Electrons on the move — from OIL RIG and half equations to oxidation numbers, test reagents, and real-world redox from rusting to breathalysers.
Section 1 — The Big Picture: What is Redox?
A redox reaction occurs when electrons are transferred between substances. One substance loses electrons (is oxidised) and another gains them (is reduced). They always happen together — you can never have one without the other!
Section 2 — Redox in Your Everyday Life
🦀 Rusting
Iron + oxygen + water → hydrated iron(III) oxide (Fe₂O₃·xH₂O) = rust. Iron is oxidised by oxygen. Salt water accelerates rusting because dissolved ions increase electrical conductivity, speeding up electron transfer.
🍎 Browning of Cut Fruit
Enzymes in fruit cells oxidise phenolic compounds to dark melanins. Lemon juice (Vitamin C / ascorbic acid) acts as a reducing agent — it is oxidised preferentially, slowing the browning of the fruit.
🧴 Bleaches
Chlorine bleaches (NaClO) and oxygen bleaches (H₂O₂) are oxidising agents that remove stains by oxidising coloured dye molecules to their colourless form. The bleach is reduced in the process.
🚗 Breathalyser Test
Acidified potassium dichromate(VI) (K₂Cr₂O₇) tests for alcohol. Ethanol in breath reduces the orange Cr₂O₇²⁻ ion to the green Cr³⁺ ion. Orange → Green = positive test!
Section 3 — Half Equations
Any redox reaction can be split into two half equations — one showing oxidation, one showing reduction. Each half equation must balance in both atoms AND charge.
Worked Example — Zn + CuSO₄
Oxidation Half (Zn loses electrons)
Zn(s) → Zn²⁺(aq) + 2e⁻
Zn starts at 0, ends at +2 → oxidation number INCREASES → oxidised. Zn is the reducing agent.
Reduction Half (Cu²⁺ gains electrons)
Cu²⁺(aq) + 2e⁻ → Cu(s)
Cu starts at +2, ends at 0 → oxidation number DECREASES → reduced. Cu²⁺ is the oxidising agent.
Adding the half equations (electrons cancel):
Worked Example — Breathalyser (Acidified K₂Cr₂O₇)
Oxidation Half (Ethanol oxidised)
C₂H₅OH → CH₃COOH + 4H⁺ + 4e⁻
Ethanol (C=−1 average) → Ethanoic acid (C=0 average). Ethanol is oxidised. It is the reducing agent.
Reduction Half (Dichromate reduced)
Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O
Cr: +6 → +3 (decrease) → reduced. K₂Cr₂O₇ is the oxidising agent. Orange → Green colour change.
Section 4 — Oxidation Numbers
An oxidation number (oxidation state) is the theoretical charge an atom would have if all its bonds were ionic. It lets us track electron transfer and identify what has been oxidised or reduced.
Rules for Assigning Oxidation Numbers
Worked Examples
Section 5 — Testing for Oxidising & Reducing Agents
These colour changes must be memorised — they come up constantly in CSEC exams!
Reagent: KI(aq) (potassium iodide)
Colourless → Brown
I⁻ ions are oxidised to I₂ (brown). The unknown oxidising agent takes electrons from I⁻.
Reagent: FeSO₄(aq) (iron(II) sulfate)
Pale green → Yellow-brown
Fe²⁺ oxidised to Fe³⁺ by the unknown oxidising agent.
Reagent: Acidified KMnO₄
Purple → Colourless
MnO₄⁻ (Mn: +7) reduced to Mn²⁺ (Mn: +2) by the unknown reducing agent. Purple disappears.
Reagent: Acidified K₂Cr₂O₇
Orange → Green
Cr₂O₇²⁻ (Cr: +6) reduced to Cr³⁺ (Cr: +3) by the unknown reducing agent.
| Reagent | Tests For | Colour Change | What Happens |
|---|---|---|---|
| Potassium iodide KI(aq) | Oxidising agent | Colourless → Brown | I⁻ oxidised to I₂ |
| Iron(II) sulfate FeSO₄(aq) | Oxidising agent | Pale green → Yellow-brown | Fe²⁺ oxidised to Fe³⁺ |
| Acidified KMnO₄ (purple) | Reducing agent | Purple → Colourless | MnO₄⁻ (+7) reduced to Mn²⁺ (+2) |
| Acidified K₂Cr₂O₇ (orange) | Reducing agent | Orange → Green | Cr₂O₇²⁻ (+6) reduced to Cr³⁺ (+3) |
Section 6 — Substances That Act as Both
Some substances can be oxidising agents OR reducing agents depending on their reaction partner. The key is comparing relative strengths.
Sulfur Dioxide (SO₂)
Usually a REDUCING AGENT (S is at +4, can go up to +6):
Acts as OXIDISING AGENT with H₂S (stronger reducer):
Hydrogen Peroxide (H₂O₂)
Usually an OXIDISING AGENT:
Acts as REDUCING AGENT with stronger oxidisers:
Section 7 — Resources & Simulations
Section 8 — CSEC Practice Questions
a In terms of electrons: Oxidation = loss of electrons. Reduction = gain of electrons.
b In terms of oxidation numbers: Oxidation = increase in oxidation number. Reduction = decrease in oxidation number.
1 Find oxidation number of Fe: In Fe₂O₃, Fe = +3. In Fe(s), Fe = 0. Fe: +3 → 0 (decrease) → REDUCED. Fe₂O₃ is the oxidising agent.
2 Find oxidation number of C: In CO, C = +2. In CO₂, C = +4. C: +2 → +4 (increase) → OXIDISED. CO is the reducing agent.
a H₂SO₄: 2(+1) + S + 4(−2) = 0 → 2 + S − 8 = 0 → S = +6
b NH₃: N + 3(+1) = 0 → N = −3
c ClO₄⁻: Cl + 4(−2) = −1 → Cl − 8 = −1 → Cl = +7 (perchlorate = chlorate(VII))
d MnO₄⁻: Mn + 4(−2) = −1 → Mn − 8 = −1 → Mn = +7 (manganate(VII) ion)
a KMnO₄ is decolourised → it has been reduced. A reducing agent must be present to donate electrons to the MnO₄⁻ ions.
b Half-equation for reduction of manganate(VII) in acid:
MnO₄⁻(aq) + 8H⁺(aq) + 5e⁻ → Mn²⁺(aq) + 4H₂O(l)
Check: Mn changes from +7 to +2 ✓ (decrease = reduction ✓)
c Acidified K₂Cr₂O₇ — if the solution changes from orange to green, it confirms a reducing agent is present.
a Fe²⁺ loses 1 electron to form Fe³⁺ (oxidation number +2 → +3, increase = oxidised):
Fe²⁺(aq) → Fe³⁺(aq) + e⁻
b Each Cl atom in Cl₂ gains 1 electron (+0 → −1, decrease = reduced):
Cl₂(g) + 2e⁻ → 2Cl⁻(aq)
c Multiply (a) by 2 so electrons cancel: 2Fe²⁺ → 2Fe³⁺ + 2e⁻, then add to (b):
2Fe²⁺(aq) + Cl₂(g) → 2Fe³⁺(aq) + 2Cl⁻(aq)