Topic 8 · page 8 of 29

Disproportionation and classifying reactions

In disproportionation one element in one species is oxidised and reduced at the same time. Write the oxidation numbers on both sides of any equation to decide whether it is not redox, redox, or disproportionation.

12 min read8.78.8

Read first: Full ionic equations

Disproportionation 8.7

The test is strict: one reactant species, one element, an increase and a decrease in its oxidation number.

Worked exampleChlorine and water

ClX2+HX2O⇌HCl+HClO\ce{Cl2 + H2O <=> HCl + HClO}

  1. Chlorine in Cl₂ is 0.
  2. In HCl chlorine is −1: its number has decreased, so it is reduced.
  3. In HClO (H +1, O −2) chlorine is +1: its number has increased, so it is oxidised.
  4. One element (Cl) in one species (Cl₂) is both oxidised and reduced: disproportionation.

Worked exampleHydrogen peroxide

2 HX2OX2→2 HX2O+OX2\ce{2H2O2 -> 2H2O + O2}

  1. In H₂O₂ oxygen is −1.
  2. In H₂O oxygen is −2: reduced.
  3. In O₂ oxygen is 0: oxidised.
  4. The same element in the same species goes both ways: disproportionation.

Worked exampleCopper(I) ions

2 CuX+(aq)→Cu(s)+CuX2+(aq)\ce{2Cu+(aq) -> Cu(s) + Cu^2+(aq)}

  1. Cu⁺ is +1. In Cu the number is 0: reduced.
  2. In Cu²⁺ it is +2: oxidised.
  3. Half-equations: CuX++eX−→Cu\ce{Cu+ + e- -> Cu} and CuX+→CuX2++eX−\ce{Cu+ -> Cu^2+ + e-}. Disproportionation.

Classifying a reaction 8.8

Oxidation numbers make the decision objective. A colour change, a gas or a heat change does not decide whether a reaction is redox. The numbers do.

MethodRedox, disproportionation or neither?

  1. Write the oxidation number of every element in every species, on both sides.
  2. If nothing changes, the reaction is not redox. (Acid–base, precipitation, most thermal decomposition of carbonates.)
  3. If something changes, it is redox.
  4. If one element in one reactant both rises and falls, it is a disproportionation.
Reaction Oxidation number changes Type
HCl+NaOH→NaCl+HX2O\ce{HCl + NaOH -> NaCl + H2O} none not redox (neutralisation)
AgX++ClX−→AgCl\ce{Ag+ + Cl- -> AgCl} none not redox (precipitation)
CaCOX3+2 HCl→CaClX2+HX2O+COX2\ce{CaCO3 + 2HCl -> CaCl2 + H2O + CO2} none: Ca +2, C +4, O −2, H +1, Cl −1 throughout not redox
CaCOX3→CaO+COX2\ce{CaCO3 -> CaO + CO2} none not redox (thermal decomposition)
Mg+2 HCl→MgClX2+HX2\ce{Mg + 2HCl -> MgCl2 + H2} Mg 0 → +2; H +1 → 0 redox
2 NaNOX3→2 NaNOX2+OX2\ce{2NaNO3 -> 2NaNO2 + O2} N +5 → +3; O −2 → 0 redox
ClX2+2 NaOH→NaCl+NaClO+HX2O\ce{Cl2 + 2NaOH -> NaCl + NaClO + H2O} Cl 0 → −1 and 0 → +1 disproportionation
2 HX2OX2→2 HX2O+OX2\ce{2H2O2 -> 2H2O + O2} O −1 → −2 and −1 → 0 disproportionation

Check yourself

  1. Define disproportionation.
  2. Show, using oxidation numbers, that ClX2+2 NaOH→NaCl+NaClO+HX2O\ce{Cl2 + 2NaOH -> NaCl + NaClO + H2O} is a disproportionation.
  3. Classify each as not redox, redox, or disproportionation: (a) Zn+CuSOX4→ZnSOX4+Cu\ce{Zn + CuSO4 -> ZnSO4 + Cu} (b) MgCOX3+2 HNOX3→Mg(NOX3)X2+HX2O+COX2\ce{MgCO3 + 2HNO3 -> Mg(NO3)2 + H2O + CO2} (c) 3 ClX2+6 NaOH→5 NaCl+NaClOX3+3 HX2O\ce{3Cl2 + 6NaOH -> 5NaCl + NaClO3 + 3H2O} (d) 2 KNOX3→2 KNOX2+OX2\ce{2KNO3 -> 2KNO2 + O2}
  4. Explain why Mg+2 HCl→MgClX2+HX2\ce{Mg + 2HCl -> MgCl2 + H2} is not a disproportionation.
  5. Hydrogen peroxide decomposes: 2 HX2OX2→2 HX2O+OX2\ce{2H2O2 -> 2H2O + O2}. State the oxidation number of oxygen in each species and say which are oxidised and reduced.
  6. A student says "All reactions that give off a gas are redox". Give two reactions from this page that disprove this, with the reason.

Answers cover these until you have tried the questions

  1. A reaction in which an element in a single species is simultaneously oxidised and reduced.
  2. Cl in Cl₂ is 0. In NaCl it is −1 (reduced); in NaClO it is +1 (Na +1, O −2). One element in one reactant (Cl₂) is both oxidised (0 → +1) and reduced (0 → −1): disproportionation.
  3. (a) Zn 0 → +2, Cu +2 → 0: redox. (b) No changes (Mg +2, C +4, O −2, N +5, H +1): not redox. (c) Cl 0 → −1 and 0 → +5: disproportionation. (d) N +5 → +3 and O −2 → 0: redox, but not disproportionation (two different elements).
  4. Mg is oxidised and H is reduced. They are different elements, in different species. For disproportionation, a single element in a single species must both rise and fall.
  5. O is −1 in H₂O₂, −2 in H₂O (reduced) and 0 in O₂ (oxidised). One element, one species: disproportionation.
  6. Calcium carbonate with hydrochloric acid (CaCOX3+2 HCl→CaClX2+HX2O+COX2\ce{CaCO3 + 2HCl -> CaCl2 + H2O + CO2}) gives CO₂, and heating calcium carbonate (CaCOX3→CaO+COX2\ce{CaCO3 -> CaO + CO2}) also gives CO₂. In both, no element changes oxidation number, so neither is redox.