻, Br⁻, I⁻) in concentrated solution → halogen. Sulfate or nitrate → oxygen.
5. Work out what stays in solution from the ions not discharged.

The four named electrolytes

Electrolyte	Cathode	Anode	Observations	Left behind
Molten lead(II) bromide	Lead	Bromine	Bulb lights only once molten; silvery lead; red-brown fumes	Nothing (compound used up)
Concentrated sodium chloride solution	Hydrogen	Chlorine	Bubbles at both; pale green gas bleaches damp litmus	Sodium hydroxide (alkaline)
Dilute sulfuric acid	Hydrogen	Oxygen	Bubbles at both; hydrogen volume twice oxygen volume	Sulfuric acid, more concentrated
Copper(II) sulfate solution	Copper	Oxygen	Pink-brown deposit; bubbles at anode; blue colour fades	Sulfuric acid (acidic)

Lead(II) bromide must be heated in a fume cupboard: bromine is toxic. Very dilute sodium chloride solution gives oxygen as well at the anode.

Half-equations (1.59C)

Reaction	Half-equation	Type
Lead forms	Pb²⁺ + 2e⁻ → Pb	Reduction
Copper forms	Cu²⁺ + 2e⁻ → Cu	Reduction
Silver forms	Ag⁺ + e⁻ → Ag	Reduction
Hydrogen forms	2H⁺ + 2e⁻ → H₂	Reduction
Bromine forms	2Br⁻ → Br₂ + 2e⁻	Oxidation
Chlorine forms	2Cl⁻ → Cl₂ + 2e⁻	Oxidation
Oxygen forms	4OH⁻ → O₂ + 2H₂O + 4e⁻	Oxidation

Method in steps – writing a half-equation

1. Ion on the left, product on the right (gases as molecules: H₂, O₂, Cl₂, Br₂).
2. Balance atoms.
3. Add electrons to the more positive side until total charge is equal on both sides.
4. Electrons on the left = gain = reduction (cathode). Electrons on the right = loss = oxidation (anode).

OIL RIG: Oxidation Is Loss, Reduction Is Gain (of electrons).

Linking the two electrodes. Electrons lost at the anode = electrons gained at the cathode. Scale the two half-equations to the same number of electrons, then read off the mole ratio of products.

Electrolyte	Electrons	Mole ratio of products
Dilute sulfuric acid	4e⁻	2 H₂ : 1 O₂
Copper(II) sulfate	4e⁻	2 Cu : 1 O₂
Sodium chloride solution	2e⁻	1 H₂ : 1 Cl₂
Molten lead(II) bromide	2e⁻	1 Pb : 1 Br₂

Equal moles of gas have equal volumes at the same temperature and pressure; at rtp, 1 mol = 24 dm³ = 24 000 cm³.

Small worked reminders

Balancing the oxygen half-equation. Start with OH⁻ → O₂. Two O atoms are needed in O₂, but the extra H atoms must go somewhere, so water forms: 4OH⁻ → O₂ + 2H₂O. Charge on the left is 4−, on the right 0, so add 4e⁻ to the right: 4OH⁻ → O₂ + 2H₂O + 4e⁻.

Sodium chloride solution, equal volumes. Both half-equations use 2e⁻, so moles of H₂ = moles of Cl₂. If 40 cm³ of hydrogen is collected, 40 cm³ of chlorine is expected. Less is usually measured, because some chlorine dissolves in the solution.

Silver: one electron. Ag⁺ has a single positive charge, so Ag⁺ + e⁻ → Ag. For every 4 electrons, 4 Ag form at the cathode and 1 O₂ at the anode in silver nitrate solution.

Molten vs aqueous at a glance

Compound	Molten: cathode / anode	Aqueous, concentrated: cathode / anode
Sodium chloride	Sodium / chlorine	Hydrogen / chlorine
Potassium bromide	Potassium / bromine	Hydrogen / bromine
Potassium iodide	Potassium / iodine	Hydrogen / iodine

The anode product is the same in each pair because the halide wins in concentrated solution. Only the cathode product changes, because all three metals are more reactive than hydrogen. A copper salt would give copper either way.

The practical (1.60C)

• Graphite electrodes, low-voltage d.c. supply, solution in a cell or beaker.
• Small test tubes full of solution inverted over each electrode to collect gas.
• Record observations at each electrode, then test gases:
  • Hydrogen: lighted splint, squeaky pop.
  • Oxygen: glowing splint relights.
  • Chlorine: damp blue litmus turns red then white.
• Rinse electrodes between solutions. Ventilate if chlorine is produced.
• Change one thing at a time (the solution, or its concentration) and keep the rest the same.

Must-know distinctions

• Molten vs aqueous sodium chloride. Molten: sodium + chlorine. Aqueous (concentrated): hydrogen + chlorine.
• Anode vs cathode. Anode positive, oxidation, anions arrive. Cathode negative, reduction, cations arrive.
• Ions vs electrons. Ions move in the electrolyte; electrons move in the external circuit.
• “Cannot move” vs “no ions”. Solid ionic compounds have ions; they are fixed.
• Copper(II) sulfate vs dilute sulfuric acid. Both give oxygen at the anode. Copper sulfate gives copper at the cathode; the acid gives hydrogen.
• Concentrated vs very dilute chloride solution. Concentrated gives chlorine at the anode; very dilute gives oxygen too.

Quick self-test

1. Name the two kinds of mobile charged particle that allow conduction.
2. Explain why solid potassium bromide does not conduct.
3. State the products at each electrode when molten calcium chloride is electrolysed.
4. State the products at each electrode when copper(II) chloride solution (concentrated) is electrolysed with inert electrodes.
5. State the products when sodium nitrate solution is electrolysed with inert electrodes.
6. Write the half-equation for iodide ions at the anode.
7. Write the half-equation for aluminium ions at the cathode, and state whether it is oxidation or reduction.
8. During electrolysis of dilute sulfuric acid, 50 cm³ of hydrogen is collected. What volume of oxygen forms at the same temperature and pressure?
9. Copper(II) sulfate solution is electrolysed with inert electrodes and 1.27 g of copper is deposited. Calculate the volume of oxygen at rtp. (A_r Cu = 63.5)
10. In the electrolysis of molten lead(II) bromide, 0.0100 mol of electrons pass. Calculate the mass of lead formed. (A_r Pb = 207)
11. Why does copper(II) sulfate solution become acidic during electrolysis with inert electrodes?

Answers

1. Ions and delocalised electrons.
2. The ions are held in fixed positions in the lattice, so they cannot move to carry charge.
3. Cathode: calcium. Anode: chlorine.
4. Cathode: copper (less reactive than hydrogen). Anode: chlorine (halide in concentrated solution).
5. Cathode: hydrogen (sodium more reactive than hydrogen). Anode: oxygen (nitrate is not discharged).
6. 2I⁻ → I₂ + 2e⁻
7. Al³⁺ + 3e⁻ → Al; reduction (gain of electrons).
8. 25 cm³ (H₂ : O₂ = 2 : 1).
9. n(Cu) = 1.27 ÷ 63.5 = 0.0200 mol; n(O₂) = 0.0200 ÷ 2 = 0.0100 mol; volume = 0.0100 × 24 000 = 240 cm³.
10. Pb²⁺ + 2e⁻ → Pb, so n(Pb) = 0.0100 ÷ 2 = 0.00500 mol; mass = 0.00500 × 207 = 1.035 g (1.04 g to 3 s.f.).
11. Cu²⁺ and OH⁻ ions are discharged; H⁺ and SO₄²⁻ ions remain, which is sulfuric acid.

Where marks are usually lost

• Writing “electrons flow through the solution” instead of “ions move”.
• Writing “the solid has no ions” when the point is that the ions cannot move.
• Giving sodium (or potassium) as the cathode product for an aqueous solution.
• Forgetting H⁺ and OH⁻ from water when listing the ions in a solution.
• Writing Cl or Br (atoms) instead of Cl₂ or Br₂ in half-equations.
• Electrons on the wrong side, or charges that do not balance, in an anode half-equation.
• Calling the cathode reaction oxidation, or explaining it as “loss of oxygen” instead of gain of electrons.
• Stating the gas ratio for dilute sulfuric acid the wrong way round (it is twice as much hydrogen).
• Describing chlorine’s test as “turns litmus red” without the bleaching.
• Leaving out the fume cupboard or the reason for it in the lead(II) bromide experiment.

Official syllabus

Pearson Edexcel International GCSE in Chemistry (4CH1), Specification, Issue 3, September 2024, published by Pearson Education Limited. Topic 1: Principles of chemistry, sub-topic (i) Electrolysis, statements 1.55C–1.60C.