What is electrolysis and how is it defined in the Edexcel IGCSE Chemistry syllabus?

Electrolysis is a chemical process that uses electricity to break down ionic compounds into their constituent elements. According to the Edexcel IGCSE Chemistry syllabus, it involves the movement of ions towards electrodes, where oxidation and reduction reactions occur, allowing the decomposition of substances.

What are the key components required for electrolysis in the context of Edexcel IGCSE Chemistry?

The key components required for electrolysis include an electrolyte, which is an ionic compound in molten or aqueous form, electrodes (anode and cathode), and a power source. The electrolyte conducts electricity, allowing ions to move towards the electrodes where chemical reactions take place.

How does the process of electrolysis differ between molten and aqueous solutions in Edexcel IGCSE Chemistry?

In molten electrolysis, the ionic compound is melted, allowing ions to move freely to the electrodes. In aqueous electrolysis, the ionic compound is dissolved in water, which introduces additional ions from the water itself. This can lead to different products being formed at the electrodes due to the presence of water.

What are some common applications of electrolysis that students should know for Edexcel IGCSE Chemistry?

Common applications of electrolysis include electroplating, extraction of metals such as aluminum and sodium, and the production of chlorine and hydrogen gas. These applications demonstrate the practical uses of electrolysis in industry and everyday life.

How can students predict the products of electrolysis in an exam setting for Edexcel IGCSE Chemistry?

To predict the products of electrolysis, students should identify the ions present in the electrolyte and determine which ions will be discharged at the electrodes. This involves understanding the reactivity series and the preferential discharge of ions, which is a key concept in the Edexcel IGCSE Chemistry curriculum.

Why is "Writing 'Na' as the product at the cathode for aqueous NaCl" a common mistake in Electrolysis?

Why it happens: Forgetting that aqueous solutions also contain H⁺ from water. How to avoid it: If the metal is MORE reactive than H (K, Na, Ca, Mg, Al), H₂ is produced at the cathode, NOT the metal. Only molten salts give Na metal at the cathode. Source: 4CH1 Examiner Reports 2022-2024

Why is "Omitting electrons from half-equations" a common mistake in Electrolysis?

Why it happens: Treating them like full balanced equations. How to avoid it: ALL half-equations must include electrons. Cathode: electrons on the LEFT (reduction). Anode: electrons on the RIGHT (oxidation). e.g. 2Cl⁻ → Cl₂ + 2e⁻ (not just '2Cl⁻ → Cl₂'). Source: 4CH1 Examiner Reports 2022-2024

Why is "Swapping the labels anode and cathode" a common mistake in Electrolysis?

Why it happens: Memorising the wrong way round. How to avoid it: Use the mnemonic **PANIC**: Positive Anode Negative Is Cathode. So anode = positive; cathode = negative. ANIONS go to ANODE; CATIONS go to CATHODE (both start with the same letter).

Why is "Using minutes (not seconds) in Q = It" a common mistake in Electrolysis?

Why it happens: Rushing the substitution. How to avoid it: Q = I × t needs SECONDS. Always convert: minutes × 60 = seconds; hours × 3600 = seconds. 32 min 10 s = 32 × 60 + 10 = 1930 s. Source: 4CH1 Examiner Reports 2023-2024

Why is "Using a 1:1 ratio of moles of electrons to moles of metal" a common mistake in Electrolysis?

Why it happens: Ignoring the charge of the metal ion. How to avoid it: Use the half-equation. For Cu²⁺ + 2e⁻ → Cu, you need 2 mol e⁻ per 1 mol Cu (so divide moles e⁻ by 2). For Al³⁺ + 3e⁻ → Al, divide by 3. For Ag⁺ + e⁻ → Ag, divide by 1.

Why is "Forgetting that with COPPER (active) electrodes, the products change for CuSO₄" a common mistake in Electrolysis?

Why it happens: Confusing inert and active electrodes. How to avoid it: INERT electrodes (C, Pt) in CuSO₄(aq): cathode Cu, anode O₂. ACTIVE copper electrodes in CuSO₄(aq): cathode Cu, anode dissolves as Cu²⁺ (no O₂). This is the basis for copper purification and electroplating.

Principles of ChemistryPearson Edexcel IGCSE Chemistry 4CH1 Higher

Electrolysis

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Electrolysis — Pearson Edexcel International GCSE Chemistry 4CH1 Study Notes (2026 onwards, Spec Issue 4)

Using a DC current to decompose ionic compounds (molten or aqueous) into elements. Half-equations at the cathode (reduction) and anode (oxidation), rules for predicting products of aqueous electrolysis, three major industrial applications (aluminium extraction, copper purification, electroplating), and Paper 2 quantitative calculations using Q = It and Faraday's constant.

At a glance

Electrolysis = decomposing an ionic compound using a DC current. Compound must be molten or aqueous (ions free to move).
Cathode (−): cations attracted → gain electrons (REDUCTION). Cu²⁺ + 2e⁻ → Cu.
Anode (+): anions attracted → lose electrons (OXIDATION). 2Cl⁻ → Cl₂ + 2e⁻.
OIL RIG: Oxidation Is Loss; Reduction Is Gain (of electrons).
Molten salt → metal at cathode, non-metal at anode.
Aqueous cathode rule: if metal more reactive than H → H₂ produced; otherwise metal produced.
Aqueous anode rule: halide present (Cl⁻, Br⁻, I⁻) → halogen produced; otherwise O₂ from OH⁻.
Quantitative (Paper 2): Q = I × t; moles of e⁻ = Q / 96500; use half-equation to find moles of product.
Aluminium: Al₂O₃ in molten cryolite (lowers mp from 2050 °C to 950 °C); C anode burns away as CO₂.

What you’ll learn

Mapped to the Cambridge IGCSE 4CH1 syllabus (2026 onwards).

1.73 — Understand that electrolysis is the decomposition of an ionic substance, when molten or in aqueous solution, by the passage of electric current.
1.74 — Identify in an electrolysis cell: the electrolyte (molten or aqueous), the electrodes (anode positive, cathode negative), the power supply (DC).
1.75 — Write half-equations for reactions at electrodes during electrolysis.
1.76 — Predict the products of electrolysis of molten ionic compounds, including binary salts like PbBr₂.
1.77 — Predict the products of electrolysis of aqueous solutions, using the rules: at the cathode, hydrogen is discharged if the metal is more reactive than hydrogen; at the anode, the halogen is discharged if a halide ion is present in concentrated solution, otherwise oxygen.
1.78 — Understand the terms 'oxidation' and 'reduction' in terms of electron loss / gain at the electrodes.
1.80 — Describe the manufacture of aluminium from purified aluminium oxide by electrolysis, including the role of cryolite, the carbon anode, and the half-equations.
1.81 — Describe the purification of copper by electrolysis with a copper(II) sulfate electrolyte; explain the half-equations and the formation of anode sludge.
1.82 — Describe the electroplating of objects with a metal (e.g. silver, copper, chromium) for appearance or corrosion resistance.
1.83P — (Paper 2 ONLY) Use the equation Q = It and Faraday's constant (96,500 C/mol of electrons) to calculate the amount of substance produced at an electrode.

What is electrolysis and what's needed (spec 1.73, 1.74)

DC current + ions free to move = decomposition.

Definition. Electrolysis is the process of decomposing an ionic compound (the electrolyte) into its constituent elements by passing an electric current through it.

Requirements for electrolysis.

An electrolyte — an ionic compound that is molten (heated above its melting point) OR dissolved in water (aqueous). In both cases the ions are free to move and can carry charge.
Two electrodes: the CATHODE (negative) and the ANODE (positive). They are usually made of an inert material (graphite or platinum) so they do not react themselves. In some processes (Cu purification, electroplating) the anode is intentionally an active metal that dissolves.
A DC (direct current) power supply — a battery or rectified mains. DC is essential because the polarity must be FIXED — alternating current would swap the electrode roles every half-cycle.

Why solid ionic compounds don't electrolyse. In a solid, the ions are fixed in the lattice and cannot move. Electrolysis only works when ions can migrate to the electrodes — hence molten or aqueous.

How current flows.

Inside the electrolyte: ions move. Cations (+) drift toward the cathode (−), anions (−) drift toward the anode (+).
At the electrodes: ions are converted to neutral atoms by gaining or losing electrons.
In the external wire: electrons flow from anode (through power supply) to cathode. Together this forms a complete circuit.

Cations drift to the negative cathode (reduction); anions drift to the positive anode (oxidation).

Useful mnemonic — PANIC. Positive Anode Negative Is Cathode.

Useful mnemonic — ANIONS to ANODE, CATIONS to CATHODE (the same first letters help you remember which goes where).

Useful mnemonic — OIL RIG. Oxidation Is Loss, Reduction Is Gain (of electrons). At the anode: oxidation (loss). At the cathode: reduction (gain).

Electrolysis = decomposing ionic compound using DC current.
Electrolyte must be molten or aqueous (ions free to move).
Cathode = negative; Anode = positive (mnemonic PANIC).
Cathode → reduction; Anode → oxidation (mnemonic OIL RIG).

See the full worked example for electrolysis →

Half-equations (spec 1.75, 1.78)

Show electrons explicitly. Cathode: e⁻ on left; anode: e⁻ on right.

A half-equation describes the reaction at ONE electrode, INCLUDING the electrons transferred. It must balance for atoms AND for charge.

Rules for writing a cathode (reduction) half-equation.

Start with the cation that's discharged (e.g. Pb²⁺, Cu²⁺, H⁺).
Add the electrons on the LEFT (e⁻ on the same side as the cation).
Number of electrons = charge magnitude of the cation.
The product (a neutral atom or molecule) appears on the right.

Examples:

$\text{Na}^+ + e^- \rightarrow \text{Na}$ (from molten NaCl)
$\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$ (from CuSO₄ solution or molten CuCl₂)
$\text{Al}^{3+} + 3e^- \rightarrow \text{Al}$ (from molten Al₂O₃ in cryolite)
$2\text{H}^+ + 2e^- \rightarrow \text{H}_2$ (from any aqueous solution where H⁺ is discharged)

Rules for writing an anode (oxidation) half-equation.

Start with the anion that's discharged (e.g. Br⁻, Cl⁻, O²⁻, OH⁻).
Add the electrons on the RIGHT (e⁻ as products).
The product is a neutral element or compound (often gaseous: Cl₂, Br₂, O₂).
Watch the stoichiometry — diatomic halogens need TWO anions per molecule.

Examples:

$2\text{Br}^- \rightarrow \text{Br}_2 + 2e^-$
$2\text{Cl}^- \rightarrow \text{Cl}_2 + 2e^-$
$2\text{O}^{2-} \rightarrow \text{O}_2 + 4e^-$ (from molten Al₂O₃; each O²⁻ loses 2 e⁻, total 4)
$4\text{OH}^- \rightarrow 2\text{H}_2\text{O} + \text{O}_2 + 4e^-$ (from aqueous OH⁻ — note H₂O byproduct)

Quick check — balance. Atoms must balance on both sides AND the total charge must balance. For $2\text{Br}^- \rightarrow \text{Br}_2 + 2e^-$ : left = 2− total charge, right = 0 + 2(−) = 2− total. ✓

Reduction / oxidation labels. Cathode reactions are always REDUCTION (gain of electrons). Anode reactions are always OXIDATION (loss of electrons). You can be asked to LABEL each half-equation with the reaction type for marks.

Cathode half-equation: e⁻ on the LEFT.
Anode half-equation: e⁻ on the RIGHT.
Number of e⁻ = charge magnitude × number of ions.
Must balance for atoms AND for charge.

See the full worked example for electrolysis →

Predicting products: molten vs aqueous electrolytes (spec 1.76, 1.77)

Molten = simple (cation + anion of salt). Aqueous = reactivity series + halide rule.

Case 1 — MOLTEN ionic compound.

The only ions present are the metal cation and the non-metal anion of the salt. The metal is discharged at the cathode; the non-metal at the anode.

Molten electrolyte	Cathode	Anode
NaCl(l)	Na	Cl₂
KBr(l)	K	Br₂
PbBr₂(l)	Pb	Br₂
Al₂O₃(l) (in cryolite)	Al	O₂

This is straightforward — there are no competing ions.

Case 2 — AQUEOUS solution.

In water, the electrolyte ions are joined by H⁺ and OH⁻ from the slight self-ionisation of water:

$\text{H}_2\text{O} \rightleftharpoons \text{H}^+ + \text{OH}^-$

So at the cathode there are TWO cations competing (the metal cation and H⁺); at the anode there are TWO anions competing (the salt's anion and OH⁻).

Cathode rule — competition between metal cation and H⁺.

If the metal is MORE reactive than hydrogen (K, Na, Ca, Mg, Al) → H₂ is produced. The reactive metal cation 'prefers' to stay in solution.
If the metal is LESS reactive than hydrogen (Cu, Ag, Au, Pb, Sn, Fe) → the METAL is produced. Less reactive metals readily gain electrons.

A useful way to remember: in the reactivity series, K > Na > Ca > Mg > Al > C > Zn > Fe > Sn > Pb > H > Cu > Hg > Ag > Au > Pt. Everything ABOVE H stays as ions; H₂ is discharged. Everything BELOW H is deposited as the metal.

Anode rule — competition between halide and OH⁻.

If a HALIDE ion (Cl⁻, Br⁻, I⁻) is present (especially in concentrated solution) → the halogen is produced (Cl₂, Br₂, I₂).
If there is NO halide (e.g. the anion is SO₄²⁻ or NO₃⁻ — both are 'oxyanions' that resist oxidation) → O₂ is produced from the OH⁻ (water): $4\text{OH}^- \rightarrow 2\text{H}_2\text{O} + \text{O}_2 + 4e^-$ .

Worked predictions.

Aqueous electrolyte	Cathode	Anode	Reasoning
NaCl(aq) (brine)	H₂	Cl₂	Na > H → H₂; Cl⁻ present → Cl₂
CuSO₄(aq) (inert electrodes)	Cu	O₂	Cu < H → Cu; SO₄²⁻ not halide → O₂ from OH⁻
H₂SO₄(aq) (dilute)	H₂	O₂	only H⁺ and OH⁻ available
NaBr(aq)	H₂	Br₂	Na > H → H₂; Br⁻ present → Br₂
AgNO₃(aq)	Ag	O₂	Ag < H → Ag; NO₃⁻ not halide → O₂
CuCl₂(aq)	Cu	Cl₂	Cu < H → Cu; Cl⁻ present → Cl₂

Exception case — copper electrodes in CuSO₄(aq). If the electrodes are ACTIVE copper (not inert graphite), the anode reaction CHANGES: the Cu anode itself dissolves rather than OH⁻ being oxidised:

Anode: Cu(s) → Cu²⁺(aq) + 2e⁻
Cathode: Cu²⁺(aq) + 2e⁻ → Cu(s)

This is the basis of copper purification and electroplating.

Molten = metal at cathode + non-metal at anode (simple).
Aqueous cathode: metal > H → H₂; metal < H → metal.
Aqueous anode: halide present → halogen; no halide → O₂ from OH⁻.
ACTIVE electrodes (e.g. Cu in CuSO₄) change the anode reaction — anode dissolves.

See the full worked example for electrolysis →

Quantitative electrolysis (Paper 2 — spec 1.83P)

Q = It → moles of e⁻ → moles of product → mass / volume.

For Paper 2, you must be able to calculate the amount (mass of metal or volume of gas) produced at an electrode given the current and time. The full chain of calculation is:

$Q = I \times t \quad \rightarrow \quad n_e = \dfrac{Q}{F} \quad \rightarrow \quad n_{\text{product}} = \dfrac{n_e}{z} \quad \rightarrow \quad m = n \times A_r$

Standard route, step by step.

Convert time to seconds. $t = \text{minutes} \times 60 + \text{seconds}$ .
Charge passed. $Q = I \times t$ in coulombs (C).
Moles of electrons. $n_e = Q / F$ where $F = 96{,}500$ C/mol (Faraday's constant — the charge of 1 mole of electrons).
Use the cathode half-equation to find the moles of product. For Cu²⁺ + 2e⁻ → Cu, 2 mol e⁻ make 1 mol Cu, so $n_{Cu} = n_e / 2$ . For Ag⁺ + e⁻ → Ag, divide by 1. For Al³⁺ + 3e⁻ → Al, divide by 3.
Mass of metal = $n \times A_r$ (g).
Volume of gas = $n \times 24$ dm³ (at rtp).

Worked example — copper deposition.

Pass 1.50 A for 32 min 10 s through CuSO₄(aq). What mass of Cu deposits at the cathode?

$t = 32 \times 60 + 10 = 1930\ \text{s}$ .
$Q = I \times t = 1.50 \times 1930 = 2895\ \text{C}$ .
$n_e = 2895 / 96500 = 0.0300\ \text{mol}$ .
Cu²⁺ + 2e⁻ → Cu → $n_{Cu} = 0.0300 / 2 = 0.0150\ \text{mol}$ .
$m_{Cu} = 0.0150 \times 63.5 = 0.953\ \text{g}$ .

Worked example — hydrogen gas at the cathode.

Pass 0.50 A for 10 minutes through dilute H₂SO₄. What volume of H₂ at rtp?

$t = 600\ \text{s}$ ; $Q = 0.50 \times 600 = 300\ \text{C}$ .
$n_e = 300 / 96500 = 3.11 \times 10^{-3}\ \text{mol}$ .
2H⁺ + 2e⁻ → H₂ → $n_{H_2} = n_e / 2 = 1.56 \times 10^{-3}\ \text{mol}$ .
$V = n \times 24 = 1.56 \times 10^{-3} \times 24 = 0.0373\ \text{dm}^3 = 37.3\ \text{cm}^3$ .

Common pitfalls.

Time MUST be in seconds (× 60 from minutes).
Always check the half-equation: how many electrons per ion.
Don't forget to divide by the charge of the ion in step 4.
Quote 3 sig figs unless the data forces fewer.

$Q = I \times t$ (charge in C; time in seconds).
Moles of electrons = $Q / 96500$ .
Use the half-equation: moles of product = $n_e \div$ (charge of ion).
Mass = $n \times A_r$ ; volume at rtp = $n \times 24$ dm³.

See the full worked example for electrolysis →

Industrial electrolysis: aluminium extraction (spec 1.80)

Al₂O₃ in molten cryolite; carbon anode burns as CO₂.

Aluminium is more reactive than carbon, so it CANNOT be extracted from its ore by reduction with C (unlike iron). Instead, electrolysis of molten aluminium oxide is used. Bauxite is the natural ore — once purified, it gives Al₂O₃.

Problem with pure Al₂O₃. Pure Al₂O₃ has a melting point of ~2050 °C — extremely costly to maintain.

Solution — cryolite. Cryolite (Na₃AlF₆) is a sodium-aluminium fluoride mineral that dissolves Al₂O₃ to give a mixture with a much LOWER melting point (~950 °C). Cryolite acts as a solvent, NOT a reactant — it lets the Al₂O₃ ions move freely while reducing the energy and cost required.

Set-up of the electrolysis cell.

A steel tank lined with carbon (graphite) — the carbon lining is the cathode.
Several large carbon blocks dipped from above into the molten mixture — the anode.
The mixture inside is molten Al₂O₃ dissolved in cryolite at ~950 °C.
A very large DC current (~100,000 A) is passed.

Cryolite dissolves the Al₂O₃ so electrolysis runs far cooler; the carbon anodes react with O₂ and must be replaced.

Half-equations.

Cathode (reduction): Al³⁺ + 3e⁻ → Al. Molten aluminium is denser than the cryolite mixture and collects at the bottom — tapped off periodically.
Anode (oxidation): 2O²⁻ → O₂ + 4e⁻. Oxygen gas is evolved at the anode.

Why the carbon anode is consumed. The carbon anode operates at ~950 °C and is exposed to the O₂ produced. Hot carbon reacts with oxygen:

$\text{C(s)} + \text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)}$

(Some CO is also formed.) The carbon anode is gradually eaten away as CO₂ → it gets thinner and must be replaced regularly. This is one reason aluminium production is expensive.

Why aluminium is expensive.

Huge amounts of electrical energy needed (high mp, high current).
Carbon anodes need replacing.
Cryolite is a relatively rare mineral.
The process must run continuously (cooling lets the mix solidify).

Why we do it anyway. Aluminium has a unique combination of low density, strength (especially in alloys like duralumin), corrosion resistance and good electrical conductivity. It is the second-most-used metal after iron.

Al₂O₃ is the source; cryolite (Na₃AlF₆) is the solvent.
Cryolite lowers operating temperature ~2050 °C → ~950 °C — saves energy.
Cathode: Al³⁺ + 3e⁻ → Al (molten Al taps from bottom).
Anode: 2O²⁻ → O₂ + 4e⁻ (O₂ evolved).
Carbon anode + O₂ → CO₂ → anode burns away → must replace.

See the full worked example for electrolysis →

Copper purification and electroplating (spec 1.81, 1.82)

Active electrodes — the anode dissolves; the cathode grows.

In the previous sections we've assumed INERT electrodes. With active (reactive) electrodes, the anode reaction changes — instead of OH⁻ being oxidised, the electrode itself dissolves. Two important applications use this trick.

Application 1 — Purification of copper.

Copper extracted from ore is typically ~99 % pure; for electrical wiring we need ≥ 99.99 % because impurities greatly increase electrical resistance.

Cathode: a thin sheet of PURE copper. Cu²⁺ ions in the solution are reduced and deposit as pure Cu metal: Cu²⁺ + 2e⁻ → Cu.
Anode: a block of IMPURE copper. Cu atoms in the anode are oxidised and enter the solution as Cu²⁺: Cu → Cu²⁺ + 2e⁻.
Electrolyte: CuSO₄(aq), often acidified with H₂SO₄.

Net effect: pure copper is transferred from the impure anode to the pure cathode.

Anode sludge. Impurities behave according to their reactivity:

Less reactive than Cu (Au, Ag, Pt): NOT oxidised → fall to the bottom as a metallic sludge (the anode sludge or 'anode mud'). Very valuable — the gold and silver recovered offset the cost of the electrolysis.
More reactive than Cu (Fe, Zn): ARE oxidised at the anode and enter solution as Fe²⁺, Zn²⁺. But they are NOT reduced at the cathode (Cu²⁺ is reduced preferentially) → they stay in solution.

Application 2 — Electroplating.

Electroplating coats one metal (the object) with a thin layer of another metal (the coating). The set-up:

Cathode: the OBJECT to be plated (e.g. a steel spoon).
Anode: a block of the plating metal (e.g. pure silver).
Electrolyte: a solution containing ions of the plating metal (e.g. AgNO₃(aq) for silver-plating; CuSO₄(aq) for copper-plating; CrO₃ for chrome-plating).

Half-equations (silver-plating example):

Anode: Ag → Ag⁺ + e⁻ (silver atoms dissolve).
Cathode: Ag⁺ + e⁻ → Ag (silver deposits onto the spoon).

The Ag⁺ concentration in the electrolyte stays roughly constant.

Why electroplate?

Appearance. Decorative finishes — silver-plated cutlery, chrome-plated bumpers, gold-plated jewellery look more expensive than the base metal.
Corrosion resistance. Tin-plated cans (food cans); zinc-plated 'galvanised' steel (nails, sheets); chrome-plated taps.
Hardness / wear resistance. Chrome plating gives a hard, scratch-resistant surface.
Electrical conductivity. Gold plating on connectors (low resistance + corrosion-free).

Copper purification: impure Cu anode dissolves; pure Cu cathode grows; CuSO₄ electrolyte.
Anode sludge contains Au, Ag, Pt (less reactive than Cu) — valuable byproduct.
Electroplating: object = cathode; plating metal = anode; electrolyte has plating-metal ions.
Uses: decoration, corrosion resistance, hardness, conductivity.

See the full worked example for electrolysis →

How it’s examined

Electrolysis is one of the most-tested 4CH1 topics — worth 8-15 marks per session combined across both papers. Paper 1 (4CH1/1C) items: (a) predict products of molten or aqueous electrolysis and write half-equations (3-6 marks); (b) explain industrial aluminium extraction including cryolite role (5-6 marks); (c) define electrolysis, identify electrodes and explain ion movement (3-5 marks). Paper 2 (4CH1/2C) BOLD 'P' items: (d) quantitative Q = It calculations to find mass of metal or volume of gas produced (5-6 marks); (e) copper purification and electroplating with half-equations (5-6 marks). Electrolysis ALWAYS appears on Paper 1, AND the quantitative version appears on most Paper 2 sittings.

Sources: Pearson Edexcel International GCSE Chemistry 4CH1 specification — Issue 4, September 2024 (valid for 2026 onwards); Pearson Edexcel 4CH1 Examiner Reports 2022-2024; 4CH1/1C May/Jun 2024 question paper and mark scheme; 4CH1/1C January 2024 question paper and mark scheme; 4CH1/1CR January 2024 question paper and mark scheme; 4CH1/2C May/Jun 2024 question paper and mark scheme; 4CH1/2C January 2024 question paper and mark scheme. Last reviewed 2026-05-12.

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Step-by-step worked examples — Electrolysis

Step-by-step solutions to past-paper-style questions on electrolysis, written exactly the way a tutor would explain them at the board.

1Electrolysis of molten lead(II) bromide (5 marks, Paper 1)
Core• Adapted from 4CH1/1C Jan 2024 Q6• electrolysis, molten, half-equations, spec-1.73, spec-1.74
Question
Molten lead(II) bromide, PbBr₂(l), is electrolysed using inert graphite electrodes. Predict the product at EACH electrode and write the half-equation for the reaction at each electrode. State the type of reaction (oxidation or reduction) at each electrode. (5 marks)
Step-by-step solution
1. Step 1
  Identify the ions present (1 M). Molten PbBr₂ contains Pb²⁺ cations and Br⁻ anions, all free to move.
2. Step 2
  Cathode — cations gain electrons (1 A). Cations are attracted to the cathode (negative electrode). Pb²⁺ gains 2 electrons to form Pb metal.
  $\text{Cathode (reduction):}\ \text{Pb}^{2+} + 2e^- \rightarrow \text{Pb}$
3. Step 3
  Reduction at cathode (1 B). Gain of electrons is REDUCTION (OIL RIG: Reduction Is Gain).
4. Step 4
  Anode — anions lose electrons (1 A). Anions are attracted to the anode (positive electrode). Two Br⁻ ions lose 1 electron each, combining to form Br₂ molecules.
  $\text{Anode (oxidation):}\ 2\text{Br}^- \rightarrow \text{Br}_2 + 2e^-$
5. Step 5
  Oxidation at anode (1 B). Loss of electrons is OXIDATION (OIL RIG: Oxidation Is Loss). Br₂ gas (orange-brown vapour) is observed at the anode; silvery lead metal collects at the cathode.
Answer
Cathode: Pb²⁺ + 2e⁻ → Pb (REDUCTION; lead metal formed). Anode: 2Br⁻ → Br₂ + 2e⁻ (OXIDATION; bromine gas evolved).
Examiner tip
Edexcel 5-mark mark scheme: 1 for each half-equation (must include electrons and balance); 1 for each correct type of reaction; 1 for state symbols / observation. The cathode half-equation MUST start with the cation; the anode must START with the anion. ECF applies if a candidate swaps anode and cathode.
2Electrolysis of sodium chloride solution (brine) (6 marks, Paper 1)
Core• Adapted from 4CH1/1C May/Jun 2024 Q9• electrolysis, brine, aqueous, spec-1.76, spec-1.77, spec-1.78
Question
Concentrated sodium chloride solution (brine) is electrolysed using inert electrodes. Predict the product at the cathode and the anode, write the half-equations, and explain why H₂ is produced at the cathode rather than sodium. State ONE industrial use of EACH product. (6 marks)
Step-by-step solution
1. Step 1
  Ions present (1 M). Brine contains Na⁺ and Cl⁻ from NaCl, AND H⁺ and OH⁻ from water (water self-ionises slightly: $\text{H}_2\text{O} \rightleftharpoons \text{H}^+ + \text{OH}^-$ ).
2. Step 2
  Cathode — choice between Na⁺ and H⁺ (1 A). Two cations are attracted to the cathode. The rule: the LESS REACTIVE element is preferentially discharged. Sodium is HIGHER in the reactivity series than hydrogen, so H⁺ is reduced and H₂ gas is produced. Sodium stays in solution as Na⁺.
  $\text{Cathode:}\ 2\text{H}^+ + 2e^- \rightarrow \text{H}_2$
3. Step 3
  Why H₂ not Na (1 B). Na is more reactive than H — it would rather stay as Na⁺ (its tendency to lose electrons is stronger). H⁺ is more easily reduced than Na⁺, so hydrogen is discharged preferentially.
4. Step 4
  Anode — choice between Cl⁻ and OH⁻ (1 A). Two anions are attracted to the anode. The rule: if a HALIDE ion is present (Cl⁻, Br⁻, I⁻) and the solution is concentrated, the halide is discharged in preference to OH⁻. Here Cl⁻ is discharged as Cl₂ gas.
  $\text{Anode:}\ 2\text{Cl}^- \rightarrow \text{Cl}_2 + 2e^-$
5. Step 5
  Leftover ions form NaOH (1 B). Na⁺ stays in solution; the cathode reaction also leaves OH⁻ behind (because water has been split). So the solution becomes NaOH(aq) — a useful by-product.
6. Step 6
  Industrial uses (1 B). H₂: making ammonia (Haber process), hardening vegetable oils into margarine, fuel cells. Cl₂: making bleach, sterilising drinking water and swimming pools, manufacturing PVC and HCl. NaOH: making soaps, detergents, paper.
Answer
Cathode: 2H⁺ + 2e⁻ → H₂ (Na more reactive than H, so H₂ discharged). Anode: 2Cl⁻ → Cl₂ + 2e⁻ (chloride discharged in preference to OH⁻). Remaining Na⁺ + OH⁻ → NaOH(aq) as a useful by-product. Uses: H₂ for Haber process / margarine; Cl₂ for bleach / water sterilisation; NaOH for soaps / paper.
Examiner tip
Edexcel 6-mark mark scheme: 1 mark for each correct cation/anion identification, half-equation, reasoning (reactivity vs halide rule), and use. Most-frequent error: writing 'Na metal' at the cathode — this would only happen for MOLTEN NaCl, not aqueous. For AQUEOUS, the reactivity-series rule applies.
3Calculating mass of copper deposited from Q = It (6 marks, Paper 2)
Extended• Adapted from 4CH1/2C May/Jun 2024 Q6• calculation, Faraday, spec-1.83P
Question
Copper(II) sulfate solution is electrolysed using inert electrodes. A current of 1.50 A is passed for 32 minutes 10 seconds. Calculate the mass of copper deposited at the cathode. Faraday's constant $F = 96{,}500$ C/mol. ( $A_r$ Cu = 63.5). (6 marks)
Step-by-step solution
1. Step 1
  Convert time to seconds (1 M). 32 min 10 s = 32 × 60 + 10 = 1930 s.
2. Step 2
  Charge passed Q = I × t (1 M).
  $Q = I \times t = 1.50 \times 1930 = 2895\ \text{C}$
3. Step 3
  Moles of electrons (1 A). Each mole of electrons carries 96,500 C of charge (Faraday's constant).
  $n_e = \dfrac{Q}{F} = \dfrac{2895}{96500} = 0.0300\ \text{mol}$
4. Step 4
  Half-equation at the cathode (1 B). Cu²⁺ + 2e⁻ → Cu. 2 moles of electrons → 1 mole of Cu. So moles of Cu = moles of electrons ÷ 2.
  $n(\text{Cu}) = \dfrac{n_e}{2} = \dfrac{0.0300}{2} = 0.0150\ \text{mol}$
5. Step 5
  Mass of Cu = n × Ar (1 B).
  $m(\text{Cu}) = 0.0150 \times 63.5 = 0.953\ \text{g}$
6. Step 6
  Final answer with sig figs and unit (1 B). Mass of copper deposited = 0.953 g (3 sig fig).
Answer
Mass of Cu deposited = 0.953 g (≈ 0.95 g).
Examiner tip
Edexcel 6-mark mark scheme for Paper 2 quantitative electrolysis: 1 each for (1) time in seconds; (2) Q = It; (3) moles of electrons from Q/F; (4) cathode half-equation / mole ratio; (5) mass calculation; (6) final answer with unit. ECF applies through each step. Always quote 3 sig figs unless the data dictates otherwise.
4Extraction of aluminium by electrolysis (6 marks, Paper 1)
Core• Adapted from 4CH1/1CR Jan 2024 Q8• aluminium, industrial electrolysis, cryolite, spec-1.80, spec-1.81
Question
Aluminium is extracted from its ore (bauxite, Al₂O₃) by electrolysis. (a) State the role of cryolite in the process. (b) Write the half-equation at each electrode. (c) Explain why the carbon anode must be replaced regularly. (6 marks)
Step-by-step solution
1. Step 1
  (a) Role of cryolite (1 M). Aluminium oxide has a very high melting point (~2050 °C) — too costly to melt directly. Cryolite (Na₃AlF₆) is added as a solvent: Al₂O₃ dissolves in molten cryolite, and the mixture has a much lower melting point (~950 °C). This saves energy and makes the process economic.
2. Step 2
  (b) Ions in the melt (1 M). Dissolved Al₂O₃ provides Al³⁺ and O²⁻ ions, free to move in the molten mixture.
3. Step 3
  (b) Cathode half-equation (1 A). Al³⁺ migrates to the negative cathode and is reduced.
  $\text{Al}^{3+} + 3e^- \rightarrow \text{Al}$
4. Step 4
  (b) Anode half-equation (1 A). O²⁻ migrates to the positive anode and is oxidised.
  $2\text{O}^{2-} \rightarrow \text{O}_2 + 4e^-$
5. Step 5
  (c) Anode wears away (1 B). The anode is made of CARBON (graphite). The operating temperature inside the cell is high (~950 °C) and oxygen is produced at the anode.
6. Step 6
  (c) Carbon + oxygen → CO₂ (1 B). Hot carbon reacts with the oxygen evolved at the anode: $\text{C} + \text{O}_2 \rightarrow \text{CO}_2$ . The carbon anode is gradually 'eaten away' as carbon dioxide → it gets thinner over time and must be replaced regularly. (CO is also formed in smaller amounts.)
Answer
(a) Cryolite is a solvent for Al₂O₃; lowers the operating temperature from ~2050 °C to ~950 °C → energy and cost saving. (b) Cathode: Al³⁺ + 3e⁻ → Al. Anode: 2O²⁻ → O₂ + 4e⁻. (c) The hot carbon anode reacts with the O₂ produced: C + O₂ → CO₂. The anode is gradually consumed → must be replaced.
Examiner tip
Edexcel 6-mark mark scheme: 1 for cryolite as solvent + low temperature; 2 for the two correct balanced half-equations; 1 for cathode reaction type / Al³⁺ → Al; 1 for stating C + O₂ → CO₂; 1 for explaining anode wearing away. Don't forget to balance the electrons: 4 in the anode equation (because 2 × O²⁻ each loses 2 electrons).

Key Formulae — Electrolysis

The formulae you need to memorise for electrolysis on the Pearson Edexcel IGCSE 4CH1 paper, with every variable defined in plain English and a note on when to use it.

Electric charge (spec 1.83P, Paper 2)
$Q = I \times t$
$Q$
Charge passed, in coulombs (C)
$I$
Current, in amperes (A)
$t$
Time, in seconds (s)
When to use
Calculate the total electric charge passed during an electrolysis. Time MUST be in seconds (convert minutes × 60). Charge in coulombs is the starting point for all quantitative electrolysis calculations.
Example
A current of 2.0 A flows for 5.0 minutes. $Q = 2.0 \times 300 = 600\ \text{C}$ .
Moles of electrons (Faraday's constant) (Paper 2)
$n_e = \dfrac{Q}{F}$
$n_e$
Moles of electrons (mol)
$Q$
Charge in coulombs (C)
$F$
Faraday's constant = 96,500 C/mol — the charge of 1 mole of electrons
When to use
Convert charge passed to moles of electrons transferred at an electrode. From here, use the cathode (or anode) half-equation to get the moles of product, then mass or volume.
Example
If $Q = 2895\ \text{C}$ , then $n_e = 2895/96500 = 0.0300\ \text{mol}$ .
Moles of product from a half-equation
$n_{\text{product}} = \dfrac{n_e}{z}$
$n_e$
Moles of electrons
$z$
Number of electrons per ion (charge magnitude) — 1 for Ag⁺, 2 for Cu²⁺, 3 for Al³⁺
When to use
Use the cathode half-equation to identify the moles of electrons needed PER mole of product. For Cu²⁺ + 2e⁻ → Cu, 2 mol of e⁻ make 1 mol of Cu, so divide moles of electrons by 2. For Ag⁺ + e⁻ → Ag, divide by 1. For Al³⁺ + 3e⁻ → Al, divide by 3.
Example
For Cu, $n(\text{Cu}) = n_e / 2 = 0.0300/2 = 0.0150\ \text{mol}$ .
Mass of metal deposited
$m = n \times A_r$
$m$
Mass of metal deposited (g)
$n$
Moles of metal (mol)
$A_r$
Relative atomic mass of the metal
When to use
Standard mass-from-moles step. Use AFTER finding moles of metal from the half-equation. Multiply moles by $A_r$ in g/mol. Always state the unit (g).
Example
For 0.0150 mol Cu ( $A_r$ = 63.5): $m = 0.0150 \times 63.5 = 0.953\ \text{g}$ .
Volume of gas produced (at rtp)
$V_{\text{gas (dm}^3\text{)}} = n_{\text{gas}} \times 24$
$V$
Volume of gas at rtp (dm³)
$n_{gas}$
Moles of gas
When to use
When a GAS is produced at an electrode (H₂, Cl₂, O₂), use $V = n \times 24$ dm³ at rtp. For O₂ from 4OH⁻ → O₂ + 2H₂O + 4e⁻, $n_{O_2} = n_e/4$ . For H₂ from 2H⁺ + 2e⁻ → H₂, $n_{H_2} = n_e/2$ .
Example
Pass 0.020 mol e⁻ through dilute H₂SO₄: $n(H_2) = 0.020/2 = 0.010\ \text{mol} \rightarrow V = 0.010 \times 24 = 0.24\ \text{dm}^3$ (240 cm³).

Model Answers — Electrolysis

High-scoring sample answers for electrolysis on the Pearson Edexcel IGCSE 4CH1 paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
4CH1/1C-style — electrolysis definition + apparatus5 marks
Define electrolysis. Describe the apparatus used to electrolyse molten lead(II) bromide, name the electrodes, and explain what happens to the ions when an electric current is passed. (5 marks)
Model answer
Definition of electrolysis. Electrolysis is the process of decomposing an ionic compound (the electrolyte — molten or in aqueous solution) into its constituent elements by passing an electric current through it.

Apparatus for electrolysis of molten PbBr₂.

The electrolyte is molten PbBr₂(l), obtained by heating solid PbBr₂ above its melting point (~370 °C) in a crucible.

Two inert electrodes (rods of graphite or platinum that do not react with the products) are dipped into the molten liquid.

The electrodes are connected by wires to the terminals of a direct current (d.c.) power supply (battery or rectified mains). The d.c. is essential — alternating current would reverse the products every half-cycle.

The electrode connected to the POSITIVE terminal is the ANODE; the one connected to the NEGATIVE terminal is the CATHODE.

What happens when the current is passed.

The molten electrolyte contains positive ions (Pb²⁺) and negative ions (Br⁻), all free to move.

Cations (Pb²⁺) are attracted to the negative cathode. At the cathode, each cation gains electrons (REDUCTION) and is converted to neutral atoms. The metal Pb is formed: Pb²⁺ + 2e⁻ → Pb.

Anions (Br⁻) are attracted to the positive anode. At the anode, each anion loses electrons (OXIDATION). The neutral atoms pair up to form Br₂ molecules: 2Br⁻ → Br₂ + 2e⁻.

The electrons released at the anode travel through the external circuit (the wires + power supply) to the cathode, completing the circuit. The "current" inside the electrolyte is carried by the moving ions.

Result. The compound PbBr₂ is decomposed into its elements: lead (silvery liquid at the bottom of the crucible) and bromine (orange-brown vapour rising from the anode).
Why this scores
Edexcel 5-mark mark scheme: 1 for the definition (decomposition of ionic compound using electricity); 1 for naming electrolyte / electrodes / d.c. power supply; 1 for ANODE = positive, CATHODE = negative; 1 for movement of ions (cations to cathode, anions to anode); 1 for what happens AT each electrode (loss vs gain of electrons). The mnemonic 'PANIC' (Positive Anode Negative Is Cathode) helps avoid confusion.
Question 2
4CH1/2C-style — rules for aqueous electrolysis6 marks
Explain the rules used to predict the products of electrolysis when the electrolyte is an aqueous solution rather than a molten salt. Illustrate with TWO examples that give different products at the cathode and anode. (6 marks)
Model answer
Why aqueous is different. Molten ionic compounds contain only the cation and anion of the salt. Aqueous solutions ALSO contain a small amount of H⁺ and OH⁻ from the slight self-ionisation of water ( $\text{H}_2\text{O} \rightleftharpoons \text{H}^+ + \text{OH}^-$ ). So there are TWO cations competing for discharge at the cathode and TWO anions competing at the anode. The element actually discharged is determined by the reactivity / ease-of-discharge rules below.

Rule for the CATHODE (cations).

If the metal cation is MORE reactive than hydrogen (in the reactivity series — i.e. K, Na, Ca, Mg, Al), then H₂ is produced at the cathode (water / H⁺ is reduced).

If the metal cation is LESS reactive than hydrogen (i.e. Cu, Ag, Pb), then the METAL is produced at the cathode (the metal cation is reduced).

Rule for the ANODE (anions).

If a HALIDE ion (Cl⁻, Br⁻, I⁻) is present in concentrated solution, the halogen (Cl₂, Br₂, I₂) is produced at the anode.

If there is NO halide (e.g. SO₄²⁻, NO₃⁻ — oxyanions only), then O₂ is produced from the discharge of OH⁻ (water): 4OH⁻ → 2H₂O + O₂ + 4e⁻.

Example 1: Copper(II) sulfate solution, CuSO₄(aq), with inert (graphite) electrodes.

Cu is LESS reactive than H → Cu²⁺ is discharged → Cu deposited at the cathode. Half-eqn: Cu²⁺ + 2e⁻ → Cu.

No halide is present (SO₄²⁻ is an oxyanion) → OH⁻ is discharged → O₂ at the anode. Half-eqn: 4OH⁻ → 2H₂O + O₂ + 4e⁻.

Example 2: Sodium chloride solution (brine), NaCl(aq).

Na is MORE reactive than H → H₂ is discharged → H₂ at the cathode. Half-eqn: 2H⁺ + 2e⁻ → H₂.

Halide (Cl⁻) present in concentrated solution → Cl₂ at the anode. Half-eqn: 2Cl⁻ → Cl₂ + 2e⁻.

Summary table.

Electrolyte Cathode product Anode product
Molten NaCl Na Cl₂
NaCl(aq) (brine) H₂ Cl₂
CuSO₄(aq) (inert) Cu O₂
H₂SO₄(aq) (dilute) H₂ O₂
Why this scores
Edexcel 6-mark mark scheme: 2 for cathode rule (reactivity vs H); 2 for anode rule (halide vs OH⁻); 2 for worked examples with correct half-equations. Most-frequent error in 4CH1 papers: candidates writing 'Na' at the cathode for aqueous NaCl. Memorise the chant 'reactive metal → H₂'.
Question 3
4CH1/2C-style — copper purification6 marks
Explain how impure copper can be purified by electrolysis. State what the cathode and anode are made of, name the electrolyte, write the half-equations, and explain the role of the 'anode sludge'. (6 marks)
Model answer
Purpose. Copper extracted by smelting is typically ~99 % pure but contains valuable impurities (Au, Ag, Pt) and some less-valuable ones (Fe, Zn, Pb). For electrical wiring, ≥ 99.99 % pure copper is needed because even small impurities greatly increase electrical resistance.

Apparatus.

Anode: a block of IMPURE copper (the material to be purified).

Cathode: a thin sheet of PURE copper (it will get thicker as pure Cu deposits on it).

Electrolyte: copper(II) sulfate solution, CuSO₄(aq) (sometimes acidified with dilute H₂SO₄ to improve conductivity).

DC power supply connecting the electrodes.

Half-equations.

At the anode (oxidation): the impure copper anode DISSOLVES — copper atoms lose 2 electrons and enter the solution as Cu²⁺ ions.

$\text{Cu(s)} \rightarrow \text{Cu}^{2+}(aq) + 2e^-$

At the cathode (reduction): Cu²⁺ ions in solution are reduced and deposited as pure copper atoms onto the cathode.

$\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu(s)}$

Net effect. Copper is transferred from the impure anode to the pure cathode. The concentration of CuSO₄(aq) stays approximately constant (dissolving at one electrode balances depositing at the other).

Role of the impurities — 'anode sludge'.

Less reactive than Cu (Ag, Au, Pt): these atoms cannot easily lose electrons, so they are NOT oxidised at the anode. As the copper around them dissolves, they fall to the bottom of the cell as a metallic sludge (the anode sludge or 'anode mud'). These are VERY valuable — silver and gold can be extracted and sold, often offsetting the cost of the electrolysis.

More reactive than Cu (e.g. Zn, Fe): these atoms DO lose electrons at the anode and enter solution as Zn²⁺, Fe²⁺. However, they are NOT reduced at the cathode (because Cu²⁺ is more easily reduced than Zn²⁺ or Fe²⁺) — they remain in solution and are eventually filtered out when the electrolyte is replaced.

Result. Cathode is now 99.99 % pure copper; anode sludge contains recoverable precious metals.
Why this scores
Edexcel 6-mark mark scheme: 1 each for naming anode (impure Cu) and cathode (pure Cu); 1 for electrolyte (CuSO₄(aq)); 1 each for anode and cathode half-equations; 1 for explaining anode sludge (Au, Ag, Pt collect at the bottom). High-frequency Paper 2 topic.
Question 4
4CH1/2C-style — electroplating5 marks
Explain the process of electroplating (e.g. silver-plating a steel spoon). Describe what is used at the cathode, the anode and the electrolyte, and give TWO reasons why objects are electroplated. (5 marks)
Model answer
What is electroplating? Electroplating is using electrolysis to deposit a thin layer of one metal onto the surface of another.

Set-up for silver-plating a steel spoon.

Cathode: the object to be plated — in this case, the steel spoon. It is the negative electrode; positive Ag⁺ ions migrate towards it.

Anode: a piece of the pure plating metal (silver). It is the positive electrode and dissolves into the solution to replenish the Ag⁺ ions.

Electrolyte: a solution containing ions of the plating metal — silver nitrate, AgNO₃(aq), or potassium argentocyanide.

Half-equations.

Anode (oxidation): the silver anode dissolves to give Ag⁺ ions: $\text{Ag} \rightarrow \text{Ag}^+ + e^-$ .

Cathode (reduction): Ag⁺ ions in solution are reduced and deposited as silver metal on the spoon: $\text{Ag}^+ + e^- \rightarrow \text{Ag}$ .

The Ag is transferred from the anode to the cathode (the spoon). After several minutes the spoon has a thin, even coating of silver. The concentration of Ag⁺ in the electrolyte is maintained.

Two reasons for electroplating.

Appearance. Cheap base metals (steel, copper) can be plated with attractive metals (silver, gold, chromium) to look more expensive. Examples: silver-plated cutlery, chrome-plated taps, gold-plated jewellery.

Corrosion resistance. A layer of an unreactive metal protects the underlying metal from oxidising. Examples: tin-plated steel cans (food cans), chromium-plated steel bumpers, zinc-plated nails (called 'galvanised' steel — protects from rusting).

Other reasons (any one accepted): increase electrical conductivity of contacts; improve scratch resistance; reduce friction; thicker coatings for hardness.
Why this scores
Edexcel 5-mark mark scheme: 1 for cathode = object to be plated; 1 for anode = plating metal; 1 for electrolyte containing the plating metal's ions; 1 each for TWO valid reasons. Common error: candidates swap the cathode and anode — remember, the OBJECT is the CATHODE (the one being negatively charged, attracting positive metal ions).

Electrolyte	Cathode product	Anode product
Molten NaCl	Na	Cl₂
NaCl(aq) (brine)	H₂	Cl₂
CuSO₄(aq) (inert)	Cu	O₂
H₂SO₄(aq) (dilute)	H₂	O₂

Key Definitions and Keywords — Electrolysis

Definitions to memorise and the exact keywords mark schemes credit for electrolysis answers — sharpened from recent examiner reports for the 2026 Pearson Edexcel IGCSE 4CH1 sitting.

Electrolysis
Examiner keyword
The decomposition of an ionic compound (the electrolyte — molten or in aqueous solution) into its constituent elements by passing an electric current through it.
Electrolyte
Examiner keyword
A substance that conducts electricity when molten or dissolved in water — and is decomposed in the process. Always an ionic compound (or an ionic solution).
Anode
Examiner keyword
The POSITIVE electrode in electrolysis. ANIONS (negative ions) are attracted to it and are OXIDISED (lose electrons).
Cathode
Examiner keyword
The NEGATIVE electrode in electrolysis. CATIONS (positive ions) are attracted to it and are REDUCED (gain electrons).
Oxidation (in electrolysis)
Examiner keyword
The LOSS of electrons by a species. Occurs at the ANODE. Mnemonic OIL: Oxidation Is Loss.
Reduction (in electrolysis)
Examiner keyword
The GAIN of electrons by a species. Occurs at the CATHODE. Mnemonic RIG: Reduction Is Gain.
Half-equation
Examiner keyword
An equation showing the reaction at a single electrode, including the electrons transferred. Cathode example: Cu²⁺ + 2e⁻ → Cu. Anode example: 2Br⁻ → Br₂ + 2e⁻. Electrons appear on the LEFT for reduction, RIGHT for oxidation.
Inert electrode
Examiner keyword
An electrode that does NOT take part in the reactions of electrolysis — only conducts the current. Examples: graphite (carbon), platinum.
Faraday's constant ( $F$ )
Examiner keyword
The total electric charge carried by one mole of electrons. $F = 96{,}500\ \text{C/mol}$ . Used in Paper 2 quantitative electrolysis calculations.
Electroplating
Examiner keyword
Using electrolysis to coat an object (the cathode) with a thin layer of another metal. The anode is made of the plating metal; the electrolyte contains ions of the plating metal.

Common Mistakes and Misconceptions — Electrolysis

The traps other students keep falling into on electrolysis questions — taken from recent Pearson Edexcel IGCSE 4CH1 examiner reports and mark schemes — and how to avoid them.

✕Writing 'Na' as the product at the cathode for aqueous NaCl
4CH1 Examiner Reports 2022-2024
Why it happens
Forgetting that aqueous solutions also contain H⁺ from water.
How to avoid it
If the metal is MORE reactive than H (K, Na, Ca, Mg, Al), H₂ is produced at the cathode, NOT the metal. Only molten salts give Na metal at the cathode.
✕Omitting electrons from half-equations
4CH1 Examiner Reports 2022-2024
Why it happens
Treating them like full balanced equations.
How to avoid it
ALL half-equations must include electrons. Cathode: electrons on the LEFT (reduction). Anode: electrons on the RIGHT (oxidation). e.g. 2Cl⁻ → Cl₂ + 2e⁻ (not just '2Cl⁻ → Cl₂').
✕Swapping the labels anode and cathode
Why it happens
Memorising the wrong way round.
How to avoid it
Use the mnemonic PANIC: Positive Anode Negative Is Cathode. So anode = positive; cathode = negative. ANIONS go to ANODE; CATIONS go to CATHODE (both start with the same letter).
✕Using minutes (not seconds) in Q = It
4CH1 Examiner Reports 2023-2024
Why it happens
Rushing the substitution.
How to avoid it
Q = I × t needs SECONDS. Always convert: minutes × 60 = seconds; hours × 3600 = seconds. 32 min 10 s = 32 × 60 + 10 = 1930 s.
✕Using a 1:1 ratio of moles of electrons to moles of metal
Why it happens
Ignoring the charge of the metal ion.
How to avoid it
Use the half-equation. For Cu²⁺ + 2e⁻ → Cu, you need 2 mol e⁻ per 1 mol Cu (so divide moles e⁻ by 2). For Al³⁺ + 3e⁻ → Al, divide by 3. For Ag⁺ + e⁻ → Ag, divide by 1.
✕Forgetting that with COPPER (active) electrodes, the products change for CuSO₄
Why it happens
Confusing inert and active electrodes.
How to avoid it
INERT electrodes (C, Pt) in CuSO₄(aq): cathode Cu, anode O₂. ACTIVE copper electrodes in CuSO₄(aq): cathode Cu, anode dissolves as Cu²⁺ (no O₂). This is the basis for copper purification and electroplating.

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

Get a report showing which sub-topics you've nailed and which ones still need work.

Video lesson

Short walkthrough of the concepts students most often get stuck on.

Electrolysis — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Electrolysis

Short Study Notes in the form of Slides

Short Study Notes — Electrolysis

Detailed Notes

What you’ll learn

How it’s examined

1Electrolysis of molten lead(II) bromide (5 marks, Paper 1)

2Electrolysis of sodium chloride solution (brine) (6 marks, Paper 1)

3Calculating mass of copper deposited from Q = It (6 marks, Paper 2)

4Extraction of aluminium by electrolysis (6 marks, Paper 1)

Electric charge (spec 1.83P, Paper 2)

Moles of electrons (Faraday's constant) (Paper 2)

Moles of product from a half-equation

Mass of metal deposited

Volume of gas produced (at rtp)

Electrolysis

Electrolyte

Anode

Cathode

Oxidation (in electrolysis)

Reduction (in electrolysis)

Half-equation

Inert electrode

Faraday's constant (FFF)

Electroplating

✕Writing 'Na' as the product at the cathode for aqueous NaCl

✕Omitting electrons from half-equations

✕Swapping the labels anode and cathode

✕Using minutes (not seconds) in Q = It

✕Using a 1:1 ratio of moles of electrons to moles of metal

✕Forgetting that with COPPER (active) electrodes, the products change for CuSO₄

Practice questions

Past paper style quiz

Assess your understanding

Video lesson

Electrolysis — frequently asked questions

Faraday's constant ( $F$ )