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6.2.2Additional lessons

Half Equations — IB Chemistry SL Revision Notes & Topical Questions

5 min read 6 parts Lesson 85 of 98

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Part 1 of 6Balancing a Half Equation in Acid

6.2.2 Half Equations

This method works for most redox reactions in aqueous acid.

1
Write the species that change, for example MnO₄⁻ → Mn²⁺.
2
Balance every atom except oxygen and hydrogen, such as the Mn or Cr atoms.
3
Balance oxygen by adding H₂O to the side that needs it.
4
Balance hydrogen by adding H⁺ to the opposite side.
5
Balance the charge by adding electrons, then check the oxidation number change matches the electrons.

Part 2 of 6Key Half Equations

Key Half Equations
Manganate(VII)
  • MnO₄⁻(aq) + 8H⁺(aq) + 5e⁻ → Mn²⁺(aq) + 4H₂O(l); purple solution becomes almost colourless.
Dichromate(VI)
  • Cr₂O₇²⁻(aq) + 14H⁺(aq) + 6e⁻ → 2Cr³⁺(aq) + 7H₂O(l); orange becomes green.
Iron(II)
  • Fe²⁺(aq) → Fe³⁺(aq) + e⁻, which is an oxidation step.
Iodide
  • 2I⁻(aq) → I₂(aq) + 2e⁻, giving a brown solution.

Part 3 of 6Combining Half Equations

Equalise electrons
  • Multiply each half equation so the electrons lost equal the electrons gained.
Add and cancel
  • Add the two equations and cancel electrons and any species appearing on both sides.
Check
  • Confirm atoms and charge balance on each side of the final equation.
Example
  • MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺ has charge +17 on each side.
TitrationReactionIndicator
Manganate(VII) with Fe²⁺MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺Self-indicating, first permanent pink
Thiosulfate with iodine2S₂O₃²⁻ + I₂ → S₄O₆²⁻ + 2I⁻Starch, blue-black to colourless

Part 4 of 6Redox Titrations in Practice

Redox Titrations in Practice
Manganate(VII) end point
  • Potassium manganate(VII) is purple, so the first permanent pale pink shows excess oxidant and no indicator is needed.
Acid medium
  • The solution is acidified with dilute sulfuric acid, because the half equation needs H⁺ ions.
Thiosulfate titrations
  • Iodine released by an oxidising agent is titrated with thiosulfate,
    • and starch added near the end gives a sharp blue-black to colourless change.
Calculation route
  • Convert the titre to moles, use the mole ratio from the combined equation, then find concentration or percentage.

Part 5 of 6Worked Example

Question

  • A 25.00 cm³ sample of acidified Fe²⁺(aq) requires 20.00 cm³ of 0.0200 mol dm⁻³ KMnO₄ for a permanent pink colour.
  • Calculate [Fe²⁺].

  1. n(MnO₄⁻) = 0.0200 × 0.02000 = 4.00 × 10⁻⁴ mol.
  2. The equation shows 1 MnO₄⁻ reacts with 5 Fe²⁺, so n(Fe²⁺) = 5 × 4.00 × 10⁻⁴ = 2.00 × 10⁻³ mol.
  3. [Fe²⁺] = 2.00 × 10⁻³ / 0.02500 = 0.0800 mol dm⁻³.

Answer [Fe²⁺] = 0.0800 mol dm⁻³.

Part 6 of 6Exam Checklist & Tips

6.2.2 – Key Points for the Exam

  • Half equation An equation showing only the oxidation or the reduction part of a redox reaction.
  • Spectator ion An ion that does not change and is absent from the ionic equation.
  • Redox titration A titration in which an oxidising agent reacts with a reducing agent.
  • Self-indicating A titration in which a reagent itself shows the end point by its colour.

Common Mistakes

  • Balancing the charge by adding electrons before balancing hydrogen.
  • Placing electrons on the wrong side: reduction half equations have electrons on the left.
  • Forgetting to multiply the half equations so electrons cancel.
  • Acidifying a manganate(VII) titration with hydrochloric acid,
    • since chloride ions can be oxidised by manganate(VII); dilute sulfuric acid is used instead.

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