Year 11 Chemistry Module 3 ⏱ ~35 min Lesson 8 of 12

Redox Reactions & Oxidation States

The bleach under your sink works by ripping electrons from the molecules in stains and bacteria — oxidation doesn’t require oxygen at all, and understanding why changes everything about how you read a chemical equation.

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1

Understand the core concepts covered in this lesson.

2

Apply your knowledge to solve problems and explain phenomena.

3

Evaluate and analyse scientific information and data.

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Think First

You’ve probably heard that iron rusts because it “oxidises” — reacts with oxygen. That makes sense. But bleach (sodium hypochlorite, NaOCl) kills bacteria and removes stains without any oxygen gas involved at all. And in a car battery, lead reacts with sulfuric acid — also called oxidation — again, no oxygen gas.

If “oxidation” doesn’t always involve oxygen, what do you think oxidation actually is? Write your best definition before reading on.

Type your definition of oxidation below — you will revisit this at the end of the lesson.

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Key Patterns — This Lesson

OIL RIG: Oxidation Is Loss (of electrons)  |  Reduction Is Gain (of electrons)
Oxidation and reduction always occur simultaneously
$\text{M}(s) \rightarrow \text{M}^{n+}(aq) + ne^-$  (oxidation)
Electrons are products (right side) in an oxidation half-equation
$\text{M}^{n+}(aq) + ne^- \rightarrow \text{M}(s)$  (reduction)
Electrons are reactants (left side) in a reduction half-equation
ON rules: element $= 0$  |  O $= -2$  |  H $= +1$  |  $\sum\text{ON} = 0$ (neutral) or $=$ charge (ion)
Exceptions: O in peroxides $= -1$  |  H in metal hydrides $= -1$
Key Terms — scan these before reading
OIL RIGIf this feels backwards, it is — commit OIL RIG and the oxidant/reductant definitions to memory as a pair.
redox reactionA reaction involving the transfer of electrons between chemical species.
reductantOxidised — these relationships feel counterintuitive and must be memorised precisely.
oxidantReduced; the reductant is oxidised — these relationships feel counterintuitive and must be memorised precisely.
oxidisedThe species that is oxidised (loses electrons) is called the reductant — it provides electrons to the other species.
Synthesis reactionA reaction where two or more reactants combine to form a single product.

📚 Content

02

Redefining Oxidation and Reduction — The Electron Transfer Definition

The original definition of oxidation as “reaction with oxygen” is a special case of a much more powerful and general definition — one that works even when no oxygen is involved at all.

The modern, general definition: Oxidation is the loss of electrons. Reduction is the gain of electrons. The mnemonic OIL RIG encodes this: Oxidation Is Loss, Reduction Is Gain.

These two processes always occur simultaneously — one species loses electrons and another gains them. A reaction in which electron transfer occurs is called a redox reaction. The species that is oxidised (loses electrons) is called the reductant — it provides electrons to the other species. The species that is reduced (gains electrons) is called the oxidant — it accepts electrons.

Example: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

Must Know: Always identify BOTH the oxidant AND the reductant in a redox reaction. The oxidant is reduced; the reductant is oxidised — these relationships feel counterintuitive and must be memorised precisely.
Common Error: Students confuse oxidant/reductant with oxidised/reduced. The reductant is the species that gets oxidised (it reduces the other by giving it electrons). The oxidant is the species that gets reduced (it oxidises the other by taking its electrons). If this feels backwards, it is — commit OIL RIG and the oxidant/reductant definitions to memory as a pair.
Insight: The original definition — oxidation as reaction with oxygen — is still correct for combustion and rusting. But it is a subset of the electron transfer definition. When iron reacts with oxygen, iron loses electrons to oxygen. Oxygen is the oxidant in both the old and new definitions.
OXIDATION Loss of electrons Zn → Zn²⁺ + 2e⁻ Species is OXIDISED Species is the REDUCTANT (donates e⁻ to other species) OIL REDUCTION Gain of electrons Cu²⁺ + 2e⁻ → Cu Species is REDUCED Species is the OXIDANT (accepts e⁻ from other species) RIG e⁻
03

Oxidation Numbers — Tracking Electron Shift

Oxidation numbers are a bookkeeping tool — they assign a fictitious charge to each atom in a compound to track which atoms have lost or gained electron density during a reaction.

Rules applied in order of priority:

  1. Elements in their standard form have oxidation number 0: Na(s) = 0, O₂(g) = 0, Fe(s) = 0.
  2. Monatomic ions have oxidation number equal to their charge: Na⁺ = +1, Fe³⁺ = +3, Cl⁻ = −1.
  3. Oxygen in compounds is usually −2, except in peroxides (e.g. H₂O₂) where it is −1.
  4. Hydrogen in compounds is usually +1, except in metal hydrides (e.g. NaH) where it is −1.
  5. The sum of oxidation numbers in a neutral compound equals zero.
  6. The sum of oxidation numbers in a polyatomic ion equals the charge of the ion.
Compound/IonAtomOxidation NumberReasoning
H₂OO−2Rule 3
H₂OH+1Rule 4
SO₄²⁻S+6Sum = −2; 4×(−2) = −8; S = −2−(−8) = +6
MnO₄⁻Mn+7Sum = −1; 4×(−2) = −8; Mn = −1−(−8) = +7
Fe₂O₃Fe+3Sum = 0; 3×(−2) = −6; 2Fe = +6; Fe = +3
NH₃N−3Sum = 0; 3×(+1) = +3; N = −3
Cr₂O₇²⁻Cr+6Sum = −2; 7×(−2) = −14; 2Cr = +12; Cr = +6

To identify a redox reaction: calculate oxidation numbers on both sides; if any atom’s oxidation number changes, a redox reaction has occurred. An increase in oxidation number = oxidation. A decrease = reduction.

Must Know: When calculating the oxidation number of a central atom in a polyatomic ion, set up the algebraic equation clearly: (number of atoms × oxidation number) + (central atom oxidation number) = charge of ion. Show this working in HSC extended response questions.
Common Error: Students apply Rule 3 (O = −2) and Rule 4 (H = +1) in peroxides and metal hydrides respectively. In H₂O₂, oxygen is −1, not −2. In NaH, hydrogen is −1, not +1. These exceptions appear in HSC questions specifically to test whether you know the priority order of the rules.
04

Writing Half-Equations

A half-equation isolates either the oxidation step or the reduction step, making explicit exactly how many electrons are transferred and in which direction.

Steps to write a balanced half-equation in aqueous solution:

  1. Write the species being oxidised or reduced on the left; write the product on the right.
  2. Balance the main element.
  3. Balance oxygen by adding H₂O molecules.
  4. Balance hydrogen by adding H⁺ ions.
  5. Balance charge by adding electrons (e⁻).
  6. Verify: atoms AND charge must both balance.

Example — oxidation of Fe²⁺ to Fe³⁺:
Fe²⁺(aq) → Fe³⁺(aq) + e⁻
Charge check: left = +2; right = +3 + (−1) = +2. ✓

Example — reduction of MnO₄⁻ to Mn²⁺ in acidic solution:
MnO₄⁻(aq) + 8H⁺(aq) + 5e⁻ → Mn²⁺(aq) + 4H₂O(l)
Atom check: 1Mn, 4O, 8H each side. ✓   Charge check: −1 + 8 − 5 = +2; right = +2. ✓

Must Know: Always check BOTH atoms and charge when verifying a half-equation. A half-equation that balances atoms but not charge is wrong. Charge balance is the most commonly missed check in HSC responses.
Common Error: Students place electrons on the wrong side. In an oxidation half-equation, electrons are products (right side) because the species is losing them. In a reduction half-equation, electrons are reactants (left side) because the species is gaining them. Draw OIL RIG above each half-equation before writing it.
05

Combining Half-Equations to Give the Overall Equation

The overall redox equation is constructed by combining the two half-equations so that all electrons cancel — the number of electrons lost must exactly equal the number gained.

Steps:

  1. Write oxidation and reduction half-equations separately.
  2. Multiply one or both half-equations so the number of electrons is equal in each.
  3. Add the two half-equations, cancelling electrons and any species appearing identically on both sides.
  4. Verify: check atoms and overall charge balance.

Example — Fe²⁺ oxidation with MnO₄⁻ in acid:
Oxidation: Fe²⁺(aq) → Fe³⁺(aq) + e⁻    (×5 to match 5 electrons in reduction)
Reduction: MnO₄⁻(aq) + 8H⁺(aq) + 5e⁻ → Mn²⁺(aq) + 4H₂O(l)
Combined: 5Fe²⁺(aq) + MnO₄⁻(aq) + 8H⁺(aq) → 5Fe³⁺(aq) + Mn²⁺(aq) + 4H₂O(l)
Charge check: left = 5(+2) + (−1) + 8(+1) = 10 − 1 + 8 = +17. Right = 5(+3) + (+2) = 15 + 2 = +17. ✓

Must Know: The number of electrons must be identical in both half-equations before you add them. If one involves 2 electrons and the other involves 3 electrons, multiply the first by 3 and the second by 2 (giving 6 electrons each) then cancel.
Common Error: Students cancel H⁺ and H₂O from the final equation when they appear on the same side. Only cancel species that appear identically on opposite sides of the combined equation.
Choose a reaction pair · watch electrons flow from reductant to oxidant · oxidation numbers change in real time Interactive
06

Bleach Chemistry — Oxidation Without Oxygen

Household bleach is an aqueous solution of sodium hypochlorite (NaOCl). The active species is the hypochlorite ion (OCl⁻). In solution, OCl⁻ acts as a powerful oxidant — it accepts electrons from organic molecules (stains, bacterial cell walls) and oxidises them, breaking chemical bonds and destroying colour-producing molecules (chromophores).

The reduction half-equation for hypochlorite in alkaline solution:
OCl⁻(aq) + H₂O(l) + 2e⁻ → Cl⁻(aq) + 2OH⁻(aq)

Verify oxidation number of Cl in OCl⁻: charge = −1; O = −2; so Cl + (−2) = −1 → Cl = +1. In Cl⁻, Cl = −1. Change: +1 → −1 (decrease of 2) → reduction confirmed (2e⁻ gained). ✓

Must Know: Bleach is the real-world anchor for this lesson. The short answer question will ask you to identify OCl⁻ as the oxidant, write its reduction half-equation, and explain why hypochlorite is described as an oxidising agent even though no O₂ is involved.
Insight: The effectiveness of bleach decreases at low pH (acidic conditions). Mixing bleach with acidic cleaners produces toxic chlorine gas (Cl₂) or chloramine gases. Never mix bleach with acidic cleaners or ammonia-based products. The chemistry: in acidic conditions OCl⁻ is converted to Cl₂ rather than remaining as the hypochlorite oxidant.

📐 Worked Examples

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Example 1 — Assigning Oxidation Numbers and Identifying Redox

Problem: For the reaction 2Fe(s) + 3Cl₂(g) → 2FeCl₃(s): assign oxidation numbers to all atoms, identify which species is oxidised and which is reduced, and name the oxidant and reductant.
Step 1 — Assign on the left

Fe(s): elemental form → oxidation number = 0.
Cl₂(g): elemental form → oxidation number = 0.

Step 2 — Assign on the right

FeCl₃: Cl = −1 (Rule 2 for ionic compound). Sum = 0; 3(−1) + Fe = 0 → Fe = +3.

Step 3 — Track changes and write half-equations

Fe: 0 → +3 (increase) → oxidised → Fe is the reductant.
Cl: 0 → −1 (decrease) → reduced → Cl₂ is the oxidant.
Oxidation: Fe(s) → Fe³⁺ + 3e⁻ (×2)
Reduction: Cl₂(g) + 2e⁻ → 2Cl⁻(aq) (×3)
Electrons balanced at 6e⁻ each: 2Fe + 3Cl₂ → 2Fe³⁺ + 6Cl⁻ = 2FeCl₃(s). ✓

Final Answer

Fe is oxidised (0→+3); Cl₂ is reduced (0→−1). Reductant: Fe. Oxidant: Cl₂.

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Example 2 — Writing and Balancing Half-Equations

Problem: Write balanced half-equations for: (a) the oxidation of zinc to zinc ions, and (b) the reduction of copper(II) ions to copper metal. Then combine them to give the overall ionic equation.
Step 1 — (a) Oxidation of Zn

Zn(s) → Zn²⁺(aq) + 2e⁻
Atom check: 1Zn each side. ✓
Charge check: left = 0; right = +2 + (−2) = 0. ✓

Step 2 — (b) Reduction of Cu²⁺

Cu²⁺(aq) + 2e⁻ → Cu(s)
Atom check: 1Cu each side. ✓
Charge check: left = +2 + (−2) = 0; right = 0. ✓

Step 3 — Combine (electrons already equal at 2e⁻)

Add half-equations and cancel 2e⁻:
Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)
Final charge check: left = 0 + (+2) = +2. Right = +2 + 0 = +2. ✓

Final Answer

Oxidation: Zn(s) → Zn²⁺(aq) + 2e⁻
Reduction: Cu²⁺(aq) + 2e⁻ → Cu(s)
Overall: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

Oxidation State Rules Free elements Oxidation state = 0 (e.g. Na, O₂, Cl₂) Monatomic ions Oxidation state = charge (e.g. Na⁺ = +1, Cl⁻ = −1) Oxygen Usually −2 (except peroxides and with fluorine) Hydrogen Usually +1 (except metal hydrides: −1) Compounds Sum of oxidation states = 0. For polyatomic ions, sum = overall charge.

✏️ Activities

Activity 1 — Oxidation Numbers

Calculate the oxidation number of the underlined atom in each species. Show your working using the algebraic method (set up the equation, solve for unknown).

  1. S in H₂SO₄
  2. N in NO₃⁻
  3. Mn in KMnO₄
  4. Cl in HClO₄

Type your working and answer for each.

Show your working in your workbook.

✏️ Show all working in your workbook
Activity 2 — Half-Equations and Overall Equation

For the reaction between magnesium metal and dilute hydrochloric acid:
Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)

Question A: Assign oxidation numbers to every atom on both sides. Identify which species is oxidised and which is reduced. Name the oxidant and reductant.

Question B: Write the balanced oxidation half-equation (Mg → Mg²⁺) and balanced reduction half-equation (2H⁺ → H₂). Verify both with atom and charge checks.

Question C: Combine the two half-equations to give the overall ionic equation.

Type your working for A, B, and C.

Show all working in your workbook.

✏️ Show all working in your workbook
Interactive: Redox Half-Equation Builder
Challenge: Redox Balancer
Revisit Your Initial Thinking

Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?

Misconceptions to Fix

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Wrong: Oxidation always involves oxygen.

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Right: The modern definition defines oxidation as loss of electrons, which may or may not involve oxygen. Mg → Mg²⁺ + 2e⁻ is oxidation with no oxygen involved. OIL RIG: Oxidation Is Loss, Reduction Is Gain — of electrons.

MC

Multiple Choice

5 random questions from a replayable lesson bank — feedback shown immediately

✍️ Short Answer

07

Extended Questions

UnderstandBand 3

8. (4 marks) For the reaction: Fe(s) + 2AgNO₃(aq) → Fe(NO₃)₂(aq) + 2Ag(s), (a) assign oxidation numbers to every atom on both sides of the equation, (b) identify which species is oxidised and which is reduced, and (c) name the oxidant and the reductant. (2 + 1 + 1 marks)

Type your working for all three parts.

Show all working in your book.

✏️ Show all working in your book
ApplyBand 4

9. (4 marks) Write balanced half-equations for the following and verify each with atom and charge checks: (a) the oxidation of iron(II) ions (Fe²⁺) to iron(III) ions (Fe³⁺) in aqueous solution, and (b) the reduction of chlorine gas (Cl₂) to chloride ions (Cl⁻). Then combine them to give the overall ionic equation for the reaction between Fe²⁺ and Cl₂. (1 + 1 + 2 marks)

Type your half-equations and combined overall equation.

Show all working in your book.

✏️ Show all working in your book
EvaluateBand 5

10. (5 marks) Household bleach contains sodium hypochlorite (NaOCl). The active oxidising species is the hypochlorite ion (OCl⁻). (a) Calculate the oxidation number of chlorine in OCl⁻ and in Cl⁻. (b) Write the balanced reduction half-equation for OCl⁻ in alkaline solution, showing the formation of Cl⁻ and OH⁻. Verify with atom and charge checks. (c) Using your understanding of oxidation and reduction, explain why bleach is described as an oxidising agent even though it contains no molecular oxygen (O₂). (1 + 2 + 2 marks)

Type your full extended response.

Write your full response in your book.

✏️ Write your full response in your book
Revisit Your Thinking

Earlier you were asked: Can you now write the precise modern definition of oxidation? Can you explain why bleach oxidises without O₂ involved?

The key insight: Oxidation is loss of electrons (OIL RIG). It does not require oxygen gas. When bleach (containing OCl⁻) oxidises a stain, the OCl⁻ ion accepts electrons from the stain molecules. The stain loses electrons → it is oxidised. The oxidising agent (OCl⁻) is itself reduced in the process. This is the modern electron-transfer definition.

Now revisit your initial response. What did you get right? What has changed in your thinking?

Look back at your initial response in your book. Annotate it with what you now understand differently.

📝 Record your prediction in your workbook

Activity 1 Answers

1. S in H₂SO₄: H = +1 (2H = +2); O = −2 (4O = −8); sum = 0. S + (+2) + (−8) = 0 → S = +6.

2. N in NO₃⁻: O = −2 (3O = −6); sum = −1. N + (−6) = −1 → N = +5.

3. Mn in KMnO₄: K = +1; O = −2 (4O = −8); sum = 0. +1 + Mn + (−8) = 0 → Mn = +7.

4. Cl in HClO₄: H = +1; O = −2 (4O = −8); sum = 0. +1 + Cl + (−8) = 0 → Cl = +7.

Activity 2 Answers

A. Left: Mg = 0; H = +1 (in HCl); Cl = −1. Right: Mg = +2 (in MgCl₂); Cl = −1 (in MgCl₂); H = 0 (in H₂). Mg: 0 → +2 (increase) → oxidised → Mg is the reductant. H: +1 → 0 (decrease) → reduced → H⁺ (from HCl) is the oxidant.

B. Oxidation: Mg(s) → Mg²⁺(aq) + 2e⁻. Atom: 1Mg each side ✓. Charge: 0 = +2 − 2 = 0 ✓. Reduction: 2H⁺(aq) + 2e⁻ → H₂(g). Atom: 2H each side ✓. Charge: +2 − 2 = 0 = 0 ✓.

C. Electrons equal (2e⁻): Mg(s) + 2H⁺(aq) → Mg²⁺(aq) + H₂(g). Charge: left = +2; right = +2 ✓.

MC Feedback

Q1 B: Mg: 0 → +2 (increase) = oxidised = reductant. O: 0 → −2 (decrease) = reduced = O₂ is the oxidant.

Q2 C: Sum = −2; 4O = −8; S + (−8) = −2 → S = +6.

Q3 B: Balance Cr (2 each side ✓). Balance O: 7O → 7H₂O ✓. Balance H: 14H → 14H⁺ ✓. Balance charge: −2+14−6 = +6 = 2(+3) ✓. Option A uses O²⁻ (doesn’t exist in solution). Option C uses OH⁻ (alkaline conditions). Option D has wrong stoichiometry.

Q4 C: Cl₂ + 2e⁻ → 2Cl⁻ is the reduction of Cl₂ (Cl: 0 → −1, decrease = reduction). Since Cl₂ is gaining electrons, it is the oxidant and is reduced.

Q5 A: Fe²⁺ loses electrons (0+2 → +3, increase) = oxidised = reductant. Mn in MnO₄⁻ (+7) gains electrons (reduces to Mn²⁺, +2) = reduced = oxidant. H⁺ is a spectator in terms of oxidation states (H stays +1 in H⁺ and H₂O).

Q6 D: In an oxidation, the species loses electrons. Electrons are products (right side) in an oxidation half-equation. The student wrote electrons as reactants, which would mean Cu is gaining electrons (reduction, not oxidation). Correct: Cu(s) → Cu²⁺(aq) + 2e⁻.

Q7 C: The reduction half-equation involves 1e⁻ and the oxidation involves 2e⁻. Multiply reduction ×2: 2Fe³⁺ + 2e⁻ → 2Fe²⁺. Add: Zn + 2Fe³⁺ + 2e⁻ → Zn²⁺ + 2Fe²⁺ + 2e⁻. Cancel 2e⁻: Zn(s) + 2Fe³⁺(aq) → Zn²⁺(aq) + 2Fe²⁺(aq). Charge: 0+6=6; 2+4=6 ✓.

Short Answer Guides

Q8: (a) Fe: 0 (left) → +2 (in Fe(NO₃)₂, right); Ag: +1 (in AgNO₃) → 0 (Ag(s) right); N: +5 both sides (spectator); O: −2 both sides (spectator). (b) Fe is oxidised (0→+2); Ag⁺ is reduced (+1→0). (c) Oxidant: Ag⁺ (gains electrons, is reduced). Reductant: Fe (loses electrons, is oxidised).

Q9: (a) Fe²⁺(aq) → Fe³⁺(aq) + e⁻. Atom: 1Fe each side ✓. Charge: +2 = +3 − 1 = +2 ✓. (b) Cl₂(g) + 2e⁻ → 2Cl⁻(aq). Atom: 2Cl each side ✓. Charge: 0 − 2 = −2 = 2(−1) ✓. Combining: multiply (a) ×2 to get 2e⁻ each. 2Fe²⁺ + Cl₂ + 2e⁻ → 2Fe³⁺ + 2e⁻ + 2Cl⁻. Cancel 2e⁻: 2Fe²⁺(aq) + Cl₂(g) → 2Fe³⁺(aq) + 2Cl⁻(aq). Charge: 2(+2)+0=+4; 2(+3)+2(−1)=6−2=+4 ✓.

Q10: (a) OCl⁻: charge = −1; O = −2; Cl + (−2) = −1 → Cl = +1. Cl⁻: charge = −1 → Cl = −1. (b) OCl⁻(aq) + H₂O(l) + 2e⁻ → Cl⁻(aq) + 2OH⁻(aq). Atom check: 1Cl, 1O(OCl)→1Cl; 1O(H₂O)+1O(OCl)=2O on left, 2O in 2OH on right; 2H each side. ✓ Charge check: −1+0−2=−3; −1+2(−1)=−3. ✓ (c) An oxidising agent accepts electrons (it is reduced). Bleach (OCl⁻) accepts 2 electrons from the molecule being oxidised, according to OIL RIG: the species receiving electrons is being reduced, so OCl⁻ is the species being reduced = the oxidant = the oxidising agent. Oxidation does not require O₂; it requires electron loss by the target species, and gain of those electrons by OCl⁻.

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