Pearson Edexcel International GCSE Chemistry 4CH1: Electrolysis – Study Guide

This guide teaches sub-topic 1(i) Electrolysis of the Pearson Edexcel International GCSE Chemistry (4CH1) specification, Issue 3 (September 2024): statements 1.55C to 1.60C. The qualification is untiered. Every statement in 1(i) is printed in bold with a C reference, so the whole sub-topic is Paper 2 only. Chemistry Paper 2 (1 hour 15 minutes, 70 marks) assesses all the content; Paper 1 assesses only the content that is not in bold.

Course links: 4CH1 Chemistry hub, printable checklist and the free all-topics diagnostic. For the rest of Topic 1, use the Principles of chemistry study guide.

What this sub-topic covers

Statement	What you must be able to do	Paper
1.55C	Explain why covalent compounds do not conduct electricity	Paper 2 only
1.56C	Explain why ionic compounds conduct only when molten or in aqueous solution	Paper 2 only
1.57C	Use the terms anion (negative ion) and cation (positive ion)	Paper 2 only
1.58C	Describe electrolysis experiments with inert electrodes on molten compounds (including lead(II) bromide) and aqueous solutions (including sodium chloride, dilute sulfuric acid and copper(II) sulfate), and predict the products	Paper 2 only
1.59C	Write ionic half-equations for the electrode reactions and explain why they are oxidation or reduction	Paper 2 only
1.60C	Practical: investigate the electrolysis of aqueous solutions	Paper 2 only

Three non-bold statements feed into this sub-topic and can be tested on either paper: 1.43 (ionic compounds conduct when molten or in solution, not when solid), 1.51 (covalent compounds do not usually conduct) and 2.20 (oxidation and reduction as loss and gain of electrons). The gas tests in 2.44 are what you use to identify the products.

Why some compounds conduct and others don’t (1.55C, 1.56C)

An electric current is a flow of charge. A substance conducts only if it contains charged particles that are free to move: delocalised electrons (as in metals and graphite) or ions.

Covalent compounds (1.55C). A covalent compound is made of neutral molecules, or a giant covalent structure, held together by shared pairs of electrons. There are no ions, and the electrons are held in bonds, so they are not free to move through the substance. With no mobile charged particles, there is nothing to carry the current. So a solution of glucose or ethanol in water does not conduct.

Ionic compounds (1.56C). An ionic compound is made of positive and negative ions.

• Solid: the ions are held in fixed positions in the giant lattice by strong electrostatic attractions. They can vibrate but cannot move from place to place, so the solid does not conduct.
• Molten: the lattice has broken down. The ions are free to move, so the liquid conducts.
• Aqueous solution: the ions separate and spread through the water. They are free to move, so the solution conducts.

It is the ions that move, not electrons.

Anions, cations and the electrodes (1.57C)

• A cation is a positive ion, e.g. Na⁺, Cu²⁺, Pb²⁺, H⁺.
• An anion is a negative ion, e.g. Cl⁻, Br⁻, OH⁻, SO₄²⁻.

Electrolysis is the breakdown of an ionic compound, molten or in aqueous solution, by passing a direct current through it. The liquid that conducts is the electrolyte. The cathode is the negative electrode and the anode is the positive electrode.

Opposite charges attract, so cations move to the cathode and anions move to the anode. Inert electrodes, usually graphite (carbon) or platinum, carry the current but do not react themselves.

Electrolysis of a molten compound: lead(II) bromide (1.58C)

Method. Put solid lead(II) bromide in a crucible with two graphite electrodes dipping into it. Connect the electrodes to a d.c. supply with a bulb (or ammeter) in series. Heat strongly in a fume cupboard.

Observations.

• While the solid is cold, the bulb stays off: the ions cannot move.
• Once the solid melts, the bulb lights.
• At the anode, red-brown fumes of bromine are given off.
• At the cathode, silvery lead forms and collects at the bottom.

The fume cupboard is needed because bromine vapour is toxic, and lead compounds are toxic too.

A molten binary ionic compound always gives the metal at the cathode and the non-metal at the anode. There is no water, so there are only two kinds of ion to choose from.

Electrolysis of aqueous solutions (1.58C)

In an aqueous solution, water provides a small number of extra ions:

H₂O ⇌ H⁺ + OH⁻

So each electrode has a choice of ions. Use these two rules to predict the products with inert electrodes.

At the cathode (negative):

• If the metal is less reactive than hydrogen (e.g. copper, silver), the metal is deposited.
• If the metal is more reactive than hydrogen (e.g. sodium, potassium, magnesium), hydrogen gas is given off instead.

At the anode (positive):

• If a halide ion (Cl⁻, Br⁻, I⁻) is present in a concentrated solution, the halogen is formed.
• Otherwise (sulfate, nitrate) oxygen is given off from the hydroxide ions. Sulfate and nitrate ions are not discharged.

The three solutions named in the specification

Electrolyte (inert electrodes)	Cathode product	Anode product	What is left in solution
Sodium chloride solution (concentrated)	Hydrogen	Chlorine	Sodium hydroxide (solution becomes alkaline)
Dilute sulfuric acid	Hydrogen	Oxygen	Sulfuric acid, slightly more concentrated
Copper(II) sulfate solution	Copper	Oxygen	Sulfuric acid (solution becomes acidic)

Sodium chloride solution. Sodium is more reactive than hydrogen, so H⁺ ions are discharged at the cathode and hydrogen bubbles off. Chloride ions are discharged at the anode as chlorine, which bleaches damp litmus paper. Na⁺ and OH⁻ ions remain, so the solution near the cathode turns universal indicator blue or purple. If the solution is very dilute, oxygen is also given off at the anode.

Dilute sulfuric acid. The ions are H⁺, SO₄²⁻ and OH⁻. Hydrogen forms at the cathode and oxygen at the anode. Water is being removed overall, so the acid slowly becomes more concentrated. The volume of hydrogen is twice the volume of oxygen (worked example 2 shows why).

Copper(II) sulfate solution. Copper is less reactive than hydrogen, so a pink-brown coating of copper forms on the cathode. Oxygen bubbles off at the anode. As Cu²⁺ ions are removed, the blue colour fades. H⁺ and SO₄²⁻ ions are left behind, so the solution becomes acidic.

Half-equations, oxidation and reduction (1.59C)

A half-equation shows what happens to one ion at one electrode, with electrons written in.

Steps:

1. Write the ion and the product. Diatomic products are written as molecules: Cl₂, Br₂, H₂, O₂.
2. Balance the atoms.
3. Add electrons so the charges balance on both sides.

Electrode	Half-equation
Cathode (lead)	Pb²⁺ + 2e⁻ → Pb
Cathode (copper)	Cu²⁺ + 2e⁻ → Cu
Cathode (hydrogen)	2H⁺ + 2e⁻ → H₂
Anode (bromine)	2Br⁻ → Br₂ + 2e⁻
Anode (chlorine)	2Cl⁻ → Cl₂ + 2e⁻
Anode (oxygen)	4OH⁻ → O₂ + 2H₂O + 4e⁻

Why these are oxidation or reduction. Use OIL RIG: Oxidation Is Loss of electrons, Reduction Is Gain.

• At the cathode, cations gain electrons, so the cathode reaction is always reduction.
• At the anode, anions lose electrons, so the anode reaction is always oxidation.

Electrons flow from the anode, round the external circuit, to the cathode. So in any given time, the number of electrons lost at the anode equals the number gained at the cathode. That fact links the amounts of the two products.

Worked example 1: half-equations for a new electrolyte

Molten zinc chloride is electrolysed. Write both half-equations and identify the oxidation.

• Ions present: Zn²⁺ and Cl⁻ only (no water).
• Cathode: zinc ions gain electrons. Zn²⁺ + 2e⁻ → Zn. Charge check: left 2+ and 2−, total 0; right 0.
• Anode: chloride ions lose electrons. 2Cl⁻ → Cl₂ + 2e⁻. Charge check: left 2−; right 0 + 2− = 2−.
• The anode reaction is oxidation, because chloride ions lose electrons.

Worked example 2: the 2 : 1 gas ratio

During the electrolysis of dilute sulfuric acid, 36 cm³ of hydrogen is collected. Calculate the volume of oxygen formed at the same time (same temperature and pressure).

• Cathode: 2H⁺ + 2e⁻ → H₂. Multiply by 2: 4e⁻ give 2H₂.
• Anode: 4OH⁻ → O₂ + 2H₂O + 4e⁻. 4e⁻ give 1O₂.
• The same 4 electrons pass through both electrodes, so mol H₂ : mol O₂ = 2 : 1.
• Equal moles of gas occupy equal volumes, so volume O₂ = 36 ÷ 2 = 18 cm³.

Worked example 3: linking the products of copper(II) sulfate

Copper(II) sulfate solution is electrolysed with inert electrodes. 0.254 g of copper is deposited. Calculate the volume of oxygen given off at rtp. (A_r Cu = 63.5; molar volume = 24 000 cm³ at rtp.)

• Moles of copper = 0.254 ÷ 63.5 = 0.00400 mol
• Cu²⁺ + 2e⁻ → Cu, so moles of electrons = 2 × 0.00400 = 0.00800 mol
• 4OH⁻ → O₂ + 2H₂O + 4e⁻, so moles of O₂ = 0.00800 ÷ 4 = 0.00200 mol
• Volume of O₂ = 0.00200 × 24 000 = 48 cm³

The mole and gas-volume methods here are from 1(e); see the formulae, equations and calculations practice if they feel shaky.

The practical: electrolysis of aqueous solutions (1.60C)

Apparatus. A beaker or electrolysis cell containing the solution, two graphite electrodes, a low-voltage d.c. supply, connecting wires, and a small test tube filled with the solution and inverted over each electrode to collect any gas.

Method.

1. Fill the cell with the solution so that both electrodes are covered.
2. Fill two small test tubes with the same solution and invert one over each electrode without letting air in.
3. Switch on and record what you see at each electrode: bubbles, colour of any solid, colour change in the solution.
4. When a tube is partly full, remove it with your thumb over the end and test the gas.
5. Repeat with each solution, rinsing the electrodes with water between runs.

Identifying the products (tests from 2.44).

Product	Test and result
Hydrogen	Lighted splint: burns with a squeaky pop
Oxygen	Glowing splint: relights
Chlorine	Damp blue litmus paper: turns red, then white (bleached)
Copper	Pink-brown solid coating on the cathode

Safety. Use low voltage. Chlorine is toxic: use small volumes, good ventilation, and stop the current once the test is done.

Possible variables to investigate. Which solution is used, its concentration (e.g. concentrated vs very dilute sodium chloride), or the current, while keeping electrode material and the other conditions the same.

Worked example 4: predicting from the rules

Predict the products at each electrode, with inert electrodes, for (a) silver nitrate solution, (b) concentrated potassium iodide solution.

(a) Cations Ag⁺ and H⁺. Silver is less reactive than hydrogen, so silver at the cathode. Anions NO₃⁻ and OH⁻; no halide, so oxygen at the anode.

(b) Cations K⁺ and H⁺. Potassium is more reactive than hydrogen, so hydrogen at the cathode. Anions I⁻ and OH⁻; iodide is a halide, so iodine at the anode (the solution near the anode turns brown).

Common errors

• Writing that electrons move through the electrolyte. Ions carry the charge there; electrons carry it in the wires.
• Saying a solid ionic compound has “no ions”. It has ions; they just cannot move.
• Swapping the electrodes. The cathode is negative and attracts cations.
• Writing Cl instead of Cl₂, or putting electrons on the wrong side of an anode half-equation.
• Predicting sodium at the cathode for sodium chloride solution. Sodium only forms from molten sodium chloride.
• Forgetting the extra H⁺ and OH⁻ ions from water when an electrolyte is aqueous.
• Calling the cathode reaction oxidation. Gain of electrons is reduction.
• Getting the hydrogen-to-oxygen ratio upside down. Hydrogen has the larger volume.

Next steps

Condense this with the electrolysis revision notes, then test yourself with the electrolysis practice questions. For the rest of the topic, use the Principles of chemistry revision notes and Principles of chemistry practice.

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.