Disproportionation and Comproportionation Reactions

In a disproportionation reaction a single element in one oxidation state is both oxidized and reduced, forming two products with higher and lower oxidation numbers; comproportionation is the reverse, in which two species containing the same element in different oxidation states react to give one product with an intermediate oxidation number. In brief: in Cl2 + 2OH− → Cl− + ClO− + H2O, chlorine goes from 0 to −1 and +1 (disproportionation); in 2H2S + SO2 → 3S + 2H2O, sulfur goes from −2 and +4 to 0 (comproportionation).

These two reaction types show how one element can serve as both oxidizing and reducing agent. They are important in the chemistry of chlorine, hydrogen peroxide, copper, manganese and sulfur. This article defines them, shows examples in a table, works through balancing and calculations, explains how reduction potentials predict disproportionation, mentions practical uses and gives some history.

What is disproportionation?

A species with an intermediate oxidation number is unstable if it can change into two species, one with a higher and one with a lower oxidation number. The same element is oxidized and reduced. Examples are the decomposition of hydrogen peroxide (O: −1 → −2 and 0) and the reaction of chlorine with cold base (Cl: 0 → −1 and +1).

Cl2​ 0 Cl−​ −1 ClO−​ +1 +e−​ −e−​ H2​O2​ −1 H2​O −2 O2​ 0 +e−​ −e−​

What is comproportionation?

Comproportionation (also called synproportionation) is the reverse: two compounds of the same element, in a higher and a lower oxidation state, give a single product with an intermediate state. The reaction of iodate with iodide in acid gives iodine (I: +5 and −1 → 0).

IO3​−​ +5 I−​ −1 I2​ 0 H2​S −2 SO2​ +4 S 0

Examples of both types

Disproportionation and comproportionation
TypeBalanced equationOxidation numbers of the element
Disproportionation2H2O2 → 2H2O + O2O: −1 → −2 and 0
DisproportionationCl2 + 2OH− → Cl− + ClO− + H2OCl: 0 → −1 and +1
Disproportionation (hot)3Cl2 + 6OH− → 5Cl− + ClO3− + 3H2OCl: 0 → −1 and +5
Disproportionation2Cu+ → Cu + Cu2+Cu: +1 → 0 and +2
Disproportionation3MnO42− + 4H+ → 2MnO4− + MnO2 + 2H2OMn: +6 → +7 and +4
Comproportionation2H2S + SO2 → 3S + 2H2OS: −2 and +4 → 0
ComproportionationIO3− + 5I− + 6H+ → 3I2 + 3H2OI: +5 and −1 → 0
ComproportionationPb + PbO2 + 2H2SO4 → 2PbSO4 + 2H2OPb: 0 and +4 → +2
ComproportionationFe + 2Fe3+ → 3Fe2+Fe: 0 and +3 → +2

How do potentials predict disproportionation?

A species disproportionates if the reduction potential for its reduction to the lower state is greater than that for the oxidation to the higher state, so that it can oxidize itself. For copper(I): Cu+ + e− → Cu has E° = +0.52 V and Cu2+ + e− → Cu+ has E° = +0.16 V. E°cell = 0.52 − 0.16 = +0.36 V, which is positive, so 2Cu+ → Cu + Cu2+ is spontaneous. That is why copper(I) salts are unstable in water. Hydrogen peroxide (E° = +1.78 V for H2O2 → H2O, and +0.70 V for O2 → H2O2) decomposes with E°cell = 1.08 V, though the reaction is slow without a catalyst.

Cu2+​ 0.16 V Cu+​ 0.52 V Cu 0.52 V > 0.16 V O2​ 0.70 V H2​O2​ 1.78 V H2​O 1.78 V > 0.70 V

Worked examples

  1. Identify. 3Br2 + 6OH− → 5Br− + BrO3− + 3H2O. Bromine: 0 → −1 (gain 1 × 5 atoms = 5 electrons) and 0 → +5 (lose 5 electrons). Disproportionation.
  2. Balance the chlorine reaction. Reduction: Cl2 + 2e− → 2Cl−. Oxidation (base): Cl2 + 4OH− → 2ClO− + 2H2O + 2e−. Add: 2Cl2 + 4OH− → 2Cl− + 2ClO− + 2H2O, divide by 2: Cl2 + 2OH− → Cl− + ClO− + H2O.
    Cl2​ + 2e−​ → 2Cl−​ Cl2​ + 4OH−​ → 2ClO−​ + 2H2​O + 2e−​ 2Cl2​ + 4OH−​ → 2Cl−​ + 2ClO−​ + 2H2​O ·2 Cl2​ + 2OH−​ → Cl−​ + ClO−​ + H2​O
  3. Bleach. 1.00 mol of Cl2 gives 1.00 mol of NaClO (74.44 g) and 1.00 mol of NaCl: the “active” hypochlorite is only half of the chlorine. From 71.0 g of Cl2, 74.4 g of NaClO forms, since 1 mol of Cl2 = 70.9 g.
  4. Hydrogen peroxide decomposition. 34.0 g of H2O2 (1.00 mol) gives 0.500 mol of O2 = 16.0 g (12.2 L at 25 °C and 1 atm).
  5. Comproportionation: H2S and SO2. S: −2 → 0 (loses 2) and +4 → 0 (gains 4). 2 H2S : 1 SO2. How much sulfur forms from 3.40 g H2S (0.100 mol)? 0.100 × 3 ÷ 2 = 0.150 mol, or 4.81 g.
    3.40 g H2​S → 0.100 mol H2​S 0.100 mol × 3/2 = 0.150 mol S 0.150 mol × 32.06 g/mol = 4.81 g S
  6. Iodine from iodate and iodide. IO3− + 5I− + 6H+ → 3I2 + 3H2O. From 0.0100 mol of IO3−: 0.0300 mol of I2 = 7.61 g. This reaction is used to standardize thiosulfate solutions in iodometric titrations.

Where are these reactions important?

How were they discovered?

The reaction of chlorine with alkali to form bleach was exploited by Claude Berthollet in the 1780s, and the term disproportionation was later used for reactions in which an intermediate oxidation state is unstable. The comproportionation of sulfur compounds was used in the 19th century to recover sulfur, in the Claus process (1883).

Which mistakes are common?

Mistake 1: Looking for two different elements

Both oxidation and reduction happen to the same element.

Mistake 2: Confusing the two terms

Disproportionation splits one species into two. Comproportionation combines two into one.

Mistake 3: Forgetting that conditions decide the products

Chlorine in cold base gives ClO−, in hot base ClO3−.

Mistake 4: Assuming that a spontaneous reaction is fast

H2O2 is thermodynamically unstable but can be stored for months without a catalyst.