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Revision notes for AQA GCSE Chemistry Oxidation and reduction in terms of electrons (HT only). Open the guide for explanations and worked examples. Written against the AQA GCSE Chemistry (8462) specification, so the content matches what's examinable rather than general Chemistry background.

Oxidation and reduction in terms of electrons (HT only)

What you'll learn

  • What oxidation and reduction mean in terms of electrons.
  • How metal displacement reactions involve electron transfer.
  • How to write ionic equations by removing spectator ions.
  • How to use half equations to identify what is oxidised and reduced.

Before you start: atoms, ions and charges

An atom is neutral overall: it has the same number of negative electrons as positive protons.

An ion is charged because it has gained or lost electrons. Metals usually form positive ions by losing electrons. For example, a magnesium atom can lose two electrons to form Mg²⁺.

Because electrons are negative:

  • losing electrons makes a particle more positive
  • gaining electrons makes a particle more negative, or less positive
Definition

Ions and species

An ion is a charged particle made when an atom or group of atoms gains or loses electrons. A species is any atom, ion or molecule shown in a chemical reaction.

Example

Working out the charge after electron loss

  1. A magnesium atom starts neutral, so its total charge is zero.

  2. It loses two electrons. Removing negative charge leaves the particle with an overall 2+ charge.

  3. The ion formed is Mg²⁺, so magnesium has changed from Mg(s) to Mg²⁺(aq).

Oxidation and reduction

This Higher Tier section explains reactions by tracking electrons.

Oxidation and reduction always happen together in a redox reaction. One species loses electrons, and another species gains those electrons.

Definition

Oxidation and reduction

Oxidation is the loss of electrons. Reduction is the gain of electrons.

Key Idea

OIL RIG

Oxidation Is Loss, Reduction Is Gain — of electrons.

A helpful way to remember this is OIL RIG. The “loss” and “gain” refer only to electrons, not to oxygen or mass.

Electron transfer in magnesium displacing copper ions

Common Mistake

Do not rely on oxygen

In this topic, define oxidation and reduction using electrons. Some redox reactions involve oxygen, but many do not.

Example

Classifying electron transfer

  1. In Mg(s) → Mg²⁺(aq) + 2e⁻, electrons are on the product side, so magnesium has lost electrons.

  2. Loss of electrons is oxidation, so Mg is oxidised.

  3. In Cu²⁺(aq) + 2e⁻ → Cu(s), electrons are on the reactant side, so copper(II) ions gain electrons.

  4. Gain of electrons is reduction, so Cu²⁺ ions are reduced.

Displacement reactions as redox reactions

A displacement reaction happens when a more reactive element takes the place of a less reactive element in a compound.

In the reactivity of metals topic, this often means:

  • a more reactive metal atom loses electrons
  • ions of a less reactive metal gain electrons
  • the more reactive metal forms ions
  • the less reactive metal is made as a solid

For example, magnesium is more reactive than copper. So magnesium can displace copper from copper(II) sulfate solution:

Mg(s) + CuSO₄(aq) → MgSO₄(aq) + Cu(s)

The visible change is usually that a brown/orange copper solid forms, while the blue copper(II) sulfate solution becomes paler.

Key Idea

What is really happening

In a metal displacement reaction, electrons are transferred from the more reactive metal atoms to the less reactive metal ions.

Ionic equations

A normal symbol equation often includes ions that are present but not actually changed during the reaction. To focus on the redox change, chemists write an ionic equation.

Definition

Ionic equation and spectator ions

An ionic equation shows only the reacting particles that change in a reaction. Spectator ions are ions that are present but unchanged, so they are left out of the final ionic equation.

How to write an ionic equation

Use this method for metal displacement reactions in solution:

  1. Write the balanced symbol equation with state symbols.
  2. Split aqueous ionic compounds into their separate ions.
  3. Keep solids, liquids and gases unchanged.
  4. Cancel ions that appear unchanged on both sides.
  5. Check that atoms and total charge balance.
Example

Writing an ionic equation for magnesium and copper(II) sulfate

  1. Start with the full balanced equation:

    Mg(s) + CuSO₄(aq) → MgSO₄(aq) + Cu(s)

  2. Split the aqueous ionic compounds into ions. Keep the solid metals unchanged:

    Mg(s) + Cu²⁺(aq) + SO₄²⁻(aq) → Mg²⁺(aq) + SO₄²⁻(aq) + Cu(s)

  3. Identify the spectator ion. SO₄²⁻(aq) appears unchanged on both sides, so it is not involved in the electron transfer.

  4. Cancel the spectator ion to get the ionic equation:

    Mg(s) + Cu²⁺(aq) → Mg²⁺(aq) + Cu(s)

  5. Check the equation: there is one Mg and one Cu on each side, and the total charge is 2+ on both sides.

Common Mistake

Cancelling formulas instead of ions

Do not cancel whole compounds like CuSO₄(aq) and MgSO₄(aq). Split aqueous ionic compounds first, then cancel the unchanged ion: SO₄²⁻(aq).

Half equations

A half equation shows the electron transfer for just one species. It focuses on either oxidation or reduction.

For oxidation, electrons are usually on the right:

Mg(s) → Mg²⁺(aq) + 2e⁻

For reduction, electrons are usually on the left:

Cu²⁺(aq) + 2e⁻ → Cu(s)

A correct half equation must balance:

  • the atoms
  • the total charge
Definition

Half equation

A half equation shows the electrons lost or gained by one species during oxidation or reduction.

Tip

Electron side shortcut

In a half equation, electrons on the right mean electrons are lost, so it is oxidation. Electrons on the left mean electrons are gained, so it is reduction.

Example

Balancing an oxidation half equation

Complete the half equation for aluminium forming aluminium ions.

  1. Start with the species changing:

    Al(s) → Al³⁺(aq)

  2. The atoms are already balanced: there is one aluminium atom on each side.

  3. The left side is neutral, but the right side has a 3+ charge from Al³⁺. Add three electrons to the right so the total charge becomes zero.

  4. The balanced half equation is:

    Al(s) → Al³⁺(aq) + 3e⁻

  5. The electrons are on the right, so aluminium has lost electrons and has been oxidised.

Identifying what is oxidised and reduced

In an exam, you might be given a word equation, symbol equation, ionic equation or half equation.

The safest method is to ask: where have the electrons gone?

If a species becomes more positive, it has lost electrons, so it has been oxidised.

If a species becomes less positive, neutral, or more negative, it has gained electrons, so it has been reduced.

Example

Identifying oxidation and reduction in a displacement reaction

For the ionic equation:

Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

  1. Compare zinc before and after the reaction. Zn(s) is neutral, but Zn²⁺(aq) has a 2+ charge, so zinc has lost two electrons.

  2. Write the zinc half equation:

    Zn(s) → Zn²⁺(aq) + 2e⁻

    Electrons are on the right, so zinc is oxidised.

  3. Compare copper before and after the reaction. Cu²⁺(aq) has a 2+ charge, but Cu(s) is neutral, so copper(II) ions have gained two electrons.

  4. Write the copper half equation:

    Cu²⁺(aq) + 2e⁻ → Cu(s)

    Electrons are on the left, so copper(II) ions are reduced.

  5. Therefore, Zn is oxidised and Cu²⁺ ions are reduced.

Common Mistake

Only if the reaction happens

A less reactive metal will not displace a more reactive metal ion. Only write a displacement ionic equation if the solid metal is more reactive than the metal ion in solution.

Putting the language together

Try to be precise with the species you name.

For this reaction:

Mg(s) + Cu²⁺(aq) → Mg²⁺(aq) + Cu(s)

A strong answer would be:

  • Magnesium atoms are oxidised because they lose electrons to form Mg²⁺ ions.
  • Copper(II) ions are reduced because they gain electrons to form copper atoms.

Avoid saying only “copper is reduced” if the reacting species is actually Cu²⁺. The examiner wants to see that you understand which particle gained the electrons.

Exam technique

In the exam

  1. For a displacement reaction, write the full balanced equation with state symbols, then split aqueous ionic compounds into ions.

  2. Cancel only spectator ions: ions that appear unchanged on both sides.

  3. Use OIL RIG to identify redox: electron loss is oxidation, electron gain is reduction.

  4. Name the species carefully, such as “Mg atoms are oxidised” and “Cu²⁺ ions are reduced”.

Self review

Check yourself

  • In Mg(s) + Zn²⁺(aq) → Mg²⁺(aq) + Zn(s), which species loses electrons?

  • Write the ionic equation for Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s).

  • In Fe³⁺(aq) + e⁻ → Fe²⁺(aq), is Fe³⁺ oxidised or reduced?

Reactivity of metals

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