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2.3.3Organic Chemistry

Halogen Redox Reactions — Edexcel AS Chemistry Revision Notes & Topical Questions

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Section 1 of 5Disproportionation

2.3.3 Halogen Redox Reactions
Definition
  • Disproportionation is a redox reaction in which the same element is both oxidised and reduced.
Chlorine in water
  • Chlorine goes from 0 to −1 in HCl and from 0 to +1 in HClO, so it is oxidised and reduced at once.
Equation
  • Cl₂(g) + H₂O(l) ⇌ HCl(aq) + HClO(aq), which is an equilibrium rather than a complete reaction.
Chloric(I) acid
  • HClO is the active chemical in the mixture, and it dissociates to give the chlorate(I) ion, ClO⁻.
Evidence
  • Damp blue litmus paper turns red from the acids and then bleaches white as HClO destroys the dye.

Section 2 of 5Water Treatment

Water Treatment
Benefit
  • Chlorate(I) ions kill bacteria and other microorganisms, so treated water carries far less risk of disease.
Risk: toxicity
  • Chlorine gas is toxic and corrosive, so storage and handling must be carefully controlled.
Risk: by-products
  • Chlorine can react with organic matter to form chlorinated hydrocarbons, some of which may be harmful.
Judgement
  • The advantage of preventing widespread disease is generally considered to outweigh these small risks.
Pool water
  • Sodium chlorate(I) solution also supplies ClO⁻ ions to disinfect swimming pools.

Section 3 of 5Reaction with Sodium Hydroxide

1
Chlorine bubbled into cold, dilute aqueous sodium hydroxide gives sodium chloride, sodium chlorate(I) and water.
2
Equation: Cl₂(g) + 2NaOH(aq) → NaCl(aq) + NaClO(aq) + H₂O(l).
3
Chlorine changes from 0 to −1 in NaCl and from 0 to +1 in NaClO, so this is also disproportionation.
4
The resulting solution of NaClO is household bleach, used for disinfection and for whitening fabrics and paper.
5
With hot concentrated alkali the oxidised product is chlorate(V) instead: 3Cl₂ + 6NaOH → 5NaCl + NaClO₃ + 3H₂O.

Section 4 of 5Worked Example

Question

  • Chlorine (7.10 g) is passed into an excess of cold, dilute sodium hydroxide solution.
  • Calculate the mass of sodium chlorate(I), NaClO, formed.
  • (Ar: Na 23.0, Cl 35.5, O 16.0.)

  1. Mr of Cl₂ = 71.0, so moles of Cl₂ = 7.10 ÷ 71.0 = 0.100 mol.
  2. Equation Cl₂ + 2NaOH → NaCl + NaClO + H₂O shows 1 mol Cl₂ gives 1 mol NaClO, so moles of NaClO = 0.100 mol.
  3. Mr of NaClO = 23.0 + 35.5 + 16.0 = 74.5.
  4. Mass = 0.100 × 74.5 = 7.45 g.

Answer 7.45 g of NaClO

Section 5 of 5Exam Checklist & Tips

2.3.3 – Key Points for the Exam

  • Disproportionation A reaction in which one element is simultaneously oxidised and reduced.
  • Oxidation number A number showing the charge an atom would have if its bonds were fully ionic.
  • Chlorate(I) The ion ClO⁻, in which chlorine has oxidation number +1.
  • Bleach A solution of sodium chlorate(I) that removes colour and kills microorganisms.
  • Equilibrium A reversible reaction in which forward and reverse rates are equal and concentrations stay constant.

Common Mistakes

  • Writing that chlorine is only oxidised in these reactions, when it is both oxidised and reduced.
  • Using NaClO₃ for bleach, which contains chlorate(I) NaClO rather than chlorate(V).
  • Forgetting the equilibrium arrow for chlorine in water, which does not go to completion.
  • Missing the temperature conditioncold dilute alkali gives chlorate(I) but hot concentrated alkali gives chlorate(V).

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