Key exam idea: chlorine can disproportionate

· Disproportionation = the same element is both oxidised and reduced in the same reaction.
· In these reactions, chlorine starts in oxidation number 0 in Cl₂.
· Some chlorine atoms are reduced to Cl⁻, oxidation number −1.
· Other chlorine atoms are oxidised to either ClO⁻ or ClO₃⁻.
· Always show disproportionation by comparing oxidation numbers of chlorine before and after reaction.

The image shows the bonding arrangement in ClO⁻, one of the products when chlorine reacts with cold aqueous sodium hydroxide. In ClO⁻, chlorine has oxidation number +1, showing that part of the chlorine has been oxidised. Source

Chlorine with cold aqueous sodium hydroxide

· Conditions: cold, dilute aqueous NaOH.
· Products: sodium chloride, NaCl, sodium chlorate(I), NaClO, and water.
· Full equation: Cl₂(aq) + 2NaOH(aq) → NaCl(aq) + NaClO(aq) + H₂O(l).
· Ionic equation: Cl₂(aq) + 2OH⁻(aq) → Cl⁻(aq) + ClO⁻(aq) + H₂O(l).
· Oxidation number changes:
· Cl₂: 0 → Cl⁻: −1, so chlorine is reduced.
· Cl₂: 0 → ClO⁻: +1, so chlorine is oxidised.
· This is a disproportionation reaction because chlorine is simultaneously oxidised and reduced.
· Exam wording: cold NaOH forms chlorate(I), ClO⁻, not chlorate(V), ClO₃⁻.

This page visually summarises the key CIE reactions of chlorine with cold and hot aqueous sodium hydroxide. It is useful for comparing the products and oxidation number changes in both disproportionation reactions. Source

Chlorine with hot aqueous sodium hydroxide

· Conditions: hot, concentrated aqueous NaOH.
· Products: sodium chloride, NaCl, sodium chlorate(V), NaClO₃, and water.
· Full equation: 3Cl₂(aq) + 6NaOH(aq) → 5NaCl(aq) + NaClO₃(aq) + 3H₂O(l).
· Ionic equation: 3Cl₂(aq) + 6OH⁻(aq) → 5Cl⁻(aq) + ClO₃⁻(aq) + 3H₂O(l).
· Oxidation number changes:
· Cl₂: 0 → Cl⁻: −1, so chlorine is reduced.
· Cl₂: 0 → ClO₃⁻: +5, so chlorine is oxidised.
· This is also a disproportionation reaction.
· Exam wording: hot NaOH forms chlorate(V), ClO₃⁻, while cold NaOH forms chlorate(I), ClO⁻.

Sodium hypochlorite contains the ClO⁻ ion and is the key product linked to bleach chemistry. This helps connect the cold NaOH reaction to a real chemical use. Source

Chlorine in water purification

· Chlorine is used in water purification because it produces species that kill bacteria.
· In water, chlorine reacts to form hypochlorous acid, HOCl:
· Cl₂(aq) + H₂O(l) ⇌ HCl(aq) + HOCl(aq).
· HOCl can dissociate to form hypochlorite ions, ClO⁻:
· HOCl(aq) ⇌ H⁺(aq) + ClO⁻(aq).
· The active species are HOCl and ClO⁻.
· These species kill bacteria, making water safer to drink.
· Exam wording: chlorine does not just “kill bacteria directly”; it reacts with water to produce HOCl and ClO⁻, which are the active sterilising species.

Oxidation number summary

· Cl₂: chlorine oxidation number = 0.
· Cl⁻ in NaCl: chlorine oxidation number = −1.
· ClO⁻ in NaClO: oxygen is −2, so chlorine is +1.
· ClO₃⁻ in NaClO₃: three oxygens total −6, ion charge −1, so chlorine is +5.
· Reduction = decrease in oxidation number: 0 → −1.
· Oxidation = increase in oxidation number: 0 → +1 or 0 → +5.

Common exam traps

· Do not confuse chlorate(I), ClO⁻, with chlorate(V), ClO₃⁻.
· Cold NaOH → ClO⁻; hot NaOH → ClO₃⁻.
· Always state that chlorine is both oxidised and reduced in disproportionation.
· Use oxidation numbers to prove disproportionation, not just the word “redox”.
· For water purification, include the equation forming HOCl and mention HOCl and ClO⁻ kill bacteria.