Electrolysis

What this lesson covers in the specification

Incandio is aligned to this specification. It is not published by, endorsed by or affiliated with Pearson, and it reproduces none of Pearson's wording — the statement numbers are given so you can check every lesson against your own copy.

• 1.58 — Electrolysis of molten compounds and aqueous solutions with inert electrodes, and predicting the products (bold C statement — Paper 2 only)
• 1.59 — Writing ionic half-equations for the electrode reactions, and classifying them as oxidation or reduction (bold C statement — Paper 2 only)
• 1.60 — Practical: investigate the electrolysis of aqueous solutions (bold C statement — Paper 2 only) (required practical)

1 · Understand it

No exam language yet. The only question this section answers is: do I actually understand what is happening?

ELECTROLYSIS is the decomposition of an ionic compound by passing an electric current through it, when the compound is molten or dissolved in water. The substance being broken down is the ELECTROLYTE, and the current enters and leaves through two ELECTRODES.

The compound must be molten or dissolved for the reason the last page established: the ions have to be free to move. In a solid nothing can travel, so nothing happens. Once they are mobile, the electrodes do the sorting: the negative CATHODE attracts the positive CATIONS and the positive ANODE attracts the negative ANIONS.

What happens when an ion reaches its electrode

1. A CATION reaching the negative cathode is attracted to it because the cathode is rich in electrons.
2. It GAINS electrons there, which cancels its positive charge and turns it back into a neutral atom. Gaining electrons is REDUCTION.
3. An ANION reaching the positive anode is attracted to it because the anode is short of electrons.
4. It LOSES its extra electrons there, which cancels its negative charge and turns it back into a neutral atom. Losing electrons is OXIDATION.
5. So reduction always happens at the cathode and oxidation always happens at the anode, in every electrolysis without exception.

Think of it like a lost-property office running in both directions

Some people arrive having lost something and are given it back; others arrive carrying something that is not theirs and hand it over. Both counters are busy, and what happens at each is fixed by which kind of person turns up. Electrolysis is exactly this with electrons. Positive ions arrive at the cathode having lost electrons and are handed them back; negative ions arrive at the anode carrying extra ones and give them up. Neither counter chooses — the ion's charge decides where it goes, and where it goes decides whether it gains or loses. That is why reduction is always at the cathode and never a matter of which compound you happen to be using.

The reactions are written as IONIC HALF-EQUATIONS, statement 1.59, which show what happens at one electrode with the electrons written in. At the cathode the electrons appear on the LEFT, because they are being gained. At the anode they appear on the RIGHT, because they are being lost.

Molten lead(II) bromide, both electrodes

Molten lead(II) bromide, PbBr₂, is electrolysed with inert electrodes. Give the products and the half-equation at each electrode, and classify each as oxidation or reduction.

1. The ions present are Pb²⁺ and Br⁻.
2. CATHODE (negative): Pb²⁺ is attracted here and gains 2 electrons. Pb²⁺ + 2e⁻ → Pb. Molten lead forms. Gaining electrons is REDUCTION.
3. ANODE (positive): Br⁻ is attracted here and loses its extra electron. Two are needed because bromine is diatomic: 2Br⁻ → Br₂ + 2e⁻. Brown bromine vapour forms. Losing electrons is OXIDATION.
4. Check the balance: charges must balance as well as atoms. On the anode side, 2 × (1−) = 2− on the left, and 0 + 2 × (1−) = 2− on the right. Balanced.
5. Check the electrons: two gained at the cathode and two lost at the anode, which must always match.

Answer: Cathode: Pb²⁺ + 2e⁻ → Pb, reduction. Anode: 2Br⁻ → Br₂ + 2e⁻, oxidation.

Electrolysing a SOLUTION is harder, and the reason is that water is present and water itself supplies a small number of H⁺ and OH⁻ ions. So there are now two candidates at each electrode, and a rule is needed to decide which one reacts.

The rules for an aqueous solution

1. AT THE CATHODE, the choice is between the metal ion and H⁺. If the metal is MORE REACTIVE than hydrogen, hydrogen gas is produced instead of the metal. If the metal is LESS reactive than hydrogen, the metal is produced.
2. So electrolysing copper sulfate solution deposits copper, but electrolysing sodium chloride solution gives hydrogen rather than sodium.
3. AT THE ANODE, if a HALIDE ion is present — chloride, bromide or iodide — that halogen is produced.
4. If no halide is present, OXYGEN is produced instead, from the hydroxide ions supplied by the water.
5. So sodium chloride solution gives chlorine at the anode, while copper sulfate solution gives oxygen, since sulfate is not a halide.

The word INERT in statement 1.58 matters and is easy to skim past. Inert electrodes, usually graphite or platinum, take no part in the reaction — they simply deliver and remove electrons. If the electrodes are made of something reactive, such as copper, the anode itself dissolves instead, and the products are entirely different. Every prediction on this page assumes inert electrodes.

Copper sulfate solution, applying both rules

Predict the products of electrolysing copper(II) sulfate solution with inert electrodes, and give both half-equations.

1. The ions present are Cu²⁺ and SO₄²⁻ from the salt, and H⁺ and OH⁻ from the water.
2. CATHODE: the choice is Cu²⁺ or H⁺. Copper is LESS reactive than hydrogen, so copper is deposited: Cu²⁺ + 2e⁻ → Cu. Reduction.
3. ANODE: the choice is SO₄²⁻ or OH⁻. Sulfate is not a halide, so oxygen is produced from the hydroxide ions.
4. The half-equation is 4OH⁻ → O₂ + 2H₂O + 4e⁻. Oxidation.
5. Observation: a pink-brown coating of copper forms on the cathode and bubbles of gas appear at the anode.

Answer: Copper at the cathode, oxygen at the anode — because copper is below hydrogen and sulfate is not a halide.

2 · Grade 9 Notes

A different job from the section above. You have already understood it; this is the precise set of things to LEARN — definitions to reproduce word for word, processes in order, equations with units, and the answers that score full marks.

Learn this definition · Electrolysis

The decomposition of an ionic compound by passing an electric current through it when molten or in aqueous solution, so that its ions are free to move to the electrodes.

The vocabulary of the cell

Electrolyte

The molten or dissolved ionic compound being decomposed.

Cathode

The NEGATIVE electrode. Cations go there, GAIN electrons, and are REDUCED.

Anode

The POSITIVE electrode. Anions go there, LOSE electrons, and are OXIDISED.

Inert electrode

Graphite or platinum — takes no part in the reaction, only delivering and removing electrons.

Writing an ionic half-equation

1. Identify which ion is attracted to that electrode: cations to the cathode, anions to the anode.
2. At the CATHODE, put the electrons on the LEFT — they are being gained.
3. At the ANODE, put the electrons on the RIGHT — they are being lost.
4. Use the number of electrons that cancels the ion's charge exactly.
5. Balance the atoms, remembering that hydrogen, oxygen and the halogens form DIATOMIC molecules, and check the charges balance too.

Statement 1.58 — predicting the products of an aqueous solution

• CATHODE — metal MORE reactive than hydrogen → HYDROGEN is produced
• CATHODE — metal LESS reactive than hydrogen → THE METAL is produced
• ANODE — a HALIDE ion present (Cl⁻, Br⁻, I⁻) → that HALOGEN is produced
• ANODE — no halide present → OXYGEN is produced from the hydroxide ions
• All of this assumes INERT electrodes; a reactive anode dissolves instead

Pb²⁺ + 2e⁻ → Pb

Conditions: At the cathode in molten lead(II) bromide. Electrons GAINED, so this is REDUCTION.

2Br⁻ → Br₂ + 2e⁻

Conditions: At the anode in molten lead(II) bromide. Electrons LOST, so this is OXIDATION. Bromine is diatomic, so two ions are needed.

Required practical 1.60 — electrolysis of aqueous solutions

1. Half-fill a small beaker with the solution to be electrolysed, such as copper(II) sulfate or sodium chloride solution.
2. Clamp two clean graphite electrodes in the solution so that they do not touch each other.
3. Connect the electrodes to a d.c. power supply, noting which is joined to the negative terminal — that one is the cathode.
4. If a gas is expected, invert a small water-filled test tube over each electrode to collect it.
5. Switch on the supply and observe both electrodes, recording any coating, colour change or bubbling.
6. Test any gas collected: a lit splint pops for hydrogen, a glowing splint relights in oxygen, and damp litmus paper is bleached white by chlorine.
7. Switch off, and repeat with the other solutions so that the products can be compared.

Variables

Independent (changed) — The solution being electrolysed
Dependent (measured) — The products formed at the cathode and at the anode

Sources of error

Model answer [4 marks]

Molten zinc chloride is electrolysed with inert electrodes. Give the half-equation at each electrode and state which is oxidation. [4]

At the cathode, zinc ions gain electrons: Zn²⁺ + 2e⁻ → Zn, and molten zinc is formed. At the anode, chloride ions lose electrons: 2Cl⁻ → Cl₂ + 2e⁻, and chlorine gas is formed. The reaction at the anode is oxidation, because the chloride ions lose electrons there. The reaction at the cathode is reduction, because the zinc ions gain electrons.

Model answer [4 marks]

Predict the products of electrolysing sodium chloride solution with inert electrodes, and explain your reasoning. [4]

Hydrogen is produced at the cathode and chlorine at the anode. At the cathode the choice is between sodium ions and the hydrogen ions supplied by the water, and because sodium is more reactive than hydrogen it stays in solution and hydrogen gas is produced instead. At the anode the choice is between chloride ions and hydroxide ions, and because a halide ion is present the halogen is produced, so chlorine gas is formed rather than oxygen.

Not this: The cathode is positive, because it attracts negative ions.

This: The CATHODE IS NEGATIVE and attracts positive CATIONS. The anode is positive and attracts anions. Each electrode attracts the ion of the OPPOSITE charge, which is why the names are worth learning as a pair.

Mark-losing trap. REDUCTION at the CATHODE, OXIDATION at the ANODE — always, in every electrolysis.

Mark-losing trap. Cathode half-equations have electrons on the LEFT; anode half-equations on the RIGHT.

Mark-losing trap. In solution, a metal MORE reactive than hydrogen gives HYDROGEN at the cathode, not the metal.

Mark-losing trap. Halogens, hydrogen and oxygen are DIATOMIC — the half-equation needs two ions, not one.

3 · Prove it — the five questions

The five questions climb Grade 6 → Grade 7 → Grade 8 → Grade 9 → Grade 9 challenge, and are marked inside Incandio on your own device, by rule, with an authored diagnosis of the mistake you actually made. The mark schemes stay in the app so that the practice is worth doing; the questions themselves are here.

1. Grade 6 · State [1 mark] — During electrolysis, which electrode do positive ions move towards?
2. Grade 7 · Write [2 marks] — Write the ionic half-equation for the reaction at the cathode when molten magnesium chloride is electrolysed.
3. Grade 8 · Predict [3 marks] — Predict the product at the cathode when sodium chloride solution is electrolysed with inert electrodes, and explain why.
4. Grade 9 · Explain [6 marks] — Select every statement that belongs in a full-mark explanation of what happens at each electrode when molten lead(II) bromide is electrolysed with inert electrodes.
5. 9+ · Analyse [6 marks] — A student electrolyses copper(II) sulfate solution and correctly obtains copper at the cathode. They then electrolyse sodium sulfate solution expecting sodium at the cathode, and instead obtain a gas that pops with a lit splint. They conclude the experiment failed. Select every statement that belongs in a full-mark analysis.

The people behind this science

Two ways into the same idea — the one who made electrolysis quantitative, and the one who did it first and could not yet explain it. Inside Incandio each of them answers knowing exactly which lesson you have just finished.

Michael Faraday — the one who made electrolysis quantitative

This page names the process and its parts, all of which Faraday defined, but his deepest result is one the specification only implies. He established that the mass of a substance released at an electrode is proportional to the quantity of electricity passed, and that the amounts for different substances stand in simple whole-number ratios related to their charges. That is a very strong hint that electricity itself comes in fixed indivisible units — an inference he was careful about and which was confirmed sixty years later when the electron was identified. He is the right person to ask how a careful measurement can point beyond what its author is willing to claim.

• “How much electricity does it take to release a given mass?”
• “Why do different substances need different amounts?”
• “Did you think electricity came in fixed units?”
• “How did you measure the quantity of charge that had passed?”
• “Why were you so cautious about what your results implied?”

Humphry Davy — the one who did it first and could not yet explain it

Davy is the first person to have performed the reaction on this page to any real purpose, and he did it without any of the vocabulary or theory it now comes with. He had no ions, no electrons, no half-equations — only a conviction that electricity was the force holding compounds together, and the largest battery in the world with which to test it. Six elements came out of that conviction in about eighteen months. He is worth asking about the gap between doing something and understanding it, particularly since the person who closed that gap was his own assistant.

• “What did you expect to appear at the electrodes?”
• “Why did you believe electricity would break the compound apart?”
• “How did you know the metal was a new element?”
• “What could you not explain about what you had done?”
• “How did it feel when Faraday explained your own experiments?”

Then defend it

On Incandio a lesson is not finished when the questions come out right. You teach the idea back to Ember, an AI apprentice who asks the awkward question, and then you argue it against Michael Faraday in a structured debate marked against descriptors you can read before you enter. Learn it, teach it, then defend it — all three happen on this page once the app loads.

Carry on through the course

• Previous lesson: Why Substances Conduct — the Four Structures Compared
• Next lesson: Group 1 — the Alkali Metals
• The Three States of Matter — What the particles are doing in a solid, a liquid and a gas — and what really changes when ice melts.
• Pure Substances, Mixtures and How to Separate Them — Why a pure substance melts at one temperature, and which separation technique to reach for when.
• Inside the Atom — Protons, neutrons and electrons — where they are, what they weigh, what they carry, and how isotopes give an element a decimal mass.
• Electronic Configurations and the Periodic Table — Why the table has the shape it has: the group number IS the outer-shell electron count, and everything follows from that.
• Solubility and Solubility Curves — Why a solution can be full, why hot water holds more, and how one curve tells you exactly how many crystals you will get back.
• All of Chemistry · Incandio Science