Key Definitions
OIL
Oxidation Is Loss (of electrons)
RIG
Reduction Is Gain (of electrons)
Oxidising agent = accepts electrons (is itself reduced). Reducing agent = donates electrons (is itself oxidised).
Three Ways to Define Redox
| Perspective | Oxidation | Reduction |
|---|---|---|
| Electrons | Loss of e⁻ | Gain of e⁻ |
| Oxidation state | Increase in ox. Number | Decrease in ox. Number |
| Oxygen / Hydrogen | Gain of O / Loss of H | Loss of O / Gain of H |
Oxidation States (Numbers)
An oxidation state is a formal charge assigned to an atom, assuming all bonds are completely ionic. It lets us track electron transfer in complex molecules.
Rules for Assigning Oxidation States
- Free elements = 0 (e.g. Na, O₂, S₈, P₄)
- Monoatomic ion = charge (e.g. Na⁺ = +1, Cl⁻ = −1)
- Fluorine = always −1
- Oxygen = −2. Exceptions: peroxides (−1, e.g. H₂O₂), OF₂ (+2)
- Hydrogen = +1. Exception: metal hydrides (−1, e.g. NaH)
- Group 1 metals = always +1; Group 2 = always +2; Al = always +3
- Halogens (Cl, Br, I) = usually −1. Exception: positive when bonded to O or F
- Sum in a neutral compound = 0; in a polyatomic ion = ion charge
Worked Example: Find the oxidation state of Mn in MnO₄⁻
Let Mn = x. Each O = −2 (there are 4). Overall charge = −1.
x + 4(−2) = −1 → x − 8 = −1 → x = +7
Mn has an oxidation state of +7 in the permanganate ion.
Identifying Redox Reactions
Assign oxidation states to every element on both sides. If any element's oxidation number changes, it is a redox reaction. If no elements change, it is not redox (e.g. Acid-base neutralisation).
Half-Equations
Half-equations show oxidation and reduction separately. Every redox reaction can be split into two half-equations.
Example: Mg + CuSO₄
Oxidation: Mg → Mg²⁺ + 2e⁻ (Mg loses electrons. Reducing agent)
Reduction: Cu²⁺ + 2e⁻ → Cu (Cu²⁺ gains electrons. Oxidising agent)
Overall: Mg + Cu²⁺ → Mg²⁺ + Cu
Balancing Complex Half-Equations (acidic conditions)
- Balance atoms other than O and H first
- Balance oxygen by adding H₂O to the side that needs O
- Balance hydrogen by adding H⁺ to the side that needs H
- Balance charge by adding electrons (e⁻) to the more positive side
- Combine: multiply half-equations so electrons cancel, then add
Worked Example: Balance MnO₄⁻ → Mn²⁺
Step 1: Mn is already balanced.
Step 2 (balance O): MnO₄⁻ → Mn²⁺ + 4H₂O
Step 3 (balance H): MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O
Step 4 (balance charge): Left = −1 + 8 = +7. Right = +2. Add 5e⁻ to the left.
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
Disproportionation
A reaction where a single element is simultaneously oxidised and reduced.
Example: Cl₂ + H₂O → HClO + HCl
Cl goes from 0 (in Cl₂) to both +1 (in HClO) and −1 (in HCl).
⚠ Common Exam Mistakes
Notation: Oxidation states have the sign before the number (+2), ionic charges have it after (2+).
Per-atom calculation: In CaCl₂, the oxidation state of Cl is −1 (not −2). Always calculate per single atom.
Naming agents: Refer to the whole substance, not just the element. Say "permanganate ion" not "manganese".
Roman numerals: Always use them for transition metals with variable oxidation states. Iron(II) oxide vs iron(III) oxide.
Confusing agents: The reducing agent is oxidised (it donates electrons). The oxidising agent is reduced (it accepts electrons).
Think About It
In the reaction: 2Fe²⁺ + Cl₂ → 2Fe³⁺ + 2Cl⁻, identify the oxidising and reducing agents.
Cl₂ is the oxidising agent (it gains electrons, is reduced). Fe²⁺ is the reducing agent (it loses electrons, is oxidised: +2 → +3).