Redox reactions

Redox (oxidation–reduction) reactions are reactions in which electrons pass from one atom or ion to another, so that the oxidation states of elements change. Combustion, respiration, photosynthesis, corrosion, metallurgy, electrolysis and the working of batteries are all redox processes. Exam questions most often ask you to assign oxidation states, identify the oxidising and reducing agents, balance an equation by the electron-balance or half-equation method and find the sum of coefficients, and predict products depending on the medium. This topic underpins "Electrolysis", "Halogens", "Nitrogen and phosphorus" and "d-block metals".

§1. Oxidation states

An oxidation state is the notional charge an atom would have if all bonds in the compound were ionic. It is found by giving the shared electrons entirely to the more electronegative atom.

The highest positive state usually equals the group number; the lowest negative state of a non-metal is group number − 8. In organic compounds each carbon is treated separately: a C–H bond contributes −1, a C–O or C–N bond +1, and C–C bonds are ignored. For example, carbon is −4 in methane and +4 in CO₂.

Worked example

Find the oxidation state of chromium in K₂Cr₂O₇.

1. Potassium +1, oxygen −2, chromium x.
2. 2(+1) + 2x + 7(−2) = 0 → 2x = 12.
3. x = +6.

§2. Oxidising and reducing agents

• A reducing agent gives electrons and is itself oxidised — its oxidation state rises.
• An oxidising agent takes electrons and is itself reduced — its oxidation state falls.

An atom in its highest oxidation state can only be an oxidising agent (N⁺⁵ in HNO₃, Mn⁺⁷ in KMnO₄, S⁺⁶ in H₂SO₄); in its lowest state only a reducing agent (S⁻² in H₂S, N⁻³ in NH₃, I⁻¹ in HI); in an intermediate state it can be either (SO₂, H₂O₂, HNO₂).
Typical oxidising agents: KMnO₄, K₂Cr₂O₇, HNO₃, conc. H₂SO₄, O₂, O₃, Cl₂, H₂O₂. Typical reducing agents: metals, H₂, C, CO, H₂S, NH₃, HI, , salts.

§3. The electron-balance method

1. Assign oxidation states to all atoms and find those that change.
2. Write the electron-loss and electron-gain schemes.
3. Use the LCM to make the electrons equal and insert the main coefficients.
4. Balance the remaining atoms (metal, acid residue, then H and O) and check the oxygen.

Worked example

Balance Cu + HNO₃(conc.) → Cu(NO₃)₂ + NO₂ + H₂O and find the sum of coefficients.

1. Cu⁰ − 2e⁻ → Cu⁺² | ×1; N⁺⁵ + 1e⁻ → N⁺⁴ | ×2.
2. 1 Cu and 2 NO₂; 2 more HNO₃ for the salt → 4 acid in total.
3. Cu + 4HNO₃ → Cu(NO₃)₂ + 2NO₂ + 2H₂O; sum 1 + 4 + 1 + 2 + 2 = 10.

Worked example

Balance K₂Cr₂O₇ + KI + H₂SO₄ → Cr₂(SO₄)₃ + I₂ + K₂SO₄ + H₂O.

1. 2Cr⁺⁶ + 6e⁻ → 2Cr⁺³ | ×1; 2I⁻ − 2e⁻ → I₂⁰ | ×3.
2. K₂Cr₂O₇ + 6KI → Cr₂(SO₄)₃ + 3I₂; potassium: 2 + 6 = 8 → 4K₂SO₄; sulfate: 3 + 4 = 7 → 7H₂SO₄; water 7H₂O.
3. K₂Cr₂O₇ + 6KI + 7H₂SO₄ → Cr₂(SO₄)₃ + 3I₂ + 4K₂SO₄ + 7H₂O; sum 1 + 6 + 7 + 1 + 3 + 4 + 7 = 29.

§4. The half-equation (ion–electron) method

This suits reactions in solution: species are written as they really exist (strong electrolytes as ions, weak ones as molecules). In acid, surplus oxygen is combined with H⁺ to form water and missing oxygen comes from water; in alkali, OH⁻ and water are used. Oxidation states are not needed, so the method is very convenient for reactions of organic compounds.

Worked example

Balance CH₃CH₂COCH₂CH₃ + MnO₄⁻ + H⁺ → CH₃CH₂COO⁻ + CH₃COO⁻ + Mn²⁺ + H₂O and find the sum of coefficients.

1. Oxidation: C₅H₁₀O + 3H₂O → C₃H₅O₂⁻ + C₂H₃O₂⁻ + 8H⁺ + 6e⁻.
2. Reduction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O.
3. LCM 30: first ×5, second ×6; after cancelling 5C₅H₁₀O + 6MnO₄⁻ + 8H⁺ → 5C₃H₅O₂⁻ + 5C₂H₃O₂⁻ + 6Mn²⁺ + 9H₂O.
4. Sum: 5 + 6 + 8 + 5 + 5 + 6 + 9 = 44.

§5. The effect of the medium

The same oxidising agent is reduced to different products in different media. For manganate(VII):

• Acidic: 2KMnO₄ + 5K₂SO₃ + 3H₂SO₄ → 2MnSO₄ + 6K₂SO₄ + 3H₂O — the purple solution is decolourised.
• Neutral: 2KMnO₄ + 3K₂SO₃ + H₂O → 2MnO₂ + 3K₂SO₄ + 2KOH.
• Alkaline: 2KMnO₄ + K₂SO₃ + 2KOH → 2K₂MnO₄ + K₂SO₄ + H₂O.

Dichromate in acid: Cr₂O₇²⁻ (orange) → Cr³⁺ (green). Hydrogen peroxide also depends on its partner: with KI it is an oxidising agent (H₂O₂ + 2KI + H₂SO₄ → I₂ + K₂SO₄ + 2H₂O), with KMnO₄ a reducing agent (O₂ is released).

§6. Types of redox reaction

• Intermolecular — oxidising and reducing agents are different substances: Zn + CuSO₄ → ZnSO₄ + Cu.
• Intramolecular — both are in one molecule, as different elements: 2KClO₃ → 2KCl + 3O₂ (Cl⁺⁵ is reduced, O⁻² oxidised).
• Disproportionation — one element is both oxidised and reduced: Cl₂ + 2NaOH → NaCl + NaClO + H₂O; 3HNO₂ → HNO₃ + 2NO + H₂O.
• Comproportionation — one element in two states ends in a single state: SO₂ + 2H₂S → 3S + 2H₂O.

Redox reactions matter enormously: in living things the oxidation of glucose supplies energy, and in photosynthesis water is oxidised to release oxygen; in industry metals are reduced from their ores and ammonia and nitric acid are synthesised; in cells and batteries redox energy becomes electrical energy. So writing redox equations is one of the most important skills in chemistry.

Check yourself

1. What are the oxidation states of the nitrogen atoms in NH₄NO₃?

2. Which substance can only be an oxidising agent: SO₂, H₂SO₄, H₂S, S?

3. Which substance is the reducing agent in Fe₂O₃ + 3CO → 2Fe + 3CO₂?

4. Balance MnO₂ + HCl → MnCl₂ + Cl₂ + H₂O.

5. What product forms when KMnO₄ is reduced in neutral solution?

6. What two roles can H₂O₂ play?

7. What mass of KMnO₄ oxidises 0.5 mol of K₂SO₃ in acid?

8. What type of redox reaction is Cl₂ + 2NaOH → NaCl + NaClO + H₂O?

Common mistakes

• Mixing up oxidation and reduction: the reducing agent is oxidised!
• Taking oxygen as −2 in H₂O₂ and peroxides.
• Forgetting the acid used to form the salt: some of the HNO₃ only forms the salt.
• Writing the reduction product of manganate(VII) without considering the medium.
• Not checking the oxygen atoms after balancing.