Oxidation States and Redox — AQA A-level Chemistry

SPEC 3.1.7 - OXIDATION, REDUCTION AND REDOX EQUATIONS

Oxidation States, Redox Reactions & Half-Equations

Assigning oxidation states, identifying redox processes, and writing balanced half-equations

Why Redox Matters

Redox reactions are among the most important reactions in chemistry. They explain how batteries work, why iron rusts, how we extract metals from ores, and even how our cells generate energy through respiration.

The term "redox" combines reduction and oxidation because these two processes always occur together. Whenever one species loses electrons, another must gain them.

Key Concept

In any redox reaction, electrons are transferred from one species to another. The total number of electrons lost must equal the total number gained - electrons cannot be created or destroyed.

To track these electron transfers systematically, chemists developed the concept of oxidation states (also called oxidation numbers). These tell us the "electron status" of each atom in a compound.

In This Lesson

You will learn to assign oxidation states, identify what gets oxidised and reduced, and write balanced half-equations that show the electron transfer explicitly.

Oxidation State Rules

Definition

Oxidation state is the charge that would exist on an individual atom if the bonding were completely ionic. It represents the electronic "status" of an element and is written as a signed integer (e.g., +2, -1) or Roman numeral (II, III).

Think of oxidation state as a "bookkeeping" system. We assign electrons to the more electronegative atom in each bond, then count how each atom's electron count compares to when it was neutral.

RuleSpeciesOxidation State

1Uncombined elements (e.g., Na, O₂, S₈)0

2Simple ions (e.g., Na⁺, Cl⁻, Al³⁺)Same as charge: +1, -1, +3

3Oxygen (in most compounds)-2

4Hydrogen (in most compounds)+1

Why These Values?

Oxygen is -2 because it's highly electronegative - it attracts electron density towards itself in bonds. Gaining two electrons fills its outer shell. Hydrogen is +1 because it's less electronegative than most elements it bonds with, so we assign the bonding electrons to the other atom.

Important Exceptions

Oxygen: In peroxides (e.g., H₂O₂, Na₂O₂), oxygen is -1. In OF₂, oxygen is +2 (fluorine is more electronegative).

Hydrogen: In metal hydrides (e.g., NaH, CaH₂), hydrogen is -1.

Examiner Tip

Oxidation state and oxidation number are used interchangeably. Single atoms always have integer oxidation states (you can't have half an electron!), but average values in complex ions can appear fractional.

More Oxidation State Rules

RuleSpeciesOxidation State

5Fluorine (always)-1

6Chlorine (usually)-1 (except with O or F)

7Sum in neutral compound= 0

8Sum in polyatomic ion= ion charge

Electronegativity Determines Sign

In compounds with two non-metals, the more electronegative element gets the negative oxidation state. Electronegativity increases across a period and decreases down a group. Fluorine is always -1 because it's the most electronegative element.

Metals

Always positive in compounds. Usually equals group number (Al = +3). Transition metals have multiple values (Fe = +2 or +3, Mn = +2, +4, +6, +7).

Non-Metals

Usually negative (based on typical ion). Can be positive up to group number when with more electronegative elements (Cl can be +1, +3, +5, +7).

Priority Order When Rules Conflict

Apply rules in this order: F (always -1) → O (usually -2) → H (usually +1) → Halogens (usually -1). Example: In OF₂, oxygen is +2 because fluorine takes priority.

Roman Numerals in Names

When elements can have multiple oxidation states, Roman numerals specify which one is present in the compound.

Metal compounds:

Iron(II) chloride, FeCl₂ → Fe = +2

Iron(III) chloride, FeCl₃ → Fe = +3

Copper(II) sulfate, CuSO₄ → Cu = +2

"-ate" compounds:

Sulfuric(VI) acid → S = +6

Sulfurous(IV) acid → S = +4

Nitric(V) acid → N = +5

"-ate" Ions Contain Oxygen

Ions ending in "-ate" contain oxygen plus another element. If that element has variable oxidation states, use Roman numerals to distinguish them.

Ion NameOxidation State of Non-MetalFormula

Sulfate(VI)+6SO₄²⁻

Sulfate(IV) / Sulfite+4SO₃²⁻

Nitrate(V)+5NO₃⁻

Nitrate(III) / Nitrite+3NO₂⁻

Chlorate(V)+5ClO₃⁻

Chlorate(I) / Hypochlorite+1ClO⁻

Worked Example01

Determine the oxidation state of sulfur in H₂SO₄ (sulfuric acid).

1

Assign known oxidation states: H = +1, O = -2

2

Write equation: 2(+1) + S + 4(-2) = 0

3

Simplify: +2 + S - 8 = 0

4

Solve: S = +6

Answer: Sulfur has oxidation state +6

Worked Example02

Determine the oxidation state of chromium in the dichromate ion, Cr₂O₇²⁻.

1

Assign known: O = -2. Let Cr = x.

2

Sum must equal ion charge: 2(x) + 7(-2) = -2

3

Simplify: 2x - 14 = -2

4

Solve: 2x = 12, so x = +6

Answer: Chromium has oxidation state +6

Worked Example03

Determine the chemical formula of copper(II) nitrate.

1

Identify oxidation states from the name: Copper(II) means Cu = +2

2

The nitrate ion (NO₃⁻) has a charge of -1

3

Balance charges: Cu²⁺ needs two NO₃⁻ ions to make neutral compound

4

(+2) + 2(-1) = 0 ✓

Answer: Cu(NO₃)₂

Worked Example04

Determine the systematic name for the ClO₂⁻ ion.

1

Assign known: Oxygen = -2. There are 2 oxygen atoms, so total = -4

2

Let chlorine = x. Sum must equal ion charge: x + (-4) = -1

3

Solve: x = -1 + 4 = +3

4

Chlorine is +3, and it's an "-ate" ion (contains oxygen)

Answer: Chlorate(III)

Defining Oxidation & Reduction

There are three definitions of oxidation and reduction used in different branches of chemistry. All are correct and describe the same underlying process.

DefinitionOxidationReduction

OxygenGain of oxygen (e.g., 2Mg + O₂ → 2MgO)Loss of oxygen (e.g., CuO + C → Cu + CO)

HydrogenLoss of hydrogen (e.g., CH₃OH → CH₂O + H₂)Gain of hydrogen (e.g., C₂H₄ + H₂ → C₂H₆)

ElectronsLoss of electrons (e.g., Al → Al³⁺ + 3e⁻)Gain of electrons (e.g., F₂ + 2e⁻ → 2F⁻)

OIL RIG

Oxidation Is Loss (of electrons) — Reduction Is Gain (of electrons)

The electron definition is the most fundamental. When an atom loses electrons, its charge becomes more positive (oxidation state increases). Gaining electrons makes it more negative (oxidation state decreases).

Visualising Electron Transfer

In any redox reaction, electrons flow from the species being oxidised to the species being reduced. Consider the reaction between magnesium and oxygen:

Key Insight

Magnesium's oxidation state increases from 0 to +2 (it is oxidised). Oxygen's oxidation state decreases from 0 to -2 (it is reduced). The 2 electrons lost by Mg are gained by O.

05Worked Example

In the reaction: 2Fe + 3Cl₂ → 2FeCl₃

Identify which species is oxidised and which is reduced.

1

Assign oxidation states to reactants: Fe = 0, Cl₂ = 0 (elements)

2

Assign oxidation states in FeCl₃: Cl = -1, so Fe = +3

3

Track changes: Fe goes 0 → +3 (increase = oxidised)

4

Cl goes 0 → -1 (decrease = reduced)

Answer: Iron is oxidised; chlorine is reduced

Oxidising & Reducing Agents

In a redox reaction, we identify which species causes the oxidation or reduction of another. For a reaction to be redox, there must be both an oxidising and reducing agent present.

Oxidising Agent

A substance that oxidises another species by causing it to lose electrons.

The oxidising agent accepts electrons and is itself reduced.

Its oxidation state decreases.

Reducing Agent

A substance that reduces another species by causing it to gain electrons.

The reducing agent donates electrons and is itself oxidised.

Its oxidation state increases.

The Key Relationship

The oxidising agent gets reduced. The reducing agent gets oxidised. Think about it: to oxidise something else, you must take its electrons - which means you gain electrons and are reduced!

Dual Role Substances

Some substances can act as both oxidising and reducing agents depending on what they react with. For example, H₂O₂ can oxidise Fe²⁺ to Fe³⁺, but can also reduce MnO₄⁻ to Mn²⁺.

Common Exam Error

Students often confuse "oxidised" with "oxidising agent". The species that IS oxidised is the reducing agent. The species that IS reduced is the oxidising agent.

06Worked Example

In the reaction: Zn + CuSO₄ → ZnSO₄ + Cu

Identify the oxidising agent and reducing agent.

1

Track Zn: 0 → +2 (oxidised, loses electrons)

2

Track Cu: +2 → 0 (reduced, gains electrons)

3

Zn is oxidised → Zn is the reducing agent

4

Cu²⁺ is reduced → Cu²⁺ (or CuSO₄) is the oxidising agent

Answer: Oxidising agent = Cu²⁺ ions (CuSO₄); Reducing agent = Zn

Metal + Acid Reactions

Reactions between metals and acids are classic examples of redox reactions. Metal atoms lose electrons (oxidised) and hydrogen ions gain electrons (reduced).

Example: Magnesium + Hydrochloric Acid

Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)

Oxidation: Mg: 0 → +2 (loses 2e⁻)

Reduction: H: +1 → 0 (gains electrons)

Reducing agent: Mg (is oxidised)

Oxidising agent: H⁺ (is reduced)

Spectator Ions

Chloride ions remain as Cl⁻ throughout - their oxidation state doesn't change. Species whose oxidation state remains constant are called spectator ions and are not included in the ionic equation:

Mg(s) + 2H⁺(aq) → Mg²⁺(aq) + H₂(g)

What Are Half-Equations?

Definition

Half-equations are ionic equations that show either the oxidation or reduction process separately, with electrons written explicitly.

Every redox reaction can be split into two half-equations: one showing oxidation (electrons on the right as products) and one showing reduction (electrons on the left as reactants).

Half-Equation Format

Oxidation: Species → Products + electrons (e⁻ on right)

Reduction: Species + electrons → Products (e⁻ on left)

Example - Zinc in copper sulfate solution:

ProcessHalf-Equation

Oxidation (Zn)Zn → Zn²⁺ + 2e⁻

Reduction (Cu²⁺)Cu²⁺ + 2e⁻ → Cu

Notice how the electrons lost by zinc (2e⁻) exactly match the electrons gained by copper ions (2e⁻). This must always be true!

Writing Half-Equations

For simple ionic reactions, half-equations are straightforward. For reactions involving covalent molecules or complex ions, you may need to balance with H₂O and H⁺.

At A-Level

Most half-equations you write will be in acidic solution, so you can use H⁺ ions. For alkaline solutions (not required at this level), you would use OH⁻ instead.

Worked Example07

Write the half-equation for the reduction of chlorine molecules to chloride ions.

1

Write species: Cl₂ → Cl⁻

2

Balance Cl atoms: Cl₂ → 2Cl⁻

3

No O or H to balance

4

Balance charge: Left = 0, Right = -2. Add 2e⁻ to left.

Cl₂ + 2e⁻ → 2Cl⁻

This is reduction (electrons on the left, being gained).

Worked Example08

Write the half-equation for the reduction of manganate(VII) ions, MnO₄⁻, to Mn²⁺ ions in acidic solution.

1

Write species: MnO₄⁻ → Mn²⁺

2

Mn is balanced (1 on each side)

3

Balance O with H₂O: MnO₄⁻ → Mn²⁺ + 4H₂O

4

Balance H with H⁺: 8H⁺ + MnO₄⁻ → Mn²⁺ + 4H₂O

5

Balance charge: Left = +8-1 = +7, Right = +2. Add 5e⁻ to left.

MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

Combining Half-Equations

To get the overall redox equation, we combine the oxidation and reduction half-equations. The key rule is that electrons must cancel out - they cannot appear in the final equation.

Method for Combining Half-Equations

Write both half-equations with electrons shown

Multiply one or both so the electron numbers are equal

Add the equations together

Cancel electrons and any species appearing on both sides

Check: atoms balance AND charges balance

Example: Combining Zn and Cu²⁺ half-equations

StepEquation

OxidationZn → Zn²⁺ + 2e⁻

ReductionCu²⁺ + 2e⁻ → Cu

CombinedZn + Cu²⁺ → Zn²⁺ + Cu

Here, both half-equations have 2 electrons, so they cancel directly when added.

Final Check

Always verify: (1) atoms balance on both sides, (2) charges balance on both sides, (3) no electrons remain in the final equation. If charges don't balance, you may need to add H⁺ or OH⁻ ions.

09Worked Example

Combine the following half-equations to give the overall equation:

Fe²⁺ → Fe³⁺ + e⁻Cl₂ + 2e⁻ → 2Cl⁻

1

Oxidation has 1e⁻, reduction has 2e⁻

2

Multiply oxidation by 2: 2Fe²⁺ → 2Fe³⁺ + 2e⁻

3

Add equations: 2Fe²⁺ + Cl₂ + 2e⁻ → 2Fe³⁺ + 2e⁻ + 2Cl⁻

4

Cancel 2e⁻ from both sides

2Fe²⁺ + Cl₂ → 2Fe³⁺ + 2Cl⁻

Check: Fe balanced (2=2), Cl balanced (2=2), charge balanced (+4 = +4)

10

Worked Example

Write the redox equation for the reaction between sodium and sulfur to produce sodium sulfide.

1

Write oxidation half-equation: Na → Na⁺ + e⁻

2

Write reduction half-equation: S + 2e⁻ → S²⁻

3

Balance electrons (1 vs 2): Multiply oxidation by 2: 2Na → 2Na⁺ + 2e⁻

4

Add: 2Na + S + 2e⁻ → 2Na⁺ + 2e⁻ + S²⁻

5

Cancel 2e⁻ and combine ions to form compound

2Na + S → Na₂S

Check: Na balanced (2=2), S balanced (1=1), charge balanced (0 = 0)

11

Worked Example

Write the redox equation for the reaction between aluminium and chlorine to produce aluminium chloride.

1

Write oxidation half-equation: Al → Al³⁺ + 3e⁻

2

Write reduction half-equation: Cl₂ + 2e⁻ → 2Cl⁻

3

Electrons: 3e⁻ vs 2e⁻. LCM = 6. Multiply accordingly.

4

2Al → 2Al³⁺ + 6e⁻ and 3Cl₂ + 6e⁻ → 6Cl⁻

5

Add and cancel 6e⁻

2Al + 3Cl₂ → 2AlCl₃

Disproportionation Reactions

Definition

A disproportionation reaction is a redox reaction in which the same element is simultaneously oxidised and reduced.

In a disproportionation reaction, atoms of one element in a single oxidation state split into two different oxidation states - one higher (oxidation) and one lower (reduction).

Classic Example: Chlorine in Dilute Alkali

Cl₂ + 2OH⁻ → ClO⁻ + Cl⁻ + H₂O

Chlorine (0) is both oxidised to +1 in ClO⁻ and reduced to -1 in Cl⁻.

How to Recognise Disproportionation

Look for the same element appearing in products with different oxidation states - one higher and one lower than in the reactant.

Practice

Question 1: Determine the oxidation state of nitrogen in each of the following: (a) NH₃, (b) NO₂, (c) HNO₃

(a)

NH₃: 3(+1) + N = 0, so N = -3

(b)

NO₂: N + 2(-2) = 0, so N = +4

(c)

HNO₃: (+1) + N + 3(-2) = 0, so N = +5

Practice

Question 2: In the thermite reaction: Fe₂O₃ + 2Al → 2Fe + Al₂O₃, identify what is oxidised, what is reduced, the oxidising agent, and the reducing agent.

1

Al: 0 → +3 (oxidation state increases) → Oxidised

2

Fe: +3 → 0 (oxidation state decreases) → Reduced

3

Al is oxidised, so Al is the reducing agent

4

Fe³⁺ is reduced, so Fe₂O₃ is the oxidising agent

Note: Oxygen stays at -2 throughout (spectator).

Practice

Question 3: Write the half-equation for the oxidation of iodide ions, I⁻, to iodine, I₂.

1

Write species: I⁻ → I₂

2

Balance I atoms: 2I⁻ → I₂

3

Balance charge: Left = -2, Right = 0. Add 2e⁻ to right.

2I⁻ → I₂ + 2e⁻

This is oxidation (electrons on the right, being lost).

Summary

Oxidation State Rules

SpeciesOxidation State

Uncombined elements0

Simple ionsSame as charge

O (usually) / in peroxides-2 / -1

H (usually) / in metal hydrides+1 / -1

F (always)-1

Sum in compound / ion= 0 / = ion charge

Key Definitions

Oxidation: Loss of electrons / increase in oxidation state / gain of oxygen / loss of hydrogen

Reduction: Gain of electrons / decrease in oxidation state / loss of oxygen / gain of hydrogen

Oxidising agent: Causes oxidation, is itself reduced (ox. state decreases)

Reducing agent: Causes reduction, is itself oxidised (ox. state increases)

Disproportionation: Same element is both oxidised and reduced

Exam Tips

Always show working when assigning oxidation states. When writing half-equations, check atoms AND charges balance. Remember OIL RIG! For disproportionation, look for the same element in products with different oxidation states.