What this chapter covers6.1
Electrochemistry begins with a piece of bookkeeping. Before anything can be said about cells, electrode potentials or electrolysis, you have to be able to look at an equation and say which substance gave electrons away and which one took them — and to say it with a number rather than an impression.
That is the whole of topic 6.1, and it is worth more marks than its length suggests, because almost every other part of the syllabus quietly depends on it. Group 2 and Group 17 chemistry is redox. The reactions of nitrogen and sulfur are redox. Half the transition-metal chemistry at A Level is redox. Every one of those questions assumes you can assign an oxidation number in a few seconds and get it right.
The five things 6.1 asks you to do
1. Calculate oxidation numbers of elements in compounds and ions.
2. Use changes in oxidation number to help balance chemical equations.
3. Explain and use redox, oxidation, reduction and
disproportionation, both in terms of electron transfer and in terms of changes in
oxidation number.
4. Explain and use oxidising agent and reducing agent.
5. Use a Roman numeral to indicate the magnitude of the oxidation number of an element.
Two definitions of the same thing run through everything here. Redox as electron transfer is the physical picture: electrons leave one species and arrive at another. Redox as a change in oxidation number is the accounting version: a rule-based count that works even when no ion ever forms and no electron is fully transferred. Part one of this chapter builds the first, part two builds the second, and the point at which they are shown to be the same statement is the centre of the topic.
What is deliberately not here
Electrode potentials, E⊖ values, electrochemical cells, the Nernst equation and electrolysis with Faraday's constant are all A Level electrochemistry, taught as separate units later. They are built entirely on 6.1: a standard electrode potential is a number that says how strongly a half-equation pulls electrons, and it means nothing until half-equations are second nature. Get this chapter solid and that one becomes arithmetic.
A note on spelling before anything else. Cambridge uses oxidation number and oxidation state interchangeably, and so does this page; if a question says "state the oxidation state of chromium", it wants the same number as "calculate the oxidation number of chromium". The older British spellings — oxidise, oxidised, oxidising — are used throughout, and the mark schemes accept the American -ize forms as well.
Oxidation and reduction as electron transfer6.1.3
Sodium burning in chlorine is the cleanest case there is. A sodium atom loses one electron and becomes Na⁺; a chlorine atom gains one and becomes Cl⁻. Nothing else happens. One species lost electrons, the other gained them, and the two events are inseparable — the electron that left sodium is the electron that arrived at chlorine.
The definitions to quote
Oxidation is the loss of electrons.
Reduction is the gain of electrons.
A redox reaction is one in which one substance is reduced and another is oxidised. The
two always happen together: electrons have to come from somewhere and go somewhere.
The mnemonic is OIL RIG — Oxidation Is Loss, Reduction Is Gain — and it is worth the two seconds it takes to recall, because the word "reduction" pulls in the wrong direction. Reduction sounds like something being taken away, and what is taken away is charge, not electrons: gaining a negative particle reduces the oxidation number. That is where the name comes from.
Why the older definitions are still taught
Before electrons were known, oxidation meant reaction with oxygen, and reduction meant having oxygen removed. Iron oxide is reduced to iron in a blast furnace; magnesium is oxidised when it burns. Those definitions are not wrong, they are just narrow — and they extend naturally to hydrogen as well, because adding hydrogen and removing oxygen do the same thing to the electron count.
| Definition | Oxidation is | Reduction is | Where it fails |
|---|---|---|---|
| oxygen | gaining oxygen | losing oxygen | most redox reactions contain no oxygen at all |
| hydrogen | losing hydrogen | gaining hydrogen | same problem, and it inverts the oxygen rule, which confuses people |
| electrons | losing electrons | gaining electrons | nothing — but in a covalent molecule the transfer is partial, which is why oxidation numbers exist |
Check the three against one reaction. In Fe₂O₃ + 3CO → 2Fe + 3CO₂, iron loses oxygen, so it is reduced on the oldest definition; iron goes from Fe³⁺ to metallic iron, gaining three electrons each, so it is reduced on the electron definition too. Carbon monoxide gains oxygen and is oxidised. All three accounts agree, and they agree because they are three views of the same electron movement.
The trap in that last statement
"When a substance burns, it is reduced" sounds plausible because burning destroys things. It is oxidation: the substance combines with oxygen and loses electrons to it. And "the aluminium ions are oxidised when aluminium is extracted from aluminium oxide" is backwards for the same reason — Al³⁺ gains three electrons to become Al, which is reduction. Extracting a metal from its ore is always a reduction of the metal, whatever is doing it.
What actually moves
In sodium chloride the electron is genuinely handed over: two ions form, and their charges are real and measurable. In hydrogen chloride nothing of the sort happens. HCl is a covalent molecule, and the bonding pair is shared — pulled towards chlorine, certainly, but shared. No ion forms, no whole electron moves, and yet H₂ + Cl₂ → 2HCl is unmistakably a redox reaction: hydrogen and chlorine both start as neutral elements and end up with an unequal share of the electrons between them.
This is the gap the second half of the chapter fills. Oxidation number is a way of counting electrons as if every bond were ionic, so that partial transfers can be handled with the same arithmetic as complete ones. It is a convention, not a measurement — and it works.
Half-equations6.1.3
A full equation hides the electrons. Mg + CuSO₄ → MgSO₄ + Cu is balanced and correct, and it says nothing at all about what moved. Splitting it in two fixes that:
A half-equation shows one half of the electron transfer
Mg → Mg2+ + 2e− the oxidation
Cu2+ + 2e− → Cu the reduction
Electrons on the right means they were lost: oxidation. Electrons on the left means they were gained: reduction. That single reading is the fastest way to label any half-equation you are given.
Two rules govern every half-equation, and both are checkable in a couple of seconds. The atoms must balance, and the charges must balance — total charge on the left equal to total charge on the right. Mg → Mg²⁺ + 2e⁻ has zero on the left and (+2) + (2 × −1) = 0 on the right, so it passes. An unbalanced charge is the single most reliable sign that a half-equation is wrong, and it is much easier to spot than a missing atom.
Why the electrons are the useful part
Every redox calculation you will ever do turns on one number: how many electrons the half-equation carries. It fixes the ratio in which two substances react, which fixes the titration answer. Get the species right and the electron count wrong and everything downstream is out by a factor — usually a tidy, plausible-looking factor, which is why the error survives to the end of the answer.
Combining half-equations6.1.2
Two half-equations add up to the full ionic equation, and the rule for adding them has one condition: the electrons must cancel exactly. If one half releases two electrons and the other takes five, neither is wrong — they just have to be scaled until both handle ten, the lowest number they both divide into.
Worked example — iron(II) with manganate(VII)
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O (takes 5)
Fe²⁺ → Fe³⁺ + e⁻ (releases 1)
Scale. Five electrons against one, so multiply the iron half by 5 — every species in it, not just the electrons.
5Fe²⁺ → 5Fe³⁺ + 5e⁻
Add and cancel. Five electrons on each side disappear:
MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺
The 1 : 5 ratio that falls out is the whole reason for doing this: one mole of manganate(VII) oxidises five moles of iron(II), and no titration calculation can be done without that number.
Exam alert
When you scale a half-equation, multiply every species in it. The commonest version of this mistake is to double the electrons and forget the H⁺ and the water, which leaves an equation that still looks reasonable and no longer balances for hydrogen. Check the charges at the end: if the total charge on the left does not equal the total on the right, something was left behind.
Notice what the combined equation does not contain: no sulfate, no potassium, no chloride from whatever salts these ions arrived in. Those are spectator ions, and section 8 explains why leaving them out is not a simplification but a more accurate description of the reaction.
Oxidising agents and reducing agents6.1.4
These two terms cause more lost marks than anything else in the topic, and for one reason: they are named after what a substance does to its partner, not after what happens to itself.
The definitions, with the consequence attached
An oxidising agent oxidises something else, by taking electrons from it. So the oxidising agent is itself reduced.
A reducing agent reduces something else, by giving electrons to it. So the reducing agent is itself oxidised.
Both halves are worth stating in an answer. "Chlorine is the oxidising agent because it is reduced from 0 to −1" scores where "chlorine is the oxidising agent" often does not.
How to get it right every time under pressure
Find the species that is reduced — the one whose oxidation number falls. That species is the oxidising agent. Then the other one is the reducing agent. Working from "reduced → oxidising agent" is one inversion to remember instead of two, and it never comes out backwards.
The agent is the whole species, not the atom inside it. In an acidified manganate(VII) reaction, manganese is the element that changes, but the oxidising agent is the MnO₄⁻ ion — that is what is added from the burette and that is what is written in the equation. Answers that say "manganese is the oxidising agent" lose the mark.
The agents worth knowing on sight6.1.4
A small number of substances appear again and again as oxidising or reducing agents across the whole syllabus. Knowing the products they turn into is what lets you write the half-equation without being told it.
| Oxidising agent | Becomes | Electrons | Conditions and what you see |
|---|---|---|---|
| MnO₄⁻, manganate(VII) | Mn²⁺ | 5 | acidified with dilute H₂SO₄; purple to colourless |
| Cr₂O₇²⁻, dichromate(VI) | 2Cr³⁺ | 6 | acidified; orange to green |
| MnO₂, manganese(IV) oxide | Mn²⁺ | 2 | warmed with concentrated HCl, making chlorine |
| Cl₂ and the halogens | 2Cl⁻ | 2 | oxidising power falls down the group |
| Fe³⁺ | Fe²⁺ | 1 | a mild one; yellow-brown to pale green |
| H₂O₂ | 2H₂O | 2 | its usual role, but see below |
| concentrated H₂SO₄ | SO₂ (or S, or H₂S) | 2 or more | hot; how far it goes depends on the reducing agent |
| dilute HNO₃ | NO | 3 | with copper; the colourless NO browns in air |
| Reducing agent | Becomes | Electrons | Where it turns up |
|---|---|---|---|
| reactive metals — Na, Mg, Zn, Fe | the metal ion | 1–3 | displacement, and reactions with acid |
| Fe²⁺ | Fe³⁺ | 1 | the standard substance in redox titrations |
| I⁻ and the other halide ions | I₂ | 2 | reducing power increases down the group |
| S₂O₃²⁻, thiosulfate | S₄O₆²⁻ | 2 per pair | titrated against iodine, with starch |
| C₂O₄²⁻, ethanedioate | 2CO₂ | 2 | titrated warm against manganate(VII) |
| SO₂ and sulfites | SO₄²⁻ | 2 | decolourises acidified dichromate(VI) |
| H₂, C, CO | H₂O, CO₂ | 2 | industrial reduction of metal oxides |
| H₂O₂ | O₂ | 2 | only against a stronger oxidising agent |
Hydrogen peroxide in both tables
Oxygen in H₂O₂ is at −1, which is halfway between the 0 of O₂ and the −2 of water — so peroxide can go either way. Against iodide it is the oxidising agent and is reduced to water. Against manganate(VII), which is a stronger oxidising agent, it is the reducing agent and is oxidised to oxygen gas. A species sitting at an intermediate oxidation state can always be pushed in either direction, and that is also the condition for disproportionation in section 16.
Exam alert
"Which is the better oxidising agent?" is not answered by how vigorous the reaction looks. At AS level the evidence is a displacement result: if A oxidises B's ion but B does not oxidise A's, A is the stronger oxidising agent. At A Level the same ordering comes out of E⊖ values with a number attached, which is what makes electrode potentials worth learning.
Metals with acids6.1.3
Magnesium in dilute hydrochloric acid fizzes, the metal disappears and hydrogen comes off. It is the first reaction most people meet, and it is redox from beginning to end.
Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)
Written like that, the chloride is in the way. Strip it out and the reaction is:
Mg(s) + 2H⁺(aq) → Mg²⁺(aq) + H₂(g)
The two halves
Mg → Mg²⁺ + 2e⁻ magnesium is oxidised, 0 → +2
2H⁺ + 2e⁻ → H₂ hydrogen ions are reduced, +1 → 0
So the magnesium is the reducing agent and the H⁺ ions — that is, the acid — are the oxidising agent. The chloride ions never change: −1 before, −1 after.
Any metal above hydrogen in the reactivity series does this; any metal below it does not. Copper and silver sit below hydrogen, which is why copper does not dissolve in dilute hydrochloric or sulfuric acid however long you leave it — Cu holds its electrons more tightly than H₂ does, so there is nothing to drive the transfer.
Copper and nitric acid: a different oxidising agent
Copper does dissolve in dilute nitric acid, and students often read that as the H⁺ finally winning. It is not. The oxidising agent is the nitrate ion, not the hydrogen ion:
3Cu + 8HNO₃ → 3Cu(NO₃)₂ + 2NO + 4H₂O
Nitrogen falls from +5 in HNO₃ to +2 in NO; copper rises from 0 to +2. No hydrogen gas is produced at all, which is the observation that gives the game away. And note that only two of the eight nitrogens are reduced — the other six leave as nitrate in the copper(II) nitrate, still at +5.
Why hot concentrated sulfuric acid behaves differently too
Dilute sulfuric acid with magnesium gives hydrogen: the H⁺ is the oxidising agent, as usual. Hot concentrated sulfuric acid is a different reagent altogether, because at that concentration the sulfate itself is a powerful oxidising agent — sulfur is at +6 and has a long way to fall. With a mild reducing agent it goes to SO₂ (+4); with a strong one such as iodide it can go all the way to H₂S (−2), a change of eight per sulfur. Recognising which species is doing the oxidising is the whole question.
Exam alert
When you are asked to write the ionic equation for a metal with an acid, the acid appears as H⁺(aq), and the number of them is set by the charge on the metal ion: two for Mg²⁺, three for Al³⁺. The state symbols are worth including — the metal is (s), the hydrogen is (g), and the fact that a gas leaves the solution is part of why the reaction goes to completion.
Displacement and spectator ions6.1.3
Drop a strip of zinc into blue copper(II) sulfate solution and three things happen at once: the zinc gets coated in pink-brown copper, the blue colour fades, and the solution warms up. All three are consequences of one electron transfer.
Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s)
What the halves say
Zn → Zn²⁺ + 2e⁻ 0 → +2, oxidised, the reducing agent
Cu²⁺ + 2e⁻ → Cu +2 → 0, reduced, the oxidising agent
Two electrons cross for each zinc atom. The blue fades because the colour belongs to Cu²⁺(aq), and those ions are being removed from the solution; the coating is the copper they turn into.
The sulfate is a spectator
Sulfate appears on both sides of the equation, unchanged, at every stage. Sulfur is +6 in SO₄²⁻ before and +6 after; nothing has happened to it. Ions like that are called spectator ions, and the ionic equation leaves them out:
Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)
Leaving them out is not a shortcut
The ionic equation is the more truthful description. Zinc in copper(II) nitrate does exactly the same thing, at the same rate, with the same colour change — because the anion was never involved. Writing the equation without it says so. It also makes the electron count visible, which the full equation hides.
Which pairs react
A metal displaces the ions of any metal below it in the reactivity series, and no others. That ordering is a statement about how readily each metal gives its electrons away: magnesium parts with them easily, silver does not. Zinc is above copper, so zinc reduces Cu²⁺; copper is below zinc, so copper in zinc sulfate solution does nothing at all — no colour change, no coating, no reaction to write.
One animation from this sequence could not be recovered
Its data is incomplete in the source file, which no emulator can repair. It belonged with the displacement material, so the interactive grid above replaces it and goes further: it works out the direction of every metal–ion pair from the reactivity order, then derives the half-equations, the electron count and the expected observation rather than showing a fixed set of results.
Exam alert
"Explain why no reaction occurs" is a redox question, not a solubility one. The answer is that the metal is below the metal in the solution in the reactivity series, so it cannot give electrons to those ions — and the observation to quote is that the solution keeps its colour and the metal stays clean.
Why oxidation number exists6.1.1
Electron transfer is a clean idea in ionic reactions and an awkward one everywhere else. Burning methane is a redox reaction — nobody doubts that — but no ions form at any point, and no electron is ever handed over completely. So how much was carbon oxidised by?
The definition
The oxidation number of an atom in a compound is the charge that atom would have if the compound consisted only of separate ions — that is, if every bonding pair were given outright to the more electronegative of the two atoms sharing it.
The phrase to hold onto is would have. Oxidation number is a deliberate fiction: it pretends every bond is ionic so that the electrons can be counted. In NaCl the fiction is exactly true and the numbers +1 and −1 are real charges. In H₂O it is not true at all — water is covalent, and the oxygen does not carry a real −2 charge — but the count still works, and it is the count that lets you say water is more oxidised than hydrogen gas.
Working out a value the long way, once
Take water. Each O–H bond is a shared pair, and oxygen is more electronegative than hydrogen, so hand both electrons of each pair to the oxygen. Oxygen started with six outer electrons and now notionally holds eight: two more than it owns, so its oxidation number is −2. Each hydrogen started with one and now has none: one fewer than it owns, so +1. The three numbers add to zero, because the molecule is neutral.
Where the sign convention comes from
A negative oxidation number means the atom has notionally gained electrons — it won the tug-of-war. A positive one means it lost them. That is the same convention as ionic charge, which is why the two agree for simple ions and why an oxidation number that goes up means the atom lost electrons, which is oxidation.
Doing this from electronegativities every time would be unbearable, and in a large molecule it is not even practical. So the fiction is packaged as a short list of rules, applied in order — and the whole of 6.1.1 is those rules and the arithmetic that follows from them.
The rules, in the order you apply them6.1.1
The rules are not a list to be scanned; they are a sequence. Everything with a fixed value is filled in first, and whatever is left over is found by arithmetic at the end. Applied in that order, almost every question in this topic takes about fifteen seconds.
| Order | Rule | Examples |
|---|---|---|
| 1 | An element on its own is 0, however many atoms are joined together | Na, O₂, Cl₂, S₈, P₄ — all zero |
| 2 | A simple monatomic ion takes the charge of the ion | Na⁺ is +1, S²⁻ is −2, Fe³⁺ is +3 |
| 3 | Fluorine is always −1 in compounds | no exceptions at all — it is the most electronegative element |
| 4 | Group 1 is +1, Group 2 is +2, aluminium is +3 | the metal is never negative in a compound |
| 5 | Hydrogen is +1 — except −1 in metal hydrides | +1 in HCl and H₂O; −1 in NaH and CaH₂ |
| 6 | Oxygen is −2 — except −1 in peroxides and positive with fluorine | −2 in H₂O; −1 in H₂O₂; +2 in OF₂ |
| 7 | The other halogens are −1 unless combined with oxygen or a more electronegative halogen | −1 in NaCl; +1 in ClO⁻; +1 for iodine in ICl |
| 8 | Everything sums to 0 in a neutral compound, or to the charge in an ion | this is the step that finds the unknown |
Worked example — sulfur in H₂SO₄
Two hydrogens at +1 give +2 (rule 5). Four oxygens at −2 give −8 (rule 6). The compound is neutral, so everything must add to zero (rule 8):
(+2) + x + (−8) = 0 → x = +6
Sulfur is +6. Note what was not needed: nothing about the structure of sulfuric acid, nothing about which atom is bonded to which. The rules do not care.
Worked example — chromium in Cr₂O₇²⁻
Seven oxygens at −2 give −14. The ion carries −2, so the two chromiums must supply +12:
2x + (−14) = −2 → 2x = +12 → x = +6
Each chromium is +6, not +12. Forgetting the final division is the most frequent single error in this topic, and it is invisible in the working unless you write the "2x" step down.
Exam alert
The charge on an ion belongs to the whole ion, never to one atom in it. MnO₄⁻ is a 1− ion, and manganese in it is +7; those two numbers are answers to different questions. Read the question carefully: "the oxidation number of manganese in the manganate(VII) ion" wants +7.
The awkward cases6.1.1
Every exception in the list above exists because of one idea: electronegativity decides who notionally keeps the electrons, and the usual answer occasionally changes.
Peroxides
In H₂O₂ the structure is H–O–O–H. The O–O bond joins two identical atoms, so its pair is split evenly and neither oxygen gains from it. Each oxygen gains only from its bond to hydrogen: one electron, so −1. The rule "oxygen is −2" assumes oxygen is bonded only to something less electronegative, and in a peroxide it is bonded to itself.
Metal hydrides
Hydrogen is more electronegative than sodium, so in NaH the hydrogen takes the pair and becomes H⁻, at −1. This is not a curiosity: sodium hydride and calcium hydride react violently with water precisely because H⁻ is a powerful reducing agent looking to get back to 0.
Oxygen with fluorine
OF₂ is the one compound where oxygen is positive. Fluorine outranks everything, takes both pairs, and oxygen is left at +2. This appears in exams roughly as often as it appears in laboratories, which is to say rarely — but it is the standard test of whether a candidate has learned the rule order or just the rule.
Fractions, and what they mean
Fe₃O₄ gives iron an oxidation number of +8/3, which is not a charge any atom can have. The fraction is the answer to the question actually asked — the average over the three irons — and it is a signal that the atoms are not all alike. Fe₃O₄ is really FeO·Fe₂O₃: one iron at +2 and two at +3, averaging +8/3. The same happens for sulfur in S₄O₆²⁻, where the +2.5 average comes from an S–S bond in the middle of the ion.
The two-site trap
Ammonium nitrate, NH₄NO₃, has two nitrogens. Apply the rules to the whole formula and you get an average of +1, which is arithmetically correct and chemically useless — neither nitrogen is at +1. Split the salt into its ions first: nitrogen is −3 in NH₄⁺ and +5 in NO₃⁻. Whenever an element appears in two different environments in one formula, deal with the ions separately.
Organic compounds get an average too
Carbon in ethanol, C₂H₅OH, comes out at −2 on average, and in ethanoic acid at 0 — even though the two carbon atoms in each molecule are in quite different environments. Cambridge does not ask you to assign oxidation numbers to individual carbons at AS level, but the direction of change is still useful: oxidising ethanol to ethanoic acid raises the average carbon oxidation number, which is exactly why acidified dichromate(VI) is the reagent that does it.
Oxidation number and the Periodic Table6.1.1
The values an element can reach are not arbitrary. They are set by how many outer electrons it has and how many it needs, which is why the pattern runs straight across a period.
| Element | Outer electrons | Highest | Lowest | Seen in |
|---|---|---|---|---|
| Na | 1 | +1 | 0 | Na₂O, NaCl |
| Mg | 2 | +2 | 0 | MgO, MgCl₂ |
| Al | 3 | +3 | 0 | Al₂O₃, AlCl₃ |
| Si | 4 | +4 | −4 | SiO₂, SiH₄ |
| P | 5 | +5 | −3 | P₄O₁₀, PH₃ |
| S | 6 | +6 | −2 | SO₃ and H₂SO₄, H₂S |
| Cl | 7 | +7 | −1 | ClO₄⁻, NaCl |
Two patterns, both worth stating in an answer. The highest oxidation number equals the number of outer electrons, because that is everything the atom has to give: sulfur reaches +6 and cannot go further. The lowest is the number of electrons needed to fill the outer shell, counted negative: chlorine needs one, so it stops at −1.
Why sodium is never +2
Removing a second electron from sodium means breaking into a full 2p shell, and the second ionisation energy is nearly ten times the first. No chemical reaction supplies that much energy, so the compound never forms. Oxidation numbers are constrained by ionisation energies, which is the link back to topic 1.
Exam alert
An impossible oxidation number is the best self-check there is. If your arithmetic gives sulfur +8 or chlorine +9, you have not made a chemistry mistake, you have made an arithmetic one — most often by using 0 instead of the ion's charge on the right-hand side. Go back to the sum rather than to the rules.
Roman numerals in names6.1.5
Iron forms two chlorides. Calling them both "iron chloride" is useless, and the older names — ferrous and ferric — require you to know which is which. The systematic name puts the oxidation number in brackets, as a Roman numeral, immediately after the element it belongs to.
What the numeral means
The Roman numeral gives the magnitude of the oxidation number of the element it follows. It carries no sign, so iron(III) means +3; there is no notation for a negative oxidation number in a name, because these names are only used where the element is positive.
| Name | Formula | Why |
|---|---|---|
| iron(II) chloride | FeCl₂ | Fe²⁺ needs two Cl⁻ |
| iron(III) chloride | FeCl₃ | Fe³⁺ needs three Cl⁻ |
| copper(I) oxide | Cu₂O | two Cu⁺ balance one O²⁻ |
| iron(III) sulfate | Fe₂(SO₄)₃ | two at +3 against three at −2 |
| lead(IV) oxide | PbO₂ | +4 against two oxides at −2 |
| manganese(IV) oxide | MnO₂ | the reagent used to make chlorine from HCl |
The numeral inside an ion
The same convention names oxoanions, and there the numeral belongs to the central element rather than to a metal. Potassium manganate(VII) is KMnO₄, in which manganese is +7. Potassium dichromate(VI) is K₂Cr₂O₇, in which each chromium is +6 — "di" counts the chromiums, "(VI)" gives their oxidation number, and the two numbers are unrelated.
This is also how the older names are being replaced. What used to be sodium chlorate, sodium hypochlorite and sodium perchlorate are chlorate(V), chlorate(I) and chlorate(VII) — three names that had to be memorised individually, replaced by three that can be worked out.
The commonest formula error
Writing chromium(III) sulfate as Cr₃(SO₄)₂, by swapping the charges over into the wrong positions. The subscripts come from making the charges cancel, not from copying the numbers across: Cr³⁺ with SO₄²⁻ needs two chromiums (+6) and three sulfates (−6), giving Cr₂(SO₄)₃. Check any formula you write by adding up the charges and requiring zero.
Redox restated: change in oxidation number6.1.3
Everything in part one can now be said again, in numbers, and this version works on any equation at all — ionic or covalent, aqueous or gaseous.
The second pair of definitions
Oxidation is an increase in oxidation number.
Reduction is a decrease in oxidation number.
A reaction is redox if any element changes its oxidation number. If nothing changes,
nothing has been oxidised or reduced, whatever else may have happened.
The two accounts agree because they are the same count. Losing an electron makes an atom one unit more positive, so oxidation number rises by one; gaining an electron drops it by one. A change of +2 means two electrons left, and in a half-equation that is exactly the number written.
| Reaction | Electron account | Oxidation-number account |
|---|---|---|
| 2Na + Cl₂ → 2NaCl | Na loses 1e⁻, Cl gains 1e⁻ | Na 0 → +1, Cl 0 → −1 |
| H₂ + Cl₂ → 2HCl | no ions form; the pair is pulled towards Cl | H 0 → +1, Cl 0 → −1 |
| CH₄ + 2O₂ → CO₂ + 2H₂O | hard to state — nothing is transferred outright | C −4 → +4, O 0 → −2 |
The third row is the point. Nobody can sensibly say how many electrons carbon "lost" in burning methane, but its oxidation number rises by eight, and that is a definite, checkable statement.
Three reaction types that are never redox
Neutralisation: H⁺ + OH⁻ → H₂O, and every element keeps its value. Precipitation: Ag⁺ + Cl⁻ → AgCl — the ions change partners, not oxidation numbers. Thermal decomposition of a carbonate: CaCO₃ → CaO + CO₂, with calcium at +2, carbon at +4 and oxygen at −2 on both sides. Recognising these instantly saves time in a multiple-choice paper.
Finding what is oxidised and what is reduced6.1.3
A reliable four-step method, worth practising until it is automatic:
The method, on 2Al + Fe₂O₃ → Al₂O₃ + 2Fe
1. Assign every element on both sides. Left: Al 0, Fe +3, O −2. Right: Al +3, O −2, Fe 0.
2. Ignore everything unchanged. Oxygen is −2 throughout; it is not involved in the transfer at all, despite being the thing that changes hands.
3. Name the direction. Al goes 0 → +3, an increase: oxidised. Fe goes +3 → 0, a decrease: reduced.
4. Convert to agents. Fe₂O₃ is reduced, so Fe₂O₃ is the oxidising agent. Al is oxidised, so aluminium is the reducing agent.
Step 2 does more work than it looks. In the thermite reaction the oxygen moves from iron to aluminium, so the oldest definition of oxidation — gaining oxygen — gives the right answer for aluminium. But oxygen itself is unchanged at −2, which is what the numbers say and what the electron account confirms: the electrons went from aluminium to iron, and the oxide ions simply came along.
Exam alert
When a question asks "by how much does the oxidation number of nitrogen change?", check whether all the nitrogen changes. In 3Cu + 8HNO₃ → 3Cu(NO₃)₂ + 2NO + 4H₂O, only two of the eight nitrogens are reduced, from +5 to +2; the other six are still +5 in the nitrate. The answer is "a decrease of 3", for those two atoms — the average across all eight is not what is being asked.
Disproportionation6.1.3
Usually two different elements share the work: one is oxidised, the other reduced. Occasionally one element does both at once.
Definition
Disproportionation is a reaction in which the same element is simultaneously oxidised and reduced. It starts from a single oxidation state and ends up in two — one higher, one lower.
The standard example is chlorine in cold dilute sodium hydroxide:
Cl₂ + 2NaOH → NaCl + NaClO + H₂O
Chlorine starts at 0 in Cl₂. In NaCl it is −1, reduced. In NaClO it is +1, oxidised. Half of it went each way, so chlorine is its own oxidising agent and its own reducing agent.
The same reagents, hotter
Chlorine with hot concentrated sodium hydroxide disproportionates further:
3Cl₂ + 6NaOH → 5NaCl + NaClO₃ + 3H₂O
Now the upward branch goes all the way to +5 in the chlorate(V) ion. The electron bookkeeping explains the coefficients: one chlorine rising by 5 has to be matched by five chlorines each falling by 1, which is why five chlorides appear for every chlorate(V). The equation is not something to memorise — it is something the oxidation numbers construct.
Three more worth recognising
Chlorine in water: Cl₂ + H₂O → HCl + HClO, which is why chlorinated water is disinfectant — the chlorate(I) does the work.
Hydrogen peroxide decomposing: 2H₂O₂ → 2H₂O + O₂, with oxygen going from −1 to −2 and to 0. It is the standard example that does not involve a halogen.
Copper(I) in solution: 2Cu⁺ → Cu + Cu²⁺, +1 going to 0 and +2. This is why copper(I) salts are unstable in water while copper(I) oxide is not.
What makes it possible
An element can only disproportionate from an intermediate oxidation state — one with room to move in both directions. Chlorine at 0 has −1 below and +1, +5, +7 above. Chloride at −1 is already at the bottom and can only be oxidised; chlorate(VII) at +7 can only be reduced. Neither can disproportionate, and being asked to explain why is a common exam question.
Not disproportionation
Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂ is redox, and chlorine changes, and it can look like the same thing. It is not: chlorine only falls, from 0 to −1, and bromine only rises. Two elements, one direction each — an ordinary redox reaction. The test is whether one element goes both ways.
Balancing with oxidation numbers6.1.2
Trial and error balances simple equations quickly and redox equations slowly, if at all. The oxidation numbers give a direct route, because they carry the one constraint that matters: the total rise must equal the total fall. Every electron lost by something was gained by something else, so the bookkeeping has to come out even.
Worked example — iron(III) oxide reduced by carbon monoxide
Fe₂O₃ + CO → Fe + CO₂, unbalanced.
1. Assign. Fe +3 → 0, a fall of 3 per iron. C +2 → +4, a rise of 2 per carbon.
2. Match the totals. The lowest common multiple of 3 and 2 is 6, so two irons must fall (2 × 3 = 6) for every three carbons that rise (3 × 2 = 6).
3. Put the numbers in. One Fe₂O₃ already contains two irons, so it needs three CO:
Fe₂O₃ + 3CO → 2Fe + 3CO₂
4. Check the atoms. Iron 2 = 2. Carbon 3 = 3. Oxygen 3 + 3 = 6 on the left, 6 on the right. Balanced — and the oxygens balanced themselves, which is what usually happens once the electron count is right.
Worked example — the harder shape
Cu + HNO₃ → Cu(NO₃)₂ + NO + H₂O, with dilute acid.
Assign the changes. Cu 0 → +2, a rise of 2. N +5 → +2 in the NO, a fall of 3. Nothing else changes — the nitrogen in the copper nitrate is still +5.
Match. LCM of 2 and 3 is 6: three coppers rising 2 each against two nitrogens falling 3 each. So 3Cu and 2NO.
Complete the rest by inspection. Three coppers need 3Cu(NO₃)₂, which uses six unchanged nitrates; with the two reduced nitrogens that is eight HNO₃ in total, and the eight hydrogens make four waters:
3Cu + 8HNO₃ → 3Cu(NO₃)₂ + 2NO + 4H₂O
Exam alert
"Use changes in oxidation number to balance the equation" is a specific instruction, and the marks are for the working: state the change for each element, state the ratio the changes force, then write the balanced equation. An unexplained correct equation can score less than a partly wrong one with the reasoning shown.
Building a half-equation in acid6.1.2
Given only "manganate(VII) is reduced to manganese(II) in acid solution", you can construct the whole half-equation from four steps, in a fixed order. The order matters: each step disturbs only what comes after it.
The four steps
1. Balance the element being oxidised or reduced.
2. Balance the oxygen by adding H₂O to the side that needs it.
3. Balance the hydrogen by adding H⁺ — including the hydrogens the
water just introduced.
4. Balance the charge by adding electrons to the more positive side.
The four steps on MnO₄⁻ → Mn²⁺
1. One manganese each side already: MnO₄⁻ → Mn²⁺
2. Four oxygens on the left, none on the right, so add four waters on the right:
MnO₄⁻ → Mn²⁺ + 4H₂O
3. Eight hydrogens on the right now, none on the left, so add eight H⁺ on the left:
MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O
4. Left: (−1) + (+8) = +7. Right: +2. Add five electrons to the left to bring +7 down
to +2:
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
And the check: manganese fell from +7 to +2, a change of 5, matching the five electrons exactly. If those two numbers disagree, the equation is wrong.
Why acid, and what changes in alkali
The H⁺ ions are not decoration: they are consumed by the reaction, which is why these reagents only work in acidic solution and why a titration must be acidified. In alkaline conditions the same job is done with OH⁻ ions and water on the other side, but 9701 sets its half-equation questions in acid, so the four steps above are the ones to learn.
Electrons on the wrong side
Adding electrons to make the charges balance is arithmetic, and it is easy to add them to the side that makes the sum work while contradicting the chemistry. Use the oxidation-number change as the independent check: manganese was reduced, reduction is gain of electrons, so the electrons must be on the left, with the manganate. Two routes to the same answer, and they should always agree.
Using the equation: redox titrations6.1.2
Everything so far has been qualitative or definitional. This section is where the balanced equation earns its keep, because a redox titration calculation is an ordinary moles calculation with one extra step — and that step is the ratio the electrons fixed.
Worked example — finding the concentration of an iron(II) solution
25.0 cm³ of an iron(II) sulfate solution, acidified with dilute sulfuric acid, needs 24.60 cm³ of 0.0200 mol dm⁻³ potassium manganate(VII) to reach the first permanent pink.
1. The equation, from the two half-equations:
MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺
2. Moles from the burette. n(MnO₄⁻) = 0.0200 × 24.60/1000 = 4.92 × 10⁻⁴ mol
3. Across the ratio. 1 : 5, so n(Fe²⁺) = 5 × 4.92 × 10⁻⁴ = 2.46 × 10⁻³ mol
4. Back to a concentration. 2.46 × 10⁻³ ÷ (25.0/1000) = 0.0984 mol dm⁻³
Exam alert — the three details that carry marks
Acidify with dilute sulfuric acid. Hydrochloric acid would itself be oxidised by manganate(VII), giving chlorine and a titre that is too high; nitric acid is an oxidising agent in its own right and attacks the iron(II) before the titration starts.
No indicator for manganate(VII). The purple colour is discharged as long as iron(II) remains; the end point is the first permanent pale pink. For dichromate(VI) an indicator is needed, because orange fading into green gives no sharp moment.
Iodine–thiosulfate needs starch, added late. Starch is added when the solution is already straw-coloured, near the end point — added early it binds the iodine too strongly and the end point comes late.
The iodine–thiosulfate titration, which is two reactions
This one appears constantly because it measures oxidising agents indirectly. An unknown oxidising agent is added to excess potassium iodide, liberating iodine:
2I⁻ → I₂ + 2e⁻
The iodine released is then titrated against standard sodium thiosulfate:
I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻
So the thiosulfate measures the iodine, and the iodine measures the original oxidising agent. Two ratios are used in sequence, and losing track of which one applies where is the usual failure. Notice that the thiosulfate–iodine ratio is 2 : 1, and that in the second equation sulfur goes from +2 to +2.5 — a change of half a unit per sulfur, which comes out as one electron for every two thiosulfate ions.
Why a fractional oxidation number is not a problem here
Tetrathionate averages +2.5 because two of its four sulfurs are joined to each other. The electron count is still a whole number: four sulfurs rising by half a unit each is two electrons, released by the two thiosulfate ions that combined. Fractional oxidation numbers never produce fractional electrons.
Checking an answer for sense
Three quick tests before moving on. Is the concentration in a plausible range — school solutions are usually 0.01 to 1 mol dm⁻³, so 84 mol dm⁻³ means a slipped factor of 1000? Did the ratio move the number in the right direction — more iron(II) than manganate(VII), so the flask concentration should come out larger than the burette one here? And has the volume been divided by 1000 exactly once, in each of the two places it appears?
Self-test6.1
Thirty-four questions covering the whole of 6.1, in random order, each with the reasoning rather than just the answer. Several are built around the specific errors described earlier in the chapter — the charge on an ion used as an atom's oxidation number, the agent named after the wrong half, the ratio left out of a titration.
Definitions to learn6.1
These are the wordings that earn the mark. Each is short enough to write out in an exam and specific enough that a paraphrase usually loses something.
| Term | Definition |
|---|---|
| oxidation | loss of electrons, or an increase in oxidation number |
| reduction | gain of electrons, or a decrease in oxidation number |
| redox reaction | a reaction in which one species is oxidised and another is reduced; electrons are transferred |
| oxidation number | the charge an atom would have if the compound consisted only of separate ions |
| oxidising agent | a species that takes electrons from another species, and is itself reduced |
| reducing agent | a species that gives electrons to another species, and is itself oxidised |
| half-equation | an equation showing either the oxidation or the reduction part of a redox reaction, including the electrons |
| disproportionation | a reaction in which the same element is simultaneously oxidised and reduced |
| spectator ion | an ion present in the reaction mixture that is unchanged, and so is omitted from the ionic equation |
| Roman numeral in a name | the magnitude of the oxidation number of the element it follows |
Two phrasings that lose marks
"An oxidising agent is a substance that is reduced" is true but incomplete — it does not say what the agent does, which is the point of the word "agent". Say that it oxidises another species by taking electrons from it, and add that it is itself reduced.
"Disproportionation is when a substance is oxidised and reduced" leaves out the crucial word. It is the same element, from a single oxidation state, going both ways.
Data used on this page6.1
Every oxidation number printed on this page, in the tables and inside the models, is computed from the formula by the rules in section 10 — none of them is a stored value. The tables below give the other data the page relies on.
| Most reactive | → | Least reactive | |||||||
|---|---|---|---|---|---|---|---|---|---|
| K | Na | Ca | Mg | Al | Zn | Fe | Pb | Cu | Ag |
| Couple | Electrons | Where it is used |
|---|---|---|
| MnO₄⁻ → Mn²⁺ | 5 | titration against iron(II) and ethanedioate |
| Cr₂O₇²⁻ → 2Cr³⁺ | 6 | titration against iron(II); oxidising alcohols |
| MnO₂ → Mn²⁺ | 2 | making chlorine from concentrated HCl |
| H₂O₂ → 2H₂O | 2 | peroxide as an oxidising agent |
| H₂O₂ → O₂ | 2 | peroxide as a reducing agent |
| NO₃⁻ → NO | 3 | copper with dilute nitric acid |
| SO₄²⁻ → SO₂ | 2 | hot concentrated sulfuric acid |
| Cl₂ → 2Cl⁻ | 2 | halogen displacement |
| Fe³⁺ → Fe²⁺ and Fe²⁺ → Fe³⁺ | 1 | the standard one-electron couple |
| I₂ → 2I⁻ and 2I⁻ → I₂ | 2 | iodine–thiosulfate titrations |
| 2S₂O₃²⁻ → S₄O₆²⁻ | 2 | thiosulfate as the reducing agent |
| C₂O₄²⁻ → 2CO₂ | 2 | ethanedioate, titrated warm |
| SO₂ → SO₄²⁻ | 2 | sulfur dioxide as a reducing agent |
| Zn → Zn²⁺ | 2 | a metal as the reducing agent |
These are reference values, not the official data booklet
The couples and the reactivity order above are the standard AS-level set and agree with the usual textbook treatments, but they are not reproduced from the Cambridge data booklet, and the booklet issued with your paper is the authority in an exam. The half-equations themselves are derived on this page from the formulae by the four-step procedure, so they are as good as that procedure — which is exactly why the page checks each one for atom and charge balance before showing it.