C4 is a big, varied topic, but most of it sorts into three areas: the reactivity series, the chemistry of acids, and electrolysis. They look quite different on the page, yet underneath each one the same thing is happening. Atoms are being rearranged into new substances, whether a metal is pulled from its ore, an acid is neutralised by a base, or a compound is split apart by electricity.
Oxidation and reduction run through the whole topic, so it’s worth getting them clear early. Learn them first in terms of oxygen gained or lost, then, if you’re on Higher Tier, in terms of electrons.
C4 is a long topic and not all of it is for everyone — check what your tier needs:
- Everyone — the reactivity series and reactions with water and acids, displacement, extracting metals with carbon, acids with metals/bases/carbonates, making soluble salts, the pH scale and neutralisation, and the products of electrolysis (including the extraction of aluminium).
- Higher Tier H — oxidation and reduction in terms of electrons, ionic and half equations (for displacement, acid–metal reactions and electrolysis), and strong vs weak acids.
- Triple (Chemistry only) T — titrations (the method is Triple; the calculations are also Higher Tier).
1Metal Oxides & Oxidation
Most metals react with oxygen to form a metal oxide. How quickly and how vigorously they do this is our first clue to how reactive a metal is — magnesium burns with a brilliant white flame, iron slowly rusts, and gold never tarnishes at all.
metal + oxygen → metal oxide
For example, magnesium burns in air to make magnesium oxide:
2Mg + O2 → 2MgO
Oxidation is the gain of oxygen. Reduction is the loss of oxygen.
When a metal reacts with oxygen it gains oxygen, so the metal is oxidised. Forming a metal oxide is always an oxidation reaction.
A handy way to remember which way round it goes:
- A substance that gains oxygen has been OXIDISED.
- A substance that loses oxygen has been REDUCED.
So when copper is heated in air it is oxidised to black copper oxide:
2Cu + O2 → 2CuO
This oxygen definition is all that Foundation Tier needs. Higher Tier students also learn to describe oxidation and reduction in terms of electrons — that comes up in Displacement Reactions and again in electrolysis.
Discovery activity
How did chemists decide which metal is most reactive?
Below are eight key metals used to build the experimental order, displayed in a random order. This is the challenge that faced early chemists — by looking at them you cannot tell which is most reactive. How would you find out?
You need to do experiments and compare how vigorously each metal reacts. The more vigorous the reaction, the more reactive the metal.
Two experiments give us the evidence we need. The length of each bar shows how vigorously the metal reacts — the longer the bar, the more reactive.
Experiment 1 — reactions with cold water
Experiment 2 — reactions with dilute acid
K, Na, Li and Ca are too dangerously reactive to test with acid — water already tells us their order.
Water separates the most reactive metals; dilute acid separates the middle ones. Together, the two experiments give us enough information to rank all of them.
Combining the evidence from both experiments — and adding the non-metals carbon and hydrogen as reference points — lets us build the reactivity series:
A few more metals — aluminium, silver and gold — slot into the full series too. See the complete diagram below.
2The Reactivity Series
The reactivity series is a league table of metals, listed in order from the most reactive at the top to the least reactive at the bottom. We build it by comparing how metals react with water and with dilute acid.
When a metal reacts, its atoms lose electrons to form positive ions. The reactivity of a metal is its tendency to form positive ions — the more easily a metal loses its outer electrons, the more reactive it is.
Reactions with water
The most reactive metals react with cold water to give a metal hydroxide (an alkaline solution) and hydrogen gas:
metal + water → metal hydroxide + hydrogen
For example, sodium fizzes across the surface of water:
2Na + 2H2O → 2NaOH + H2
Potassium, sodium and lithium (the alkali metals) and calcium all react with cold water. How vigorously they fizz tells us their order: potassium reacts most violently, then sodium, then lithium, then calcium more gently. Less reactive metals react only very slowly — or not at all — which makes water a poor test for telling them apart.
At GCSE these reactions are limited to room temperature: you are not expected to know the reactions of metals with steam.
Pick a metal to drop onto the water and watch how vigorously it reacts.
Reactions with dilute acid
To compare the metals that barely react with water, we use dilute acid instead. Metals more reactive than hydrogen react with acid to give a salt and hydrogen, and the rate of fizzing (effervescence of hydrogen) shows how reactive each one is:
metal + acid → salt + hydrogen
Magnesium fizzes rapidly, zinc steadily, iron slowly, and copper not at all — placing them in the order Mg > Zn > Fe > Cu.
Each reactive metal gives off hydrogen, which fizzes off as bubbles — the faster the fizzing, the more reactive the metal (Mg > Zn > Fe > Cu). Copper lies below hydrogen in the reactivity series, so it does not react with dilute acid at all.
The reactivity series. The non-metals carbon and hydrogen are included as reference points — their positions decide how a metal can be extracted and whether it reacts with acid.
Carbon and hydrogen are non-metals, but they are placed in the series as reference points. A metal’s position relative to carbon decides how it is extracted (see §4), and its position relative to hydrogen decides whether it reacts with acids (see §5).
🧪 Exam-style questions
The reactivity of a metal is best described as its tendency to…
A student adds four metals to dilute hydrochloric acid. Metal W fizzes very fast, X fizzes slowly, Y fizzes steadily and Z does not react. What is the order of reactivity, most reactive first?
3Displacement Reactions
The core idea here is for everyone; the ionic and half equations at the end are Higher Tier only.
A displacement reaction is one in which a more reactive metal displaces a less reactive metal from a compound (usually a solution of its salt).
If you put a more reactive metal into a solution of a less reactive metal’s salt, the more reactive metal “pushes out” the less reactive one and takes its place. For example, magnesium is more reactive than copper, so it displaces copper from copper sulfate solution:
magnesium + copper sulfate → magnesium sulfate + copper
Mg + CuSO4 → MgSO4 + Cu
You can see it happen: the blue colour of the copper sulfate solution fades, and a coating of brown copper forms on the magnesium. If the added metal is less reactive than the one in the compound, nothing happens — silver cannot displace copper, for instance.
Use the reactivity series: the reaction only happens if the metal you add is higher than the metal in the compound. Displacement is the basis of the whole reactivity series — ranking metals by which can displace which gives exactly the same order as ranking them by their reactions with water and acid.
🧪 Test a displacement reaction
Pick a salt solution, then a metal to drop in. Both metals light up in the reactivity series so you can compare them — then see whether a reaction happens.
Ionic equations and redox Higher
In a displacement reaction, electrons are transferred from one metal to the other. One metal is oxidised (loss of electrons) while the other is reduced (gain of electrons), so it is a redox reaction (reduction and oxidation happening in one reaction). Higher Tier defines these in terms of electrons:
Oxidation Is Loss of electrons · Reduction Is Gain of electrons.
Let’s build the ionic equation for magnesium displacing copper from copper(II) sulfate solution, one step at a time.
Step 1 — Write the normal balanced equation. Magnesium is more reactive than copper, so it takes copper’s place in the compound:
Mg(s) + CuSO4(aq) → MgSO4(aq) + Cu(s)
Step 2 — Split the dissolved ionic compounds into their ions. Any ionic compound that is dissolved in water (aq) has had its lattice broken up, so its ions are pulled apart and float about separately in the solution. We rewrite each aqueous ionic compound as its free ions. (The magnesium and copper metals are solid elements, not in solution, so they stay exactly as they are.)
Mg(s) + Cu2+(aq) + SO42−(aq) → Mg2+(aq) + SO42−(aq) + Cu(s)
Step 3 — Cancel the spectator ions. The sulfate ion (SO42−) is identical on both sides — it starts and ends as the same free aqueous ion and takes no part in the reaction. An ion like this is called a spectator ion, and we cancel it from both sides:
Mg(s) + Cu2+(aq) + SO42−(aq) → Mg2+(aq) + SO42−(aq) + Cu(s)
What is left is the ionic equation — it shows only the species that actually change:
Mg(s) + Cu2+(aq) → Mg2+(aq) + Cu(s)
Half equations — oxidation and reduction separately Higher
A half equation goes one step further: it shows the oxidation or reduction of just one element on its own, including the electrons that are transferred. We can split the ionic equation above into its two halves.
Start with “what becomes what”. Magnesium atoms turn into magnesium ions:
Mg → Mg2+
The charge of a simple ion comes from its group number (see C2 Ionic Bonding). Metals lose electrons to form positive ions, and the number of electrons lost equals the group number — so a Group 2 metal like magnesium forms a 2+ ion. Transition metals don’t follow this rule; their charge is given to you in roman numerals — copper(II) means Cu2+.
Now finish the half equation with a two-point checklist:
- Balance the atoms. One Mg on each side — already balanced.
- Balance the charge using electrons. The left side has a total charge of 0; the right side is 2+. Each electron is 1−, so add electrons to the more positive side until both sides match — here, 2 electrons on the right:
Mg → Mg2+ + 2e− (oxidation — magnesium loses electrons)
The copper does the opposite — copper ions gain those electrons to become copper atoms:
Cu2+ + 2e− → Cu (reduction — copper ions gain electrons)
The more reactive metal is always oxidised (it loses electrons to become an ion); the less reactive metal ion is always reduced (it gains electrons to become an atom). The electrons lost by one are exactly the electrons gained by the other.
Marks are lost on ionic and half equations even when the chemistry is understood. Watch for: impossible species such as Zn2− (a metal forms a positive ion, Zn2+); writing full formulae instead of separate ions (use Cu2+, not CuSO4); a charge that does not balance on the two sides; and missing or incorrect state symbols. For zinc displacing copper the full mark answer is Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s). Check in order: species → balancing → charge → state symbols.
🧪 Build a half equation
Pick a displacement reaction and one of its metals, then build that metal’s half equation: choose which side the atom and the ion go on, and add electrons until the atoms and charges balance.
4Extracting Metals: Reduction with Carbon
A few very unreactive metals — gold, silver and platinum — are found in the Earth as the metal itself (“native” metals), because they are so unreactive they have stayed uncombined for billions of years. Most metals, though, are found locked inside compounds in rocks called ores, and have to be chemically “pulled apart” to get the metal out.
An ore is a rock that contains enough of a metal (or its compound) to make it economical to extract.
Most metals are found as compounds inside ores. Crushing and concentrating the ore is physical — the metal is only freed by a chemical reduction step, using carbon or electrolysis.
Many metals occur as oxides. To get the metal, the oxygen has to be removed — in other words the oxide must be reduced. How we do this depends on the metal’s position relative to carbon in the reactivity series.
A metal less reactive than carbon can be extracted from its oxide by reduction with carbon. The carbon displaces the metal because it is more reactive, taking the oxygen for itself.
A metal more reactive than carbon (aluminium, magnesium, calcium, sodium, potassium) cannot — it has to be extracted by electrolysis instead.
The same reactivity series, regrouped by extraction method. A metal’s position relative to carbon decides how it is extracted: more reactive → electrolysis; less reactive → reduction with carbon; unreactive → found native as the element.
For example, iron oxide is heated with carbon in a blast furnace. The carbon gains the oxygen (it is oxidised) and the iron loses it (it is reduced):
2Fe2O3 + 3C → 4Fe + 3CO2
The same happens with copper oxide and lead oxide:
2CuO + C → 2Cu + CO2
2PbO + C → 2Pb + CO2
| In a reduction-with-carbon reaction… | Gains / loses oxygen | Oxidised or reduced? |
|---|---|---|
| The metal oxide | loses oxygen | reduced |
| The carbon | gains oxygen | oxidised |
🧪 Exam-style questions
Which of these metals cannot be extracted by heating its oxide with carbon?
In the reaction 2CuO + C → 2Cu + CO2, name the substance that is reduced and explain your choice.
Show answer
- Copper oxide is reduced. 1 mark
- It has lost oxygen (to the carbon), and loss of oxygen is reduction. 1 mark
5Acids + Metals
The reaction and salt-naming here are for everyone; the redox explanation at the end is Higher Tier only.
Acids react with metals that are more reactive than hydrogen to produce a salt and hydrogen gas:
acid + metal → salt + hydrogen
For example:
Mg + 2HCl → MgCl2 + H2
The bubbles are hydrogen gas — you can test for it with a lit splint, which gives a squeaky pop. (AQA limits these reactions to magnesium, zinc and iron with hydrochloric and sulfuric acids.)
The first part of the salt’s name comes from the metal; the second part comes from the acid:
- Hydrochloric acid (HCl) → chloride salts
- Sulfuric acid (H2SO4) → sulfate salts
- Nitric acid (HNO3) → nitrate salts
So zinc + sulfuric acid gives zinc sulfate; iron + hydrochloric acid gives iron chloride.
Why these are redox reactions Higher
Hydrogen sits in the reactivity series too. A metal above hydrogen can displace it from the acid: the metal loses electrons (oxidised) and the hydrogen ions gain them (reduced). So acid–metal reactions are redox reactions. Taking zinc and hydrochloric acid:
Zn + 2HCl → ZnCl2 + H2
Zn → Zn2+ + 2e− (oxidation)
2H+ + 2e− → H2 (reduction)
Remember that hydrogen is diatomic (H2), so you need two H+ ions to make one molecule — that is why the half equation starts with 2H+ and gains 2e−.
Copper is below hydrogen, so it cannot displace it — which is why copper does not react with dilute acids.
🧪 Exam-style questions
Which salt is produced when zinc reacts with sulfuric acid?
Copper does not react with dilute hydrochloric acid. Use the reactivity series to explain why.
Show answer
- Copper is less reactive than hydrogen (it is below hydrogen in the series). 1 mark
- So it cannot displace hydrogen from the acid, and no reaction occurs. 1 mark
6Neutralisation & Making Salts
Acids are also neutralised by bases. A base is any compound that neutralises an acid; the products always include a salt and water.
A base is a substance that neutralises an acid. The bases you meet here are metal oxides, metal hydroxides and metal carbonates.
An alkali is a base that is soluble in water (for example sodium hydroxide). Most metal oxides and carbonates are insoluble bases.
All bases neutralise acids. Alkalis are the soluble ones that dissolve in water; insoluble bases stay as solid powders — the kind you add in excess to make a salt.
There are three neutralisation patterns to know. The salt is named exactly as before — from the metal and the acid.
acid + metal oxide → salt + water
acid + metal hydroxide → salt + water
acid + metal carbonate → salt + water + carbon dioxide
Carbonates are the odd one out — they also give off carbon dioxide, which you would see as effervescence. For example:
CuO + H2SO4 → CuSO4 + H2O
CaCO3 + 2HCl → CaCl2 + H2O + CO2
Working out the formula of a salt
To write the formula of a salt, balance the charges of the positive ion (from the metal) and the negative ion (from the acid) so the overall charge is zero. Some common ions to know:
| Positive ions | Negative ions |
|---|---|
| Na+, K+, Li+, H+, NH4+ | Cl−, NO3−, OH− |
| Mg2+, Ca2+, Cu2+, Zn2+, Fe2+ | O2−, SO42−, CO32− |
| Al3+, Fe3+ |
So magnesium (Mg2+) and chloride (Cl−) need two chlorides to balance one magnesium: MgCl2. Sodium (Na+) and sulfate (SO42−) need two sodiums: Na2SO4.
🧪 Build a salt formula
Pick a positive ion and a negative ion from the table, then add ions of each until the total positive charge exactly cancels the total negative charge. The formula appears when the charges balance in the simplest ratio.
Prepare a soluble salt (e.g. copper sulfate) by neutralising an acid with an insoluble base (e.g. copper oxide):
- Gently warm the dilute acid (e.g. sulfuric acid) with a Bunsen burner — this speeds up the reaction.
- Add the insoluble base (copper oxide) a little at a time, stirring, until no more dissolves and some solid remains. This excess guarantees all the acid has been neutralised, giving the maximum yield of salt.
- Filter to remove the unreacted excess base. The filtrate is a solution of the pure salt.
- Pour the filtrate into an evaporating basin and crystallise: heat to evaporate off some water, then leave the solution to cool so crystals form slowly.
- Pat the crystals dry between filter paper (or leave to dry) to get a pure, dry sample.
7The pH Scale & Neutralisation
What makes a solution acidic or alkaline comes down to two ions:
- Acids release hydrogen ions, H+, in aqueous solution — the more H+, the more acidic.
- Alkalis release hydroxide ions, OH−, in aqueous solution — the more OH−, the more alkaline.
The pH scale runs from 0 to 14 and measures how acidic or alkaline a solution is. pH 7 is neutral; below 7 is acidic; above 7 is alkaline. You can measure pH with universal indicator (which changes colour) or, more precisely, with a pH probe.
The pH scale. Universal indicator turns red in strong acids, green at neutral, and purple in strong alkalis.
When an acid neutralises an alkali, the hydrogen ions from the acid react with the hydroxide ions from the alkali to make water. This is true for every acid–alkali neutralisation:
H+(aq) + OH−(aq) → H2O(l)
The leftover metal and non-metal ions stay in solution as the dissolved salt — which you can recover by crystallisation.
🧪 Exam-style questions
A solution turns universal indicator purple. Which row is correct?
8Titrations Triple Only Higher
A titration finds the exact volumes of acid and alkali that react together. This topic is Triple (Chemistry only). The method is for all triple students; the calculations are Higher Tier only.
- Use a pipette to measure a fixed volume of alkali into a conical flask, and add a few drops of a suitable indicator. Stand the flask on a white tile so the colour change shows up clearly.
- Fill a burette with the acid and record the starting reading (from the bottom of the meniscus — see below).
- Do a rough (trial) titration first: run the acid in fairly quickly, swirling, until the indicator just changes colour. This gives an approximate idea of how much acid you will need.
- Repeat carefully: run the acid in quickly at first, then dropwise as you approach the end-point, until the indicator just changes colour. Record the final reading.
- Repeat until you get concordant results (within 0.10 cm3) and take a mean (ignoring the rough titre).
Use a single-colour-change indicator such as phenolphthalein (pink→colourless) or methyl orange (yellow→red) — not universal indicator, whose gradual rainbow of colours gives no sharp end-point.
A pipette measures one fixed volume (e.g. 25.0 cm3) — it delivers the same amount every time, which is why it is used for the alkali. A burette measures variable volumes — you read off exactly how much liquid has run out, which is why it is used for the acid whose volume you are trying to find (the titre).
The required practical in two stages: first pipette a fixed volume of alkali into the flask. Add a few drops of indicator. Then run acid in from the burette drop by drop until the colour just changes.
Reading the burette
The surface of the liquid in a burette (or pipette) curves into a meniscus. Always take the reading from the bottom of the meniscus, with your eye level with it to avoid a parallax error. Burette readings are taken to the nearest 0.05 cm3.
The meniscus curves up at the glass walls and dips in the middle. Line your eye up with the lowest point and read from there — reading the higher edges gives a smaller, wrong value.
Titration calculations Higher
Turning your mean titre into a concentration is a calculation, taught in full — with the three-step method, a worked example and an interactive table — in C3: Quantitative Chemistry. In short: find the moles of the solution you know (moles = concentration × volume in dm3), use the balanced equation to get the moles of the other solution, then divide by its volume to find its concentration in mol/dm3. To express that concentration in g/dm3, multiply by the Mr — Higher Tier titration questions can ask for either unit.
🧪 Exam-style questions
A student titrates a sodium hydroxide solution against citric acid solution. This is part of the method the student used.
2. Add a few drops of indicator to the flask.
3. Fill a burette with citric acid solution.
Describe how the student would complete the titration.
Show answer
- Add the citric acid to the flask until there is a (permanent) colour change. 1 mark
- Measure / record the volume of citric acid added — i.e. take the final (and initial) burette reading. 1 mark
- Plus any one good-technique point: swirl the flask · stand it on a white tile · add the acid dropwise (slowly) near the end-point · repeat and calculate a mean. 1 mark
The examiner ignores any mention of which colours the indicator turns — just that there is a colour change at the end-point.
Give two reasons why a burette is used for the citric acid solution.
Show answer
- It can add the citric acid in small increments (drop by drop / slowly). 1 mark
- It can measure variable volumes (it has a scale). 1 mark
- It is more accurate than a measuring cylinder. 1 mark
Any two of these reasons score the 2 marks — both are about reading off the exact titre precisely, which a pipette (one fixed volume) or measuring cylinder cannot do.
9Strong & Weak Acids Higher
This whole section is Higher Tier only. The key is to separate two ideas that sound similar but mean different things: strong vs weak, and concentrated vs dilute.
A strong acid is completely ionised in aqueous solution — every molecule splits up to release its H+ ions. Examples: hydrochloric, nitric and sulfuric acids.
A weak acid is only partially ionised — just a small fraction of the molecules release their H+ ions at any moment. Examples: ethanoic, citric and carbonic acids.
Strong / weak is about the degree of ionisation — what fraction of molecules split up. Concentrated / dilute is about the amount of acid dissolved in a given volume (the number of moles per dm3). You can have a dilute strong acid or a concentrated weak acid — they are independent ideas.
For solutions of the same concentration, a stronger acid releases more H+ ions, so it has a lower pH.
The pH scale is logarithmic: as the pH decreases by one unit, the hydrogen-ion concentration increases by a factor of 10.
So an acid at pH 2 has 10× the H+ concentration of one at pH 3, and 100× that of one at pH 4.
Because the scale is logarithmic, each whole pH unit is a factor of 10, not an equal step. For a change of two pH units the H+ concentration changes by 10 × 10 = 100× — so a fall of 2 pH units means you multiply H+ by 100, not divide by 100 (and never by 2). Decide the direction first: lower pH = more H+.
🧪 Strong & weak acids — ionisation, concentration & pH
Both acids start at the same concentration. Notice the strong acid is far more acidic. Then drag its slider to dilute it — and see how far you have to go before it only matches the weak acid’s pH.
● H⁺ hydrogen ion · ● A⁻ acid anion (Cl⁻ or CH₃COO⁻) · H–A un-ionised acid molecule
≈100% ionised · picture schematic
· picture schematic
pH scale — compare the two acids
🧪 Exam-style questions
Which statement describes a weak acid?
Acid A has pH 1 and acid B has pH 4. How does the H+ concentration of A compare with B?
10Introduction to Electrolysis
Electrolysis uses electricity to break down an ionic compound into its elements. It is how we extract the most reactive metals (like aluminium), and how we make chlorine and hydrogen from salt water.
Recall which metals this matters for: the ones more reactive than carbon — at the top of the reactivity series — cannot be extracted by reduction with carbon, so electrolysis is the only way to get them.
The reactivity series, regrouped by extraction method. Metals above carbon can only be extracted by electrolysis — that is what this topic is about.
An electrolyte is an ionic compound that is molten or dissolved in water, so its ions are free to move and carry charge. Two electrodes are dipped into it and connected to a power supply:
- the cathode is the negative electrode;
- the anode is the positive electrode.
The electrodes are usually made of graphite (a form of carbon). Recall from C2 Carbon Allotropes that in graphite each carbon atom forms only three covalent bonds, leaving its fourth outer electron delocalised and free to move — so graphite conducts electricity, which any electrode must do. It also has a very high melting point, so it stays solid in hot molten electrolytes, and it is inert (unreactive), so it doesn’t normally take part in the reactions at the electrodes.
In a solid ionic compound the ions are locked in a lattice and cannot move, so it does not conduct. Only when molten or dissolved are the ions free to move and carry charge — so only then can it be electrolysed.
Once a current flows, the ions move towards the oppositely-charged electrode, because opposite charges attract:
- positive ions (cations) move to the negative cathode;
- negative ions (anions) move to the positive anode.
At the electrodes the ions are discharged — turned back into neutral atoms — producing the elements.
Inside the electrolyte, charge is carried by moving ions — not by electrons travelling through the liquid. Electrons only flow through the wires and electrodes. Also, the ions are already present in the molten or dissolved compound; electrolysis does not create them, it discharges them at the electrodes.
The ions are free to move in the molten or dissolved electrolyte.
11Changes at the Electrodes
The discharge of ions and the products are for everyone; the half equations are Higher Tier only.
The simplest case is a molten (liquid) ionic compound (one made of just two elements) — for example molten lead bromide or zinc chloride. The rule is simple:
- The metal is produced at the cathode (negative electrode) — you see solid metal deposits.
- The non-metal is produced at the anode (positive electrode).
This is a redox process. Remember OIL RIG — oxidation is loss of electrons, reduction is gain:
- At the cathode, positive metal ions gain electrons — this is reduction.
- At the anode, negative ions lose electrons — this is oxidation.
Half equations Higher
A half equation shows what happens to one ion, including the electrons (e−). For molten lead bromide, PbBr2:
Pb2+ + 2e− → Pb (cathode — reduction)
2Br− → Br2 + 2e− (anode — oxidation)
Notice the bromine forms a diatomic molecule (Br2) — the halogens always do. The number of electrons lost at the anode must equal the number gained at the cathode.
Make sure the charges balance. On the left of 2Br− the total charge is 2−; adding 2e− to the right gives a matching 2− there too, so the equation balances for both atoms and charge.
🧪 Build a half equation — molten electrolysis
Pick a molten compound and an electrode, then build that electrode’s half equation: choose which side the ion and the product go on, set how many of each, and add electrons until the atoms and charges balance.
🧪 Exam-style questions
Molten zinc chloride is electrolysed. What is produced at the cathode?
Write the half equation for the reaction at the anode when molten zinc chloride is electrolysed, and state whether it is oxidation or reduction.
Show answer
- 2Cl− → Cl2 + 2e− 1 mark
- It is oxidation — the chloride ions lose electrons. 1 mark
12Extraction of Aluminium
Aluminium is more reactive than carbon, so it cannot be extracted by reduction with carbon — it has to be extracted by electrolysis. The process and reasons are for everyone; the half equations are Higher Tier.
Bauxite is purified to aluminium oxide; because aluminium is more reactive than carbon, the oxide is then split by electrolysis to free the metal.
Aluminium’s ore is bauxite, which is dug straight out of the ground by open-cast mining. An ore is a rock that contains enough of a metal (or its compound) to make extracting it worthwhile. The bauxite is purified to aluminium oxide (Al2O3). To electrolyse it, the aluminium oxide must be molten so its ions are free to move and carry the current.
Aluminium oxide has a very high melting point (over 2000 °C), so melting it on its own would use a huge amount of energy. Instead it is dissolved in molten cryolite to form the electrolyte, which lowers the melting point to around 950 °C. This uses less energy and makes the process cheaper. Even so, this stage is expensive because it still needs a lot of heat energy and a large amount of electricity to drive the electrolysis.
The molten mixture is the electrolyte. It contains aluminium ions (Al3+) and oxide ions (O2−), which are free to move and carry the current. Both electrodes are made of carbon (graphite):
- The whole carbon-lined tank is the cathode (negative electrode). The Al3+ ions move to it and are reduced to molten aluminium. The molten metal is denser than the electrolyte, so it sinks and collects at the bottom of the tank, where it is tapped off through a tap.
- Several graphite blocks dipping into the electrolyte are the anodes (positive electrodes). The O2− ions move to them and are oxidised to oxygen gas. More than one anode is used because they steadily burn away and must be replaced (see below), so several keep the electrolysis running.
The carbon-lined tank is the cathode where Al³⁺ is reduced to molten aluminium; the graphite anodes are where O²⁻ is oxidised.
The half equations (Higher) are:
Al3+ + 3e− → Al (cathode — reduction)
2O2− → O2 + 4e− (anode — oxidation)
At the high operating temperature, the oxygen made at the anode reacts with the carbon of the electrode to form carbon dioxide:
C + O2 → CO2
This slowly burns away the carbon anodes, so they have to be continually replaced.
The graphite anode dips into the molten electrolyte. Press play to see why it wears away.
13Electrolysis of Aqueous Solutions
The products and rules are for everyone; the half equations are Higher Tier. When an ionic compound is dissolved in water, there is an extra complication: the water itself provides a small supply of hydrogen ions (H+) and hydroxide ions (OH−). So at each electrode there is a choice of which ion is discharged.
At the cathode (−): hydrogen is produced unless the metal is less reactive than hydrogen — in which case the metal is produced. (The less reactive species is discharged.)
At the anode (+): oxygen is produced (from hydroxide ions) unless the solution contains a halide ion (Cl−, Br−, I−) — in which case the halogen is produced.
| Aqueous solution | Cathode (−) | Anode (+) |
|---|---|---|
| Copper chloride | copper (Cu below H) | chlorine (halide) |
| Sodium chloride (brine) | hydrogen (Na above H) | chlorine (halide) |
| Sodium sulfate | hydrogen (Na above H) | oxygen (no halide) |
| Copper sulfate | copper (Cu below H) | oxygen (no halide) |
Electrolysing brine (concentrated sodium chloride solution) is industrially important: it produces hydrogen, chlorine, and sodium hydroxide solution — all valuable products.
Half equations Higher
At the cathode, if hydrogen is given off:
2H+ + 2e− → H2
At the anode, if a halide is present (e.g. chloride), the halogen is given off:
2Cl− → Cl2 + 2e−
At the anode with no halide, oxygen is given off from hydroxide ions:
4OH− → O2 + 2H2O + 4e−
🧪 Predict the products — aqueous electrolysis
Choose a dissolved compound. It splits into its ions — and the water adds H⁺ and OH⁻ too. Use the rules (with the reactivity series and Group 7) to decide which ion is discharged at each electrode.
Investigate the products formed when different aqueous solutions are electrolysed using inert (graphite) electrodes. Identify the gases: hydrogen gives a squeaky pop with a lit splint; oxygen relights a glowing splint; chlorine bleaches damp litmus paper white. These three gas tests are taught in full later in the course, in C8 Chemical Analysis. Predict the products first, then test your hypothesis.
Two setups covering every electrode product: a metal or hydrogen at the cathode, oxygen or a halogen at the anode.
🧪 Exam-style questions
Predict the products at each electrode when aqueous sodium sulfate is electrolysed.
Show answer
- Cathode: hydrogen — sodium is more reactive than hydrogen, so hydrogen is discharged instead. 1 mark
- Anode: oxygen — there is no halide ion present, so oxygen forms from hydroxide ions. 1 mark
Why is hydrogen, not sodium, produced at the cathode during the electrolysis of aqueous sodium chloride?
- Oxidation & reduction (oxygen): oxidation is gain of oxygen, reduction is loss of oxygen. Forming a metal oxide is oxidation.
- Reactivity series: metals ordered by their tendency to form positive ions, deduced from reactions with water and acid. Carbon and hydrogen are included as reference points.
- Displacement: a more reactive metal displaces a less reactive one from its compound. H It is redox — the more reactive metal is oxidised (loses electrons), the less reactive ion is reduced (gains electrons).
- Extracting metals: metals below carbon are extracted by reduction with carbon (heat the oxide with carbon); metals above carbon need electrolysis.
- Acids + metals: acid + metal → salt + hydrogen (only for metals above hydrogen). HCl → chlorides, H2SO4 → sulfates, HNO3 → nitrates.
- Neutralisation: acid + base → salt + water (carbonates also give CO2). Make a pure salt from an insoluble base by adding excess, filtering, then crystallising.
- pH scale: 0–14; acids release H+, alkalis release OH−. Neutralisation: H+ + OH− → H2O.
- Titration T: find reacting volumes with a burette, pipette and single-colour indicator. H Calculate concentrations using moles = concentration × volume and the mole ratio.
- Strong vs weak H: strong acids fully ionise, weak acids partially ionise. This is different from concentrated vs dilute. pH down 1 = H+ up ×10.
- Electrolysis: splits molten/dissolved ionic compounds. Cathode (−) attracts cations (reduced); anode (+) attracts anions (oxidised).
- Molten: metal at the cathode, non-metal at the anode. Aluminium is extracted from aluminium oxide dissolved in cryolite; carbon anodes burn away and are replaced.
- Aqueous: cathode gives hydrogen unless the metal is less reactive than hydrogen; anode gives the halogen if a halide is present, otherwise oxygen.