Redox Means Electrons Change Hands
Oxidation is the loss of electrons. Reduction is the gain of electrons. The memory hook is OIL RIG: Oxidation Is Loss, Reduction Is Gain.
The two never happen alone. Electrons lost by one species must be gained by another, so every oxidation is matched by a reduction; together, the process is a redox reaction.
Oxidation numbers track the transfer: the number increases in oxidation and decreases in reduction.
In Zn + Cu²⁺ → Zn²⁺ + Cu, zinc moves from 0 to +2, losing two electrons per atom: oxidised. Copper moves from +2 to 0, gaining them: reduced.
Oxidation Numbers: The Bookkeeping Rules
Free elements are 0. Na, O₂ and Cl₂ are all 0.
A monatomic ion takes its charge. Na⁺ is +1, Cl⁻ is −1, Al³⁺ is +3.
Fluorine is always −1, the most electronegative element, so it takes the negative value in every compound.
Oxygen is usually −2. In peroxides, such as H₂O₂, it is −1, because the two oxygens are bonded to each other. Bonded to fluorine it is positive: in OF₂ oxygen is +2.
Hydrogen is usually +1 with nonmetals. In metal hydrides, such as NaH, the metal is less electronegative than hydrogen, so hydrogen is −1.
The sum rule ties it together. Oxidation numbers add up to 0 in a neutral compound and to the charge on a polyatomic ion. Fixed values help: Group 1 metals are +1, Group 2 metals +2, aluminium +3.
Oxidation numbers are bookkeeping, not real charges: carbon is +4 in CO₂, but no C⁴⁺ ion exists there.
Finding an Unknown Oxidation Number
Use the sum rule as an equation: insert the unknown, fill in the known values, solve.
Sulfur in H₂SO₄: 2(+1) + S + 4(−2) = 0, so S = +6.
Nitrogen in NO₃⁻: N + 3(−2) = −1, so N = +5.
Chromium in Cr₂O₇²⁻: 2Cr + 7(−2) = −2, so 2Cr = +12 and Cr = +6.
The Agents Do the Opposite of Their Names
The oxidising agent takes electrons from its partner, causing oxidation, and is itself reduced in the process.
The reducing agent gives electrons to its partner, causing reduction, and is itself oxidised.
In 2Na + Cl₂ → 2NaCl, chlorine gains electrons, falling from 0 to −1: chlorine is reduced, so Cl₂ is the oxidising agent. Sodium loses electrons and is oxidised, so Na is the reducing agent.
The strongest oxidising agent is the species most readily reduced: the one that takes electrons most eagerly, and so oxidises its partners most powerfully.
Half-Reactions: Balancing in Acidic Solution
Splitting an equation into an oxidation half and a reduction half shows the electrons explicitly and makes balancing systematic.
In each half, balance the atoms other than oxygen and hydrogen first. Balance oxygen by adding H₂O, hydrogen by adding H⁺, which acidic solution supplies freely. Then balance charge by adding electrons: on the right of an oxidation half, on the left of a reduction half.
Multiply the halves so electrons lost equal electrons gained, add them, and the electrons cancel. Finish by checking atoms and total charge.
For basic solutions, balance as for acid first, then add OH⁻ to both sides to convert H⁺ into H₂O, and cancel duplicated water.
Disproportionation: One Species, Both Jobs
In a disproportionation reaction, the same species is both oxidised and reduced: part of it climbs to a higher oxidation number, part falls to a lower one. The element must start in an intermediate state, able to move both ways.
Decomposing hydrogen peroxide, 2H₂O₂ → 2H₂O + O₂: oxygen starts at −1, falls to −2 in H₂O (reduced) and rises to 0 in O₂ (oxidised). Chlorine in alkali does the same: Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O, chlorine moving from 0 to both −1 and +1.
Redox Titrations
A redox titration spends the electron transfer itself as the measurement. The balanced equation gives the mole ratio between titrant and analyte, and the titre at the endpoint converts that ratio into the unknown concentration, as in Topic 4.6 but with the ratio set by electrons, not protons.
Permanganate is self-indicating: MnO₄⁻ is purple, its product Mn²⁺ colourless, so no separate indicator is needed; the endpoint is the first permanent purple tinge.
The ratio is rarely 1 to 1: Fe²⁺ to MnO₄⁻ runs 5 to 1, and assuming equal moles would miss the answer by a factor of five.
Worked Examples
Example 1: redox or not, decided by the numbers. In CH₄ + 2O₂ → CO₂ + 2H₂O, carbon moves from −4 to +4: oxidised. Oxygen moves from 0 to −2: reduced. Numbers moved, so the reaction is redox. In CaCO₃ → CaO + CO₂, calcium stays +2, carbon +4 and oxygen −2 on both sides: nothing moved, so this decomposition is not redox.
Example 2: balancing MnO₄⁻ + Fe²⁺ in acid. Oxidation half: Fe²⁺ → Fe³⁺ + e⁻, atoms already balanced, one electron fixing the charge. Reduction half: MnO₄⁻ → Mn²⁺. Add 4H₂O on the right for the oxygens, 8H⁺ on the left for the hydrogens. The left now carries charge −1 + 8 = +7 against +2 on the right, so add 5e⁻ to the left: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O, matching manganese falling from +7 to +2. Multiply the oxidation half by 5 and add: MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺. Check: atoms match, charge +17 on each side.
Example 3: a titration with a 5 to 1 ratio. A 25.0 mL sample of Fe²⁺ solution needs 24.0 mL of 0.0200 mol/L KMnO₄ to reach the endpoint. Moles of MnO₄⁻ = 0.0200 × 0.0240 = 4.80 × 10⁻⁴ mol. With Fe²⁺ : MnO₄⁻ at 5 : 1, moles of Fe²⁺ = 5 × 4.80 × 10⁻⁴ = 2.40 × 10⁻³ mol. Concentration = 2.40 × 10⁻³ ÷ 0.0250 = 0.0960 mol/L.
Traps That Cost Marks in Topic 4.9
Naming the agent after its own fate. The oxidising agent is itself reduced; the reducing agent is itself oxidised. Find the electron flow first, then name each agent by what it does to its partner.
Treating oxidation numbers as real ionic charges. They are a counting scheme: sulfur is +6 in H₂SO₄ without any S⁶⁺ ion existing.
Missing the three exceptions. Free elements are 0, so oxygen in O₂ is 0, not −2; oxygen in peroxides is −1; hydrogen in metal hydrides is −1. Questions target exactly these cases.
Putting the electrons on the wrong side. Oxidation produces electrons, so they are written as products; reduction consumes them, so they are written as reactants. MnO₄⁻ → Mn²⁺ gains 5e⁻, and those electrons belong on the left.
Balancing the atoms and forgetting the charge. An equation is balanced only when atoms and total charge both match; the charge check is what fixes how many electrons a half-reaction needs.
Carrying the 1 to 1 habit into redox titrations. Ratios come from the balanced equation: 5 Fe²⁺ per MnO₄⁻, 6 Fe²⁺ per Cr₂O₇²⁻, 2 S₂O₃²⁻ per I₂.
Hunting for redox in partner swapping. Double replacement and acid-base reactions change no oxidation number, so are never redox; single replacement and combustion always are.
Key Relationships
- Oxidation is loss of electrons and the oxidation number rises; reduction is gain and the number falls; OIL RIG; the two always occur together
- Oxidising agent: gains electrons, is itself reduced, causes oxidation. Reducing agent: loses electrons, is itself oxidised, causes reduction. The strongest oxidising agent is the species most readily reduced
- Oxidation number rules: free element 0; monatomic ion equals its charge; F −1; O −2 except peroxides (−1) and with fluorine (positive); H +1 except metal hydrides (−1); sums to 0 for a neutral compound, to the charge for an ion
- Half-reaction method in acid: balance other atoms, O with H₂O, H with H⁺, charge with e⁻; scale so electrons cancel; check atoms and charge
- Disproportionation: one species is oxidised and reduced at once, its key element starting in an intermediate state, as oxygen does at −1 in H₂O₂
- The redox test: assign oxidation numbers before and after; any change means redox. Single replacement and combustion always qualify; double replacement and acid-base never do
- Redox titrations turn on the balanced equation’s mole ratio, not on 1 to 1: MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺, and purple MnO₄⁻ is its own indicator
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