HookThe metal cap that was once worth more than gold
On 6 December 1884, builders lifted a 2.85 kg pyramid of pure aluminium onto the tip of the Washington Monument — at the time the largest single piece of aluminium ever cast, and pound for pound about as costly as silver. A few decades earlier Napoleon III had reputedly kept aluminium cutlery for his most honoured guests while everyone else made do with gold. The reason a metal that makes up roughly 8% of the Earth's crust — more common than iron — was a luxury comes down to one stubborn fact: aluminium clings to its oxygen so tightly that no furnace fire could tear the two apart.
Within two years the price collapsed. In 1886 Charles Martin Hall in Ohio and Paul Héroult in France, both aged 22, independently worked out how to prise aluminium from its oxide using electricity rather than heat — electrolysis. Aluminium tumbled from precious-metal prices to the stuff of drinks cans within a generation. That reversal is the whole of C4 in miniature: some metals give up their electrons willingly and some cling on for dear life, that ordering — the reactivity series — decides how you extract each one, and where ordinary chemistry cannot win, electrolysis forces the reaction uphill with a current. Underneath every reaction in this section is a single idea: the transfer of electrons.
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 version of the idea is oxidation as the gain of oxygen: burn a metal and you get a metal oxide. Magnesium ribbon flares brilliant white as it does exactly this, \(2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}\).
The reactivity series lines metals up 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 slots them into the table on purpose — they are the two reference lines that decide how a metal is extracted and whether it reacts with acid. With cold water, only the top of the table reacts: potassium ignites with a lilac flame and skitters across the surface, sodium melts into a ball, calcium sinks and fizzes steadily, releasing hydrogen and forming a hydroxide. 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.
Gold sits at the very bottom precisely because it will not give up its electrons, which is why it survives buried for millennia. When the Staffordshire Hoard was ploughed up in 2009 its gold came out of the ground as bright as the day it was buried, while any iron beside it would long since have rusted to nothing.
MechanismExtraction and redox — reduction is just electron gain
A few 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 taken away) to release the metal. The reactivity series is the decision rule for how. Metals below carbon — zinc, iron, tin, lead, copper — can be extracted by heating the ore with carbon, because carbon is more reactive and grabs the oxygen for itself: in a blast furnace, \(2\text{Fe}_2\text{O}_3 + 3\text{C} \rightarrow 4\text{Fe} + 3\text{CO}_2\). The iron oxide is reduced; the carbon is oxidised. Metals above carbon — aluminium, magnesium, calcium, sodium, potassium — hold their oxygen too tightly for carbon to win, so they are extracted by electrolysis instead.
The deeper definition of redox, and the Higher-tier one, is written in electrons: OIL RIG — Oxidation Is Loss, Reduction Is Gain, of electrons. When magnesium burns, each atom loses two electrons, \(\text{Mg} \rightarrow \text{Mg}^{2+} + 2e^-\), and the oxygen 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.
Use half equations to explain what is oxidised and what is reduced when zinc displaces copper from copper(II) sulfate.
The overall reaction is \(\text{Zn} + \text{CuSO}_4 \rightarrow \text{ZnSO}_4 + \text{Cu}\). Split it into what each metal does:
Zinc loses two electrons: \(\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-\) — loss of electrons, so zinc is oxidised.
Copper ions gain two electrons: \(\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}\) — gain of electrons, so copper is reduced.
The sulfate is a spectator — it is \(\text{SO}_4^{2-}\) before and after and takes no part. Exactly two electrons pass from each zinc atom to each copper ion, so the electrons cancel when you add the halves, which is the check that your half equations balance. That pair of half equations, with 'oxidised' and 'reduced' named explicitly, is the full answer AQA's Higher-tier redox questions are looking for.
ModelAcids, alkalis and the pH scale
An acid is a substance that releases \(\text{H}^+\) ions in water; an alkali is a soluble base that releases \(\text{OH}^-\) ions. The pH scale runs 0–14 and measures the concentration of \(\text{H}^+\): below 7 is acidic, 7 is neutral, above 7 is alkaline, and universal indicator maps the whole range onto a colour, red through green to purple. On the Higher tier, each whole step on the pH scale is a factor of ten in \(\text{H}^+\) concentration — pH 3 holds ten times the \(\text{H}^+\) of pH 4.
Neutralisation is the single reaction hiding under all the salt-making: \(\text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O}\). There are three acid reactions to know. Acid plus a metal gives a salt plus hydrogen, \(\text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2\); on the Higher tier this is redox — the metal is oxidised and the hydrogen ions are reduced to hydrogen gas. Acid plus a base (a metal oxide or hydroxide) gives a salt plus water. Acid plus a carbonate gives a salt, water and carbon dioxide — the fizz that turns limewater milky. The salt takes its surname from the acid: hydrochloric acid makes chlorides, sulfuric acid makes sulfates, nitric acid makes nitrates.
The Higher-tier trap is strong versus weak. A strong acid — hydrochloric, sulfuric, nitric — ionises completely: every molecule splits into ions in water. A weak acid — ethanoic, citric, carbonic — only partially ionises, sitting at an equilibrium with most molecules intact. At the same concentration the strong acid therefore has more \(\text{H}^+\) and a lower pH. 'Concentrated' (a lot of acid per litre) and 'strong' (a large fraction ionised) are different ideas entirely.
CaseMaking a salt you can hold — the insoluble base method (Required practical 1)
To make a pure, dry sample of a soluble salt such as copper sulfate, you react 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 whole trick: you keep adding the solid until no more will dissolve, which guarantees every last bit of acid has been used up, then you filter off the leftover. That is why you cannot make this salt by mixing acid and alkali — with two solutions you would have no visible signal that neutralisation was complete, and any excess would contaminate the crystals.
The method for copper sulfate: warm some dilute sulfuric acid, then add copper(II) oxide a spatula at a time, stirring, until unreacted black powder settles at the bottom — the sign that all the acid has reacted, \(\text{CuO} + \text{H}_2\text{SO}_4 \rightarrow \text{CuSO}_4 + \text{H}_2\text{O}\). Filter to remove the excess oxide, leaving clear blue copper sulfate solution. Heat the solution to the point of crystallisation — evaporate until crystals just begin to form at the edge — then leave it to cool slowly so that large, well-formed crystals grow. Pat them dry between filter paper.
The errors examiners flag every year: evaporating the solution to dryness, which spoils the crystal shape and can decompose the salt; not adding the base in excess, which leaves acid behind to contaminate the product; and forgetting to filter off the excess solid before crystallising.
DataTitration — finding a concentration to three figures (Required practical 2)
A titration finds an unknown concentration precisely. You measure a known volume of alkali into a conical flask with a pipette, add a few drops of a single indicator — methyl orange or phenolphthalein, never universal indicator, whose spread of colours gives no sharp end-point — and run acid in from a burette until the indicator just changes colour. That colour change is the end-point, where the acid exactly neutralises the alkali.
You repeat until you have concordant titres agreeing within 0.10 cm³, and you average only those. Read the burette at eye level to the bottom of the meniscus, to the nearest 0.05 cm³. The Higher-tier calculation then runs on moles: convert to moles with \(n = c \times \dfrac{V}{1000}\), apply the balancing ratio from the equation, and convert back to the unknown concentration.
25.0 cm³ of sodium hydroxide of unknown concentration is titrated against 0.100 mol/dm³ hydrochloric acid, and the mean titre is 20.0 cm³. The reaction is \(\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}\), a 1:1 ratio.
Step 1 — moles of the known (the acid): \(n(\text{HCl}) = 0.100 \times \dfrac{20.0}{1000} = 0.00200\ \text{mol}\).
Step 2 — ratio: 1 mole of acid reacts with 1 mole of alkali, so \(n(\text{NaOH}) = 0.00200\ \text{mol}\).
Step 3 — concentration of the unknown: \(c = \dfrac{n}{V} = \dfrac{0.00200}{0.0250} = 0.0800\ \text{mol/dm}^3\).
To put that in g/dm³, multiply by the relative formula mass, \(M_r(\text{NaOH}) = 40\): \(0.0800 \times 40 = 3.20\ \text{g/dm}^3\). The three-step spine — moles of the known, the ratio, then the concentration of the unknown — solves every titration calculation on the paper.
MechanismElectrolysis — pushing reactions uphill with a current
Electrolysis uses electricity to break down an ionic compound, and the ions must 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 — reduction; negative ions (anions) move to the positive electrode, the anode, and lose electrons — oxidation. The line worth memorising is 'reduction at the cathode'.
With a molten ionic compound the products are obvious because only two ions are present: molten lead bromide gives lead metal at the cathode and bromine gas at the anode. Aqueous solutions are trickier, because water itself supplies \(\text{H}^+\) and \(\text{OH}^-\). 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 (copper plates out; sodium does not). At the anode you get a halogen if a halide such as chloride, bromide or iodide is present, and otherwise oxygen from the hydroxide ions. Required practical 3 is exactly this investigation — electrolysing solutions with inert graphite electrodes and identifying the products by test: hydrogen gives a squeaky pop, oxygen relights a glowing splint, chlorine bleaches damp litmus paper.
The hook returns here. Aluminium oxide from bauxite melts at around 2000°C, which is ruinously expensive, so it is dissolved in molten cryolite to bring the working temperature down to about 950°C. At the cathode, \(\text{Al}^{3+} + 3e^- \rightarrow \text{Al}\), and molten aluminium is tapped off the bottom. At the anode, \(2\text{O}^{2-} \rightarrow \text{O}_2 + 4e^-\), and that oxygen burns the carbon anodes away to carbon dioxide, so they must be replaced continually — the running cost that makes recycling aluminium, which skips electrolysis altogether, save roughly 95% of the energy.
Predict the electrode products and write the half equations for the electrolysis of molten aluminium oxide.
The only ions present are \(\text{Al}^{3+}\) and \(\text{O}^{2-}\).
Cathode (negative, reduction — ions gain electrons): \(\text{Al}^{3+} + 3e^- \rightarrow \text{Al}\).
Anode (positive, oxidation — ions lose electrons): \(2\text{O}^{2-} \rightarrow \text{O}_2 + 4e^-\).
Now balance the electrons. The cathode uses 3 electrons per aluminium and the anode releases 4 per oxygen molecule, so scale both to 12: \(4\text{Al}^{3+} + 12e^- \rightarrow 4\text{Al}\) and \(6\text{O}^{2-} \rightarrow 3\text{O}_2 + 12e^-\). The electrons now cancel, giving the overall reaction \(2\text{Al}_2\text{O}_3 \rightarrow 4\text{Al} + 3\text{O}_2\). Making the electrons balance is the step students most often drop, and it is worth 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, justify 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 turns a lucky guess into a marked answer.
Half equations are worth practising until they are automatic: put the electrons on the correct side, state plainly which species is oxidised and which is reduced, and check that the electrons cancel when the halves are combined. On aluminium extraction, examiners specifically reward the roles of cryolite (lowering the melting point to save energy) and of the carbon anode (burning away in the oxygen, so it must be replaced).
In titration calculations, always write \(n = c \times \dfrac{V}{1000}\), apply the ratio from the balanced equation, quote the unit, and give three significant figures; and remember to average only the concordant titres. For salt preparations, name the salt from its acid, add the insoluble base in excess and filter it off, then crystallise slowly rather than evaporating to dryness — each of those is a separate, creditable step.