Redox Reactions: Revision Notes

Condensed for the final weeks. For the full explanation, use the Redox Reactions study guide.

Three definitions of the same thing

	Oxidation	Reduction
Oxygen	Gain of oxygen	Loss of oxygen
Electrons	Loss of electrons	Gain of electrons
Oxidation number	Increase	Decrease

OIL RIG — Oxidation Is Loss, Reduction Is Gain (of electrons).

Oxidation and reduction always occur together: if one species loses electrons, another must gain them.

Worked example (oxygen definition). CuO + H2 -> Cu + H2O (naming the oxidising and reducing agents is (0620 Extended, 5070 required))

CuO loses oxygen  -> reduced   -> copper(II) oxide is the oxidising agent
H2  gains oxygen  -> oxidised  -> hydrogen is the reducing agent

Nothing here involves electron transfer explicitly, but the oxygen and electron definitions always agree on which species is oxidised and which is reduced — this is exactly the kind of reaction the oxygen definition was designed for, before electron transfer is introduced as the more general picture.

Agents — the reversal students get wrong

(0620 Extended, 5070 required) Defining and identifying oxidising and reducing agents is 0620 Supplement content (6.4 S11-S13).

• Oxidising agent — causes oxidation in something else, so is itself reduced and gains electrons.
• Reducing agent — causes reduction, so is itself oxidised and loses electrons.

Common oxidising agents: acidified potassium manganate(VII), potassium dichromate(VI), chlorine, oxygen. Common reducing agents: carbon, carbon monoxide, hydrogen, reactive metals, potassium iodide.

In a displacement reaction (0620 Extended, 5070 required), a more reactive metal displaces a less reactive one from solution, and this is always a redox reaction: the more reactive metal is oxidised (loses electrons to form ions), while the less reactive metal’s ions are reduced (gain electrons to form the metal). This is the same electron-transfer logic used throughout this topic, just applied to a pair of metals rather than a metal and a non-metal.

Oxidation number rules

(0620 Extended, 5070 required) Deducing oxidation numbers is 0620 Supplement (6.4 S9); 0620 Core needs only the Roman numeral in a name (6.4 C1).

Uncombined element                        0
Simple ion                    = its charge (Na+ = +1, S2- = -2)
Oxygen                                   -2   (except peroxides -1)
Hydrogen                                 +1   (except metal hydrides -1)
Group I / II                       +1 / +2
Fluorine                                 -1
Sum in a neutral compound                 0
Sum in an ion                = the ion charge

Roman numerals in a name give the oxidation number directly: iron(III) chloride contains Fe at +3.

Worked example. Find the oxidation number of Mn in KMnO₄.

K is always +1, O is normally -2, compound is neutral:
(+1) + Mn + 4(-2) = 0
Mn = 0 - 1 + 8 = +7

This matches the “(VII)” in potassium manganate(VII) — the Roman numeral and the calculated oxidation number always agree, which is a quick way to check your working once you have finished a calculation.

Half equations

(0620 Extended, 5070 required) Constructing ionic half-equations is 0620 Supplement content (4.1 S11).

Write each half separately, balance atoms, then balance charge with electrons:

Zn(s)  -> Zn2+(aq) + 2e-        oxidation (electrons on the RIGHT)
Cu2+(aq) + 2e- -> Cu(s)         reduction (electrons on the LEFT)
-----------------------------------------------------------
Zn(s) + Cu2+(aq) -> Zn2+(aq) + Cu(s)

Electrons must cancel exactly when the halves are combined — if they don’t, multiply one half through first.

The two tests to memorise

(0620 Extended, 5070 required)

Test	Reagent	Positive result	Shows
Oxidising agent	Potassium iodide solution	Colourless → brown	I⁻ oxidised to I₂
Reducing agent	Acidified potassium manganate(VII)	Purple → colourless	MnO₄⁻ reduced to Mn²⁺

Background — beyond the syllabus, not examinable: acidified potassium dichromate(VI) turning orange → green also indicates a reducing agent, but this test is not part of the 0620/5070 subject content.

Worked identification

(0620 Extended, 5070 required)

In 2Mg + O₂ → 2MgO:

Mg:  0  ->  +2   increase -> OXIDISED -> Mg is the reducing agent
O:   0  ->  -2   decrease -> REDUCED  -> O2 is the oxidising agent

Exam traps

• (0620 Extended, 5070 required) The agent is the opposite of what happens to it — the single most common error in this topic.
• (0620 Extended, 5070 required) Oxidation number is per atom, not for the whole formula.
• (0620 Extended, 5070 required) In H₂O₂ oxygen is −1, not −2.
• (0620 Extended, 5070 required) Electrons appear on the right for oxidation, the left for reduction.
• For 0620 Core, the oxygen definition (gain/loss of oxygen) is the complete requirement on its own — the electron and oxidation-number definitions are Extended/5070 only, so do not mark a Core candidate down for using oxygen alone.
• (0620 Extended, 5070 required) Displacement reactions are redox: the more reactive metal is oxidised.
• (0620 Extended, 5070 required) Forgetting that the oxygen and electron definitions must always agree — if a working shows one species gaining oxygen but also gaining electrons, at least one step has gone wrong.

Related: Redox Reactions practice questions for further worked examples, including displacement reactions and the reactivity series.

Self-test

1. In Fe₂O₃ + 3CO → 2Fe + 3CO₂, which species is reduced?
2. (0620 Extended, 5070 required) Give the oxidation number of S in H₂SO₄.
3. (0620 Extended, 5070 required) A solution turns acidified KMnO₄ from purple to colourless. What does this show?
4. (0620 Extended, 5070 required) Write the half equation for chloride ions forming chlorine.
5. (0620 Extended, 5070 required) Why is a reducing agent itself oxidised?
6. (0620 Extended, 5070 required) In CuO + H2 → Cu + H2O, identify the oxidising agent and reducing agent using the oxygen definition.
7. (0620 Extended, 5070 required) Find the oxidation number of Cr in K2Cr2O7.

Answers: 1. Fe₂O₃ — iron goes from +3 to 0, a decrease, so it is reduced (CO is the reducing agent). 2. (+1×2) + S + (−2×4) = 0 → S = +6. 3. The solution is a reducing agent; MnO₄⁻ has been reduced to Mn²⁺. 4. 2Cl⁻ → Cl₂ + 2e⁻. 5. It donates electrons to the other species — donating electrons is oxidation, so causing reduction elsewhere necessarily means being oxidised itself. 6. CuO loses oxygen and is reduced, so it is the oxidising agent; H₂ gains oxygen and is oxidised, so it is the reducing agent. 7. 2(+1) + 2Cr + 7(−2) = 0 → 2Cr = 12 → Cr = +6.