Bleach and clean water.

Six original Cambridge-style questions on the chemistry of chlorine: disproportionation, oxidation numbers, water treatment and a calculation.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
Keep these straight

One element, two directions.

01
Recall
[2 marks]

Write equations for the reaction of chlorine with (a) cold, dilute and (b) hot, concentrated aqueous sodium hydroxide.

  • (a) Cl2 + 2NaOH → NaCl + NaClO + H2O. ✓
  • (b) 3Cl2 + 6NaOH → 5NaCl + NaClO3 + 3H2O. ✓
02
Analysis
[3 marks]

Use oxidation numbers to explain why the reaction of chlorine with hot, concentrated NaOH is disproportionation.

  • Chlorine in Cl2 is 0. ✓
  • In NaCl it is −1 (reduced); in NaClO3 it is +5 (oxidised). ✓
  • The same element is oxidised and reduced in the same reaction. ✓
03
Recall
[2 marks]

Write the equation for the reaction of chlorine with water and name the product responsible for killing bacteria.

  • Cl2 + H2O ⇌ HCl + HClO. ✓
  • Chloric(I) acid, HClO. ✓
04
Explain
[3 marks]

Discuss one benefit and two risks of adding chlorine to drinking water.

  • Benefit: it kills pathogenic bacteria, preventing waterborne diseases such as cholera. ✓
  • Risk: chlorine is toxic at high concentrations. ✓
  • Risk: it can react with organic compounds to form chlorinated organic compounds that may be harmful. ✓
05
Calculation
[2 marks]

Calculate the mass of NaClO (Mr 74.5) formed when 1.42 g of Cl2 (Mr 71.0) reacts with excess cold, dilute NaOH.

  • n(Cl2) = 1.42 ÷ 71.0 = 0.0200 mol = n(NaClO). ✓
  • Mass = 0.0200 × 74.5 = 1.49 g. ✓
06
Analysis
[2 marks]

Explain how the electrons balance in the reaction of chlorine with hot, concentrated NaOH.

  • Five Cl atoms each gain one electron (0 → −1): five electrons gained. ✓
  • One Cl atom loses five electrons (0 → +5), so electrons lost equal electrons gained. ✓

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