A-Level Chemistry 9701 / Topic 6 / AS

Balance the electrons.

A redox equation is two half-equations stitched together: one losing electrons, one gaining them. Balance each half on its own, match the electrons, and even the manganate(VII) equations become routine.

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The key idea

A half-equation shows one species gaining or losing electrons. In acidic solution balance it in four steps: the element changing oxidation number, then O with H₂O, then H with H⁺, then charge with electrons. The number of electrons equals the change in oxidation number. To combine two half-equations, multiply them so the electrons lost equal the electrons gained, add them, and cancel anything that appears on both sides. No electrons may remain in the final equation.

Fig. 1: balancing the manganate(VII) half-equation in four steps

Section 01

Where the electrons go.

Watch electrons pass from zinc to copper(II) ions in the simulator, then split the reaction into its two halves.

Oxidation: Zn → Zn²⁺ + 2e⁻ (electrons on the right).
Reduction: Cu²⁺ + 2e⁻ → Cu (electrons on the left).
Add: Zn + Cu²⁺ → Zn²⁺ + Cu. The two electrons cancel.

Section 02

Half-equations in acid.

Build the manganate(VII) and dichromate(VI) half-equations in the simulator one step at a time, following Fig. 1.

Species	Half-equation	Change in oxidation number
manganate(VII)	MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O	+7 → +2, gains 5
dichromate(VI)	Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O	2 × (+6 → +3), gains 6
iron(II)	Fe²⁺ → Fe³⁺ + e⁻	+2 → +3, loses 1
hydrogen peroxide	H₂O₂ → O₂ + 2H⁺ + 2e⁻	2 × (−1 → 0), loses 2

Stage 1 · Learn

Check what the sim just showed you

Four quick checks. Each correct answer earns XP and records this skill on your profile.

Quick check+10 XP

In a reduction half-equation, the electrons appear:

Quick check+10 XP

Which is the correctly balanced half-equation for Fe²⁺ changing to Fe³⁺?

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When balancing a half-equation in acidic solution, oxygen atoms are balanced by adding:

Quick check+10 XP

How many electrons are gained when one MnO₄⁻ ion is reduced to Mn²⁺?

Section 03

Combining halves.

Find the lowest common multiple of the electrons, scale each half-equation, add and cancel.

Manganate(VII) and iron(II): 5 electrons gained, 1 lost, so multiply the iron half by 5: MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O. Mole ratio MnO₄⁻ : Fe²⁺ = 1 : 5.
Dichromate(VI) and iron(II): 6 gained, so six Fe²⁺: Cr₂O₇²⁻ + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O.
Check: atoms balance and the total charge is the same on both sides (+17 = +17 for the manganate equation).

Examiner trap

Electrons go on the left in a reduction and on the right in an oxidation, and they must all cancel in the full equation. A final equation that still shows e⁻ has not been combined properly. Always finish by checking that the charges balance as well as the atoms.

Stage 2 · Exam

Exam-style questions

Unlocks once the four checks above are done. Worth more XP, written to AS Paper 1 and Paper 2 standard.

Finish the four checks above to unlock the exam questions

Exam style+20 XP

What is the mole ratio of MnO₄⁻ to Fe²⁺ in their reaction in acidic solution?

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Which is the correctly balanced half-equation for dichromate(VI) in acid?

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Combine Cl₂ + 2e⁻ → 2Cl⁻ with Fe²⁺ → Fe³⁺ + e⁻. The overall equation is:

Exam style+20 XP

20.0 cm³ of 0.0200 mol dm⁻³ KMnO₄ reacts exactly with Fe²⁺ in acid. How many moles of Fe²⁺ reacted?

Skill unlocked

Redox equations, mastered.

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Go deeper · practice

Six original Cambridge-style questions

Writing and balancing half-equations in acid, combining them into full redox equations, balancing using oxidation number changes and using redox mole ratios.