Oxidation and reduction: oxygen, electrons and oxidation states
By the end of this lesson you will be able to
- Define oxidation and reduction in terms of oxygen and in terms of electrons, and identify what is oxidised and what is reduced in any redox equation.
- Write and balance half-equations with electrons and state symbols, and combine two half-equations into an ionic equation.
- Determine oxidation numbers with the rules, and use them to name compounds such as iron(III) oxide and to spot redox changes.
- Distinguish an oxidising agent from a reducing agent and explain the colour changes that show common agents at work.
- Analyse reacting-mass data from a reduction experiment to determine the formula of an oxide.
You met this before
- Metal atoms lose their outer electrons to form positive ions: Na → Na⁺, Mg → Mg²⁺, Al → Al³⁺. Non-metal atoms gain electrons to fill their outer shell: Cl → Cl⁻, O → O²⁻ (Year 3 ionic bonding).
- An ionic formula balances the charges: Fe³⁺ with O²⁻ needs two iron ions (+6) for three oxide ions (−6), so iron(III) oxide is Fe₂O₃.
- A symbol equation is balanced by counting atoms of each element on both sides, changing only the numbers in front of formulas: 4Fe + 3O₂ → 2Fe₂O₃.
- Kinds of reaction you already name: combustion (burning in oxygen), displacement (a more reactive element pushes a less reactive one out of its compound), decomposition, neutralisation, precipitation (Year 3).
- State symbols: (s) solid, (l) liquid, (g) gas, (aq) dissolved in water.
Give the electron arrangement of a magnesium atom and of a magnesium ion, and write the formula of the compound formed between magnesium and chlorine, with the charge on each ion.
Reveal
Mg atom 2,8,2; Mg²⁺ ion 2,8 — the atom has lost its two outer electrons. Chlorine atoms (2,8,7) each gain one electron to become Cl⁻ (2,8,8).
One Mg²⁺ balances two Cl⁻, so the formula is MgCl₂. Hold on to the idea that the magnesium lost electrons and the chlorine gained them: that is this whole lesson.
Revise it in full: Year 3 · Chemistry · Atomic structure and bonding · Lesson 3 — Ionic bonding · Year 3 · Energy and chemical change · Lesson 2 — Formulas, symbol equations and balancing · Year 3 · Energy and chemical change · Lesson 3 — Kinds of reaction
The puzzle
Hold a strip of magnesium in a Bunsen flame and it burns with a white light so bright you must not stare at it, leaving white magnesium oxide. Drop an identical strip into blue copper(II) sulfate solution and nothing so dramatic happens: the strip turns brown, the solution warms and slowly loses its colour.
There is no oxygen gas in the second beaker at all. Yet chemists say that in both beakers the magnesium was oxidised — and that in both, something else was reduced at exactly the same moment. What do the two reactions have in common that the word “oxygen” cannot explain?
In the copper sulfate beaker the magnesium ends up as Mg²⁺ ions in solution, and brown copper metal appears where Cu²⁺ ions used to be. Predict where the two electrons that each magnesium atom lost have gone, and why the blue colour fades.
Compare your answer
Each magnesium atom hands its two outer electrons to a copper(II) ion, which becomes a neutral copper atom: Mg + Cu²⁺ → Mg²⁺ + Cu. The blue colour belongs to the Cu²⁺ ions, so as they are turned into copper metal the colour fades.
In the flame the magnesium also lost two electrons each — to oxygen atoms, making O²⁻ ions. Losing electrons is what “oxidised” really means; gaining them is “reduced”.
The science — structured notes
1.1 Oxidation and reduction as gain and loss of oxygen
The words are older than atoms. Metal workers reduced an ore to its metal — the heap of rock “reduced” to a smaller lump of metal. When a metal took up oxygen from the air it was oxidised. So the first definitions are about oxygen:
- Oxidation is the gain of oxygen by a substance: 2Mg(s) + O₂(g) → 2MgO(s). Magnesium is oxidised.
- Reduction is the loss of oxygen from a substance: CuO(s) + H₂(g) → Cu(s) + H₂O(g). Copper(II) oxide is reduced; the hydrogen, which gains the oxygen, is oxidised.
Look at the second equation again. One substance cannot lose oxygen unless another takes it. Oxidation and reduction always happen together, which is why these reactions are called redox reactions (reduction + oxidation).
The blast furnace (Lesson 3) is a redox reaction on a colossal scale: Fe₂O₃ + 3CO → 2Fe + 3CO₂ — the iron oxide is reduced and the carbon monoxide is oxidised.
A few older textbooks also define oxidation as loss of hydrogen and reduction as gain of hydrogen (in the second equation hydrogen is oxidised because it becomes attached to oxygen). You will meet this mainly in biology: respiration oxidises glucose, photosynthesis reduces carbon dioxide.
For each reaction, state which substance is oxidised and which is reduced, using gain or loss of oxygen: (i) 2CuO + C → 2Cu + CO₂; (ii) Fe₂O₃ + 2Al → Al₂O₃ + 2Fe; (iii) CH₄ + 2O₂ → CO₂ + 2H₂O.
Reveal answer
(i) Copper(II) oxide loses oxygen, so it is reduced; carbon gains oxygen, so it is oxidised.
(ii) Iron(III) oxide is reduced; aluminium is oxidised — this is the thermite reaction (Lesson 2).
(iii) Methane is oxidised (its carbon gains oxygen and it loses hydrogen); oxygen is reduced — every combustion is a redox reaction.
1.2 The better definition: transfer of electrons
Now look at the magnesium strip in copper sulfate. The ionic equation, which leaves out the sulfate ions because they do not change, is:
Mg(s) + Cu²⁺(aq) → Mg²⁺(aq) + Cu(s)
No oxygen is gained or lost by anything. What has happened is that two electrons have moved from each magnesium atom to a copper(II) ion. Magnesium has done exactly what it did in the flame, where it also ended up as Mg²⁺. So chemists now define the words by electrons:
- Oxidation is the loss of electrons. Mg → Mg²⁺ + 2e⁻
- Reduction is the gain of electrons. Cu²⁺ + 2e⁻ → Cu
The memory aid is OIL RIG: Oxidation Is Loss, Reduction Is Gain (of electrons). The electron definition includes the oxygen one: when magnesium burns, each oxygen atom gains two electrons to become O²⁻, so oxygen is reduced and magnesium oxidised in both senses.
Ions that appear unchanged on both sides — here the sulfate ions, SO₄²⁻ — are spectator ions. They are there, they carry charge through the solution, but they take no part in the electron transfer, so the ionic equation leaves them out.
“Oxidation must involve oxygen, so the reaction of sodium with chlorine to make sodium chloride is not an oxidation.”
Reveal answer
False. 2Na + Cl₂ → 2NaCl is a redox reaction: each sodium atom loses one electron (Na → Na⁺ + e⁻, oxidation) and each chlorine atom gains one (Cl₂ + 2e⁻ → 2Cl⁻, reduction).
The oxygen definition is a special case; the electron definition covers every redox reaction, with or without oxygen.
1.3 Half-equations: writing each side of the transfer
A half-equation shows one half of a redox reaction — the oxidation or the reduction — with the electrons written in. It must balance in two ways: the atoms of each element, and the total charge on each side. Electrons, e⁻, each carry a charge of 1−.
| Change | Half-equation | Charge check (left → right) | Type |
|---|---|---|---|
| zinc atom to zinc ion | Zn(s) → Zn²⁺(aq) + 2e⁻ | 0 → +2 − 2 = 0 | oxidation |
| copper(II) ion to copper | Cu²⁺(aq) + 2e⁻ → Cu(s) | +2 − 2 = 0 → 0 | reduction |
| aluminium atom to ion | Al(s) → Al³⁺(aq) + 3e⁻ | 0 → +3 − 3 = 0 | oxidation |
| chlorine molecule to chloride ions | Cl₂(aq) + 2e⁻ → 2Cl⁻(aq) | −2 → −2 | reduction |
| iron(II) to iron(III) | Fe²⁺(aq) → Fe³⁺(aq) + e⁻ | +2 → +3 − 1 = +2 | oxidation |
| silver ion to silver | Ag⁺(aq) + e⁻ → Ag(s) | 0 → 0 | reduction |
To build the full ionic equation, multiply each half-equation so that the electrons lost equal the electrons gained, add the two, and cancel the electrons. The electrons never appear in the overall equation — every one released by the reducing side is taken by the oxidising side.
Aluminium foil (with its oxide layer scratched off) displaces copper from copper(II) sulfate solution. Write the two half-equations and the ionic equation with state symbols.
Oxidation half. Al(s) → Al³⁺(aq) + 3e⁻
Reduction half. Cu²⁺(aq) + 2e⁻ → Cu(s)
Match the electrons. The lowest common multiple of 3 and 2 is 6.
2Al(s) → 2Al³⁺(aq) + 6e⁻
3Cu²⁺(aq) + 6e⁻ → 3Cu(s)
Add and cancel the 6e⁻. 2Al(s) + 3Cu²⁺(aq) → 2Al³⁺(aq) + 3Cu(s)
Check. Atoms: 2 Al and 3 Cu on each side. Charge: left +6, right +6. Aluminium is oxidised, copper(II) ions are reduced.
Borehole water often contains dissolved iron(II) ions, which later turn into brown iron(III) deposits that stain baths and clog pipes. Water works add chlorine to oxidise the iron(II) on purpose, so that the iron(III) hydroxide can be filtered out. Write the ionic equation and state the role of each species.
Oxidation half. Fe²⁺(aq) → Fe³⁺(aq) + e⁻
Reduction half. Cl₂(aq) + 2e⁻ → 2Cl⁻(aq)
Match the electrons. Double the iron half: 2Fe²⁺(aq) → 2Fe³⁺(aq) + 2e⁻
Add. Cl₂(aq) + 2Fe²⁺(aq) → 2Cl⁻(aq) + 2Fe³⁺(aq)
Check. Charge left: 0 + 4 = +4; right: −2 + 6 = +4. Balanced.
Roles. Iron(II) ions are oxidised; chlorine is reduced. No oxygen appears anywhere, yet this is an oxidation — the electron definition is the one that works.
A student writes three half-equations: (i) Fe(s) + 2e⁻ → Fe²⁺(aq); (ii) Br₂(aq) + e⁻ → 2Br⁻(aq); (iii) Na(s) → Na⁺(aq) + e⁺. Find and correct every error.
Reveal answer
(i) The electrons are on the wrong side: a neutral iron atom becoming a 2+ ion must lose two electrons. Correct: Fe(s) → Fe²⁺(aq) + 2e⁻ (charge 0 → +2 − 2 = 0).
(ii) Charge does not balance: left −1, right −2. Two bromine atoms need two electrons. Correct: Br₂(aq) + 2e⁻ → 2Br⁻(aq).
(iii) There is no such particle as “e⁺” in chemistry; the electron is e⁻. Correct: Na(s) → Na⁺(aq) + e⁻.
Checking charge on both sides catches all three — do it every time.
1.4 Oxidation numbers: book-keeping for electrons
Electron transfer is easy to see in Mg → Mg²⁺. It is harder in a molecule like SO₂ or an ion like MnO₄⁻, where the electrons are shared rather than handed over. Chemists therefore give every atom an oxidation number (oxidation state) — the charge it would have if all its bonds were ionic. Then:
- Oxidation is an increase in oxidation number (electrons lost).
- Reduction is a decrease in oxidation number (electrons gained).
| Rule | Oxidation number | Examples |
|---|---|---|
| 1. An uncombined element | 0 | Mg, Fe, O₂, Cl₂, S₈ |
| 2. A simple (monatomic) ion | its charge | Na⁺ +1, Mg²⁺ +2, Cl⁻ −1, O²⁻ −2 |
| 3. The sum in a neutral compound | 0 | H₂O: 2(+1) + (−2) = 0 |
| 4. The sum in a polyatomic ion | the charge on the ion | SO₄²⁻: S + 4(−2) = −2, so S = +6 |
| 5. Oxygen in compounds | −2 (−1 in peroxides such as H₂O₂) | CO₂, Fe₂O₃; H₂O₂ |
| 6. Hydrogen in compounds | +1 (−1 in metal hydrides such as NaH) | HCl, NH₃ |
| 7. Fluorine; Group 1; Group 2 | −1; +1; +2 always in compounds | NaF, KCl, CaO |
Work out the unknown atom last, from the sum. Oxidation numbers are written with the sign first (+3, −2), to tell them apart from ionic charges, which put the sign after the number (Fe³⁺, O²⁻).
Find the oxidation number of manganese in KMnO₄.
K is in Group 1, so +1; each O is −2; the compound is neutral.
(+1) + Mn + 4(−2) = 0
Mn + 1 − 8 = 0
Mn = +7
That +7 is written as the Roman numeral in the name: potassium manganate(VII). In the same way, the iron in Fe₂O₃ is +3, so the name is iron(III) oxide; in FeO it is +2, iron(II) oxide.
Oxidation numbers make redox visible at a glance. In the combustion of methane, CH₄ + 2O₂ → CO₂ + 2H₂O, carbon goes from −4 (in CH₄) to +4 (in CO₂): oxidised, eight electrons’ worth. Oxygen goes from 0 to −2: reduced.
A reaction in which no oxidation number changes is not redox: in neutralisation, NaOH + HCl → NaCl + H₂O, every atom keeps its number.
Determine the oxidation number of sulfur in each: H₂S, S₈, SO₂, SO₃²⁻, H₂SO₄. Then state whether sulfur is oxidised or reduced when hydrogen sulfide burns to sulfur dioxide.
Reveal answer
H₂S: 2(+1) + S = 0, so S is −2.
S₈: an element, so 0.
SO₂: S + 2(−2) = 0, so S is +4.
SO₃²⁻: S + 3(−2) = −2, so S is +4.
H₂SO₄: 2(+1) + S + 4(−2) = 0, so S is +6.
H₂S → SO₂ takes sulfur from −2 to +4, an increase of 6: sulfur is oxidised (and oxygen, 0 → −2, is reduced).
1.5 Oxidising agents and reducing agents
The substance that causes an oxidation is the oxidising agent: it takes electrons from something else, so the oxidising agent is itself reduced. The substance that causes a reduction is the reducing agent: it gives electrons away, so it is itself oxidised.
In Mg + Cu²⁺ → Mg²⁺ + Cu, copper(II) ions are the oxidising agent and magnesium is the reducing agent. It sounds backwards at first; say it slowly and it becomes obvious — the agent does to the other what happens in reverse to itself.
| Common oxidising agents (are reduced) | What you see | Common reducing agents (are oxidised) | Used for |
|---|---|---|---|
| Oxygen, O₂ | burning, rusting | Reactive metals (Mg, Zn, Al) | displacement, thermite |
| Chlorine, Cl₂; sodium chlorate(I) in bleach | bleaches colours; kills microbes | Carbon and carbon monoxide | extracting iron (blast furnace) |
| Hydrogen peroxide, H₂O₂ | fizzes, bleaches hair and teeth | Hydrogen, H₂ | reducing metal oxides |
| Acidified potassium manganate(VII), KMnO₄ | purple → colourless (Mn +7 → +2) | Iron(II) ions, Fe²⁺ | analysis by titration |
| Acidified potassium dichromate(VI), K₂Cr₂O₇ | orange → green (Cr +6 → +3) | Iodide ions, I⁻ | tested with an oxidising agent: brown iodine forms |
Strong oxidising agents are hazardous in their own way: they supply oxygen or take electrons from anything that burns, so they make fires fiercer. Their hazard pictogram is a flame over a circle — which is the “oxidising” symbol a real paper asked about for hydrogen peroxide (see the study box).
Chlorine water is added to colourless potassium bromide solution and the mixture turns orange. Write the ionic equation, identify the oxidising agent and the reducing agent, and explain the colour.
Reveal answer
Cl₂(aq) + 2Br⁻(aq) → 2Cl⁻(aq) + Br₂(aq)
Chlorine takes electrons from bromide ions: chlorine is the oxidising agent (it is reduced, 0 → −1); bromide ions are the reducing agent (they are oxidised, −1 → 0). Potassium ions are spectators.
The orange colour is bromine, Br₂, formed from colourless bromide ions. Chlorine is the more reactive halogen: its outer shell is closer to the nucleus and gains an electron more readily.
A few drops of purple, acidified potassium manganate(VII) are added to a solution of iron(II) sulfate. Predict what you see and name what is oxidised and what is reduced, using oxidation numbers.
Reveal answer
The purple colour disappears as each drop is added (while iron(II) remains), and the pale green solution turns very pale yellow.
Manganese goes from +7 in MnO₄⁻ to +2 in Mn²⁺ (almost colourless): reduced — manganate(VII) is the oxidising agent. Iron goes from +2 to +3: oxidised — iron(II) is the reducing agent.
When all the iron(II) is used up, the next drop stays pink — the end-point used to measure the iron in iron tablets.
1.6 Redox in one sentence, and where it is hiding
Every redox reaction contains a species that loses electrons (oxidised; the reducing agent) and a species that gains the same number of electrons (reduced; the oxidising agent). Oxidation numbers track the change: up for oxidation, down for reduction, and the total rise equals the total fall.
- Respiration oxidises glucose to carbon dioxide (carbon 0 → +4) and reduces oxygen to water.
- Photosynthesis runs it backwards: carbon dioxide is reduced, water is oxidised to oxygen.
- Rusting oxidises iron to iron(III) (Lesson 5); extraction reduces ores to metals (Lesson 3).
- Batteries and electrolysis separate the two halves in space so the electrons flow through a wire (Lessons 4 and 5).
In the photographic darkroom of the twentieth century, light made silver bromide crystals release silver atoms, and the “developer” solution then reduced more silver ions to silver around those specks. What if the developer were replaced by an oxidising agent: would an image appear? Explain in terms of electrons.
Reveal answer
No. Developing needs Ag⁺ + e⁻ → Ag: the developer must give electrons, so it must be a reducing agent.
An oxidising agent would do the reverse — take electrons from the few silver atoms the light had made, turning them back into Ag⁺ ions — and erase the latent image. (Photographers did exactly this on purpose to “bleach” prints.)
Household bleach is made by bubbling chlorine into sodium hydroxide solution: Cl₂ + 2NaOH → NaCl + NaClO + H₂O. Determine the oxidation number of chlorine in each chlorine-containing species and explain why this reaction is unusual. Then explain, with an equation, why the label warns never to mix bleach with acidic toilet cleaners.
Reveal answer
Cl₂: 0. NaCl: −1. NaClO: +1 + Cl + (−2) = 0, so Cl = +1.
The same element is both reduced (0 → −1) and oxidised (0 → +1) in one reaction — a disproportionation. In each chlorine molecule, one atom takes an electron from the other.
With acid the reverse happens: ClO⁻(aq) + Cl⁻(aq) + 2H⁺(aq) → Cl₂(g) + H₂O(l). Chlorine at +1 and at −1 meet at 0, and toxic chlorine gas is released into a small bathroom.
Try it — electrons on the move between halogens
This lab shows a redox reaction you can measure. A halogen solution is added from a pipette to a potassium halide solution; if the added halogen is the stronger oxidising agent, it takes electrons from the halide ions and a coloured halogen appears.
A colorimeter reads the colour as an absorbance, A, at 450 nm. Cl₂ + 2I⁻ → 2Cl⁻ + I₂ makes brown iodine. With the potassium iodide in excess, the chlorine is the limiting reactant, so the iodine formed — and the absorbance — should follow the chlorine you add.
Suppose your means were: 0.25 → 0.13; 0.50 → 0.25; 1.00 → 0.50; 1.50 → 0.74; 2.00 → 1.01. Calculate A ÷ c for each, state whether the hypothesis is supported, and explain why the same absorbance appears when 1.00 mmol dm⁻³ bromine water is used instead of chlorine water.
Reveal answer
A ÷ c = 0.52, 0.50, 0.50, 0.49, 0.51 dm³ mmol⁻¹ — constant within scatter, and 0.50 → 1.00 mmol dm⁻³ doubles A from 0.25 to 0.50. The hypothesis is supported: the absorbance is proportional to the chlorine added.
Bromine is also a stronger oxidising agent than iodine: Br₂ + 2I⁻ → 2Br⁻ + I₂. Each Br₂ makes one I₂, just as each Cl₂ did, so the same amount of iodine forms and the absorbance is the same. The colorimeter measures the product, iodine, not the halogen that was added.
Common mistakes to avoid
- OIL RIG backwards — oxidation is loss of electrons; if a species’ charge goes up (Fe²⁺ → Fe³⁺) it has been oxidised.
- Mixing up the agent and what happens to it — the oxidising agent is reduced; the reducing agent is oxidised.
- Half-equations that do not balance charge — count the charge on both sides, including every electron as 1−.
- Electrons left in the overall equation — multiply the half-equations until the electrons match, then cancel them.
- Leaving spectator ions in the ionic equation — sulfate, nitrate, potassium and sodium ions usually do not change.
- Writing oxidation numbers like charges — +3, not 3+, and every atom gets its own number, even in a molecule.
- Guessing formulas from names — iron(III) oxide is Fe₂O₃; build it from Fe³⁺ and O²⁻.
- Missing state symbols — the May 2026 report found students struggled to write half-equations with them; metals (s), ions (aq), halogens in water (aq), gases (g).
Study smart: how this is examined
Study routine. Today: write OIL RIG, the agent rule and the seven oxidation-number rules on one card, then redo both half-equation EXAMPLES from memory. In two days: assign oxidation numbers to ten species chosen from this page without looking, and write the half-equations for Zn/Cu²⁺, Cl₂/Br⁻ and Fe²⁺/Cl₂.
In a week: run the halogen lab as a full investigation, then the Think task and the quiz.
How we found out — from phlogiston to electrons
For most of the eighteenth century chemists explained burning with phlogiston, a “fire principle” that substances were supposed to release when they burned or rusted. The theory had an awkward fact to swallow: metals got heavier when they turned into their “calx” (what we call the oxide). Some chemists even gave phlogiston negative weight.
In 1774 Joseph Priestley in England heated red mercury calx with sunlight focused through a large lens and collected a gas in which a candle burned brilliantly and a mouse lived longer than in ordinary air. He called it “dephlogisticated air”. Carl Wilhelm Scheele in Sweden had made the same gas a little earlier, but his book appeared only in 1777.
Antoine Lavoisier in Paris saw what the gas meant. He heated mercury in a retort connected to a bell jar of air for twelve days: red calx formed on the mercury and the air shrank by about a sixth (the oxygen did not all react). Then he heated the calx strongly and recovered the same volume of gas, and the mercury.
The metal had combined with part of the air. He named the gas oxygène, and in his Traité élémentaire de chimie of 1789 burning, rusting and breathing all became combination with oxygen — oxidation. The balance, not the theory, won the argument.
“Reduction” kept its metal-workers’ meaning — getting the metal back from its calx — until the electron was discovered by J. J. Thomson in 1897. Once chemists understood that atoms form ions by losing and gaining electrons, both words were redefined in the early twentieth century as the loss and gain of electrons.
By the 1930s chemists such as Wendell Latimer in California were tabulating the oxidation states of the elements and how strongly each pair of states attracts electrons — the same idea that ranks metals in the reactivity series in Lesson 2.
Lavoisier’s mercury gained mass as it became calx, and the air lost volume. Explain why these two observations together disproved the phlogiston theory, and write the modern equation for forming the calx, with oxidation numbers.
Reveal answer
If burning released phlogiston, the calx should weigh less than the metal and the air should gain something. Instead the metal gained mass and exactly the gas the air lost could be recovered from the calx. The mass gained came from the air: the metal had combined with oxygen, and mass was conserved.
2Hg(l) + O₂(g) → 2HgO(s). Mercury 0 → +2 (oxidised); oxygen 0 → −2 (reduced).
Where it is used now
- Safe drinking water. Water utilities chlorinate treated water — in Botswana as in cities from Nairobi to Lima — because chlorine and hypochlorite oxidise the proteins of bacteria and viruses. The same oxidation removes dissolved iron(II) from borehole water.
- Bleaches and hair dye. Hydrogen peroxide oxidises the coloured pigment melanin in hair, and the stain molecules on teeth; sodium chlorate(I) bleach oxidises the coloured compounds in stains.
- Breath tests for alcohol. Early breathalysers used orange acidified dichromate, which turned green as it oxidised ethanol; today’s handheld units oxidise the ethanol in a small fuel cell and measure the current.
- Iron tablets and blood tests. Pharmacies’ quality-control laboratories measure iron(II) in tablets by titrating with manganate(VII) until the purple colour stays.
- Industrial chemicals. Nitric acid for fertilisers is made by oxidising ammonia step by step (Ostwald process), the staircase in the nitrogen diagram.
Science now — and the scientist you need to be
What is happening in 2026
Redox chemistry is at the centre of the energy transition. Vanadium redox flow batteries, which store energy by shifting vanadium between oxidation states +2, +3, +4 and +5 in two tanks of solution, are being built for power grids; one of the largest, in Dalian, China, was connected in 2022 at 100 MW and 400 MWh.
Steelmakers are replacing carbon with hydrogen as the reducing agent for iron ore: the Swedish HYBRIT project delivered its first hydrogen-reduced “fossil-free” steel in 2021, and plants on a much larger scale are under construction.
Artificial intelligence is now part of the search for new redox materials. In January 2024 Microsoft and the US Pacific Northwest National Laboratory reported using AI screening to narrow some 32 million candidate materials to a handful for a battery electrolyte, and made one that uses less lithium.
The results still had to be tested in a real laboratory — the machine suggests; the experiment decides. Chemists are also working on electrochemical cells that reduce carbon dioxide back into fuels, which is photosynthesis done with electricity.
The mindshift
What you need to be able to do
Look at any equation and say which species loses electrons, which gains them, and by how much, using oxidation numbers where charges do not help. Write each half with its electrons and state symbols, and join them without leftover electrons. These skills are needed in every remaining lesson of this unit.
See an example
Calcium hypochlorite, Ca(OCl)₂: calcium +2, oxygen −2, so each chlorine is +1. Chlorine at +1 is a strong oxidising agent because it can take two electrons and fall to −1.
A correct half-equation: ClO⁻(aq) + H₂O(l) + 2e⁻ → Cl⁻(aq) + 2OH⁻(aq). Charge: left −1 − 2 = −3; right −1 − 2 = −3. If the assistant wrote only one electron, its equation does not balance — the kind of check you now make yourself.
Key terms
| Oxidation | Loss of electrons; gain of oxygen; an increase in oxidation number. |
| Reduction | Gain of electrons; loss of oxygen; a decrease in oxidation number. |
| Redox reaction | A reaction in which oxidation and reduction happen together, electrons passing from one species to another. |
| Half-equation | An equation for the oxidation or the reduction alone, with electrons shown; it balances atoms and charge. |
| Ionic equation | An equation showing only the species that change, with spectator ions left out. |
| Spectator ion | An ion present in the mixture that is unchanged by the reaction. |
| Oxidation number (state) | The charge an atom would have if all its bonds were ionic, assigned by rules; written with the sign first. |
| Oxidising agent | A species that takes electrons from another and is itself reduced. |
| Reducing agent | A species that gives electrons to another and is itself oxidised. |
| Disproportionation | A redox reaction in which one element is both oxidised and reduced. |
Think like a scientist
Criterion C Black copper oxide can be reduced to copper by heating it in a stream of natural gas (methane) in a hard-glass tube; the unused gas is burned at a small hole at the end. A student wants to decide whether the black oxide is copper(II) oxide, CuO, or copper(I) oxide, Cu₂O.
She weighs five different samples of the oxide, heats each in the gas until it has all turned pink-brown, lets it cool in the gas stream and weighs the copper. Her hypothesis: “The black oxide is CuO, so the mass of copper will be 0.799 times the mass of oxide, because Cu has a relative atomic mass of 63.55 and CuO a formula mass of 79.55.”
| Mass of copper oxide / g | 1.00 | 1.50 | 2.00 | 2.50 | 3.00 |
|---|---|---|---|---|---|
| Mass of copper left / g | 0.80 | 1.20 | 1.60 | 2.10 | 2.39 |
(a) Calculate the ratio mass of copper ÷ mass of oxide for each sample, to three significant figures. [2]
(b) Identify the anomalous result and suggest two specific causes, stating the direction each would push the mass. [2]
(c) Describe the graph you would draw to find the ratio, and state its gradient. [3]
(d) Use the 2.00 g sample to determine the ratio of moles of copper to moles of oxygen, and hence the formula of the oxide. [3]
(e) State whether the hypothesis is supported, with data, and identify the oxidising and reducing agents in 4CuO + CH₄ → 4Cu + CO₂ + 2H₂O, the reaction in the tube. [2]
(f) Evaluate one step of the method and suggest one improvement. [2]
Model answer
(a) Ratios: 0.800, 0.800, 0.800, 0.840, 0.797.
(b) The 2.50 g sample, ratio 0.840, is anomalous — well above the other four (0.797–0.800).
Cause 1: incomplete reduction — some black oxide left inside the sample, so the “copper” still contains oxygen and weighs too much.
Cause 2: the gas was turned off before the tube had cooled, so hot copper reacted with air again and gained oxygen — again too heavy. Both push the mass up, which is the direction observed.
(c) Mass of copper (vertical axis, g) against mass of oxide (horizontal axis, g), both from zero, points plotted as crosses, the anomaly circled and ignored.
A straight line of best fit through the origin and the other four points.
Gradient = Δ(mass of copper) ÷ Δ(mass of oxide), read from the line
Gradient = 2.39 g ÷ 3.00 g
Gradient = 0.797, about 0.80 (accept 0.797–0.800; no unit, g ÷ g)
(d) Mass of oxygen = 2.00 g − 1.60 g
Mass of oxygen = 0.40 g
n(Cu) = 1.60 ÷ 63.55
n(Cu) = 0.0252 mol
n(O) = 0.40 ÷ 16.00
n(O) = 0.0250 mol
Ratio Cu : O = 0.0252 : 0.0250, which is 1.01 : 1, so 1 : 1 — the formula is CuO. (Cu₂O would give 2 : 1 and a copper ratio of 0.888.)
(e) Supported: the four reliable ratios are 0.797–0.800, matching the predicted 0.799 and far from the 0.888 of Cu₂O.
In 4CuO + CH₄ → 4Cu + CO₂ + 2H₂O, copper(II) oxide is the oxidising agent (copper +2 → 0, reduced) and methane is the reducing agent (carbon −4 → +4, oxidised).
(f) Deciding that reduction is complete “when it has turned pink-brown” is a judgement by eye, and a black core can hide inside; improvement: heat, cool in the gas and weigh, then heat again for 2 minutes and reweigh, repeating until two masses agree to 0.01 g (heating to constant mass). Also acceptable: keep the gas flowing until the tube is cool to stop re-oxidation.
Probe deeper — exam-style questions
Attempt every question before opening the answer. Marks in brackets show what a full answer is worth. Relative atomic masses: Fe 55.85, O 16.00, Mn 54.94.
Define oxidation and reduction in terms of electrons, and state which one happens to a chlorine molecule when it becomes chloride ions.
Reveal answer
Oxidation is the loss of electrons; reduction is the gain of electrons [1].
Cl₂ + 2e⁻ → 2Cl⁻: chlorine gains electrons, so it is reduced [1].
Determine the oxidation number of: chromium in Cr₂O₇²⁻; nitrogen in NH₄⁺; iron in FeCl₃.
Reveal answer
2Cr + 7(−2) = −2
2Cr = +12, so Cr is +6 [1]
N + 4(+1) = +1, so N = −3 [1]
Fe + 3(−1) = 0, so Fe = +3 — iron(III) chloride [1]
Silver ions are displaced from silver nitrate solution by copper metal. Write the oxidation half-equation, the reduction half-equation and the ionic equation, all with state symbols.
Reveal answer
Cu(s) → Cu²⁺(aq) + 2e⁻ [1]
Ag⁺(aq) + e⁻ → Ag(s) [1]
Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s) [1] (the silver half doubled so that two electrons are gained for the two lost).
In the blast furnace, Fe₂O₃ + 3CO → 2Fe + 3CO₂. Using oxidation numbers, identify the oxidising agent and the reducing agent.
Reveal answer
Iron: +3 in Fe₂O₃ → 0 in Fe, a decrease, so iron(III) oxide is reduced and is the oxidising agent [1].
Carbon: +2 in CO → +4 in CO₂, an increase, so carbon monoxide is oxidised and is the reducing agent [1].
Iron in a paint reacts with oxygen: 4Fe + 3O₂ → 2Fe₂O₃. Calculate the amount, in moles, of oxygen that reacts with 0.600 g of iron.
Reveal answer
n(Fe) = m ÷ M [1]
n(Fe) = 0.600 g ÷ 55.85 g mol⁻¹
n(Fe) = 0.01074 mol [1]
n(O₂) = 0.01074 × 3 ÷ 4 [1]
n(O₂) = 8.06 × 10⁻³ mol [1] (the November 2022 markscheme accepted 0.0080–0.0083 mol).
Explain why the reaction Cl₂(aq) + 2Br⁻(aq) → 2Cl⁻(aq) + Br₂(aq) is described as a redox reaction even though no oxygen is involved.
Reveal answer
Bromide ions lose electrons (oxidation number −1 → 0): they are oxidised [1].
Chlorine gains those electrons (0 → −1): it is reduced; oxidation and reduction occur together, so the reaction is redox [1].
7. In acid, manganate(VII) is reduced: MnO₄⁻(aq) + 8H⁺(aq) + 5e⁻ → Mn²⁺(aq) + 4H₂O(l). Combine this with the oxidation of iron(II) to iron(III) to give the ionic equation. An iron tablet dissolved in acid needs 25.0 cm³ of 0.0200 mol dm⁻³ KMnO₄ to reach the end-point. Calculate the mass of iron in the tablet.
Reveal answer
Iron half × 5: 5Fe²⁺ → 5Fe³⁺ + 5e⁻
MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O [1]
n(MnO₄⁻) = 0.0250 dm³ × 0.0200 mol dm⁻³
n(MnO₄⁻) = 5.00 × 10⁻⁴ mol [1]
n(Fe²⁺) = 5 × 5.00 × 10⁻⁴
n(Fe²⁺) = 2.50 × 10⁻³ mol [1]
m = 2.50 × 10⁻³ mol × 55.85 g mol⁻¹
m = 0.140 g, which is 140 mg of iron [1]
8. Hydrogen peroxide decomposes: 2H₂O₂(aq) → 2H₂O(l) + O₂(g). Use oxidation numbers to show that hydrogen peroxide is both oxidised and reduced in this reaction, and explain why it can act as an oxidising agent towards iodide ions but as a reducing agent towards manganate(VII).
Reveal answer
Oxygen in H₂O₂ is −1 (peroxide rule) [1].
In H₂O it is −2 (decrease: reduced); in O₂ it is 0 (increase: oxidised) — a disproportionation [1].
Because its oxygen sits in the middle at −1, it can go either way. With iodide, a reducing agent, H₂O₂ takes electrons and falls to −2 (H₂O₂ + 2I⁻ + 2H⁺ → 2H₂O + I₂) [1].
With manganate(VII), a much stronger oxidising agent, H₂O₂ is forced to give electrons and rises to 0, releasing O₂ [1].
Summary
- Oxidation is loss of electrons (gain of oxygen, rise in oxidation number); reduction is gain of electrons (loss of oxygen, fall in oxidation number) — OIL RIG — and the two always happen together.
- Half-equations show each side with its electrons and balance atoms and charge; combine them by matching electrons lost and gained, then cancel the electrons.
- Oxidation numbers are assigned by rules (elements 0, simple ions their charge, O −2, H +1, sums 0 or the ion’s charge) and give the Roman numerals in names such as iron(III) oxide, Fe₂O₃.
- The oxidising agent takes electrons and is reduced; the reducing agent gives electrons and is oxidised. Colour changes (purple manganate(VII), orange dichromate(VI), halogens) show agents at work.
- Burning, rusting, respiration, photosynthesis, displacement, extraction, electrolysis and batteries are all redox — the rest of this unit.
Next time → The reactivity series and displacement: why magnesium beats zinc, zinc beats copper and copper beats silver, how carbon and hydrogen fit into the league table, and a lab in which the temperature rise measures how hard each metal pushes its electrons away.