Redox Reactions & Oxidation States
In 2015, the Tianjin chemical warehouse explosions killed 173 people when sodium cyanide (NaCN) stockpiles were exposed to floodwater. The NaCN reacted with water to produce hydrogen cyanide gas — a reaction driven entirely by electron transfer: CN⁻ reduced water while itself being oxidised. Oxidation states tell you exactly what transferred electrons and what didn’t.
Practise this lesson
Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
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.
Key facts
- The modern (electron-transfer) definition of oxidation and reduction — OIL RIG: Oxidation Is Loss, Reduction Is Gain (of electrons)
- The rules for assigning oxidation numbers in priority order: elements = 0; monatomic ions = charge; O = −2 (−1 in peroxides); H = +1 (−1 in hydrides); sum = 0 (neutral) or ion charge
- The meaning of oxidant (gains electrons, is reduced) and reductant (loses electrons, is oxidised)
Concepts
- Why oxidation and reduction always occur simultaneously — electrons lost by one species must be gained by another
- How to use changes in oxidation numbers to identify which species is oxidised and which is reduced in a reaction
- How to construct and verify balanced half-equations in aqueous solution using the step-by-step method
Skills
- Assign oxidation numbers to all atoms in a compound or polyatomic ion (including exceptions)
- Identify the oxidant and reductant from a balanced equation by tracking oxidation number changes
- Write and balance half-equations in aqueous solution; combine two half-equations to give a balanced overall ionic equation
When a strip of zinc metal is dropped into blue copper sulfate solution, two visible things happen: the zinc surface turns dark and the blue colour fades to colourless. No oxygen is involved — yet something has clearly reacted. The zinc is losing electrons, the copper ions are gaining them. This is the broader picture of oxidation and reduction: electron transfer.
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)
- Zinc loses 2 electrons → zinc is oxidised → zinc is the reductant
- Copper(II) ions gain 2 electrons → copper is reduced → Cu²⁺ is the oxidant
OIL RIG: Oxidation Is Loss of electrons; Reduction Is Gain of electrons. The reductant loses electrons and is oxidised; the oxidant gains electrons and is reduced. Both half-processes always occur simultaneously. E.g. Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s): Zn is oxidised (reductant); Cu²⁺ is reduced (oxidant).
Pause — copy the highlighted definition into your book before moving on.
Quick check: In the reaction Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), which statement correctly identifies the oxidant and reductant?
We just saw that redox reactions involve electron transfer tracked by OIL RIG. That raises a question: how do we systematically identify which atoms have been oxidised or reduced in complex compounds — especially covalent ones with no visible ions? This card answers it → oxidation numbers are an algebraic bookkeeping system that assigns fictitious charges to track 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:
- Elements in their standard form have oxidation number 0: Na(s) = 0, O₂(g) = 0, Fe(s) = 0.
- Monatomic ions have oxidation number equal to their charge: Na⁺ = +1, Fe³⁺ = +3, Cl⁻ = −1.
- Oxygen in compounds is usually −2, except in peroxides (e.g. H₂O₂) where it is −1.
- Hydrogen in compounds is usually +1, except in metal hydrides (e.g. NaH) where it is −1.
- The sum of oxidation numbers in a neutral compound equals zero.
- The sum of oxidation numbers in a polyatomic ion equals the charge of the ion.
| Compound/Ion | Atom | Oxidation Number | Reasoning |
|---|---|---|---|
| H₂O | O | −2 | Rule 3 |
| H₂O | H | +1 | Rule 4 |
| SO₄²⁻ | S | +6 | Sum = −2; 4×(−2) = −8; S = −2−(−8) = +6 |
| MnO₄⁻ | Mn | +7 | Sum = −1; 4×(−2) = −8; Mn = −1−(−8) = +7 |
| Fe₂O₃ | Fe | +3 | Sum = 0; 3×(−2) = −6; 2Fe = +6; Fe = +3 |
| NH₃ | N | −3 | Sum = 0; 3×(+1) = +3; N = −3 |
| Cr₂O₇²⁻ | Cr | +6 | Sum = −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.
Oxidation number rules: elements = 0; monatomic ions = charge; O = −2 (except −1 in peroxides); H = +1 (except −1 in metal hydrides); sum in neutral compound = 0; sum in polyatomic ion = ion charge. Increase in oxidation number = oxidised; decrease = reduced.
Add the highlighted rules to your notes before the check below.
True or false: If an atom's oxidation number increases from the left side to the right side of a chemical equation, that atom has been oxidised.
We just saw that oxidation numbers identify which atoms are oxidised or reduced. That raises a question: how do we formally write the oxidation and reduction steps as separate balanced equations, including the electrons? This card answers it → half-equations isolate each electron-transfer step; balance in order: main element, O (add H₂O), H (add H⁺), then charge (add e⁻).
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:
- Write the species being oxidised or reduced on the left; write the product on the right.
- Balance the main element.
- Balance oxygen by adding H₂O molecules.
- Balance hydrogen by adding H⁺ ions.
- Balance charge by adding electrons (e⁻).
- 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. ✓
Half-equation balancing order: (1) balance main element; (2) add H₂O for O; (3) add H⁺ for H; (4) add e⁻ to balance charge. Oxidation: e⁻ on RIGHT (lost). Reduction: e⁻ on LEFT (gained). Verify both atoms AND charge. E.g. MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O.
Pause — write the highlighted procedure into your book.
Explain it: A student is trying to write the balanced reduction half-equation for MnO₄⁻ to Mn²⁺ in acidic solution but is unsure what to add and where to put the electrons. Explain the step-by-step method they should use, including why the charge balance check is essential. (3–4 sentences)
We just saw how to write balanced half-equations for oxidation and reduction separately. That raises a question: how do we combine two half-equations into a single overall ionic equation? This card answers it → scale each half-equation so the number of electrons matches, then add and cancel the electrons completely.
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:
- Write oxidation and reduction half-equations separately.
- Multiply one or both half-equations so the number of electrons is equal in each.
- Add the two half-equations, cancelling electrons and any species appearing identically on both sides.
- 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. ✓
To combine half-equations: scale each so the electron count matches; add the two equations; cancel electrons completely (none must remain). Verify atoms and charge in the final equation. E.g. 5Fe²⁺ + MnO₄⁻ + 8H⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O (charge = +17 each side ✓).
Add the highlighted method to your notes before the check below.
Quick check: To combine the half-equations Zn → Zn²⁺ + 2e⁻ and MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O into an overall ionic equation, what is the first step?
We just saw how to combine half-equations to give a complete redox ionic equation. That raises a question: where does this skill apply in everyday chemistry — specifically, how does bleach remove stains without molecular oxygen? This card answers it → hypochlorite (OCl⁻) is the oxidant in bleach; it accepts electrons from organic chromophores, destroying their colour via redox, not combustion.
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). ✓
Bleach (NaOCl) works by OCl⁻ acting as an oxidant: OCl⁻(aq) + H₂O(l) + 2e⁻ → Cl⁻(aq) + 2OH⁻(aq). Chlorine is reduced from +1 to −1. OCl⁻ oxidises chromophore molecules (destroying colour) without molecular O₂. Never mix bleach with acidic cleaners — toxic Cl₂ gas forms.
Pause — write the highlighted half-equation into your book.
True or false: The hypochlorite ion (OCl⁻) acts as an oxidising agent in bleach because it accepts electrons from stain molecules, causing them to be oxidised — even though no molecular oxygen (O₂) is involved.
Worked examples · reveal as you go
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.
Cl₂(g): elemental form → oxidation number = 0.
Cl: 0 → −1 (decrease) → reduced → Cl₂ is the oxidant.
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.
Atom check: 1Zn each side. ✓
Charge check: left = 0; right = +2 + (−2) = 0. ✓
Atom check: 1Cu each side. ✓
Charge check: left = +2 + (−2) = 0; right = 0. ✓
Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)
Final charge check: left = 0 + (+2) = +2. Right = +2 + 0 = +2. ✓
Oxidation: Zn(s) → Zn²⁺(aq) + 2e⁻
Reduction: Cu²⁺(aq) + 2e⁻ → Cu(s)
Overall: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)
Before combining the half-equations for MnO₄⁻ + Fe²⁺ in acidic solution: predict which species will be the oxidant and which will be the reductant. Record your prediction, then reveal.
Oxidant: MnO₄⁻ — Mn goes from +7 (in MnO₄⁻) to +2 (in Mn²⁺), a decrease in oxidation number → reduction → MnO₄⁻ is the oxidant.
Reductant: Fe²⁺ — Fe goes from +2 to +3, an increase → oxidation → Fe²⁺ is the reductant.
Half-equations: Oxidation: Fe²⁺ → Fe³⁺ + e⁻ (×5); Reduction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O.
Overall: 5Fe²⁺ + MnO₄⁻ + 8H⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O.
Key Patterns — This Lesson
Common errors · the 3 traps that cost marks
Common misconception
Oxidation always involves oxygen.
Fix: 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.
Oxygen always has oxidation number −2
A student writes: "In H₂O₂, O = −2 (as always), so 2H = +2, H = +1." This is wrong — H₂O₂ is a peroxide.
Fix: In peroxides (e.g. H₂O₂, Na₂O₂), oxygen has oxidation number −1, not −2. This is the Rule 3 exception. In H₂O₂: neutral compound, H = +1 (Rule 4), so 2(+1) + 2(O) = 0 → O = −1. Always check for the peroxide linkage (O–O bond) before applying O = −2.
Subtracting electrons when combining half-equations
A student "cancels" electrons by subtracting one half-equation from the other rather than adding them after equalising electron counts.
Fix: Never subtract — multiply each half-equation by a factor so the number of electrons is equal in both, then ADD the equations and cancel identical species. Electrons must disappear completely from the final equation. Write out the multiplication step explicitly in exam responses.
Quick-fire practice · 5 reps +2 XP per reveal
Q8 (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.
Q9 (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.
Q10 (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) Explain why bleach is described as an oxidising agent even though it contains no molecular oxygen (O₂).
Assign the oxidation number of sulfur in SO₄²⁻. Show full algebraic working.
A student writes the half-equation: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. Verify this by checking both atom and charge balance.
The 2015 Tianjin disaster involved multiple redox reactions. When NaCN met water: Na⁺ oxidation state stays +1 throughout (spectator). The reactive change: CN⁻(−1 on N, +2 on C in cyanide) transferred electrons in the presence of water. More dramatically, the sodium metal stored there underwent Na → Na⁺ + e⁻ — sodium is oxidised (oxidation state 0 → +1); water is reduced (H: +1 → 0 as H₂ gas formed). Oxidation numbers let you trace every electron transfer: any species whose oxidation number increases is oxidised; any whose number decreases is reduced.
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.
Pick your answer, then rate your confidence — that tells the system what to drill next.
A student assigns oxidation numbers to H₂O₂: "O = −2 (as always), so 2H = +2, H = +1." Find and correct the error.
Error: H₂O₂ is a peroxide (contains an O–O bond). The Rule 3 exception applies: in peroxides, O = −1, not −2.
Correct working: H₂O₂ is neutral → sum = 0. H = +1 (Rule 4). So: 2(+1) + 2(O) = 0 → O = −1.
The student applied O = −2 uncritically. HSC questions set peroxide examples specifically to test knowledge of this exception.
Q1. 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)
Q2. 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)
Q3. 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)
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