HookThe night a chemist danced round his lab because a metal caught fire on water
In October 1807, at the Royal Institution in London, Humphry Davy connected the largest battery in the world to a lump of molten potash — potassium hydroxide — and watched. Tiny globules of a bright, silvery metal budded at the negative wire, skittered across the surface and burst into lilac flame. He had made potassium, a metal so reactive that no one had ever laid eyes on it, because it had always been locked away inside its compounds and no furnace could prise it loose. His cousin recorded that Davy 'danced about the room in ecstatic delight.' Within days he had sodium; within a year, calcium, magnesium, barium and strontium — every one torn free by the same trick: electrolysis, using electricity to do what no chemical reaction could.
That is the whole of C4 in one story. Metals differ enormously in how fiercely they cling to their electrons, and lining them up by that willingness — the reactivity series — decides everything that follows: how a metal reacts with water and with acid, whether it is dug up native or smelted with carbon or, like Davy's potassium, only ever won by electrolysis. Sitting underneath all of it is a single idea you will meet again and again in this section — the transfer of electrons — which is why acids reacting with metals, displacement, extraction and electrolysis are really all the same kind of event.
ModelThe reactivity series — a league table of electron-givers
Metals react by losing electrons to form positive ions, and the more readily a metal does this, the more reactive it is. The oldest way to see a metal reacting is oxidation as the gain of oxygen: heat a metal in air and you get a metal oxide. Magnesium ribbon flares a blinding white as it does exactly this, \(2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}\).
The reactivity series orders metals by how vigorously they react: potassium, sodium, lithium, calcium, magnesium, then carbon, then zinc, iron, tin, lead, then hydrogen, then copper, silver and gold. Carbon and hydrogen are non-metals, but AQA slips them into the table on purpose, because they are the two reference lines that decide how a metal is extracted and whether it reacts with acid. With cold water, only the very top reacts — potassium ignites and races across the surface, sodium melts into a ball, calcium sinks and fizzes steadily — each giving a metal hydroxide and hydrogen, for example \(2\text{Na} + 2\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2\). With dilute acid, every metal above hydrogen gives a salt plus hydrogen, the fizzing growing gentler as you move down; the metals below hydrogen — copper, silver, gold — do not react with dilute acid at all, which is exactly why gold coins survive centuries in the ground while iron rusts to crumbs beside them.
MechanismExtraction and reduction — how you win a metal from its ore
A handful of very unreactive metals — gold, sometimes copper — are found native, as the metal itself, and can simply be dug up. Everything else is locked inside a compound, usually an oxide, and must be reduced — have its oxygen removed — to release the metal. The reactivity series is the decision rule for how. Metals below carbon can be reduced by heating the ore with carbon, because carbon is more reactive and takes the oxygen for itself; the metal oxide is reduced while the carbon is oxidised. Metals above carbon — aluminium, magnesium and the rest of the top of the table — hold their oxygen far too tightly for carbon to win, so they are extracted by electrolysis instead.
You are given tin(IV) oxide and aluminium oxide and asked how each metal is extracted. Tin sits below carbon in the reactivity series, so carbon can reduce it: heat the ore with carbon and the oxygen transfers across, \[ \text{SnO}_2 + \text{C} \rightarrow \text{Sn} + \text{CO}_2 \] The tin oxide is reduced (it loses oxygen) and the carbon is oxidised (it gains oxygen). Aluminium sits above carbon, so no amount of heating with carbon will work — its oxide has to be split by electrolysis, which is why aluminium was a precious metal until the electrical method was found. Naming the reference line — 'below carbon, so reduced by carbon' — is what turns a guess into a marked answer.
MechanismRedox in electrons — OIL RIG (Higher tier)
The deeper, Higher-tier definition of oxidation and reduction is written not in oxygen but in electrons: OIL RIG — Oxidation Is Loss, Reduction Is Gain, of electrons. This is why displacement reactions and the reaction of a metal with an acid are both redox: the metal loses electrons and something else gains them. Oxidation and reduction are two halves of one event — you never get one without the other — which is why the combined reaction is called redox, and why the electrons must cancel exactly when you add the two half equations together.
Explain, using half equations, what is oxidised and what is reduced when magnesium reacts with dilute sulfuric acid.
The overall reaction is \(\text{Mg} + \text{H}_2\text{SO}_4 \rightarrow \text{MgSO}_4 + \text{H}_2\). Split it into what each part does:
Magnesium loses two electrons: \(\text{Mg} \rightarrow \text{Mg}^{2+} + 2e^-\) — loss of electrons, so magnesium is oxidised.
Hydrogen ions gain those electrons: \(2\text{H}^+ + 2e^- \rightarrow \text{H}_2\) — gain of electrons, so hydrogen is reduced.
The sulfate is a spectator: it is \(\text{SO}_4^{2-}\) before and after and takes no part. Two electrons leave each magnesium atom and two arrive at the pair of hydrogen ions, so the electrons cancel when the halves are combined — the check that your half equations are right. That pair of equations, with 'oxidised' and 'reduced' named explicitly, is the full Higher-tier answer.
ModelAcids, alkalis and the three ways to make a salt
An acid is a substance that releases hydrogen ions \((\text{H}^+)\) in water; a base is a substance that neutralises an acid, and a base that dissolves is an alkali, releasing hydroxide ions \((\text{OH}^-)\). Underneath every neutralisation is one reaction, \(\text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O}\).
There are three acid reactions you must know, and each makes a salt. Acid plus a metal gives a salt plus hydrogen, for example \(\text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2\) — and on the Higher tier this is a redox reaction, the metal oxidised and the hydrogen ions reduced. Acid plus a base (a metal oxide or hydroxide) gives a salt plus water. Acid plus a metal carbonate gives a salt, water and carbon dioxide — the fizz that turns limewater milky, for example \(\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2\). The salt takes its surname from the acid: hydrochloric acid makes chlorides, sulfuric acid makes sulfates, nitric acid makes nitrates. Get the acid right and the salt name writes itself.
ModelThe pH scale, and strong versus weak acids
The pH scale runs from 0 to 14 and measures how acidic or alkaline a solution is by tracking its hydrogen-ion concentration: below 7 is acidic, 7 is neutral, above 7 is alkaline. Universal indicator maps the whole range onto a spread of colours, red through green to purple, while a pH probe gives a precise numerical reading. On the Higher tier there is a sharper rule — each whole step down the pH scale is a tenfold rise in hydrogen-ion concentration, so a solution at pH 2 has ten times the \(\text{H}^+\) of one at pH 3, and a hundred times that of pH 4.
The Higher-tier trap is strong versus weak. A strong acid — hydrochloric, sulfuric, nitric — ionises completely: in water every molecule splits into ions. A weak acid — ethanoic, citric, carbonic — only partially ionises, sitting at an equilibrium with most of its molecules still intact. So at the same concentration a strong acid has more hydrogen ions and therefore a lower pH than a weak one. 'Concentrated' (a lot of acid dissolved per litre) and 'strong' (a large fraction of it ionised) are completely different ideas — you can have a dilute strong acid and a concentrated weak one.
CaseMaking a salt you can hold — the insoluble base method (Required practical 8)
Required practical 8 makes a pure, dry sample of a soluble salt, such as copper sulfate, by reacting an acid with an insoluble base — a metal oxide or carbonate you can add in excess and then simply filter away. Using an insoluble base is the entire trick: you keep adding the solid until no more dissolves, which guarantees every last bit of acid has reacted, then you filter off the leftover. This is why you cannot make the salt by mixing two solutions — with acid and alkali there is no visible signal that neutralisation is exactly complete, and any excess would contaminate the crystals.
The method for copper sulfate runs like this. Warm some dilute sulfuric acid, then add copper(II) oxide a spatula at a time, stirring, until unreacted black powder settles on the bottom — the sign that all the acid is used up, \(\text{CuO} + \text{H}_2\text{SO}_4 \rightarrow \text{CuSO}_4 + \text{H}_2\text{O}\). Filter to remove the excess oxide, leaving a clear blue copper sulfate solution. Heat that solution to the point of crystallisation — evaporate until crystals just start to form at the edge — then leave it to cool slowly so large, well-shaped crystals grow, and pat them dry between filter paper. The errors examiners flag every year: evaporating to dryness (which spoils the crystal shape), failing to add the base in excess (which leaves acid to contaminate the salt), and forgetting to filter off the excess solid before crystallising.
MechanismElectrolysis — splitting compounds with a current (Higher tier half equations)
Electrolysis uses an electric current to break down an ionic compound, and the ions have to be free to move, so the compound is either melted or dissolved. Positive ions (cations) move to the negative electrode, the cathode, and gain electrons — that is reduction. Negative ions (anions) move to the positive electrode, the anode, and lose electrons — that is oxidation. The single line worth memorising is 'reduction at the cathode'. With a molten compound the products are easy to predict, because only two kinds of ion are present, and the electrode reactions are written as half equations showing the electrons.
Predict the products and write the half equations for the electrolysis of molten lead(II) bromide, \(\text{PbBr}_2\).
The only ions present are \(\text{Pb}^{2+}\) and \(\text{Br}^-\).
At the cathode (negative — reduction, so ions gain electrons) the lead ions are discharged as molten lead metal: \(\text{Pb}^{2+} + 2e^- \rightarrow \text{Pb}\).
At the anode (positive — oxidation, so ions lose electrons) the bromide ions form bromine gas: \(2\text{Br}^- \rightarrow \text{Br}_2 + 2e^-\).
Each lead ion takes two electrons and each pair of bromide ions releases two, so the electrons balance exactly. Lead collects at the bottom as a molten bead and orange bromine vapour rises from the anode — Davy's method, made visible.
DataElectrolysis of solutions — predicting the products (Required practical 9)
Aqueous solutions are trickier than molten compounds, because water itself supplies a few \(\text{H}^+\) and \(\text{OH}^-\) ions, so at each electrode there is a competition. Two rules settle it. At the cathode you get hydrogen unless the metal is less reactive than hydrogen, in which case the metal is deposited instead — copper plates out, but sodium stays in solution as ions. At the anode you get a halogen if a halide such as chloride, bromide or iodide is present, and otherwise oxygen, released from the hydroxide ions. Required practical 9 is exactly this investigation: electrolysing solutions with inert graphite electrodes and identifying the products by test — hydrogen gives a squeaky pop with a lit splint, oxygen relights a glowing splint, and chlorine bleaches damp litmus paper.
Predict the products at each electrode for two solutions.
Copper(II) sulfate solution: at the cathode, copper is less reactive than hydrogen, so copper is deposited (a pink-brown coat on the electrode); at the anode there is no halide, so oxygen is given off from the hydroxide ions.
Sodium chloride solution: at the cathode, sodium is more reactive than hydrogen, so hydrogen bubbles off and the sodium stays as ions; at the anode a halide (chloride) is present, so chlorine is produced. Test each gas to confirm — the squeaky pop for hydrogen, and damp litmus bleached white for chlorine. The two rules, applied electrode by electrode, predict every case on the paper.
CaseExtracting aluminium — the hook's payoff (Higher tier half equations)
Aluminium is the third most common element in the Earth's crust, yet it was once costlier than silver, because it sits above carbon and cannot be smelted — only electrolysis will free it. Its ore, purified to aluminium oxide from bauxite, melts at around 2000°C, which would be ruinously expensive to sustain, so it is dissolved in molten cryolite to bring the working temperature down to about 950°C. At the cathode the aluminium ions are reduced, \(\text{Al}^{3+} + 3e^- \rightarrow \text{Al}\), and molten aluminium is tapped from the bottom of the cell. At the anode the oxide ions are oxidised to oxygen, \(2\text{O}^{2-} \rightarrow \text{O}_2 + 4e^-\), and that hot oxygen burns the carbon anodes away to carbon dioxide, so they must be replaced over and over — the running cost that makes recycling aluminium, which skips electrolysis entirely, save roughly 95% of the energy.
Combine the electrode half equations for aluminium extraction into the overall equation. The cathode reaction uses 3 electrons per aluminium, \(\text{Al}^{3+} + 3e^- \rightarrow \text{Al}\); the anode reaction releases 4 electrons per oxygen molecule, \(2\text{O}^{2-} \rightarrow \text{O}_2 + 4e^-\). To make the electrons match, scale both to the lowest common multiple, 12: \[ 4\text{Al}^{3+} + 12e^- \rightarrow 4\text{Al} \qquad 6\text{O}^{2-} \rightarrow 3\text{O}_2 + 12e^- \] The 12 electrons now cancel, giving the overall reaction \(2\text{Al}_2\text{O}_3 \rightarrow 4\text{Al} + 3\text{O}_2\). Balancing the electrons before you combine the halves is the step candidates most often drop, and AQA gives it its own mark.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
Whenever a question turns on the reactivity series, the AO2 marks come from justifying every claim with a comparison: a metal reacts with dilute acid because it is 'more reactive than hydrogen', and it needs electrolysis because it is 'more reactive than carbon'. Naming the reference line is what the mark scheme credits. For word and symbol equations, balance them and add state symbols where the question shows them — an unbalanced equation caps the marks.
Half equations are worth drilling until they are automatic: put the electrons on the correct side, state plainly which species is oxidised and which is reduced, and show that the electrons cancel when the halves are combined. On aluminium extraction, examiners specifically reward the role of cryolite (dissolving the oxide to lower the melting point and save energy) and of the carbon anode (burning away in the oxygen, so it must be replaced). For the electrolysis of solutions in Required practical 9, apply the two rules electrode by electrode and quote the gas test — squeaky pop, relit splint, bleached litmus.
On salt preparation (Required practical 8), each step is a separate creditable point: name the salt from its acid, add the insoluble base in excess, filter off the excess, then crystallise slowly rather than evaporating to dryness. And on the AO3 'strong versus weak' question, never let 'concentrated' stand in for 'strong' — the marks are for degree of ionisation, and on the Higher tier for the tenfold change in hydrogen-ion concentration per pH unit.