2.3.3Unit 2. Energetics, Group Chemistry, Halogenoalkanes and Alcohols
Disproportionation Reactions — IAL Edexcel Linear (Chemistry) Revision Notes & Topical Questions
5 min read 5 parts Lesson 23 of 53 Unit quiz
Part 1 of 5Defining Disproportionation
In a disproportionation reaction one species acts as both oxidising agent and reducing agent at once.
Defining Disproportionation
Same element
- A single element in a single oxidation state ends up in two different oxidation states, one higher and one lower.
Evidence
- Check oxidation numbersthe same element must increase in one product and decrease in another.
Hydrogen peroxide
- In 2H₂O₂ → 2H₂O + O₂, oxygen goes from -1 to -2 in water and from -1 to 0 in oxygen gas.
Copper(I)
- Copper(I) oxide in acid forms copper metal (0) and copper(II) ions (+2), so Cu(I) is both oxidised and reduced.
Part 2 of 5Chlorine and Aqueous Alkali
Chlorine disproportionates in alkali, and the products depend on the temperature.
Cold dilute alkali
- Cl₂(g) + 2NaOH(aq) → NaCl(aq) + NaClO(aq) + H₂O(l), where chlorine goes from 0 to -1 and +1.
Hot concentrated alkali
- 3Cl₂(g) + 6NaOH(aq) → 5NaCl(aq) + NaClO₃(aq) + 3H₂O(l), producing chloride and chlorate(V).
Chlorine in water
- Cl₂(g) + H₂O(l) ⇌ HCl(aq) + HClO(aq) is also a disproportionation, giving chloric(I) acid used to sterilise water.
Oxidation numbers
- In chlorate(V), ClO₃⁻, chlorine is +5, while in chlorate(I), ClO⁻, it is +1.
| Conditions | Products containing chlorine | Oxidation numbers of Cl |
|---|---|---|
| Cold dilute NaOH | NaCl and NaClO | -1 and +1 |
| Hot concentrated NaOH | NaCl and NaClO₃ | -1 and +5 |
Part 3 of 5Balancing by Oxidation Number Changes
1
Write the unbalanced equation and mark the oxidation numbers of the element that changes in both directions.
2
Work out each change: chlorine to chloride falls by 1 per atom, while chlorine to chlorate(V) rises by 5 per atom.
3
Multiply species so the total rise equals the total fall, so five chloride ions are needed for each chlorate(V) ion.
4
Balance the charge using hydroxide ions on the left, and then balance hydrogen and oxygen with water.
5
Check the final atom and charge totals on both sides, giving 3Cl₂ + 6OH⁻ → 5Cl⁻ + ClO₃⁻ + 3H₂O.
Part 4 of 5Worked Example
Question Show that the reaction 3Cl₂ + 6OH⁻ → 5Cl⁻ + ClO₃⁻ + 3H₂O is a disproportionation and that it is balanced.
- Chlorine in Cl₂ is 0, in Cl⁻ is -1 (reduced), and in ClO₃⁻ is +5 (oxidised).
- Five Cl atoms each fall by 1, a total fall of 5; one Cl atom rises by 5, so the changes balance.
- Atoms
- 6 Cl, 6 O and 6 H appear on both sides (O: 6 = 3 + 3;
- H: 6 = 3 × 2).
- 6 Cl, 6 O and 6 H appear on both sides (O: 6 = 3 + 3;
- Charge
- left side is 6 × (-1) = -6;
- right side is 5 × (-1) + (-1) = -6.
- left side is 6 × (-1) = -6;
Answer Chlorine is both reduced and oxidised, so it disproportionates, and the equation balances for atoms and charge.
Part 5 of 5Exam Checklist & Tips
2.3.3 – Key Points for the Exam
- Disproportionation A redox reaction in which the same element is simultaneously oxidised and reduced.
- Chlorate(I) The ion ClO⁻, in which chlorine has oxidation number +1, found in bleach.
- Chlorate(V) The ion ClO₃⁻, in which chlorine has oxidation number +5.
- Chloric(I) acid HClO, formed when chlorine dissolves in water, which kills bacteria.
Common Mistakes
- Defining disproportionation as two different elements being oxidised and reduced; it is the same element.
- Mixing up the conditions, giving chlorate(V) in cold alkali instead of hot concentrated alkali.
- Forgetting that the oxidation number changes must balance, so the ratio of Cl⁻ to ClO₃⁻ is 5 to 1.
- Leaving out state symbols or the charge on the ions in ionic equations.
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