Topic 16 · page 12 of 19

Disproportionation

In disproportionation one species is oxidised and reduced at the same time. Test it with two standard electrode potentials that both involve that species.

12 min read16.14

Read first: Predicting feasibility

What disproportionation is 16.14

You can spot it from oxidation numbers: one element appears at a higher and a lower oxidation number in the products. Chlorine in cold sodium hydroxide solution is an example you may know:

ClX2(aq)+2 OHX−(aq)→ClX−(aq)+ClOX−(aq)+HX2O(l) \begin{aligned} &\ce{Cl2(aq) + 2OH-(aq)} \\ &\ce{-> Cl-(aq) + ClO-(aq) + H2O(l)} \end{aligned}

Chlorine goes from 0 to −1 (reduced) and from 0 to +1 (oxidised) in the same reaction.

Testing a species with electrode potentials 16.14

A species can disproportionate when it appears in two couples:

Then treat it as an ordinary feasibility question.

MethodWill a species disproportionate?

  1. Find the two couples that contain the species.
  2. Decide which couple it is reduced in and which it is oxidised in.
  3. Ecell⊖E^\ominus_\text{cell} = E⊖E^\ominus(couple where it is reduced) − E⊖E^\ominus(couple where it is oxidised).
  4. If positive, it disproportionates.

Worked exampleCopper(I)

CuX2++eX−⇌CuX+\ce{Cu^2+ + e- <=> Cu+} has E⊖E^\ominus = +0.15 V. CuX++eX−⇌Cu\ce{Cu+ + e- <=> Cu} has E⊖E^\ominus = +0.52 V.

  1. Cu⁺ is on the left of the second couple: it can be reduced to Cu.
  2. Cu⁺ is on the right of the first couple: it can be oxidised to Cu²⁺.
  3. Ecell⊖E^\ominus_\text{cell} = +0.52 − (+0.15) = +0.37 V.
  4. Positive, so copper(I) does disproportionate in aqueous solution:
2 CuX+(aq)→Cu(s)+CuX2+(aq) \ce{2Cu+(aq) -> Cu(s) + Cu^2+(aq)}
Two ladders. Left: copper(I) appears in two couples, at +0.15 V and +0.52 V, the arrow runs downwards, E cell +0.37 V, so it disproportionates. Right: iron(II) with couples at minus 0.44 V and plus 0.77 V, the arrow would run upwards, E cell minus 1.21 V, so it does not.
Testing disproportionation on a ladder. The species is the reduced form of the upper couple and the oxidised form of the lower couple.

You can see this happen. Add copper(I) oxide to dilute sulfuric acid and a pink-brown solid (copper) forms in a blue solution (copper(II) sulfate).

CuX2O(s)+2 HX+(aq)→Cu(s)+CuX2+(aq)+HX2O(l) \begin{aligned} &\ce{Cu2O(s) + 2H+(aq)} \\ &\ce{-> Cu(s) + Cu^2+(aq) + H2O(l)} \end{aligned}

This is why copper(I) compounds are rare in water: the ion is unstable with respect to disproportionation.

Worked exampleIron(II) does not disproportionate

FeX2++2 eX−⇌Fe\ce{Fe^2+ + 2e- <=> Fe} has E⊖E^\ominus = −0.44 V. FeX3++eX−⇌FeX2+\ce{Fe^3+ + e- <=> Fe^2+} has E⊖E^\ominus = +0.77 V.

  1. Fe²⁺ could be reduced to Fe (−0.44 V) or oxidised to Fe³⁺ (+0.77 V).
  2. Ecell⊖E^\ominus_\text{cell} = −0.44 − (+0.77) = −1.21 V.
  3. Negative, so Fe²⁺ does not disproportionate.

The reverse is feasible: Fe + 2Fe³⁺ → 3Fe²⁺. That is why an iron nail is sometimes kept in iron(II) solutions: any Fe³⁺ that forms in the air is turned back into Fe²⁺.

Worked exampleHydrogen peroxide

Data: HX2OX2+2 HX++2 eX−⇌2 HX2O\ce{H2O2 + 2H+ + 2e- <=> 2H2O} E⊖E^\ominus = +1.77 V; OX2+2 HX++2 eX−⇌HX2OX2\ce{O2 + 2H+ + 2e- <=> H2O2} E⊖E^\ominus = +0.68 V.

  1. H₂O₂ is on the left of the first couple, so it can be reduced to water. It is on the right of the second, so it can be oxidised to oxygen.
  2. Ecell⊖E^\ominus_\text{cell} = +1.77 − (+0.68) = +1.09 V, positive.
  3. Hydrogen peroxide disproportionates:
2 HX2OX2(aq)→2 HX2O(l)+OX2(g) \ce{2H2O2(aq) -> 2H2O(l) + O2(g)}

Check yourself

  1. Define disproportionation.
  2. Use E⊖E^\ominus(Cu⁺/Cu) = +0.52 V and E⊖E^\ominus(Cu²⁺/Cu⁺) = +0.15 V to show that Cu⁺(aq) disproportionates. Write the ionic equation.
  3. Show that Fe²⁺(aq) does not disproportionate. Use −0.44 V (Fe²⁺/Fe) and +0.77 V (Fe³⁺/Fe²⁺).
  4. Describe what you would see when copper(I) oxide is added to dilute sulfuric acid, and explain.
  5. Show with oxidation numbers that Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O is a disproportionation.
  6. Use E⊖E^\ominus(Fe²⁺/Fe) = −0.44 V and E⊖E^\ominus(Fe³⁺/Fe²⁺) = +0.77 V to show that iron metal reduces Fe³⁺ to Fe²⁺.

Answers cover these until you have tried the questions

  1. The same element in one species is simultaneously oxidised and reduced.
  2. Cu⁺ is reduced to Cu (+0.52 V) and oxidised to Cu²⁺ (+0.15 V). Ecell⊖E^\ominus_\text{cell} = +0.52 − (+0.15) = +0.37 V, positive, so it disproportionates: 2 CuX+(aq)→Cu(s)+CuX2+(aq)\ce{2Cu+(aq) -> Cu(s) + Cu^2+(aq)}
  3. Fe²⁺ is reduced to Fe (−0.44 V) and oxidised to Fe³⁺ (+0.77 V). Ecell⊖E^\ominus_\text{cell} = −0.44 − (+0.77) = −1.21 V, negative, so it does not disproportionate.
  4. A pink-brown solid (copper) and a blue solution (copper(II) sulfate). Copper(I) in Cu₂O is oxidised to Cu²⁺ and reduced to Cu at the same time.
  5. Chlorine goes from 0 to −1 in Cl⁻ (reduced) and from 0 to +1 in ClO⁻ (oxidised). One element, one species, both changes: disproportionation.
  6. Fe³⁺ is reduced and Fe is oxidised. Ecell⊖E^\ominus_\text{cell} = +0.77 − (−0.44) = +1.21 V, positive, so Fe + 2Fe³⁺ → 3Fe²⁺ is feasible.