Oxidation, reduction and disproportionationEdexcel A-Level Chemistry: Revision notes
Section 1
Oxidation and reduction as electron transfer
Oxidation is loss of electrons and reduction is gain of electrons (remember OIL RIG). They always occur together, so the overall reaction is called redox.
When an element is oxidised its oxidation number increases; when it is reduced its oxidation number decreases.
Example: 2Na + Cl₂ → 2NaCl.
- Na → Na⁺ + e⁻: oxidation number 0 to +1, so Na is oxidised
- Cl₂ + 2e⁻ → 2Cl⁻: oxidation number 0 to −1, so Cl is reduced
This applies to reactions of s-block elements (Group 1 and 2 metals) and p-block elements (halogens, oxygen, sulfur).
Describe changes using both ideas: electrons lost or gained and the oxidation number change, with the species named.
Section 2
Oxidising agents and reducing agents
An oxidising agent gains electrons, so it is itself reduced, and it oxidises another species. A reducing agent loses electrons, so it is itself oxidised, and it reduces another species.
- In 2Mg + O₂ → 2MgO, Mg is the reducing agent (0 to +2) and O₂ is the oxidising agent (0 to −2)
- In Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂, Cl₂ is the oxidising agent and Br⁻ is the reducing agent
The oxidising agent is the species that is reduced.
Mixing up the agent and the change. The oxidising agent is reduced, and the reducing agent is oxidised.
Section 3
Metals and non-metals forming ions
Metals generally form positive ions by electron loss, so their oxidation number increases (Na to Na⁺, Mg to Mg²⁺, Al to Al³⁺).
Non-metals generally form negative ions by electron gain, so their oxidation number decreases (Cl to Cl⁻, O to O²⁻).
This explains why metals are usually reducing agents and why halogens and oxygen are usually oxidising agents. Down Group 7 the halogens become weaker oxidising agents, and down Group 1 and 2 the metals become stronger reducing agents.
Section 4
Disproportionation
Disproportionation is a redox reaction in which an element in a single species is simultaneously oxidised and reduced.
Example 1: Cl₂ + H₂O → HCl + HClO. Chlorine goes from 0 to −1 (reduced, in HCl) and from 0 to +1 (oxidised, in HClO).
Example 2: Cl₂ + 2NaOH → NaCl + NaClO + H₂O (cold, dilute NaOH). With hot concentrated sodium hydroxide: 3Cl₂ + 6NaOH → 5NaCl + NaClO₃ + 3H₂O. Chlorine goes from 0 to −1 and from 0 to +5.
In each case state the oxidation number of the element in the reactant and in each product, and say which has increased and which has decreased.
Disproportionation needs one element in one reactant going both up and down. A reaction where two different elements change is ordinary redox.
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Oxidation, reduction and disproportionation
- A teacher demonstrates two reactions of s- and p-block elements: magnesium ribbon burning in oxygen, 2Mg + O₂ → 2MgO, and sodium burning in chlorine, 2Na + Cl₂ → 2NaCl.Explain, in terms of electron transfer, why magnesium acts as a reducing agent in its reaction with oxygen.2 marks
- Chlorine gas is bubbled through aqueous potassium bromide. The solution turns orange as bromine forms. The ionic equation is Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂.Use oxidation numbers to show that this is a redox reaction.2 marks
- Chlorine is used to disinfect swimming pool water. When it dissolves in water it forms chloride ions and chlorate(I) ions, ClO⁻, which kill bacteria.The reaction of chlorine with water is Cl₂ + H₂O → HCl + HClO. Use oxidation numbers to explain why this is a disproportionation.3 marks
Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).