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6.2.2Additional lessons

Half Equations — IB Chemistry HL Revision Notes & Topical Questions

5 min read 5 parts Lesson 133 of 172

📚 IB Chemistry HL Hub / Topical Revision
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Part 1 of 5What a Half-Equation Shows

6.2.2 Half Equations

A half-equation describes only one of the two processes in a redox reaction, with electrons shown explicitly.

Oxidation half
  • Electrons appear on the product side, for example Fe²⁺ → Fe³⁺ + e⁻, showing electrons being lost.
Reduction half
  • Electrons appear on the reactant side, for example Cl₂ + 2e⁻ → 2Cl⁻, showing electrons being gained.
Charge balance
  • The total charge on both sides must be equal, which is checked after electrons have been added.
Atom balance
  • Every element must appear in equal numbers on each side before the charges are balanced.

Part 2 of 5Balancing in Acidic Solution

1
Write the two species containing the element that changes oxidation state, and balance that element first.
2
Balance oxygen atoms by adding water molecules to the side that needs oxygen.
3
Balance hydrogen atoms by adding H⁺ ions to the side that needs hydrogen.
4
Balance the charge by adding electrons to the more positive side until both charges are equal.
5
Check the final atom and charge totals once more before moving on.

Part 3 of 5Combining Half-Equations

Combining Half-Equations
Equalise electrons
  • Multiply each half-equation by a whole number so that electrons lost equal electrons gained.
Add and cancel
  • Add the two half-equations together and cancel electrons, plus any H⁺ or H₂O appearing on both sides.
Add state symbols
  • Include (aq), (s), (l) and (g) in the final equation, as the data booklet and mark schemes expect.
Final check
  • Confirm that total charge and atoms of every element balance in the overall equation.
Half-equationType
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂OReduction
Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂OReduction
Fe²⁺ → Fe³⁺ + e⁻Oxidation

Part 4 of 5Worked Example

Question Combine MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O with Fe²⁺ → Fe³⁺ + e⁻ to give the overall equation.

  1. Multiply the iron half-equation by 5 so that 5 electrons are lost: 5Fe²⁺ → 5Fe³⁺ + 5e⁻.
  2. Add the two half-equations and cancel the 5e⁻ on each side.
  3. Check the chargeleft is −1 + 8 + 10 = +17, and right is +2 + 15 = +17.

Answer MnO₄⁻(aq) + 8H⁺(aq) + 5Fe²⁺(aq) → Mn²⁺(aq) + 4H₂O(l) + 5Fe³⁺(aq)

Part 5 of 5Exam Checklist & Tips

6.2.2 – Key Points for the Exam

  • Half-equation An equation showing either the loss or the gain of electrons by one species.
  • Electron The particle transferred in redox, written as e⁻ with a charge of −1.
  • Acidic solution Conditions where H⁺ and H₂O are used to balance hydrogen and oxygen.
  • Spectator ion An ion that is present but does not change during the reaction.

Common Mistakes

  • Balancing charge before balancing oxygen and hydrogen, which leads to wrong electron numbers.
  • Leaving electrons in the final overall equation instead of cancelling them.
  • Multiplying only one half-equation, so that electrons lost and gained do not match.
  • Adding H⁺ on the wrong side, which gives balanced atoms but unbalanced charge.

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