How do I know what is made at each electrode in electrolysis?

Work it in two questions. First, is the substance molten or in solution? A molten binary compound gives the metal at the cathode and the non-metal at the anode, with no water to complicate it. Second, if it is aqueous, the water competes: at the cathode you get the metal only if it is less reactive than hydrogen (copper, silver), otherwise hydrogen; at the anode you get a halogen if a halide is present in reasonable concentration, otherwise oxygen. Those two questions predict almost every 0620 electrode product.

Why do I get hydrogen instead of the metal at the cathode?

Because water is present and the metal is more reactive than hydrogen. In any aqueous solution of a reactive metal (sodium, potassium, calcium, magnesium, aluminium, and the rest above hydrogen in the reactivity series), the hydrogen from water is discharged in preference to the metal, so you collect hydrogen gas at the cathode and the metal ions stay in solution. Only metals below hydrogen (copper, silver, gold) are deposited from solution.

When is the anode product oxygen and when is it a halogen?

If the solution contains a halide ion (chloride, bromide, iodide) at a reasonable concentration, the halogen is produced (chlorine, bromine, iodine). If there is no halide, or the halide is very dilute, oxygen is produced from the hydroxide ions in water instead. This is why concentrated sodium chloride solution gives chlorine at the anode but very dilute sodium chloride gives oxygen: concentration decides it.

How do I write electrolysis half-equations correctly?

Remember which electrode does which job: reduction (gain of electrons) at the cathode, oxidation (loss of electrons) at the anode. Write the ion, add or remove the right number of electrons to balance the charge, and check both sides balance for atoms and charge. Cathode example: Cu2+ + 2e⁻ → Cu. Anode example: 2Cl⁻ → Cl2 + 2e⁻. The commonest lost mark is the wrong number of electrons or putting them on the wrong side.

What is the difference between inert and active electrodes?

Inert electrodes (carbon or platinum) take no part in the reaction; they just carry current, so the products are decided purely by the ions present. Active electrodes take part: in the purification of copper, the impure copper anode dissolves (Cu → Cu2+ + 2e⁻) and pure copper deposits on the cathode. If a question uses copper electrodes in copper sulfate, the anode loses mass and the cathode gains it, which is a different answer from the same solution with carbon electrodes.

Electrolysis: How to Predict Every Electrode Product

IGCSE Chemistry electrolysis exam technique: the rules for predicting cathode and anode products, writing half-equations, and the errors examiners flag.

Published by IGCSEChemistry.com.my

Chemistry teaching team: K. S. Tan (15+ years teaching IGCSE Chemistry) and Ms Yash (10+ years teaching IGCSE Chemistry) and Ms Kartini (15+ years teaching IGCSE Chemistry).

Mapped to Cambridge IGCSE Chemistry 0620 (2026–2028). Last updated 2026-08-04.

Electrolysis loses more marks per question than almost any topic in 0620, and the reason is simple: the products are not fixed. Change the electrolyte, dilute it, or swap the electrodes and the answer changes, so a student who memorised one example gets the next one wrong. The good news is that the products are decided by a small, finite decision tree, not by luck. Learn the tree and you can predict every electrode product the exam can ask. This is the exam technique; the underlying chemistry is taught in full under electrochemistry.

First, fix the two electrodes in your head

Every electrolysis mark depends on not confusing the electrodes, so lock this down before anything else:

Cathode is the negative electrode. Cations (positive ions, including metals and hydrogen) move to it and gain electrons. Gaining electrons is reduction.
Anode is the positive electrode. Anions (negative ions, including halides and hydroxide) move to it and lose electrons. Losing electrons is oxidation.

The phrase that survives exam pressure is OIL RIG: Oxidation Is Loss, Reduction Is Gain. Cathode reduces, anode oxidises. Get this backwards and you lose the half-equation marks and the product marks together.

The decision tree for products

Ask two questions in order, every time.

Question 1: is it molten or aqueous?

A molten binary compound (one metal, one non-metal, nothing else, no water) is the easy case: the metal forms at the cathode, the non-metal forms at the anode. There is no water to compete. Electrolysing molten lead(II) bromide gives lead at the cathode and bromine at the anode, full stop.

If the compound is aqueous (dissolved in water), water is now present and competes at both electrodes. This is where most marks are lost, and it needs Question 2.

Question 2 (aqueous only): what wins at each electrode?

At the cathode, it is a contest between the metal ion and the hydrogen from water:

If the metal is more reactive than hydrogen (potassium, sodium, calcium, magnesium, aluminium, zinc, iron, and anything above hydrogen in the reactivity series), hydrogen gas is produced and the metal stays in solution.
If the metal is less reactive than hydrogen (copper, silver, gold), the metal is deposited.

At the anode, it is a contest between any halide ion and the hydroxide from water:

If a halide (chloride, bromide, iodide) is present at reasonable concentration, the halogen is produced (chlorine, bromine, iodine).
If there is no halide (for example sulfate or nitrate solutions), or the halide is very dilute, oxygen is produced from hydroxide ions.

The whole tree in one table

Electrolyte	Cathode product	Anode product	Why
Molten lead(II) bromide	Lead	Bromine	Molten binary: metal + non-metal
Concentrated sodium chloride (aq)	Hydrogen	Chlorine	Na more reactive than H; halide present
Dilute sodium chloride (aq)	Hydrogen	Oxygen	Halide too dilute, so hydroxide wins
Copper(II) sulfate (aq), carbon electrodes	Copper	Oxygen	Cu less reactive than H; no halide
Copper(II) sulfate (aq), copper electrodes	Copper	Anode dissolves	Active electrode (purification of copper)
Dilute sulfuric acid (aq)	Hydrogen	Oxygen	No metal below H, no halide (electrolysis of water)
Molten aluminium oxide	Aluminium	Oxygen	Extraction of aluminium

If you can reproduce that table from the two questions rather than memory, you can answer any electrode-product question 0620 sets.

Writing the half-equations

Once you know the product, the half-equation is mechanical. Write the ion, balance the charge with electrons on the correct side, then check atoms and charge balance.

Cathode (reduction, electrons on the left): Cu2+ + 2e⁻ → Cu ; 2H+ + 2e⁻ → H2 ; Na+ + e⁻ → Na (molten).
Anode (oxidation, electrons on the right): 2Cl⁻ → Cl2 + 2e⁻ ; 4OH⁻ → O2 + 2H2O + 4e⁻ ; 2Br⁻ → Br2 + 2e⁻.

The two reliable checks: the number of electrons must make the charges balance (Cu2+ needs 2e⁻, not 1), and electrons sit on the left at the cathode and on the right at the anode. The general technique for balancing these is in ionic equations and balancing.

The mistakes examiners flag every series

Examiner reports repeat the same electrolysis errors year after year. Avoid these five and you bank most of the marks:

Predicting the metal at the cathode of a reactive-metal solution. Electrolysing sodium chloride solution gives hydrogen, not sodium. The metal is more reactive than hydrogen, so water wins.
Giving the halogen when the solution is too dilute, or oxygen when it is concentrated. Concentration decides the anode product in halide solutions; read the question for the word “concentrated” or “dilute”.
Wrong number of electrons in the half-equation. Cu2+ + e⁻ → Cu does not balance; it needs 2e⁻.
Electrons on the wrong side. Cathode reduction has electrons as a reactant (left); anode oxidation has them as a product (right). OIL RIG fixes this.
Treating copper electrodes as inert. With copper electrodes in copper sulfate, the anode dissolves and loses mass; this is the purification of copper, a different answer from carbon electrodes. These are exactly the kind of trap catalogued in common exam mistakes.

How examiners want it phrased

Marks are awarded for precise language, not the general idea:

Name the electrode and the process: “at the cathode, copper ions are reduced” scores where “copper appears” does not.
State the product and, where asked, the observation: “bubbles of a pale green gas (chlorine)” beats “gas is made”.
In explanations, link reactivity to the outcome: “hydrogen is discharged because sodium is more reactive than hydrogen”.

Worked exam question

Q (Paper 4 style): Concentrated aqueous sodium chloride is electrolysed using inert carbon electrodes. (a) Name the product at the cathode and write the half-equation. (2) (b) Name the product at the anode and explain why it is not oxygen. (2)

Model answer: (a) Hydrogen (1). 2H+ + 2e⁻ → H2 (1). (b) Chlorine (1). The chloride ions are present in high concentration, so the halide is discharged in preference to hydroxide, giving chlorine rather than oxygen (1).

Mark-scheme logic: (a) sodium is more reactive than hydrogen, so hydrogen is discharged from water; the half-equation needs two electrons on the left and H2, not H. (b) the mark for “chlorine” is the product; the second mark is specifically for linking the answer to the high chloride concentration, the exact reasoning the scheme rewards and the exact thing a memorised “anode gives oxygen” answer misses.

The Malaysia note

In Malaysian international-school cohorts, electrolysis tends to be taught near the end of the inorganic course and then under-practised, so students meet it again only in the mock with the rules half-forgotten. Because the topic is rule-based rather than memory-heavy, it responds faster than almost any other to focused work: a single session spent building the two-question decision tree and drilling half-equations usually moves a student from guessing electrode products to predicting them. If electrolysis is one of the topics quietly costing your child marks, a trial lesson can diagnose exactly where the rule is breaking and fix it in an hour.