Magnesium ribbon burning in oxygen gives a bright white flame and a white solid, magnesium oxide. The visible change is dramatic, but the redox claim is not "something burned"; the redox claim is that electrons and oxidation numbers have changed.

For the reaction:

2Mg + O₂ -> 2MgO

the useful A level question is: what has happened to each element's oxidation number?

An increase in oxidation number is oxidation. A decrease in oxidation number is reduction. The same reaction can therefore be described in two linked ways:

• magnesium is oxidised because its oxidation number increases from 0 to +2
• oxygen is reduced because its oxidation number decreases from 0 to -2

This is stronger than the older GCSE shortcut "oxidation is gain of oxygen". That shortcut works for magnesium oxide, but Edexcel 9CH0 expects the more general model: redox is about electron transfer and oxidation-number change.

Observation To Model

When sodium reacts with chlorine:

2Na + Cl₂ -> 2NaCl

there is no oxygen in the equation, but it is still redox. Sodium atoms become Na⁺ ions in the product, so sodium's oxidation number increases from 0 to +1. Chlorine atoms in Cl₂ become Cl^- ions in sodium chloride, so chlorine's oxidation number decreases from 0 to -1.

This is a typical s-block/p-block redox pattern: an electropositive s-block metal loses electrons, while a p-block non-metal gains electrons.

Feynman Diagnostic

Explain to a younger student why Na + Cl₂ -> NaCl is redox even though oxygen is not involved. If your explanation says only "chlorine reacts with sodium", it is not enough. The mark-earning idea is that sodium's oxidation number increases as it loses electrons, while chlorine's oxidation number decreases as it gains electrons.

The two definitions of oxidation and reduction must point in the same direction:

The phrase "oxidation is loss" is useful, but A level answers should normally connect it to the chemical species in the equation. A species that loses electrons becomes more positive, so its oxidation number increases. A species that gains electrons becomes more negative, so its oxidation number decreases.

Worked Example: Calcium And Chlorine

Classify the redox changes in:

Ca + Cl₂ -> CaCl₂

Step 1: assign oxidation numbers.

• Ca in Ca is 0 because it is an element.
• Cl in Cl₂ is 0 because it is an element.
• Ca in CaCl₂ is +2 because the compound contains Ca²⁺ and two Cl^- ions.
• Cl in CaCl₂ is -1.

Step 2: compare reactant and product values.

Step 3: connect to electrons.

Calcium can be represented by the electron-change statement:

Ca -> Ca²⁺ + 2e^-

Electrons appear on the product side, so calcium has lost electrons. Chlorine can be represented by:

Cl₂ + 2e^- -> 2Cl^-

Electrons appear on the reactant side, so chlorine has gained electrons.

Chemical interpretation: calcium transfers electrons to chlorine. Calcium is oxidised; chlorine is reduced. The next lesson will make the formal construction of ionic half-equations explicit, but the electron placement already tells you the redox direction.

Guided Practice

For:

2Al + 3Br₂ -> 2AlBr₃

fill the missing words.

• Aluminium changes from 0 to ___, so aluminium is ___.
• Bromine changes from 0 to ___, so bromine is ___.
• Aluminium has ___ electrons.
• Bromine has ___ electrons.

Answer: aluminium changes from 0 to +3, so aluminium is oxidised. Bromine changes from 0 to -1, so bromine is reduced. Aluminium has lost electrons. Bromine has gained electrons.

Common Error Contrast

Do not decide oxidation and reduction from the size of a formula alone. In AlBr₃, there are three bromine atoms for each aluminium atom, but that coefficient pattern does not decide the redox change. The mark-earning comparison is oxidation number before and after reaction.

Agent language is where many correct redox answers get flipped. The agent is named for what it does to the other species, not for the change happening to itself.

So the two compact Edexcel facts are:

• oxidising agents gain electrons
• reducing agents lose electrons

That can feel backwards at first. It becomes less strange if you think of the oxidising agent as the electron acceptor: by taking electrons from another species, it makes that other species lose electrons and become oxidised.

Worked Example: Chlorine And Bromide Ions

Consider the reaction:

Cl₂ + 2Br^- -> 2Cl^- + Br₂

Oxidation-number changes:

Chlorine is reduced because it gains electrons. Therefore chlorine is the oxidising agent: it has caused bromide ions to lose electrons and become bromine.

Bromide ions are oxidised because they lose electrons. Therefore bromide ions are the reducing agent: they have caused chlorine to gain electrons and become chloride ions.

The agent labels belong to the reactants, not to the products. In this example, Cl₂ is the oxidising agent and Br^- is the reducing agent.

Feynman Diagnostic

A student says, "An oxidising agent is oxidised." Correct the sentence in one simple explanation. The clean version is: an oxidising agent oxidises another species; the oxidising agent itself gains electrons and is reduced.

A disproportionation reaction is a redox reaction with a special pattern: the same element, starting in a single species, is both oxidised and reduced.

Ordinary redox has one element going up in oxidation number and a different element going down. Disproportionation has one element splitting into two oxidation-number outcomes.

Worked Example: Hydrogen Peroxide

Hydrogen peroxide decomposes as follows:

2H₂O₂ -> 2H₂O + O₂

Assign oxidation numbers to oxygen:

• In H₂O₂, oxygen is -1. This is the peroxide exception from oxidation-number rules.
• In H₂O, oxygen is -2.
• In O₂, oxygen is 0.

Now compare oxygen before and after:

Oxygen in H₂O₂ has simultaneously decreased in oxidation number and increased in oxidation number. This is disproportionation. Hydrogen peroxide is acting as both an oxidising agent and a reducing agent because the same reactant species supplies oxygen atoms for both changes.

Second Example: Chlorine In Alkali

The detailed halogen chemistry belongs later in Topic 4, but the classification pattern is useful here:

Cl₂ + 2OH^- -> Cl^- + ClO^- + H₂O

Chlorine begins at 0 in Cl₂. It ends at -1 in Cl^- and +1 in ClO^-. The same element from the same starting species has been reduced and oxidised, so the reaction is disproportionation.

Feynman Diagnostic

Explain the difference between "redox" and "disproportionation" using one sentence. A strong answer is: disproportionation is a special type of redox where the same element in one reactant species is both oxidised and reduced.

When a Pearson question asks you to identify, deduce, explain or justify a redox classification, use a repeatable routine.

1. Assign oxidation numbers to the elements likely to change.
2. Compare each element in reactants and products.
3. Label increases as oxidation and decreases as reduction.
4. If at least one oxidation and one reduction occur, classify the reaction as redox.
5. If the same element in one reactant species is both oxidised and reduced, classify it as disproportionation.
6. Name oxidising and reducing agents from the reactant species: the electron gainer is the oxidising agent; the electron loser is the reducing agent.

Worked Classification Set

Reaction A:

Mg + 2HCl -> MgCl₂ + H₂

Magnesium changes from 0 to +2, so magnesium is oxidised. Hydrogen changes from +1 in HCl to 0 in H₂, so hydrogen ions are reduced. This is redox. Mg is the reducing agent; H⁺ from the acid is the oxidising agent. It is not disproportionation because magnesium and hydrogen are different elements.

Reaction B:

NaOH + HCl -> NaCl + H₂O

Oxidation numbers do not change: Na remains +1, Cl remains -1, H remains +1 and O remains -2. This is acid-base neutralisation, not redox. There is no oxidising agent or reducing agent in this reaction.

Reaction C:

2H₂O₂ -> 2H₂O + O₂

Oxygen changes from -1 to -2 and from -1 to 0. This is redox and also disproportionation. The useful evidence is not just that oxygen gas is produced; it is the two different oxidation-number changes for oxygen from the same reactant species.

Pearson-Aware Recap

For Edexcel 9CH0, a redox answer earns marks by making the oxidation-number or electron-transfer evidence visible. "Oxidised" means electron loss and an increase in oxidation number; "reduced" means electron gain and a decrease in oxidation number. Oxidising agents gain electrons and are reduced; reducing agents lose electrons and are oxidised. Disproportionation is the special case where one element in a single starting species is simultaneously oxidised and reduced.