OIL RIG, oxidation states, half-equations, and the mighty reactivity series — master the transfer of electrons! ⚡
In a redox reaction, oxidation and reduction always occur simultaneously — one substance loses electrons while another gains them. You cannot have one without the other.
| Process | Electrons | Oxidation State | Agent |
|---|---|---|---|
| Oxidation | LOSES electrons | Increases (becomes more positive) | The oxidised substance is the reducing agent (it causes reduction) |
| Reduction | GAINS electrons | Decreases (becomes more negative) | The reduced substance is the oxidising agent (it causes oxidation) |
The oxidation state is a number assigned to an atom representing its apparent charge in a compound. It tracks electron transfer in redox reactions.
| Rule | Value | Example |
|---|---|---|
| Uncombined element | 0 | Fe, Cl₂, O₂ → all 0 |
| Simple monoatomic ion | = ionic charge | Na⁺ = +1, Cl⁻ = −1, Fe³⁺ = +3 |
| Oxygen in compounds | −2 (usually) | H₂O: O = −2. Exception: peroxides (O = −1) |
| Hydrogen in compounds | +1 (usually) | HCl: H = +1. Exception: metal hydrides (H = −1) |
| Sum of all oxidation states in neutral compound | = 0 | H₂O: 2(+1) + (−2) = 0 ✓ |
| Sum of all oxidation states in polyatomic ion | = ionic charge | SO₄²⁻: S + 4(−2) = −2 → S = +6 |
A half-equation shows either the oxidation or reduction part of a redox reaction separately. Each half-equation must be balanced for both atoms AND charge.
| Half-Equation | Type | Notes |
|---|---|---|
| Zn → Zn²⁺ + 2e⁻ | Oxidation | Zn loses 2 electrons; oxidation state 0 → +2 |
| Cu²⁺ + 2e⁻ → Cu | Reduction | Cu²⁺ gains 2 electrons; +2 → 0 |
| Fe²⁺ → Fe³⁺ + e⁻ | Oxidation | Fe loses 1 electron; +2 → +3 |
| 2H⁺ + 2e⁻ → H₂ | Reduction | H⁺ gains electrons to form H₂ gas |
| Cl₂ + 2e⁻ → 2Cl⁻ | Reduction | Each Cl: 0 → −1 |
| 2Cl⁻ → Cl₂ + 2e⁻ | Oxidation | Each Cl: −1 → 0 |
To combine, multiply half-equations so electrons cancel, then add:
Every displacement reaction is a redox reaction — the more reactive metal is oxidised (loses electrons) while the metal ion in solution is reduced (gains electrons).
| Reaction | Oxidation half | Reduction half | Observation |
|---|---|---|---|
| Zn + CuSO₄ → ZnSO₄ + Cu | Zn → Zn²⁺ + 2e⁻ | Cu²⁺ + 2e⁻ → Cu | Blue fades; pink Cu deposits on Zn |
| Fe + CuSO₄ → FeSO₄ + Cu | Fe → Fe²⁺ + 2e⁻ | Cu²⁺ + 2e⁻ → Cu | Blue fades; Cu deposits on Fe |
| Mg + 2HCl → MgCl₂ + H₂ | Mg → Mg²⁺ + 2e⁻ | 2H⁺ + 2e⁻ → H₂ | Effervescence; Mg dissolves |
| Common Oxidising Agents | What they become (get reduced to) |
|---|---|
| Potassium manganate(VII) KMnO₄ (acidic, purple) | Mn²⁺ (colourless/pale pink) — decolourises |
| Potassium dichromate(VI) K₂Cr₂O₇ (orange) | Cr³⁺ (green) — colour change orange→green |
| Chlorine Cl₂ | Cl⁻ ions |
| Hydrogen peroxide H₂O₂ | H₂O |
| Common Reducing Agents | What they become (get oxidised to) |
|---|---|
| Iron(II) ions Fe²⁺ | Fe³⁺ |
| Sulfur dioxide SO₂ | SO₄²⁻ (sulfate) |
| Iodide ions I⁻ | I₂ (turns starch solution blue-black) |
| Hydrogen H₂ | H⁺ |
A disproportionation reaction is a special redox reaction where the same element is simultaneously oxidised AND reduced in the same reaction.
Select a reaction to watch electrons animate between atoms. Red = losing electrons (oxidised). Blue = gaining electrons (reduced).
Select a compound and identify the element to find its oxidation state step by step.
Metals higher up the reactivity series are stronger reducing agents — they lose electrons more easily. Click any bar for details.
👆 Click a bar to learn about that metal's redox behaviour
Click to flip!
Problem: Iron(II) ions are oxidised by potassium manganate(VII) (KMnO₄) in acidic solution. MnO₄⁻ is reduced to Mn²⁺. Fe²⁺ is oxidised to Fe³⁺. Write the two half-equations and combine them into a balanced ionic equation.
Fe²⁺ loses one electron to become Fe³⁺. Write the half-equation for oxidation. (Format: Fe²⁺ → Fe³⁺ + ?e⁻)
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O (given — acidic solution needs H⁺ and H₂O to balance). How many electrons does each Mn gain? Type the number.
Oxidation: Fe²⁺ → Fe³⁺ + 1e⁻. Reduction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. To cancel electrons, the Fe half-equation must be multiplied by how much?
Adding: 5Fe²⁺ + MnO₄⁻ + 8H⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O. Which species is the oxidising agent? (the one that gets reduced)
Click a term, then its match!
1 (a) Mg is oxidised (0 → +2, loses electrons). ✓ Cu²⁺ is reduced (+2 → 0, gains electrons). ✓
2 (b) Oxidation: Mg → Mg²⁺ + 2e⁻ ✓ | Reduction: Cu²⁺ + 2e⁻ → Cu ✓
3 (c) Oxidising agent: CuSO₄ / Cu²⁺ (it causes Mg to be oxidised; Cu²⁺ itself is reduced) ✓ | Reducing agent: Mg (it causes Cu²⁺ to be reduced; Mg itself is oxidised) ✓
1 (a) Cr₂O₇²⁻: 2Cr + 7(−2) = −2 → 2Cr = −2+14 = 12 → Cr = +6 ✓✓
2 (b) NO₃⁻: N + 3(−2) = −1 → N − 6 = −1 → N = +5 ✓✓
3 (c) SO₂: S + 2(−2) = 0 → S = +4 ✓✓
1 (a) Orange → green ✓
2 (b) Orange: Cr₂O₇²⁻ (dichromate ion, Cr = +6) ✓ | Green: Cr³⁺ (chromium(III) ion) ✓
3 (c) Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O ✓✓ (accept without state symbols; must balance atoms and charge)
1 Disproportionation: a reaction where the same element is simultaneously oxidised AND reduced. ✓
2 Cl₂: oxidation state of Cl = 0 ✓
3 In HCl: Cl = −1 (reduced from 0 to −1) ✓
4 In HOCl: Cl = +1 (oxidised from 0 to +1) ✓ Therefore the same element (Cl) is both oxidised and reduced → disproportionation.
Oxidation = Loss of e⁻
Reduction = Gain of e⁻
Reducing agent → oxidised
Oxidising agent → reduced
Element = 0
O in compound = −2
H in compound = +1
Sum in compound = 0
KMnO₄: purple → colourless
K₂Cr₂O₇: orange → green
I⁻ → I₂ (blue-black w/ starch)
Balance atoms → balance charge
Add e⁻ to balance charge
Combine: cancel e⁻
Same element: oxidised AND reduced
Cl₂ + H₂O → HCl + HOCl
Cl: 0 → −1 and 0 → +1
K > Na > Ca > Mg > Al > Zn
> Fe > Pb > H > Cu > Ag
Higher = stronger reducing agent