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Redox And Half Equations

Learn and apply Redox And Half Equations in the published Chemistry course sequence.

  • GCE A-Level H1 Chemistry 8873-2027
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H1 Redox and Half-equations: Orientation

Redox equations must balance two things at once: atoms and charge. Use oxidation numbers to identify what changes, then use half-equations to show where the electrons go.

H1 8873 scope
  • H1 8873 places electron transfer, oxidation-number change and half-equation construction inside Stoichiometry.
  • Electrode potentials and H2 Electrochemistry calculations are not imported into this lesson.

Definitions (Must Know)

  • Oxidation is loss of electrons or an increase in oxidation number.
  • Reduction is gain of electrons or a decrease in oxidation number.
  • An oxidising agent accepts electrons and is reduced.
  • A reducing agent donates electrons and is oxidised.
  • An oxidation number is the formal charge assigned using agreed electron-accounting rules.
  • A half-equation represents either oxidation or reduction and shows electrons explicitly.

Detailed Explanations

A. Building a half-equation in acid

For MnO₄⁻ → Mn²⁺:

  1. manganese is already balanced;
  2. add 4H₂O to the product side to balance oxygen;
  3. add 8H⁺ to the reactant side to balance hydrogen;
  4. the left charge is + 7, so add 5e⁻ to the left to make both sides + 2.

MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

B. Combining half-equations

Write both halves, scale them so electrons cancel, then add and remove identical species from both sides. Finish with an atom audit and a charge audit.

C. Connecting agents to oxidation numbers

In 2Fe²⁺ + Cl₂ → 2Fe³⁺ + 2Cl⁻, iron increases from + 2 to + 3 and is oxidised, so Fe²⁺ is the reducing agent. Chlorine decreases from 0 to -1 and is reduced, so Cl₂ is the oxidising agent.

Worked Examples

Modelled example 1

Construct a Dichromate Half-equation

Core

Problem

Balance Cr₂O₇²⁻ → Cr³⁺ in acidic solution.
Study the worked solution

  1. Balance chromium and oxygen

    Method

    Place 2 before Cr³⁺ and add 7 water molecules to the product side.

    Reason

    The dichromate ion contains two chromium atoms and seven oxygen atoms.

    Working

    Cr₂O₇²⁻ → 2Cr³⁺ + 7H₂O

  2. Balance hydrogen

    Reason

    Acidic solution supplies hydrogen ions to match the 14 hydrogen atoms in seven water molecules.

    Working

    Cr₂O₇²⁻ + 14H⁺ → 2Cr³⁺ + 7H₂O

  3. Balance charge

    Method

    Add six electrons to the reactant side.

    Reason

    The reactant charge before electrons is + 12, while the product charge is + 6.

    Working

    Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O

Guided practice 2

Combine Permanganate with Iron(II)

About 6 min

Problem

Combine MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O with Fe²⁺ → Fe³⁺ + e⁻ to write the overall equation in acidic solution.

Try this before viewing the solution

Hints

Hint 1: compare electron counts
Permanganate consumes five electrons while one iron(II) ion releases one.
Hint 2: scale the oxidation
Multiply every term in the iron half-equation by five.
View solution step by step

  1. Scale iron oxidation

    Method

    Multiply the iron half-equation by five.

    Reason

    Electron loss must equal the five-electron gain by permanganate.

    Working

    5Fe²⁺ → 5Fe³⁺ + 5e⁻

  2. Add and cancel

    Method

    Add the half-equations and remove the five electrons.

    Reason

    The electrons are transferred between the two reacting species.

    Working

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

Common misconception 3

Correct Partial Scaling

Find and correct the mistake

Learner attempt

A learner combines Zn → Zn²⁺ + 2e⁻ and Ag⁺ + e⁻ → Ag by writing the scaled reduction as 2Ag⁺ + e⁻ → 2Ag. Identify the first error and write the correct overall equation.

Find the error before correcting it

First error

View solution step by step

  1. Scale the complete half-equation

    Method

    Double silver ions, electrons and silver atoms together.

    Reason

    A half-equation is one balanced relationship; changing only selected coefficients breaks charge balance.

    Working

    2Ag⁺ + 2e⁻ → 2Ag

  2. Combine the pair

    Method

    Add and cancel two electrons.

    Reason

    Zinc releases exactly the electron amount accepted by two silver ions.

    Working

    Zn + 2Ag⁺ → Zn²⁺ + 2Ag

Examiner practice 4

Construct and Interpret an Iron–Chlorine Equation

4 marks

Problem

Combine Fe²⁺ → Fe³⁺ + e⁻ and Cl₂ + 2e⁻ → 2Cl⁻. Write the overall equation and identify the oxidising and reducing agents. [4 marks]

Try this before viewing the solution

View solution step by step

  1. Scale iron oxidation

    1 mark

    Method

    Multiply the iron half-equation by two.

    Reason

    Two iron(II) ions must supply the two electrons accepted by one chlorine molecule.

    Working

    2Fe²⁺ → 2Fe³⁺ + 2e⁻

  2. Write the overall equation

    1 mark

    Reason

    Adding the half-equations cancels the two transferred electrons.

    Working

    2Fe²⁺ + Cl₂ → 2Fe³⁺ + 2Cl⁻

  3. Identify the oxidising agent

    1 mark

    Method

    Name chlorine.

    Reason

    Chlorine gains electrons and is reduced from oxidation number 0 to -1.

    Working

    Oxidising agent: Cl₂.

  4. Identify the reducing agent

    1 mark

    Method

    Name iron(II) ions.

    Reason

    Iron(II) loses electrons and is oxidised from + 2 to + 3.

    Working

    Reducing agent: Fe²⁺.

Challenge 5

Recover a Missing Half-equation

Minimal support

Problem

The overall reaction is 2Al + 3Cu²⁺ → 2Al³⁺ + 3Cu and the copper reduction is Cu²⁺ + 2e⁻ → Cu. Infer the aluminium oxidation half-equation and show how the two halves reproduce the overall equation.

Try this before viewing the solution

Hints

Hint 1: follow aluminium
One aluminium atom changes from oxidation number 0 to + 3.
Hint 2: match the total electrons
The overall equation contains two aluminium atoms and three copper(II) ions.
View solution step by step

  1. Infer aluminium oxidation

    Method

    Write one aluminium atom releasing three electrons.

    Reason

    The oxidation-number increase from 0 to + 3 corresponds to loss of three electrons.

    Working

    Al → Al³⁺ + 3e⁻

  2. Reach a common electron count

    Reason

    Two aluminium oxidations release six electrons, while three copper reductions consume six.

    Working

    2Al → 2Al³⁺ + 6e⁻ 3Cu²⁺ + 6e⁻ → 3Cu

  3. Reconstruct the overall equation

    Method

    Add the scaled halves and cancel six electrons.

    Reason

    The reconstruction checks that the inferred half-equation is consistent with every overall coefficient.

    Working

    2Al + 3Cu²⁺ → 2Al³⁺ + 3Cu

Mind Stretchers

Attempt each unfamiliar application before opening the hint, then compare your reasoning chain with the solution.

Mind stretcher 1: Identifying both agentsExtension

Question. Combine 2I⁻ → I₂ + 2e⁻ with Cl₂ + 2e⁻ → 2Cl⁻ and identify both agents.

Show Hint

The electron counts already match; identify which reactant loses and which gains them.

Show Answer

2I⁻ + Cl₂ → I₂ + 2Cl⁻. Chlorine gains electrons and is the oxidising agent. Iodide loses electrons and is the reducing agent.

Mind stretcher 2: Auditing an incorrect equationExtension

Question. A student proposes MnO₄⁻ + 8H⁺ + 4e⁻ → Mn²⁺ + 4H₂O. Find and correct the error.

Show Hint

Compare total charge on the two sides.

Show Answer

The left charge is -1 + 8-4 = +3, but the right charge is + 2. Five electrons are required: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O.