C18: Oxidation and Reduction

Higher

Oxidation and reduction are defined in terms of electron transfer using OIL RIG. Writing balanced half equations and ionic equations allows you to identify which species are oxidised and reduced in a redox reaction.

Exam Board:

Oxidation and Reduction in Terms of Electrons

Oxidation Is Loss of electrons. Reduction Is Gain of electrons. Remember this using the mnemonic OIL RIG.

In earlier topics, you learned that oxidation is the gain of oxygen and reduction is the loss of oxygen. At Higher level, you need to understand oxidation and reduction in terms of electron transfer, which is a more general definition that applies to all redox reactions, not just those involving oxygen.

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

When a substance is oxidised, it loses electrons and its oxidation number increases. When a substance is reduced, it gains electrons and its oxidation number decreases. Oxidation and reduction always happen simultaneously in a redox reaction because the electrons lost by one species must be gained by another.

Process	Electron Transfer	Effect on Oxidation Number	Memory Aid
Oxidation	Loses electrons	Increases (becomes more positive)	OIL — Oxidation Is Loss
Reduction	Gains electrons	Decreases (becomes more negative)	RIG — Reduction Is Gain

If a species loses electrons, it is oxidised. If a species gains electrons, it is reduced. You can identify which is which by looking at the charges: if the charge becomes more positive, the species has lost electrons (oxidised); if the charge becomes more negative, the species has gained electrons (reduced).

Identifying Oxidation and Reduction

To identify oxidation and reduction, look at how the charge on each species changes. If a species becomes more positively charged, it has lost electrons and been oxidised. If a species becomes less positively charged (or more negatively charged), it has gained electrons and been reduced.

Displacement of Copper by Zinc

Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s)

In ionic terms: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

Zinc goes from Zn (0) to Zn²⁺ (+2): charge becomes more positive, so zinc has lost 2 electrons and been oxidised.

Copper goes from Cu²⁺ (+2) to Cu (0): charge becomes less positive, so copper has gained 2 electrons and been reduced.

This is a redox reaction because oxidation and reduction happen simultaneously.

Reaction of Sodium with Chlorine

2Na(s) + Cl₂(g) → 2NaCl(s)

Na → Na⁺ + e⁻: Sodium loses one electron per atom and is oxidised.

Cl₂ + 2e⁻ → 2Cl⁻: Chlorine gains one electron per atom and is reduced.

The electrons lost by sodium are gained by chlorine. This is a redox reaction.

Reaction of Iron with Copper(II) Sulfate

Fe(s) + Cu²⁺(aq) → Fe²⁺(aq) + Cu(s)

Iron: Fe → Fe²⁺ + 2e⁻ — iron loses 2 electrons and is oxidised.

Copper: Cu²⁺ + 2e⁻ → Cu — copper gains 2 electrons and is reduced.

The two electrons lost by iron are gained by copper(II) ions.

Writing Half Equations

A half equation shows the oxidation or reduction of a single species, including the electrons transferred. Oxidation half equations have electrons on the product side. Reduction half equations have electrons on the reactant side.

Steps for writing a balanced half equation:

Identify the species before and after the reaction.
Write the unbalanced equation showing the change.
Balance all atoms except oxygen and hydrogen.
Balance oxygen atoms by adding H₂O (for reactions in aqueous solution).
Balance hydrogen atoms by adding H⁺ ions.
Balance the charge by adding electrons (e⁻).
Check that both sides have the same total charge.

Half Equation for the Oxidation of Iron(II) to Iron(III)

Step 1: Fe²⁺ → Fe³⁺

Step 2: Atoms are balanced (1 Fe on each side).

Step 3: Charges: left = +2, right = +3. Add 1 electron to the right to balance charge.

Fe²⁺ → Fe³⁺ + e⁻

Check: left charge = +2, right charge = +3 + (-1) = +2. Balanced.

Iron(II) is oxidised because it loses an electron.

Half Equation for the Reduction of Manganate(VII) in Acidic Solution

Step 1: MnO₄⁻ → Mn²⁺

Step 2: Balance Mn: already balanced (1 Mn on each side).

Step 3: Balance O by adding H₂O: MnO₄⁻ → Mn²⁺ + 4H₂O

Step 4: Balance H by adding H⁺: MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O

Step 5: Balance charge: left = -1 + 8(+1) = +7, right = +2 + 0 = +2. Add 5e⁻ to the left.

MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

Check: left charge = -1 + 8 + (-5) = +2, right charge = +2. Balanced.

Manganate(VII) is reduced because it gains 5 electrons.

Half Equation for the Oxidation of Sulfite to Sulfate

Step 1: SO₃²⁻ → SO₄²⁻

Step 2: S atoms balanced. O atoms: need to add H₂O and H⁺.

Step 3: Balance O by adding H₂O to the other side: SO₃²⁻ + H₂O → SO₄²⁻

Step 4: Balance H by adding H⁺: SO₃²⁻ + H₂O → SO₄²⁻ + 2H⁺

Step 5: Balance charge: left = -2 + 0 = -2, right = -2 + 2 = 0. Add 2e⁻ to the right.

SO₃²⁻ + H₂O → SO₄²⁻ + 2H⁺ + 2e⁻

Check: left charge = -2, right charge = -2 + 2 + (-2) = -2. Balanced.

Sulfite is oxidised because it loses 2 electrons.

Ionic Equations

An ionic equation shows only the species that actually take part in a reaction. Spectator ions (ions that do not change during the reaction) are removed from the equation.

Spectator ions appear on both sides of the full ionic equation in exactly the same form. They are present in the reaction mixture but do not participate in the reaction itself.

Steps for writing an ionic equation:

Write the full balanced symbol equation.
Split all soluble ionic compounds into their individual ions.
Cancel out the spectator ions that appear on both sides.
Write the final ionic equation showing only the species that change.

Ionic Equation for Zinc Displacing Copper from Copper(II) Sulfate

Full equation: Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s)

Split into ions: Zn(s) + Cu²⁺(aq) + SO₄²⁻(aq) → Zn²⁺(aq) + SO₄²⁻(aq) + Cu(s)

Cancel spectator ions: SO₄²⁻ appears on both sides — it is a spectator ion.

Ionic equation: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

This shows clearly that zinc is oxidised (loses 2 electrons) and copper is reduced (gains 2 electrons).

Ionic Equation for Neutralisation

Full equation: HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

Split into ions: H⁺(aq) + Cl⁻(aq) + Na⁺(aq) + OH⁻(aq) → Na⁺(aq) + Cl⁻(aq) + H₂O(l)

Cancel spectator ions: Na⁺ and Cl⁻ appear on both sides.

Ionic equation: H⁺(aq) + OH⁻(aq) → H₂O(l)

This is the same for all neutralisation reactions between a strong acid and a strong alkali, regardless of which specific acid and alkali are used.

Ionic Equation for the Reaction of Magnesium with Hydrochloric Acid

Full equation: Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)

Split into ions: Mg(s) + 2H⁺(aq) + 2Cl⁻(aq) → Mg²⁺(aq) + 2Cl⁻(aq) + H₂(g)

Cancel spectator ions: Cl⁻ appears on both sides.

Ionic equation: Mg(s) + 2H⁺(aq) → Mg²⁺(aq) + H₂(g)

Magnesium is oxidised (loses 2 electrons) and hydrogen is reduced (gains 2 electrons).

State symbols are important in ionic equations. Only split compounds that are aqueous (aq). Solid (s), liquid (l), and gas (g) substances should not be split into ions because their ions are not free to move independently.

Redox Reactions

A redox reaction is a reaction in which both oxidation and reduction occur simultaneously. In any redox reaction, the number of electrons lost by the species being oxidised equals the number of electrons gained by the species being reduced.

Every redox reaction can be split into two half equations: one showing the oxidation and one showing the reduction. The overall equation is obtained by combining the half equations so that the electrons cancel out.

Combining Half Equations: Iron and Copper(II) Ions

Oxidation half equation: Fe(s) → Fe²⁺(aq) + 2e⁻

Reduction half equation: Cu²⁺(aq) + 2e⁻ → Cu(s)

The number of electrons lost (2) equals the number of electrons gained (2), so the half equations can be combined directly.

Add the two half equations together and cancel the electrons:

Fe(s) + Cu²⁺(aq) + 2e⁻ → Fe²⁺(aq) + 2e⁻ + Cu(s)

Fe(s) + Cu²⁺(aq) → Fe²⁺(aq) + Cu(s)

Combining Half Equations with Different Electron Numbers

Oxidation half equation: Zn(s) → Zn²⁺(aq) + 2e⁻

Reduction half equation: Ag⁺(aq) + e⁻ → Ag(s)

The number of electrons does not match (2 lost vs 1 gained). Multiply the silver half equation by 2:

2Ag⁺(aq) + 2e⁻ → 2Ag(s)

Now combine: Zn(s) + 2Ag⁺(aq) → Zn²⁺(aq) + 2Ag(s)

Check: left charge = 0 + 2(+1) = +2, right charge = +2 + 0 = +2. Balanced.

Combining Half Equations: Aluminium and Iron(III) Oxide (Thermite Reaction)

Oxidation half equation: Al(s) → Al³⁺(aq) + 3e⁻ (multiply by 2 = 6e⁻)

Reduction half equation: Fe³⁺(aq) + 3e⁻ → Fe(s) (multiply by 2 = 6e⁻)

Combined: 2Al(s) + 2Fe³⁺(aq) → 2Al³⁺(aq) + 2Fe(s)

Aluminium is oxidised (loses electrons) and iron is reduced (gains electrons). The thermite reaction is extremely exothermic and is used for welding railway tracks.

Oxidising and Reducing Agents

An oxidising agent is a substance that causes oxidation in another species by accepting electrons from it (the oxidising agent itself is reduced). A reducing agent is a substance that causes reduction in another species by donating electrons to it (the reducing agent itself is oxidised).

This can seem confusing at first, but remember:

An oxidising agent oxidises something else, which means it takes electrons from that substance. By gaining electrons, the oxidising agent is itself reduced.
A reducing agent reduces something else, which means it gives electrons to that substance. By losing electrons, the reducing agent is itself oxidised.

Term	What It Does	What Happens to It	Electron Transfer
Oxidising agent	Causes oxidation of another species	Gets reduced itself	Accepts electrons
Reducing agent	Causes reduction of another species	Gets oxidised itself	Donates electrons

Identifying the Oxidising and Reducing Agent

Reaction: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

Zinc is oxidised (loses electrons): Zn → Zn²⁺ + 2e⁻

Copper(II) ions are reduced (gain electrons): Cu²⁺ + 2e⁻ → Cu

Zinc causes the reduction of Cu²⁺, so zinc is the reducing agent.

Cu²⁺ causes the oxidation of Zn, so Cu²⁺ is the oxidising agent.

Common Oxidising and Reducing Agents

Common oxidising agents:

Potassium manganate(VII) (KMnO₄) — turns from purple to colourless when reduced
Potassium dichromate(VI) (K₂Cr₂O₇) — turns from orange to green when reduced
Chlorine, oxygen, concentrated sulfuric acid

Common reducing agents:

Reactive metals (sodium, magnesium, zinc, iron)
Carbon, carbon monoxide, sulfur dioxide
Hydrogen (in some reactions)

The oxidising agent is the species that gets reduced. The reducing agent is the species that gets oxidised. Think of it this way: the agent does the action to the other species. The oxidising agent causes oxidation in the other substance.

Redox in Displacement Reactions

All metal displacement reactions are redox reactions. The more reactive metal is oxidised (loses electrons) and acts as the reducing agent. The less reactive metal ion is reduced (gains electrons) and acts as the oxidising agent.

Iron Displacing Silver from Silver Nitrate

Fe(s) + 2AgNO₃(aq) → Fe(NO₃)₂(aq) + 2Ag(s)

Ionic equation: Fe(s) + 2Ag⁺(aq) → Fe²⁺(aq) + 2Ag(s)

Iron is oxidised: Fe → Fe²⁺ + 2e⁻ (iron is the reducing agent)

Silver ions are reduced: Ag⁺ + e⁻ → Ag (Ag⁺ is the oxidising agent)

The reaction shows why iron displaces silver: iron is more reactive and loses electrons more readily.

Magnesium Displacing Zinc from Zinc Sulfate

Mg(s) + ZnSO₄(aq) → MgSO₄(aq) + Zn(s)

Ionic equation: Mg(s) + Zn²⁺(aq) → Mg²⁺(aq) + Zn(s)

Magnesium is oxidised: Mg → Mg²⁺ + 2e⁻ (magnesium is the reducing agent)

Zinc ions are reduced: Zn²⁺ + 2e⁻ → Zn (Zn²⁺ is the oxidising agent)

Redox in Electrolysis

Electrolysis is always a redox process. At the anode, negative ions lose electrons (oxidation). At the cathode, positive ions gain electrons (reduction). The two half equations together describe the full redox process.

Redox in the Electrolysis of Molten Sodium Chloride

At the cathode (reduction): Na⁺(l) + e⁻ → Na(l)

At the anode (oxidation): 2Cl⁻(l) → Cl₂(g) + 2e⁻

To combine these half equations, multiply the cathode equation by 2 so the electrons balance:

2Na⁺(l) + 2e⁻ → 2Na(l)

2Na⁺(l) + 2Cl⁻(l) → 2Na(l) + Cl₂(g)

Sodium ions are reduced (gain electrons). Chloride ions are oxidised (lose electrons).

Redox in the Extraction of Aluminium

At the cathode: Al³⁺(l) + 3e⁻ → Al(l) — reduction

At the anode: 2O²⁻(l) → O₂(g) + 4e⁻ — oxidation

Multiply the cathode equation by 4 and the anode equation by 3 to balance electrons (12e⁻ on each side):

4Al³⁺(l) + 12e⁻ → 4Al(l)

6O²⁻(l) → 3O₂(g) + 12e⁻

Overall: 4Al³⁺(l) + 6O²⁻(l) → 4Al(l) + 3O₂(g)

Which simplifies to: 2Al₂O₃(l) → 4Al(l) + 3O₂(g)

Oxidation States and Redox

An oxidation state (or oxidation number) is a number assigned to an element in a compound that indicates the number of electrons lost or gained by an atom of that element compared to its uncombined state. Oxidation increases the oxidation state; reduction decreases it.

Rules for assigning oxidation states:

The oxidation state of an uncombined element is 0 (e.g. Fe, O₂, Cl₂).
The oxidation state of a monatomic ion equals the charge on the ion (e.g. Na⁺ = +1, Cl⁻ = -1).
In compounds, oxygen usually has an oxidation state of -2 (except in peroxides where it is -1).
In compounds, hydrogen usually has an oxidation state of +1 (except in metal hydrides where it is -1).
The sum of all oxidation states in a neutral compound is 0.
The sum of all oxidation states in a polyatomic ion equals the charge on the ion.

Identifying Redox Using Oxidation States

Reaction: Fe₂O₃ + 3CO → 2Fe + 3CO₂

Iron: Fe in Fe₂O₃ has oxidation state +3; Fe as a product has oxidation state 0.

Change from +3 to 0 = decrease = reduction.

Carbon: C in CO has oxidation state +2; C in CO₂ has oxidation state +4.

Change from +2 to +4 = increase = oxidation.

Iron(III) oxide is reduced (and is the oxidising agent). Carbon monoxide is oxidised (and is the reducing agent).

Oxidation States in the Reaction of Magnesium with Oxygen

2Mg(s) + O₂(g) → 2MgO(s)

Magnesium: 0 → +2 (increase, so Mg is oxidised).

Oxygen: 0 → -2 (decrease, so O₂ is reduced).

Magnesium is the reducing agent (it causes the reduction of oxygen). Oxygen is the oxidising agent (it causes the oxidation of magnesium).

Practice Questions

1. Use OIL RIG to explain why the conversion of Fe²⁺ to Fe³⁺ is an oxidation reaction.

Fe²⁺ → Fe³⁺ + e⁻. The iron(II) ion loses one electron to become iron(III). Since oxidation is loss of electrons (OIL), this is an oxidation reaction. The oxidation state increases from +2 to +3.

2. Write half equations for the reaction: Zn(s) + Pb²⁺(aq) → Zn²⁺(aq) + Pb(s). Identify which species is oxidised and which is reduced.

Oxidation half equation: Zn(s) → Zn²⁺(aq) + 2e⁻ (zinc is oxidised, it loses electrons).

Reduction half equation: Pb²⁺(aq) + 2e⁻ → Pb(s) (lead is reduced, it gains electrons).

Zinc is the reducing agent. Pb²⁺ is the oxidising agent.

3. Write an ionic equation for the reaction between magnesium and copper(II) sulfate solution. Identify the oxidising agent and the reducing agent.

Mg(s) + Cu²⁺(aq) → Mg²⁺(aq) + Cu(s) (SO₄²⁻ is the spectator ion).

Magnesium is oxidised (loses electrons) and is the reducing agent. Cu²⁺ is reduced (gains electrons) and is the oxidising agent.

4. Write a balanced half equation for the reduction of dichromate(VI) ions in acidic solution: Cr₂O₇²⁻ → Cr³⁺.

Step 1: Cr₂O₇²⁻ → Cr³⁺ (unbalanced)

Step 2: Balance Cr: Cr₂O₇²⁻ → 2Cr³⁺

Step 3: Balance O with H₂O: Cr₂O₇²⁻ → 2Cr³⁺ + 7H₂O

Step 4: Balance H with H⁺: Cr₂O₇²⁻ + 14H⁺ → 2Cr³⁺ + 7H₂O

Step 5: Balance charge: left = -2 + 14 = +12, right = 6. Add 6e⁻ to left.

Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O

Check: left = -2 + 14 + (-6) = +6, right = +6. Balanced.

5. In the thermite reaction, 2Al(s) + Fe₂O₃(s) → Al₂O₃(s) + 2Fe(s), explain which species is the oxidising agent and which is the reducing agent.

Aluminium is oxidised: Al → Al³⁺ + 3e⁻ (oxidation state increases from 0 to +3). Aluminium is the reducing agent because it causes the reduction of iron by donating electrons.

Iron is reduced: Fe³⁺ + 3e⁻ → Fe (oxidation state decreases from +3 to 0). Fe₂O₃ is the oxidising agent because it causes the oxidation of aluminium by accepting electrons.

6. Write an ionic equation for the precipitation of lead(II) iodide from lead(II) nitrate and potassium iodide. Explain why this is NOT a redox reaction.

Pb²⁺(aq) + 2I⁻(aq) → PbI₂(s)

This is not a redox reaction because no electrons are transferred. The oxidation states of all elements remain unchanged: lead remains +2 and iodine remains -1 throughout the reaction. This is a precipitation reaction, not a redox reaction.

Maths Skills Higher

Balancing Ionic and Half Equations

Writing balanced ionic equations: (1) write the full balanced symbol equation, (2) split all aqueous ionic compounds into individual ions, (3) cancel spectator ions that appear identically on both sides, (4) check that both atoms and charges balance.

Writing half equations: (1) identify the species before and after the reaction, (2) balance all atoms, (3) balance oxygen by adding H₂O (in aqueous reactions), (4) balance hydrogen by adding H⁺, (5) balance the total charge by adding electrons, (6) verify both sides have the same total charge.

Combining Half Equations

Combining half equations: ensure the electrons lost in oxidation equal the electrons gained in reduction. Multiply one or both half equations by appropriate factors, then add them together and cancel the electrons. Check that the final ionic equation is balanced for both atoms and charge.

Common Misconceptions

Oxidation and Oxygen Higher

Wrong: Oxidation always involves oxygen Correct: Oxidation is defined as the loss of electrons (OIL RIG). While the original definition involved gaining oxygen, the electron-based definition is more general and applies to all redox reactions, including those with no oxygen involved. For example, when Zn loses electrons to form Zn²⁺, it is oxidised even though no oxygen is present

Oxidising and Reducing Agents Higher

Wrong: The oxidising agent gets oxidised Correct: The oxidising agent causes oxidation in another species by accepting electrons from it — so the oxidising agent itself gets reduced. Similarly, the reducing agent causes reduction in another species by donating electrons — so the reducing agent itself gets oxidised. Think of it this way: the agent does the action to the other substance

Wrong: Reduction means getting smaller Correct: Reduction means gain of electrons (RIG). The term comes from the reduction in oxidation number, not a physical reduction in size. When Cu²⁺ gains 2 electrons to become Cu, the oxidation number decreases from +2 to 0

6-Mark Extended Question Higher

Oxidation and Reduction in Terms of Electrons

Explain oxidation and reduction in terms of electrons. Use the reaction of zinc with copper sulfate to illustrate your answer.

Oxidation is loss of electrons (OIL) and reduction is gain of electrons (RIG). Oxidation and reduction always happen simultaneously in a redox reaction because the electrons lost by one species must be gained by another. [2 marks]

In the reaction Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s), the ionic equation is Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s). Zinc is oxidised because it loses two electrons: Zn → Zn²⁺ + 2e⁻. The oxidation state of zinc increases from 0 to +2. [2 marks]

Copper is reduced because it gains two electrons: Cu²⁺ + 2e⁻ → Cu. The oxidation state of copper decreases from +2 to 0. Zinc is the reducing agent (it causes the reduction of Cu²⁺) and Cu²⁺ is the oxidising agent (it causes the oxidation of Zn). The two electrons lost by zinc are exactly the two gained by copper. [2 marks]

AO3: Analyse and Evaluate Higher

Identifying Oxidation and Reduction in a Displacement Reaction

For the reaction: Fe(s) + CuSO₄(aq) → FeSO₄(aq) + Cu(s). (a) Write the ionic equation for this reaction. (b) Identify which species is oxidised and which is reduced. (c) State which is the oxidising agent and which is the reducing agent. Write half equations for both processes.

(a) Ionic equation: Fe(s) + Cu²⁺(aq) → Fe²⁺(aq) + Cu(s). SO₄²⁻ is the spectator ion. (b) Iron is oxidised: Fe → Fe²⁺ + 2e⁻ (loses electrons, oxidation state increases from 0 to +2). Copper is reduced: Cu²⁺ + 2e⁻ → Cu (gains electrons, oxidation state decreases from +2 to 0). (c) Iron is the reducing agent because it donates electrons and causes the reduction of Cu²⁺. Cu²⁺ is the oxidising agent because it accepts electrons and causes the oxidation of Fe. The 2 electrons lost by iron are exactly the 2 gained by copper.

📝 Exam Questions by Topic

AQA Past Papers - Topic Questions -- AQA
Edexcel Past Papers - Topic Questions -- Edexcel
OCR Past Papers - Topic Questions -- OCR
Eduqas Past Papers - Topic Questions -- Eduqas
CCEA Past Papers - Topic Questions -- CCEA

🎬 Video Resources

Ready to ace your GCSE Chemistry exams?

Get the best revision books and guides to boost your grades.

📚 Shop Chemistry Books 📖 Shop Revision Guides ✅ All GCSE Guides

← Previous: Electrolysis Next →

C18: Oxidation and Reduction

Oxidation and Reduction in Terms of Electrons

Identifying Oxidation and Reduction

Writing Half Equations

Ionic Equations

Redox Reactions

Oxidising and Reducing Agents

Redox in Displacement Reactions

Redox in Electrolysis

Oxidation States and Redox

Practice Questions

Maths Skills Higher

Balancing Ionic and Half Equations

Combining Half Equations

Common Misconceptions

Oxidation and Oxygen Higher

Oxidising and Reducing Agents Higher

6-Mark Extended Question Higher

Oxidation and Reduction in Terms of Electrons

AO3: Analyse and Evaluate Higher

Identifying Oxidation and Reduction in a Displacement Reaction

📝 Exam Questions by Topic

🎬 Video Resources

Share this page

Related

Related Topics

Related Subjects

Ready to ace your GCSE Chemistry exams?