International ASChemistryEdexcelRevision Notes2. Energetics, Group Chemistry, Halogenoalkanes & AlcoholsRedox Chemistry & Acid-Base TitrationsDisproportionation Reactions

Disproportionation Reactions (Edexcel International AS Chemistry): Revision Note

Exam code: XCH11

Written by: Philippa Platt

Reviewed by: Caroline Carroll

Updated on 24 October 2026

Disproportionation

Disproportionation reactions

A disproportionation reaction is a reaction in which the same species is simultaneously oxidised and reduced

Cl2 + H2O -> HCl + HClO with oxidation numbers: chlorine is 0 in Cl2, -1 in HCl (reduced) and +1 in HClO (oxidised)

Example of a disproportionation reaction in which the same species (chlorine in this case) has been both oxidised and reduced

In Cl₂ + H₂O → HCl + HClO, chlorine goes from 0 in Cl₂ to -1 in HCl (reduced, change of -1) and to +1 in HClO (oxidised, change of +1).

Worked Example

Balancing disproportionation reactions

Balance the disproportionation reaction which takes place when chlorine is added to hot concentrated aqueous sodium hydroxide. The products are Cl^-and ClO₃^-ions and water

Answer

Step 1: Write the unbalanced equation and identify the atoms that change in oxidation number:

Step 1: Cl2 + OH- -> Cl- + ClO3- + H2O with oxidation numbers marked: Cl2 0, O -2, H +1, Cl- -1, Cl in ClO3- +5

Step 2: Deduce the oxidation number changes:

Step 2: oxidation number changes: chlorine 0 to -1 (change -1) and 0 to +5 (change +5)

Step 3: Balance the oxidation number changes:

Step 3: balancing the oxidation number changes: 5 x (-1) = -5 matches 1 x (+5) = +5, giving 5Cl-

Step 4: Balance the charges

Step 4: charge balance: total charge on the right is 6-, left has 1-, so 6 OH- ions are needed

Step 5: Balance the atoms

Summary of Steps 1 to 4, which appear only in the images:

Step 1: Cl₂ + OH^- → Cl^- + ClO₃^- + H₂O, with oxidation numbers Cl₂ = 0, Cl^- = -1 and Cl in ClO₃^- = +5
Step 2: chlorine goes from 0 to -1 (change -1, reduced) and from 0 to +5 (change +5, oxidised)
Step 3: balance the changes, 5Cl^- (5 x -1 = -5) for 1ClO₃^- (1 x +5 = +5), giving Cl₂ + OH^- → 5Cl^- + ClO₃^- + H₂O
Step 4: balance the charges; the total charge on the right is 5(-1) + (-1) = -6, so 6OH^- are needed on the left
Step 5: balance the atoms (3Cl₂ for 6 Cl; 3H₂O for 6 H): 3Cl₂ + 6OH^- → 5Cl^- + ClO₃^- + 3H₂O

Step 5: balanced equation 3Cl2 + 6OH- -> 5Cl- + ClO3- + 3H2O

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