OCR GCSE Chemistry A: Electrolysis (J248) – Revision Notes

These notes boil down C3.4 Electrolysis from the J248 specification for OCR GCSE (9–1) Chemistry A (Gateway Science), Version 4.0 (August 2026): statements C3.4a–C3.4e, mathematical learning outcome CM3.4i and the PAG C2 practical suggestions. Both tiers are assessed on every C3.4 statement. For full explanations and five worked examples, use the Electrolysis study guide.

For exam-style problems with full answers, move on to the Electrolysis practice questions. Other J248 topics are on the J248 course hub, and you can record progress on the printable checklist. Unsure where to focus? Take a free 10-minute diagnostic.

Core definitions

Term	Meaning
electrolysis	decomposition of a liquid (molten or aqueous) by passing a direct current through it
electrolyte	ionic substance, melted or in solution, that carries the current and is broken down
cathode	negative electrode; attracts cations
anode	positive electrode; attracts anions
cation	positive ion (metal ions, H⁺)
anion	negative ion (Cl⁻, Br⁻, I⁻, O²⁻, OH⁻, SO₄²⁻, NO₃⁻)
inert electrode	conducts but does not react (graphite, platinum)
non-inert electrode	takes part in the electrode reaction (for example copper)

Why melting or dissolving matters. In the solid, ions are locked in the lattice and cannot move. Melted or dissolved, the ions move, so the liquid conducts. Ions carry the charge through the electrolyte; electrons move only in the wires.

C3.4a: the one-line rule

With inert electrodes: metal or hydrogen at the cathode, non-metal at the anode. Cations go to the cathode and gain electrons. Anions go to the anode and lose electrons. (Calling these reduction and oxidation in terms of electrons is C3.3b, Higher Tier only.)

C3.4b: molten binary compounds

Only two ions are present, so there is no competition.

Method in steps

1. Write the ions in the compound.
2. Cation → cathode → metal.
3. Anion → anode → non-metal, written as a diatomic molecule for halogens and oxygen.

Quick reminders:

• molten NaCl → sodium at the cathode, chlorine at the anode
• molten KI → potassium and iodine
• molten ZnCl₂ → zinc and chlorine

C3.4c: aqueous solutions

Water adds H⁺ and OH⁻ to the solution, so ions compete at each electrode.

Electrode	Rule (inert electrodes)
cathode	metal forms only if it is less reactive than hydrogen (Cu, Ag); otherwise hydrogen forms
anode	halogen forms if Cl⁻, Br⁻ or I⁻ is present; otherwise oxygen forms from OH⁻

The two specification examples

Solution	Cathode	Anode	What is left in solution
NaCl(aq)	hydrogen	chlorine (some oxygen if very dilute)	sodium hydroxide (alkaline)
CuSO₄(aq)	copper	oxygen	sulfuric acid (acidic); blue colour fades

Observations to quote

• Copper: pink-brown solid coating the cathode.
• Hydrogen, oxygen, chlorine: bubbles. Chlorine is pale green and smells of bleach.
• Bromine: orange-brown colour near the anode. Iodine: brown colour.

C3.4d: half equations

Product	Half equation	Electrode
a metal M with a 2+ ion	M²⁺ + 2e⁻ → M	cathode
a Group 1 metal	Li⁺ + e⁻ → Li (and similar)	cathode
hydrogen	2H⁺ + 2e⁻ → H₂	cathode
a halogen	2Br⁻ → Br₂ + 2e⁻	anode
oxygen (aqueous)	4OH⁻ → O₂ + 2H₂O + 4e⁻	anode
oxygen (molten oxide)	2O²⁻ → O₂ + 4e⁻	anode
copper anode dissolving	Cu → Cu²⁺ + 2e⁻	anode

Method in steps for any half equation

1. Start with the ion as reactant and the element as product.
2. Balance atoms (two halide ions make one halogen molecule).
3. Add electrons to make the total charge equal on both sides: on the left at the cathode, on the right at the anode.

Checking a half equation: charge on the left must equal charge on the right. For 4OH⁻ → O₂ + 2H₂O + 4e⁻, the left is 4−, the right is 4−.

CM3.4i: ratio and empirical formula

Gas ratio. Match electron numbers before comparing products. Hydrogen needs 2e⁻ per molecule; oxygen releases 4e⁻ per molecule. So hydrogen and oxygen form in a 2 : 1 volume ratio. Hydrogen and chlorine (each 2e⁻) form in a 1 : 1 ratio.

Empirical formula from electrode products.

1. Divide each mass by its Ar.
2. Divide each answer by the smaller one.
3. Write the whole-number ratio as the formula.

Worked reminder: a molten bromide gives 0.48 g of magnesium (Ar 24) and 3.20 g of bromine (Ar 80). Ratios: 0.48 ÷ 24 = 0.02 and 3.20 ÷ 80 = 0.04, giving 1 : 2, so MgBr₂.

C3.4e and PAG C2: practical summary

Inert electrodes (aqueous): graphite rods in a beaker of solution, d.c. power supply, inverted test tubes of solution over each electrode to collect gas. Molten compounds are heated in a crucible; use a fume cupboard for chlorine or bromine.

Gas tests

Gas	Test	Result
hydrogen	burning splint held at the mouth	squeaky pop
oxygen	glowing splint pushed in	reignites
chlorine	moist blue litmus paper	goes red, then bleached

Non-inert copper electrodes in CuSO₄(aq):

• anode dissolves (loses mass): Cu → Cu²⁺ + 2e⁻
• cathode gains copper: Cu²⁺ + 2e⁻ → Cu
• colour stays the same because Cu²⁺ ions are replaced as fast as they are removed
• mass lost by the anode equals mass gained by the cathode

Good technique: rinse electrodes with distilled water, dry before weighing, handle the cathode gently so loose copper does not fall off.

Model explanations to adapt

Short, linked sentences gain more credit than a list of facts. Use these patterns and change the names.

Why does a molten salt conduct? “In the solid salt, every ion is locked in place in the lattice. Once the salt melts, the ions are free to move, so they can carry charge to the electrodes.”

Why does hydrogen form instead of the metal? “The solution contains potassium ions and hydrogen ions from water. Potassium is more reactive than hydrogen, so the hydrogen ions gain electrons at the cathode and hydrogen gas is produced. The potassium ions stay in solution.”

Why does oxygen form at the anode? “There are no halide ions in the solution, so hydroxide ions from water lose electrons at the anode and oxygen is formed. The nitrate (or sulfate) ions stay in solution.”

Why does the anode lose mass with copper electrodes? “Copper atoms in the anode lose electrons and enter the solution as Cu²⁺ ions, so the anode gets lighter. The same number of Cu²⁺ ions gain electrons at the cathode, so the cathode gains the same mass.”

Explaining a gas volume ratio. “Each hydrogen molecule needs two electrons and each oxygen molecule releases four, so twice as many hydrogen molecules form as oxygen molecules, and the hydrogen volume is double the oxygen volume.”

Must-know distinctions

• Molten vs aqueous: molten has one possible product per electrode; aqueous adds H⁺ and OH⁻ and the rules decide.
• Inert vs non-inert anode in CuSO₄: graphite gives oxygen and a fading colour; copper dissolves and the colour stays.
• Cathode vs anode: cathode negative (cations, metals or hydrogen); anode positive (anions, non-metals).
• Ions vs electrons: ions move through the liquid; electrons move through the wires.
• Chlorine vs chloride: chloride ions (Cl⁻) are in the electrolyte; chlorine gas (Cl₂) is the product.

Quick self-test

1. Name the electrode where non-metals form with inert electrodes.
2. What is an anion?
3. Give the products of electrolysing molten lithium chloride.
4. Why does solid potassium bromide not conduct?
5. Predict the cathode product for aqueous silver nitrate.
6. Predict both products for aqueous potassium sulfate with inert electrodes.
7. Write the half equation for the formation of iodine.
8. Complete and balance: OH⁻ → O₂ + H₂O + e⁻.
9. In electrolysis of a sulfate solution, 64 cm³ of hydrogen is collected. What volume of oxygen would you expect?
10. A copper anode goes from 18.40 g to 17.92 g in copper sulfate solution. What mass is gained by the copper cathode?
11. Why does the blue colour fade when copper sulfate is electrolysed with graphite electrodes?

Answers

1. The anode (positive electrode).
2. A negative ion.
3. Lithium at the cathode, chlorine at the anode.
4. Its ions are held in the lattice and cannot move.
5. Silver, because silver is less reactive than hydrogen.
6. Hydrogen at the cathode (potassium is more reactive than hydrogen); oxygen at the anode (no halide present).
7. 2I⁻ → I₂ + 2e⁻
8. 4OH⁻ → O₂ + 2H₂O + 4e⁻
9. 64 ÷ 2 = 32 cm³ (2 : 1 ratio).
10. 18.40 − 17.92 = 0.48 g.
11. Cu²⁺ ions are removed as copper at the cathode and are not replaced.

Where marks are usually lost

• Writing that electrons flow through the electrolyte.
• Saying an ionic solid does not conduct because electrons cannot move, instead of ions.
• Giving sodium or potassium as the cathode product of an aqueous solution.
• Naming sulfur, sulfur dioxide or “sulfate” as the anode product for a sulfate solution.
• Writing Cl, Br or O instead of Cl₂, Br₂ or O₂.
• Electrons written on the wrong side, or charge totals that do not add up.
• Comparing gas volumes without matching electron numbers first (hydrogen to oxygen is 2 : 1, not 1 : 1).
• Saying the colour fades with copper electrodes. It stays the same.
• Describing the chlorine test without saying the litmus paper is damp.

Official syllabus

Source: OCR’s specification for GCSE (9–1) Chemistry A (Gateway Science), J248, Version 4.0 (August 2026), for first assessment in 2018. Content condensed here: C3.4 Electrolysis (statements C3.4a–C3.4e), mathematical learning outcome CM3.4i and the PAG C2 practical suggestions.