Redox Reactions

Redox reactions are among the most important types of chemical reaction. They are responsible for processes ranging from combustion and corrosion to photosynthesis and respiration. The name "redox" comes from reduction-oxidation, because the two processes always occur together.

Definitions of Oxidation and Reduction

There are three equivalent ways to define oxidation and reduction. All three are valid and useful in different contexts:

In Terms of Oxygen

Oxidation: Gain of oxygen
Reduction: Loss of oxygen

In Terms of Electrons

Oxidation: Loss of electrons (OIL)
Reduction: Gain of electrons (RIG)

The mnemonic OIL RIG helps: Oxidation Is Loss, Reduction Is Gain.

In Terms of Oxidation States

Oxidation: Increase in oxidation state
Reduction: Decrease in oxidation state

All three definitions are consistent with each other. The electron transfer definition is the most general and applies to all redox reactions.

Definition	Oxidation	Reduction
Oxygen	Gain of O	Loss of O
Electrons	Loss of e⁻	Gain of e⁻
Oxidation state	Increase	Decrease

Oxidising Agents and Reducing Agents

An oxidising agent (oxidant) is a substance that causes another substance to be oxidised. It does this by accepting electrons — so the oxidising agent itself is reduced.

A reducing agent (reductant) is a substance that causes another substance to be reduced. It does this by donating electrons — so the reducing agent itself is oxidised.

This can be confusing at first. The key is: the agent causes the change in the other substance, but the agent itself undergoes the opposite change.

Common Oxidising and Reducing Agents

Oxidising agents	What they become	Reducing agents	What they become
KMnO₄ (MnO₄⁻)	Mn²⁺	Metals (Zn, Fe, Mg)	Metal ions
K₂Cr₂O₇ (Cr₂O₇²⁻)	Cr³⁺	Fe²⁺	Fe³⁺
Conc. H₂SO₄	SO₂	H₂	H⁺ / H₂O
Cl₂	Cl⁻	C (coke/charcoal)	CO / CO₂
O₂	O²⁻ / H₂O	I⁻	I₂
HNO₃	NO / NO₂	S₂O₃²⁻	S₄O₆²⁻

Example

In the reaction: CuO + H₂ → Cu + H₂O

Copper oxide (CuO) is the oxidising agent — it oxidises H₂ to H₂O. The Cu is reduced (Cu²⁺ → Cu⁰).
Hydrogen (H₂) is the reducing agent — it reduces CuO to Cu. The H is oxidised (H⁰ → H⁺).

Writing Half-Equations

Half-equations separate a redox reaction into its oxidation and reduction components. Each half-equation must be balanced for:

Atoms — every element must balance
Charge — total charge on left = total charge on right

Steps for Writing Half-Equations

flowchart TD
    A["Write species being oxidised/reduced on each side"] --> B["Balance atoms other than O and H"]
    B --> C["Balance O by adding H₂O"]
    C --> D["Balance H by adding H⁺"]
    D --> E["Balance charge by adding e⁻"]
    E --> F["Check: atoms balanced AND charge balanced"]

Worked Example 1

Write the half-equation for the reduction of MnO₄⁻ to Mn²⁺ in acidic solution.

Step 1: MnO₄⁻ → Mn²⁺

Step 2: Mn is already balanced.

Step 3: Balance O by adding 4H₂O to the right. MnO₄⁻ → Mn²⁺ + 4H₂O

Step 4: Balance H by adding 8H⁺ to the left. MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O

Step 5: Balance charge. Left: (−1) + (+8) = +7. Right: +2. Need to add 5e⁻ to the left. MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

Check: Mn = 1:1, O = 4:4, H = 8:8. Charge: +7 − 5 = +2 on left; +2 on right.

Worked Example 2

Write the half-equation for the oxidation of Fe²⁺ to Fe³⁺.

Fe²⁺ → Fe³⁺ + e⁻

Charge: +2 on left; +3 − 1 = +2 on right.

Worked Example 3

Write the half-equation for the reduction of Cr₂O₇²⁻ to Cr³⁺ in acidic solution.

Step 1: Cr₂O₇²⁻ → 2Cr³⁺ Step 2: Balance O: add 7H₂O. Cr₂O₇²⁻ → 2Cr³⁺ + 7H₂O Step 3: Balance H: add 14H⁺. Cr₂O₇²⁻ + 14H⁺ → 2Cr³⁺ + 7H₂O Step 4: Charge: Left = −2 + 14 = +12. Right = +6. Add 6e⁻ to left.

Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O

Worked Example 4

Write the half-equation for the oxidation of ethanal to ethanoic acid.

CH₃CHO → CH₃COOH

Step 1: C and H — we need to balance O. Left has 1 O; right has 2 O. Add 1 H₂O to left. CH₃CHO + H₂O → CH₃COOH

Step 2: Balance H. Left has 4 + 2 = 6 H (in CH₃CHO) + 2 H (in H₂O) = 6 H. Right has 4 H (in CH₃COOH). Wait — let us count more carefully.

Left: CH₃CHO has 4 H + H₂O has 2 H = 6 H. Right: CH₃COOH has 4 H. Add 2H⁺ to right. CH₃CHO + H₂O → CH₃COOH + 2H⁺

Step 3: Balance charge. Left = 0. Right = +2. Add 2e⁻ to right. CH₃CHO + H₂O → CH₃COOH + 2H⁺ + 2e⁻

Combining Half-Equations

To write the overall redox equation, combine the two half-equations so that the electrons cancel:

Multiply each half-equation by the appropriate factor so the number of electrons is the same.
Add the two half-equations together.
Cancel species that appear on both sides (electrons, H₂O, H⁺).

Worked Example 5

Combine the permanganate and iron half-equations:

Reduction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O Oxidation: Fe²⁺ → Fe³⁺ + e⁻ (× 5)

Overall: MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺

Worked Example 6

Combine the dichromate and iron half-equations:

Reduction: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O Oxidation: Fe²⁺ → Fe³⁺ + e⁻ (× 6)

Overall: Cr₂O₇²⁻ + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 7H₂O + 6Fe³⁺

Real-World Redox Reactions

Redox reactions are everywhere in daily life and industry:

Process	Oxidation	Reduction
Rusting of iron	Fe → Fe²⁺/Fe³⁺	O₂ → O²⁻
Burning fuels (combustion)	C/H in fuel → CO₂/H₂O	O₂ → O²⁻ (in CO₂/H₂O)
Batteries (electrochemical cells)	Anode metal dissolves	Cathode ions deposited
Photosynthesis	H₂O → O₂	CO₂ → C₆H₁₂O₆
Respiration	C₆H₁₂O₆ → CO₂	O₂ → H₂O
Bleaching with chlorine	Coloured compound → colourless	Cl₂ → Cl⁻
Extracting metals from ores	C or CO → CO₂	Metal oxide → metal

Rusting in detail: Iron rusts when exposed to water and oxygen. The overall process involves:

Fe(s) → Fe²⁺(aq) + 2e⁻ (oxidation at anodic areas)
O₂(aq) + 2H₂O(l) + 4e⁻ → 4OH⁻(aq) (reduction at cathodic areas)
Fe²⁺ is further oxidised to Fe³⁺, which combines with water to form hydrated iron(III) oxide — rust.

This is why rusting requires both water and oxygen, and why salt accelerates rusting (it increases the conductivity of the water, allowing electrons to flow more easily).

Displacement Reactions as Redox

A more reactive metal will displace a less reactive metal from a solution of its salt. These are redox reactions:

Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s)

Zn: 0 → +2 (oxidised — loses electrons)
Cu: +2 → 0 (reduced — gains electrons)

The ionic equation is: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

The reactivity series determines which displacements are possible. A metal higher in the series will displace one lower in the series.

Disproportionation Revisited

Disproportionation occurs when the same element is both oxidised and reduced in a single reaction.

Key Example for Edexcel

Cl₂(g) + 2NaOH(aq) → NaCl(aq) + NaClO(aq) + H₂O(l)

Cl₂: oxidation state 0 In NaCl: Cl = −1 (reduced, gained 1 electron) In NaClO: Cl = +1 (oxidised, lost 1 electron)

This reaction is important in water treatment — chlorine is added to water supplies and undergoes disproportionation to form hypochlorite ions (ClO⁻), which kill bacteria.

Common Mistakes in Redox

Mistake	Correction
"Oxidation is gain of electrons"	No — oxidation is LOSS of electrons (OIL RIG)
"The reducing agent is reduced"	No — the reducing agent is oxidised (it reduces something else)
Forgetting that both processes must occur	You cannot have oxidation without reduction
Not balancing charge in half-equations	Total charge must be equal on both sides
Adding electrons to the wrong side	Electrons go on the left for reduction, right for oxidation

Redox chemistry connects to nearly every topic in A-Level Chemistry — from electrochemistry and transition metals to organic oxidation reactions. A solid understanding of the fundamentals covered here will serve you throughout the course.

A-Level Deep Dive: Redox Reactions

Spec mapping

Edexcel 9CH0 specification Topic 3 — Redox I and Topic 14 — Redox II, sub-strand on redox half-equations and combined redox equations, requires students to write half-equations (balanced for atoms and charge), combine them into full ionic equations, and identify oxidising and reducing agents (refer to the official specification document for exact wording). Examined across Paper 2 (organic redox: alcohols → carbonyls; alkenes via KMnO $_4$ ) and Paper 3 (synoptic with Topic 14 electrochemistry, Topic 15 transition metals). The data sheet provides standard electrode potentials (used in Topic 14) and the periodic table; half-equations themselves must be derived from oxidation-state changes.

Worked example with full mark scheme

Question (8 marks):

(a) Construct the half-equation for the reduction of $\text{MnO}_4^-$ to $\text{Mn}^{2+}$ in acidic solution. (3)

(b) Construct the half-equation for the oxidation of $\text{Fe}^{2+}$ to $\text{Fe}^{3+}$ . (1)

(d) State the role of H $^+$ in the overall reaction. (2)

Solution with mark scheme:

(a) Step 1 — change in Mn oxidation state.

Mn: +7 → +2, gain of 5 electrons.

Step 2 — balance O with H $_2$ O.

$\text{MnO}_4^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$ (4 oxygens balanced).

Step 3 — balance H with H $^+$ (acidic).

$\text{MnO}_4^- + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$ .

Step 4 — balance charge with electrons.

LHS charge: -1 + 8(+1) = +7. RHS charge: +2. Add 5 electrons to LHS.

$\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$ .

M1 — atoms balanced. M1 — H and O balanced. A1 — full half-equation with electrons.

(b) $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-$ .

B1.

$5\text{Fe}^{2+} \rightarrow 5\text{Fe}^{3+} + 5e^-$ .

Add to Mn half-equation:

$\text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}$ .