UNIT 2 CHEMISTRY · PEARSON AAQ · C1.2.4
Half Equations and Redox Equations
FREE CHEMISTRY REVISION · TOPIC 22
Balance the atoms, then the charge, then match the electrons.
Construct and combine half equations systematically, from simple ions to increasingly demanding balances in a stated medium.
Pearson BTEC Level 3 AAQ Applied Science
Unit 2: Principles and Applications of Chemistry · C1.2.4
What you will learn
Construct oxidation and reduction half equations; balance atoms and total charge; equalise transferred electrons; add and cancel to obtain net ionic redox equations. Progress from monatomic ions to an oxygen-containing ion in a stated medium, checking every coefficient and sign.
1. What a half equation shows
A half equation isolates either electron loss or electron gain. Oxidation has electrons on the product side: Zn → Zn²⁺ + 2e⁻. Reduction has electrons on the reactant side: Cu²⁺ + 2e⁻ → Cu. The electrons show the transfer that is hidden when the two changes are combined.
Two quantities must be conserved independently: the number of atoms of each element and the total electric charge. Each e⁻ contributes −1 to the charge sum. A half equation may have a positive or negative total on both sides; it need not be neutral. For Fe²⁺ → Fe³⁺ + e⁻, the total is +2 on each side.
Keep an ion’s charge and formula fixed when balancing. Change whole-number coefficients in front of species. Writing Cu⁺ instead of Cu²⁺ changes the species; changing Cl₂ to Cl changes the given chlorine molecule rather than balancing it. Include state symbols when required.
2. Construct simple half equations systematically
1. Write the starting and finishing species, with correct formulas and charges.
2. Balance the changing element by coefficients.
3. Compare total charge on the two sides.
4. Add electrons to the more positive side until charges are equal.
5. Check that the electron position agrees with oxidation or reduction.
6. Recount both atoms and charges.
For chlorine reduction begin Cl₂ → Cl⁻. Balance atoms: Cl₂ → 2Cl⁻. The right-hand charge is −2, so add 2e⁻ to the left: Cl₂ + 2e⁻ → 2Cl⁻. For iodide oxidation the direction is 2I⁻ → I₂ + 2e⁻.
Hydrogen evolution in acidic solution is 2H⁺ + 2e⁻ → H₂. The coefficient 2 accounts for both hydrogen atoms. The two +1 charges and two electron charges sum to zero on the left, matching neutral H₂. H⁺ + e⁻ → H₂ would fail atom conservation.
3. Combine half equations only after electrons match
One half equation must supply the electrons consumed by the other. Find the lowest common multiple of the electron coefficients. Multiply every term of each half equation by the required factor; then add left sides together and right sides together.
Cancel identical electrons from opposite sides. Cancel other identical species on opposite sides only where they genuinely have the same formula, charge and state. Simplify the remaining coefficients where appropriate. A complete overall redox reaction has no electrons left over.
For Zn → Zn²⁺ + 2e⁻ and Cu²⁺ + 2e⁻ → Cu, the counts already match, giving Zn + Cu²⁺ → Zn²⁺ + Cu. For Al and Cu²⁺, use six electrons before adding. The sum must conserve both atoms and total charge.
4. Oxygen and hydrogen: use the stated medium
For an oxygen-containing species in acidic aqueous solution, balance the other element first, then O using H₂O, H using H⁺, and finally charge using e⁻. H₂O and H⁺ are balancing species available in the stated medium, not arbitrary replacements for the given reactants.
For ClO⁻ → Cl⁻ in acid: add H₂O to the right for the oxygen, then 2H⁺ to the left for its two H atoms. Before electrons, left charge is −1 + 2 = +1 and right charge is −1. Add two electrons on the left. The final reduction is ClO⁻ + 2H⁺ + 2e⁻ → Cl⁻ + H₂O.
Cl changes from +1 in ClO⁻ to −1 in Cl⁻, confirming two-electron gain. Oxygen and hydrogen must still be accounted for: oxidation numbers do not replace atom balancing.
5. An alkaline application of the same balancing principle
If the question specifies alkaline conditions, H⁺ must not remain in the final half equation. First balance as in acid, then add enough OH⁻ to both sides to convert all H⁺ into H₂O, and cancel identical water molecules.
For the ClO⁻ example, add 2OH⁻ to both sides. The left 2H⁺ + 2OH⁻ becomes 2H₂O; one water cancels against the water on the right. The alkaline form is ClO⁻ + H₂O + 2e⁻ → Cl⁻ + 2OH⁻. Each side has one Cl, two O, two H and total charge −3. Use the given medium.
These oxyanion applications extend the construction method. Pearson C1.2.4 does not prescribe a fixed list of half-equation species. Use a defensible atom-and-charge balance for the species and conditions supplied in the question.
6. From net ionic to full equations
A net ionic equation omits unchanged spectators. A sulfate ion unchanged in a Zn/Cu²⁺ displacement cancels. When a full formula equation is required, restore the appropriate spectator counter-ions in balanced quantities; do not insert extra species into half equations merely to make them neutral.
A mathematically balanced equation does not prove that a reaction occurs rapidly under every condition. Reactants, products and medium must be appropriate. Passivation and reaction conditions can matter; balancing establishes ratios once the chemical change is specified.
Worked examples
Example 1 · Oxidation of aluminium
Start Al → Al³⁺. One Al is present on each side. Add 3e⁻ to the right: Al → Al³⁺ + 3e⁻. Charge check: 0 = +3 − 3. Electron loss is oxidation.
Example 2 · Reduction of chlorine
Start Cl₂ → Cl⁻; balance Cl as Cl₂ → 2Cl⁻. Add 2e⁻ to the left: Cl₂ + 2e⁻ → 2Cl⁻. Atoms: 2 Cl each side. Charge: −2 each side.
Example 3 · Combine magnesium and acid
Mg → Mg²⁺ + 2e⁻ and 2H⁺ + 2e⁻ → H₂ already involve two electrons each. Add and cancel: Mg + 2H⁺ → Mg²⁺ + H₂. Atoms balance and total charge is +2 on each side.
Example 4 · Combine zinc and silver ions
Zn → Zn²⁺ + 2e⁻; Ag⁺ + e⁻ → Ag. Multiply the whole silver equation by 2: 2Ag⁺ + 2e⁻ → 2Ag. Add: Zn + 2Ag⁺ → Zn²⁺ + 2Ag. Charge is +2 on each side.
Example 5 · Use the lowest common multiple
Al → Al³⁺ + 3e⁻ and Cu²⁺ + 2e⁻ → Cu need six electrons. Multiply the first by 2 and second by 3. Add: 2Al + 3Cu²⁺ → 2Al³⁺ + 3Cu. Charge is +6 on each side. Multiply all terms, not just electrons.
Example 6 · One-electron iron change with a two-electron molecule
Fe²⁺ → Fe³⁺ + e⁻; Cl₂ + 2e⁻ → 2Cl⁻. Double the iron equation and add: 2Fe²⁺ + Cl₂ → 2Fe³⁺ + 2Cl⁻. Charge: +4 on the left; +6 − 2 = +4 on the right.
Example 7 · Combine an acidic oxyanion reduction
Combine ClO⁻ + 2H⁺ + 2e⁻ → Cl⁻ + H₂O with 2Fe²⁺ → 2Fe³⁺ + 2e⁻. Result: ClO⁻ + 2H⁺ + 2Fe²⁺ → Cl⁻ + H₂O + 2Fe³⁺. Each side contains Cl 1, O 1, H 2, Fe 2. Charge: −1 + 2 + 4 = +5 and −1 + 6 = +5.
Example 8 · Repair an unbalanced half equation
Fe²⁺ + e⁻ → Fe³⁺ balances Fe but gives left charge +1 and right +3. The actual change is oxidation: Fe²⁺ → Fe³⁺ + e⁻. Now both charges are +2. Atom balance alone missed the error.
Example 9 · Check the alkaline form
For ClO⁻ + H₂O + 2e⁻ → Cl⁻ + 2OH⁻, left charge is −1 − 2 = −3 and right is −1 − 2 = −3. Both sides contain Cl 1, O 2, H 2. The medium changes the H/O balancing species, not the two-electron requirement.
Common misconceptions and exam tips
“Electrons always go on the left.” They are products in oxidation and reactants in reduction.
“Charge must be zero.” It must be equal on both sides, not necessarily zero.
“Change ion charges to balance.” Keep the species fixed; change coefficients.
“Multiply only electrons.” Multiply every term.
“Matching atoms is sufficient.” Check charge separately.
“Keep spare electrons in the overall equation.” Equalise and cancel them.
“H⁺ can remain in a final alkaline equation.” Use the stated medium and the water/OH⁻ balance.
Show both half equations, the multipliers and the final combined equation. Count atoms and calculate total charge numerically. Revisit electron placement before changing any correct formula.
Your revision summary
Half equations conserve atoms and charge. Oxidation releases e⁻; reduction consumes e⁻. Equalise electron counts by multiplying entire half equations, then add and cancel. Use H₂O and the acid or alkaline species appropriate to the stated medium.
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