11.4 The reactions of chlorine
- Syllabus
- 9701–2028–2029
- Topic
- 11.4
- Level
- AS
Learning objectives
11.4.1And interpret• Describe and interpret, in terms of changes in oxidation number, the reaction of chlorine with cold and with hot aq. sodium hydroxide and recognise these as disproportionation reactions11.4.2Explain• Explain, including by use of an equation, the use of chlorine in water purification to include the production of the active species HOCl and Cl O – which kill bacteria
Chlorine disproportionates differently in cold and hot alkali
In a disproportionation reaction, the same element is both oxidised and reduced. Chlorine reacts with cold dilute sodium hydroxide to form chloride and chlorate(I), while hot concentrated alkali forms chloride and chlorate(V).
Use oxidation numbers to show the two pathways. In cold solution chlorine goes from 0 to −1 and +1; in hot solution it goes from 0 to −1 and +5. Balance atoms and charge before naming the reaction.
Cold: Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O. Hot: 3Cl₂ + 6OH⁻ → 5Cl⁻ + ClO₃⁻ + 3H₂O. The conditions determine the product.
Do not use one equation for every sodium-hydroxide condition, and do not call the reaction simple oxidation or reduction.
Chlorine forms two active bactericidal species in water
Cl2(aq)+H2O(l)⇌HCl(aq)+HOCl(aq)
HOCl(aq)⇌H+(aq)+ClO−(aq)
Hypochlorous acid, HOCl, and hypochlorite ions, ClO⁻, are the active species that kill bacteria. Chloride ions from HCl are not the disinfecting species.
HOCl is a weak acid, so both HOCl and ClO⁻ are present in water. Their relative proportions depend on pH, but both species named in the syllabus contribute to purification.
Do not write ClO₃⁻, the hot-alkali product, in the water-purification equilibrium. This card explains formation of HOCl and ClO⁻; it does not prescribe a chlorine dose or claim that unrestricted chlorine addition is safe.