Disproportionation Reactions
Disproportionation is a specific type of redox reaction where a single element is simultaneously oxidised and reduced. For the ESAT, you must be able to identify these reactions by tracking changes in oxidation states.
Part of the ESAT Chemistry syllabus — revision for the Engineering and Science Admissions Test (ESAT), the UAT-UK admissions test for Cambridge, Imperial, Oxford and UCL.
Key points
- A single element, starting in one oxidation state in a reactant, must end up in at least two different products with different oxidation states.
- In one product, the element's oxidation state will have increased (oxidation).
- In another product, the element's oxidation state will have decreased (reduction).
- This means the element must start in an intermediate oxidation state, not its highest or lowest possible state.
- Common examples include the reaction of halogens with cold or hot alkali, and the decomposition of hydrogen peroxide.
Definitions
- Disproportionation
- A redox reaction in which one element in a single chemical species is concurrently oxidised and reduced to form two or more different products.
Worked example
Nitrogen dioxide reacts with water to form nitric acid and nitrous acid. The unbalanced equation is: NO2 + H2O → HNO3 + HNO2. By assigning oxidation states, determine if this is a disproportionation reaction.
- 1
Step 1:
Identify the element that appears to change its oxidation state.
Here, Nitrogen (N) is in the reactant NO2 and appears in two different products, HNO3 and HNO2.
- 2
Step 2:
Calculate the oxidation state of Nitrogen in the reactant, NO2.
Oxygen is -2.
For the molecule to be neutral, N + 2(-2) = 0, so the oxidation state of N is +4.
- 3
Step 3:
Calculate the oxidation state of Nitrogen in the first product, HNO3.
H is +1 and O is -2.
For a neutral molecule:
(+1) + N + 3(-2) = 0, so N - 5 = 0.
The oxidation state of N is +5.
- 4
Step 4:
Calculate the oxidation state of Nitrogen in the second product, HNO2.
H is +1 and O is -2.
For a neutral molecule:
(+1) + N + 2(-2) = 0, so N - 3 = 0.
The oxidation state of N is +3.
- 5
Step 5:
Compare the changes.
Nitrogen starts at +4.
It increases to +5 (oxidation) in HNO3 and decreases to +3 (reduction) in HNO2.
- 6
Step 6:
Conclude.
Since the same element (N) from a single reactant (NO2) has been both oxidised and reduced, the reaction is an example of disproportionation.
Answer: Yes, this is a disproportionation reaction. Nitrogen is oxidised from +4 to +5 and simultaneously reduced from +4 to +3.
Common mistakes
- ×Failing to correctly calculate oxidation states is the most common error. Review the rules, especially for polyatomic ions and exceptions like peroxides.
- ×Mistaking a simple redox reaction for disproportionation. If two different elements are oxidised and reduced, it is not disproportionation.
- ×Forgetting that the element must come from a single reactant species. If an element from two different reactants combines, it is not disproportionation.
No-calculator tips
- ✓Quickly scan equations: look for an element that is in one substance on the left-hand side but two or more substances on the right. This is your primary candidate.
- ✓Memorise the standard oxidation state hierarchy (e.g., Group 1 metals are always +1, Fluorine is always -1). This allows you to solve for the element of interest quickly.
- ✓For ions, remember the sum of oxidation states equals the charge. Mentally subtract the known values from the overall charge to find the unknown.