Sec 3 Chemistry — Oxidation and Reduction

Prepared by Miss Clarissa Ng · www.clartutors.com

Part A · What the words mean
1 Oxidation and Reduction Always Happen Together
A redox reaction is a chemical reaction in which oxidation and reduction occur at the same time. Oxidation and reduction are not two different kinds of reaction — they are the two halves of one.

A substance is oxidised or reduced if it undergoes at least one of these four changes. Learn the table as four pairs of opposites:

Oxidation — the substance…Reduction — the substance…
gains oxygenloses oxygen
loses hydrogengains hydrogen
loses electronsgains electrons
shows an increase in oxidation stateshows a decrease in oxidation state
Read the table in one direction only. Oxygen and hydrogen are opposites, but note that hydrogen sits the other way round from oxygen: losing hydrogen is oxidation, while gaining oxygen is also oxidation. Both rows are really the same electron story, which is why section 2 is the one to understand properly.
Oxidation and reduction shown by oxygen — the blast furnace
Fe2O3 (s) + 3CO (g) → 2Fe (s) + 3CO2 (g)
SubstanceOxygenWhat it means
Iron(III) oxideloses oxygenIt is reduced to iron — Fe goes from +3 to 0.
Carbon monoxidegains oxygenIt is oxidised to carbon dioxide — C goes from +2 to +4.

One substance takes the oxygen and the other gives it up, so oxidation and reduction have both happened. That is what makes the extraction of iron from its ore a redox reaction.

Oxidation and reduction shown by hydrogen
CH4 (g) + Cl2 (g) → CH3Cl (g) + HCl (g)
SubstanceHydrogenWhat it means
Methaneloses hydrogenIt is oxidised to chloromethane — C goes from −4 to −2.
Chlorinegains hydrogenIt is reduced to hydrogen chloride — Cl goes from 0 to −1.
Notice the pattern: the substance that loses something is always the one oxidised, and the substance that gains it is always the one reduced. Oxygen and hydrogen are just two different ways of watching the same transfer of electrons, which is where section 2 picks up.
2 The Real Definition: Electrons Move
Oxidation is the loss of electrons. Reduction is the gain of electrons. When electrons are transferred from one substance to another, both processes happen in the same reaction.

Take a piece of zinc and put it into copper(II) sulfate solution. The zinc atoms lose two electrons each and become zinc ions, while the copper(II) ions in solution gain those electrons and come out as copper metal:

Zn (s) + CuSO4 (aq) → ZnSO4 (aq) + Cu (s)
Zinc in copper(II) sulfate: electrons move acrossone substance loses electrons, the other gains them — at the same timeZnZn atomloses 2e⁻Zn²⁺zinc ionCu²⁺copper(II) iongains 2e⁻Cucopper atomOxidation — zinc is oxidisedZn → Zn²⁺ + 2e⁻Reduction — copper(II) is reducedCu²⁺ + 2e⁻ → Cuadding the two half-equations gives Zn + Cu²⁺ → Zn²⁺ + Cu

Reactions that involve the transfer of electrons can be written as two half-equations — one for the oxidation and one for the reduction:

Half-equationWhat it shows
OxidationZn → Zn2+ + 2e−Each zinc atom loses two electrons.
ReductionCu2+ + 2e− → CuEach copper(II) ion gains two electrons.
Exam habit: the electrons must cancel when you add the two half-equations together. If one half-equation has 2e− and the other has 3e−, multiply them up to 6e− each before combining — that is exactly how the equation for a redox titration is built.
Part B · Oxidation states
3 What an Oxidation State Is
The oxidation state of an element is the charge an atom would have if it existed as an ion in the compound. If the oxidation state increases, the substance is oxidised; if it decreases, the substance is reduced.

This is the most powerful of the four views, because it works even when no oxygen, no hydrogen and no obvious ions are involved. Watch both elements in the same equation:

Following the oxidation state through an equationN₂ + 3H₂ → 2NH₃ · every element starts at 0 in its pure formN₂ (g)+3H₂ (g)→2NH₃ (g)0−3reduced: 0 → −30+1oxidised: 0 → +1both changes happen together — a redox reaction
In that reaction nitrogen goes from 0 to −3, so nitrogen is reduced, and hydrogen goes from 0 to +1, so hydrogen is oxidised. One equation, both changes — a redox reaction.
4 The Rules for Working Out Oxidation States

Use the rules in order. Start with the element you already know, and let the sum rule finish the job.

Element or situationOxidation stateNote
A pure element0Na, Al, Cl2, N2, Ar — uncombined elements are always zero.
Group 1 metal in a compound+1Including sodium and potassium.
Group 2 metal in a compound+2Including magnesium and calcium.
Aluminium in a compound+3Fixed for aluminium.
Zinc in a compound+2Fixed for zinc — useful because zinc is a common reducing agent.
Hydrogen in a compound+1Except in hydrides such as NaH, where it is −1.
Oxygen in a compound−2Except in peroxides such as H2O2, where it is −1.
Fluorine in a compound−1Always −1; it is the most electronegative element.
Sum in a neutral compound= 0The oxidation states of all the atoms add up to zero.
Sum in a polyatomic ion= the charge on the ionFor CO32− the states add up to −2, not 0.
Exam habit: the two exceptions are the marks. Hydrogen in NaH and oxygen in H2O2 are the two places where the usual values flip, and questions are written to check that you know it.
5 Worked Example — a Compound and an Ion
(a) The oxidation state of chromium in potassium dichromate, K2Cr2O7
Potassium is Group 1, so K = +1. Oxygen = −2. Let chromium be x.
Sum in a neutral compound = 0:   2(+1) + 2x + 7(−2) = 0
2x = 12  →  chromium is +6
(b) The oxidation state of sulfur in the sulfate ion, SO42−
Oxygen = −2. Let sulfur be x.
Sum in a polyatomic ion = the charge on the ion:   x + 4(−2) = −2
x = −2 + 8 = sulfur is +6
(c) The oxidation state of manganese in the manganate(VII) ion, MnO4−
x + 4(−2) = −1  →  x = −1 + 8 = manganese is +7
That last value is why the next part of the chapter works: manganese sits at +7 in manganate(VII) and drops to +2 when it acts as an oxidising agent. A fall of five explains the 5e− in its half-equation.
6 Elements with Variable Oxidation States

Some elements take different oxidation states in different compounds — most transition metals, and some non-metals such as carbon, nitrogen, sulfur and chlorine. Because the oxidation state changes, the same two elements can form several different compounds, and the Roman numeral in the name tells you which one you are holding.

FormulaNameOxidation state of the metal
FeOiron(II) oxide+2 — iron is oxidised from 0 to +2 here
Fe2O3iron(III) oxide+3 — iron is oxidised from 0 to +3 here
FeCl2iron(II) chloride+2
FeCl3iron(III) chloride+3
Exam habit: the Roman numeral is not decoration and it is not a charge on the metal ion's own — it is the oxidation state. Writing "iron(III) sulfate" without the numeral where it is needed loses the mark.
7 Redox or Not? The One Test to Apply
Non-redox reactions are reactions in which there is no change in the oxidation state of any element. To decide which kind of reaction you are looking at, compare the oxidation state of every element before and after — one element changing is enough to make it redox.
Redox — oxidation states changeNot redox — no oxidation state changes
Displacement — a more reactive element displaces a less reactive one from a solution of its ions, e.g. Zn + CuSO4Thermal decomposition of carbonates, e.g. CaCO3 → CaO + CO2
Cellular respiration — cells take in oxygen and break down glucose to release energyNeutralisation, e.g. HCl + NaOH → NaCl + H2O
Photosynthesis — plants take in carbon dioxide and water to make glucose and oxygenPrecipitation, e.g. AgNO3 + NaCl → AgCl + NaNO3
The Haber process — nitrogen and hydrogen combine to form ammonia
Disproportionation — one element in a substance is oxidised and reduced at the same time, giving two different products
Why the non-redox list is short and worth memorising: in a neutralisation or a precipitation reaction every ion simply swaps partner — nothing changes its oxidation state, so nothing is oxidised or reduced. Disproportionation is the interesting case: one element plays both parts at once.
Part C · Oxidising and reducing agents
8 The Two Agents, and How to Spot Them
An oxidising agent is a substance that causes another substance to be oxidised while it is itself reduced. It gains electrons from the other substance.

A reducing agent is a substance that causes another substance to be reduced while it is itself oxidised. It loses electrons to the other substance.
The agent does the opposite to itself. The oxidising agent is the one that is reduced; the reducing agent is the one that is oxidised. Anchor the pair with that one sentence and the names stop being confusing.

Return to the zinc and copper(II) sulfate reaction from section 2:

SubstanceWhat happens to itRole
ZincZn → Zn2+ + 2e− — loses electrons, oxidation state 0 → +2reducing agent — it causes the copper(II) ions to be reduced
Copper(II) sulfateCu2+ + 2e− → Cu — gains electrons, oxidation state +2 → 0oxidising agent — it causes the zinc to be oxidised
Exam habit: find the element whose oxidation state falls. That substance is the oxidising agent. Then find the one whose oxidation state rises — that is the reducing agent. It works every time, even when the question is written entirely in words.
9 Testing for an Oxidising Agent
Testing for an oxidising agent and for a reducing agenteach test works because of a permanent colour changeAcidified KMnO₄ + a reducing agentpurplecolourlessMnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂OKI(aq) + an oxidising agentcolourlessyellow-brown2I⁻ → I₂ + 2e⁻in both tests the colour change is permanent, so no indicator is neededMn goes +7 → +2 (reduced); I goes −1 → 0 (oxidised)

Aqueous potassium iodide, KI(aq), is the test. If an oxidising agent is present, the solution changes from colourless to yellow-brown.

2I− (aq) → I2 (aq) + 2e−     colourless → yellow-brown
SpeciesOxidation state of iodineWhat it means
Iodide ion, I−−1Before the reaction
Iodine, I20After — the oxidation state increased, so the iodide was oxidised
The iodide ions are the ones being oxidised, so the test works by catching the oxidising agent in the act: whatever you added took electrons from the iodide, and the colour tells you it happened.
10 Testing for a Reducing Agent

Acidified aqueous potassium manganate(VII), KMnO4, is the test. If a reducing agent is present, the solution changes from purple to colourless.

MnO4− (aq) + 8H+ (aq) + 5e− → Mn2+ (aq) + 4H2O (l)     purple → colourless
SpeciesOxidation state of manganeseWhat it means
Manganate(VII) ion, MnO4−+7Before the reaction — purple
Manganese(II) ion, Mn2++2After — the oxidation state decreased, so the manganate(VII) was reduced
A drop of five in the oxidation state is why the half-equation takes 5e−. Compare the two tests: iodide gives electrons away and manganate(VII) takes them, which is exactly why each one detects the opposite kind of agent.
Part D · Redox titration
11 The Set-up, and Why No Indicator Is Needed

A redox titration uses a redox reaction to find the amount of a substance in a sample. Because redox reactions involve the transfer of electrons, one substance can be titrated against the other directly.

To find the concentration of a sample of aqueous iron(II) sulfate, a fixed volume of it is titrated with a standard solution of acidified aqueous potassium manganate(VII) — one whose concentration is known exactly.

The redox titration set-upthe burette holds the standard solution of known concentrationretort standburette with standardacidified KMnO₄(aq)purpleconical flask withFeSO₄(aq)pale greenend-point: pale green → pale pink, permanent for 30 s
SpeciesColour
Fe2+ (aq)pale green
MnO4− (aq)purple
Fe3+ (aq)pale yellow
Mn2+ (aq)colourless
5Fe2+ (aq) + MnO4− (aq) + 8H+ (aq) → 5Fe3+ (aq) + Mn2+ (aq) + 4H2O (l)

During the reaction the iron(II) ions lose electrons and are oxidised to iron(III) ions, while the manganate(VII) ions gain electrons and are reduced to manganese(II) ions.

The end-point is reached when a single drop of the standard solution added from the burette causes a permanent colour change from pale green to pale pink. That means all the iron(II) ions have been oxidised and there is now an excess of manganate(VII) ions in the flask.

No indicator is needed, because the reagents themselves are coloured — one of them changes colour when the reaction is complete. The same is true of redox titrations using other transition-metal compounds.
12 Worked Example — a Titration Calculation

25.0 cm3 of aqueous iron(II) sulfate of unknown concentration is titrated with 0.0200 mol dm−3 acidified aqueous potassium manganate(VII). The average volume required to reach the end-point is 24.00 cm3. Determine the concentration of the iron(II) sulfate.

5Fe2+ (aq) + MnO4− (aq) + 8H+ (aq) → 5Fe3+ (aq) + Mn2+ (aq) + 4H2O (l)
StepWorking
1 · moles of MnO4−KMnO4 (aq) → K+ (aq) + MnO4− (aq), so the moles of MnO4− equal the moles of KMnO4.
24.00 cm3 = 0.02400 dm3    n = 0.0200 × 0.02400 = 4.80 × 10−4 mol
2 · moles of Fe2+The mole ratio of Fe2+ to MnO4− is 5 : 1, so n(Fe2+) = 5 × 4.80 × 10−4 = 2.40 × 10−3 mol
3 · concentration of FeSO4FeSO4 (aq) → Fe2+ (aq) + SO42− (aq), so the moles of FeSO4 equal the moles of Fe2+.
25.0 cm3 = 0.0250 dm3    c = 2.40 × 10−3 ÷ 0.0250 = 0.0960 mol dm−3
Exam habit: the mole ratio comes from the balanced ionic equation, not from the two substances' formulae. Write the ratio down explicitly in step 2 — "Fe2+ : MnO4− = 5 : 1" — and the arithmetic after that is routine.
13 Put It Together — Exam-Style Question
[3](a) Zinc reacts with copper(II) sulfate solution: Zn (s) + CuSO4 (aq) → ZnSO4 (aq) + Cu (s). State what happens to the zinc and to the copper(II) ions in terms of electrons, and name the oxidising agent.
[3](b) Calculate the oxidation state of the element shown in each of the following. (i) Chromium in the dichromate ion, Cr2O72−. (ii) Sulfur in SO42−. (iii) Oxygen in hydrogen peroxide, H2O2.
[2](c) Aqueous potassium iodide is added to a solution containing an oxidising agent. State the colour change and explain it in terms of oxidation states.
[2](d) Acidified aqueous potassium manganate(VII) is used to test for a reducing agent. State the colour change and explain it in terms of electrons.
[2](e) State, with a reason, whether the reaction CaCO3 (s) → CaO (s) + CO2 (g) is a redox reaction.
[3](f) 25.0 cm3 of aqueous iron(II) sulfate is titrated with 0.0200 mol dm−3 acidified aqueous potassium manganate(VII) and 24.00 cm3 is required. Calculate the concentration of the iron(II) sulfate, given that 5Fe2+ + MnO4− + 8H+ → 5Fe3+ + Mn2+ + 4H2O.
Model answers.
(a) Each zinc atom loses two electrons and is oxidised (0 → +2); each copper(II) ion gains two electrons and is reduced (+2 → 0). The oxidising agent is copper(II) sulfate, because it causes the zinc to be oxidised while it is itself reduced.
(b)(i) 2(+1) + 2x + 7(−2) = 0 gives 2x = 12, so chromium is +6. (ii) x + 4(−2) = −2 gives +6. (iii) In a peroxide oxygen is −1, not −2.
(c) Colourless to yellow-brown. The iodide ions are oxidised to iodine: the oxidation state of iodine rises from −1 in I− to 0 in I2, so the solution darkens.
(d) Purple to colourless. The manganate(VII) ions gain electrons and are reduced to manganese(II) ions, with manganese falling from +7 to +2.
(e) Not a redox reaction. Each element keeps the same oxidation state: calcium stays +2, carbon stays +4 and oxygen stays −2 on both sides.
(f) n(MnO4−) = 0.0200 × 0.02400 = 4.80 × 10−4 mol. Ratio 5 : 1, so n(Fe2+) = 2.40 × 10−3 mol, and c = 2.40 × 10−3 ÷ 0.0250 = 0.0960 mol dm−3.
★ Chapter Concept Map
Oxidation and Reduction — one reaction, two halves
Oxidationgains oxygen · loses hydrogen · loses electrons · oxidation state increases
ALWAYS
TOGETHER
Reductionloses oxygen · gains hydrogen · gains electrons · oxidation state decreases
Half-equations: write one for each, then make the electrons cancel before adding them — e.g. Zn → Zn2+ + 2e− with Cu2+ + 2e− → Cu
Oxidation states — the rulespure element 0 · Groups 1 and 2 give +1 and +2 · Al +3 · Zn +2 · H +1 (−1 in hydrides) · O −2 (−1 in peroxides) · F −1
THE SUM
RULE
Add them upneutral compound sums to 0 · a polyatomic ion sums to its charge · Roman numerals in a name give the metal's oxidation state
Oxidising agentcauses oxidation, is itself reduced, gains electrons · tested with KI(aq): colourless → yellow-brown
FIND THE
AGENT
Reducing agentcauses reduction, is itself oxidised, loses electrons · tested with acidified KMnO4: purple → colourless
Redox or not?compare every oxidation state before and after — no change anywhere means not redox (neutralisation, precipitation, carbonate decomposition)
TITRATE
Redox titrationKMnO4 is its own indicator: end-point is a permanent pale pink · moles from c × V, ratio from the balanced ionic equation, then c = n / V
Miss Clarissa Ng | www.clartutors.com