3.3-3.8 - Redox And Disproportionation

Recognising Redox

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:

$\ce{2Mg + O2 -> 2MgO}$

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

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 A level chemistry uses the more general model: redox is about electron transfer and oxidation-number change.

Observation To Model

When sodium reacts with chlorine:

$\ce{2Na + Cl2 -> 2NaCl}$

there is no oxygen in the equation, but it is still redox. Sodium atoms become $\ce{Na+}$ ions in the product, so sodium's oxidation number increases from 0 to +1. Chlorine atoms in $\ce{Cl2}$ become $\ce{Cl-}$ 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.

Electron Transfer And Oxidation Numbers

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

[DIAGRAM: asset_name: Edexcel 9Ch0 Redox Electron Transfer; asset_slug: edexcel_9ch0_redox_electron_transfer; recommended_method: deterministic_code; description: Magnesium supplies two electrons, one to each chlorine atom; oxidising and reducing agents are labelled. Pale 1600 by 900 diagram rendered with Pillow; technical geometry and Unicode chemical labels inspected.]

The phrase "oxidation is loss" is useful, but A level answers should normally connect it to the chemical species in the equation. For a monatomic ion, losing electrons makes the actual charge more positive; gaining electrons makes it more negative. In covalent substances, an oxidation-number change tracks assigned electron ownership without implying that free ions appear.

Worked Example: Calcium And Chlorine

Classify the redox changes in:

$\ce{Ca + Cl2 -> CaCl2}$

Step 1: assign oxidation numbers.

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

Step 2: compare reactant and product values.

Step 3: connect to electrons.

Calcium can be represented by the electron-change statement:

$\ce{Ca -> Ca^2+ + 2e-}$

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

$\ce{Cl2 + 2e- -> 2Cl-}$

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:

$\ce{2Al + 3Br2 -> 2AlBr3}$

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 $\ce{AlBr3}$, there are three bromine atoms for each aluminium atom, but that coefficient pattern does not decide the redox change. The key comparison is oxidation number before and after reaction.

Oxidising And Reducing Agents

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.

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:

$\ce{Cl2 + 2Br- -> 2Cl- + Br2}$

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, $\ce{Cl2}$ is the oxidising agent and $\ce{Br-}$ is the reducing agent.

Disproportionation

One reactant species can supply atoms that undergo two opposite oxidation-number changes.

Redox requires both oxidation and reduction. These may involve different elements or different starting species. Disproportionation specifically takes the same element from a single starting species to both higher and lower oxidation numbers.

Worked Example: Hydrogen Peroxide

Hydrogen peroxide decomposes as follows:

$\ce{2H2O2 -> 2H2O + O2}$

Assign oxidation numbers to oxygen:

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

Now compare oxygen before and after:

Oxygen in $\ce{H2O2}$ 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:

$\ce{Cl2 + 2OH- -> Cl- + ClO- + H2O}$

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

Redox And Non-Redox Reactions

When a 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:

$\ce{Mg + 2HCl -> MgCl2 + H2}$

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

Reaction B:

$\ce{NaOH + HCl -> NaCl + H2O}$

Oxidation numbers do not change: $\ce{Na}$ remains +1, $\ce{Cl}$ 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:

$\ce{2H2O2 -> 2H2O + O2}$

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.

Recap

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.