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Writing Redox Equations from Half-equations

Learn and apply Writing Redox Equations from Half-equations in the published Chemistry course sequence.

  • GCE A-Level H2 Chemistry 9476-2027
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Writing Redox Equations from Half-equations: Orientation

This lesson is the “turn table → equation” skill: use E⦵ to decide which half-equation stays as reduction, reverse the other for oxidation, balance electrons, and add.

Before the harder applications, review Standard Electrode Potentials and SHE and keep the Electrochemistry hub as your route map.

This page is really testing:

  • Can you identify which half-equation is reversed (oxidation)?
  • Can you balance electrons cleanly without changing species/charges incorrectly?
  • Can you present one final overall equation with no electrons left?

Definitions (Must Know)

A. Half-equations

A half-equation shows either oxidation (loss of electrons) or reduction (gain of electrons).

B. Overall redox equation

An overall redox equation is obtained by adding the oxidation and reduction half-equations after balancing electrons.

C. Oxidation-number test

Oxidation is an increase in oxidation number; reduction is a decrease. This test is useful when electron transfer is not visually explicit in the molecular equation.

Detailed Explanations

A. A reliable step-by-step method

  1. Write the two relevant reduction half-equations from the table.
  2. Identify the more positive E⦵ (this reduction happens).
  3. Reverse the other half-equation (that one becomes oxidation).
  4. Multiply half-equations so electrons cancel.
  5. Add and simplify.

Because the table lists reductions, therefore oxidation is always created by reversing one half-equation (and reversing its E⦵ sign).

Mini example: Using Cu²⁺/Cu (+0.34 V) and Zn²⁺/Zn (−0.76 V), copper stays as reduction and zinc is reversed to oxidation.

B. Worked method (mark-scheme order)

Use this order in long-structured questions:

  1. Write both data-booklet reduction half-equations.
  2. Circle the more positive E⦵ (this is the reduction that happens).
  3. Reverse the other half-equation and label it oxidation.
  4. Multiply to cancel electrons.
  5. Add, cancel electrons, then check atom and charge balance.

C. Checking redox by oxidation numbers

After constructing the overall equation, identify the atom whose oxidation number increases and the atom whose oxidation number decreases. The total increase must equal the total decrease when electron transfer is balanced. This is an independent check on coefficients and on the identities of the oxidising and reducing agents.

Worked Examples

Modelled example 1

Combine a Zinc–Copper Pair

Core

Problem

The standard half-equations are Cu²⁺(aq) + 2e⁻ ⇌ Cu(s) Zn²⁺(aq) + 2e⁻ ⇌ Zn(s) Write the overall equation for the feasible cell reaction under standard conditions.
Study the worked solution

  1. Choose the two directions

    Method

    Keep copper(II) reduction and reverse the zinc half-equation for oxidation.

    Reason

    The copper couple has the more positive reduction potential, so zinc supplies electrons to copper(II) ions.

    Working

    Cu²⁺(aq) + 2e⁻ → Cu(s) Zn(s) → Zn²⁺(aq) + 2e⁻

  2. Add and cancel electrons

    Method

    Add the half-equations and remove the two electrons appearing on opposite sides.

    Reason

    Electrons are transferred internally and must not remain in the overall redox equation.

    Working

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

Guided practice 2

Scale the Iron Half-equation

About 6 min

Problem

Given I₂(aq) + 2e⁻ ⇌ 2I-(aq) Fe³⁺(aq) + e⁻ ⇌ Fe²⁺(aq) write the overall equation for Fe³⁺ oxidising I⁻.

Try this before viewing the solution

Hints

Hint 1: identify the oxidation
Iodide is oxidised, so reverse the iodine half-equation.
Hint 2: equalise the electrons
The iron reduction transfers one electron; scale it to match the two-electron iodide oxidation.
View solution step by step

  1. Reverse the oxidation half-equation

    Method

    Write iodide losing two electrons.

    Reason

    The prompt states that iodide is oxidised.

    Working

    2I-(aq) → I₂(aq) + 2e⁻

  2. Scale the reduction

    Reason

    Two iron(III) ions must accept the two electrons released by iodide.

    Working

    2Fe³⁺(aq) + 2e⁻ → 2Fe²⁺(aq)

  3. Combine the changes

    Method

    Add the equations and cancel electrons.

    Reason

    Equal electron loss and gain gives the smallest whole-number overall equation.

    Working

    2Fe³⁺(aq) + 2I-(aq) → 2Fe²⁺(aq) + I₂(aq)

Common misconception 3

Correct Two Half-equations Pointing the Same Way

Find and correct the mistake

Learner attempt

For aluminium reacting with chlorine, a learner scales and adds these two reduction half-equations: Al³⁺ + 3e⁻ → Al Cl₂ + 2e⁻ → 2Cl⁻ 2Al³⁺ + 3Cl₂ + 12e⁻ → 2Al + 6Cl⁻ Identify the first error and construct the correct overall equation.

Find the first error

First error

View solution step by step

  1. Locate the first error

    Method

    Reverse the aluminium half-equation.

    Reason

    Aluminium is oxidised in this reaction, so it must release electrons rather than consume them.

    Working

    Al → Al³⁺ + 3e⁻

  2. Equalise electron transfer

    Reason

    The lowest common multiple of three and two electrons is six.

    Working

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

  3. Write the corrected equation

    Method

    Add and cancel the six electrons.

    Reason

    The overall equation must conserve atoms and charge without displaying transferred electrons.

    Working

    2Al + 3Cl₂ → 2Al³⁺ + 6Cl⁻

Examiner practice 4

Construct and Label a Bromine–Iron Reaction

4 marks

Problem

The standard reduction potentials are Br₂(aq) + 2e⁻ ⇌ 2Br-(aq) E⦵ = +1.07 V Fe³⁺(aq) + e⁻ ⇌ Fe²⁺(aq) E⦵ = +0.77 V Construct the feasible overall equation under standard conditions and identify the oxidising and reducing agents. [4 marks]

Try this before viewing the solution

View solution step by step

  1. Select the reduction

    1 mark

    Method

    Keep the bromine half-equation as reduction.

    Reason

    Its standard reduction potential is more positive.

    Working

    Br₂(aq) + 2e⁻ → 2Br-(aq)

  2. Reverse and scale iron

    1 mark

    Method

    Reverse the iron half-equation and multiply it by two.

    Reason

    Iron(II) is oxidised and must release the two electrons accepted by bromine.

    Working

    2Fe²⁺(aq) → 2Fe³⁺(aq) + 2e⁻

  3. State the overall equation

    1 mark

    Reason

    Adding the two directions cancels the transferred electrons.

    Working

    Br₂(aq) + 2Fe²⁺(aq) → 2Br-(aq) + 2Fe³⁺(aq)

  4. Identify the agents

    1 mark

    Method

    Name bromine as the oxidising agent and iron(II) ions as the reducing agent.

    Reason

    Bromine is reduced, while iron(II) ions are oxidised.

    Working

    Oxidising agent: Br₂; reducing agent: Fe²⁺.

Challenge 5

Combine an Acidic Oxyanion Half-equation

Minimal support

Problem

Acidified dichromate(VI) ions oxidise iodide ions. Use the half-equations Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O I₂ + 2e⁻ → 2I⁻ to construct the overall equation. Then use oxidation numbers to check the direction of both changes.

Try this before viewing the solution

Hints

Hint 1: set the iodide direction
The prompt says iodide is oxidised, so reverse its listed reduction half-equation.
Hint 2: match six electrons
Scale the reversed iodide half-equation so it releases six electrons.
View solution step by step

  1. Reverse and scale iodide oxidation

    Method

    Reverse the iodine half-equation and multiply it by three.

    Reason

    Six iodide ions must release the six electrons accepted by one dichromate(VI) ion.

    Working

    6I⁻ → 3I₂ + 6e⁻

  2. Combine the acidic half-equations

    Reason

    The six electrons now cancel without changing the H⁺ or H₂O terms in the dichromate half-equation.

    Working

    Cr₂O₇²⁻ + 14H⁺ + 6I⁻ → 2Cr³⁺ + 7H₂O + 3I₂

  3. Audit oxidation numbers

    Method

    Check that chromium decreases and iodine increases in oxidation number.

    Reason

    Equal total decrease and increase independently confirms the electron-transfer direction.

    Working

    Each Cr changes + 6 → + 3; each I changes -1 → 0. Both changes account for six electrons.

Mind Stretchers

Mind stretcher 1Extension

Why is it useful that data booklets list reduction potentials rather than a mix of reduction and oxidation potentials?

Show Hint

A common reduction convention makes every comparison use the same direction; oxidation is generated only when needed.

Show Answer

Mark scheme:

  • It keeps sign conventions consistent: you always compare “tendency to be reduced”.
  • You can decide oxidation by reversing the relevant half-equation, rather than memorising separate tables.

Mind stretcher 2: Constructing an unfamiliar acidic redox equationExtension

Question. Combine the half-equations MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O and Fe³⁺ + e⁻ → Fe²⁺ to write the equation for acidified permanganate oxidising Fe²⁺. Identify the oxidising agent.

Show Hint

Reverse the iron half-equation, scale it to five electrons, then cancel electrons before checking atoms and charge.

Show Answer

Reverse the iron half-equation and multiply it by 5: 5Fe²⁺ → 5Fe³⁺ + 5e⁻ Adding and cancelling electrons gives: MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺ MnO₄⁻ is the oxidising agent because it is reduced to Mn²⁺.