AS Chemistry: Redox Processes — Revision Notes

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

Oxidation number rules — in priority order

1  Uncombined element                          0
2  Simple ion                        = its charge
3  Fluorine                                   -1   (always)
4  Group 1 / Group 2                     +1 / +2
5  Hydrogen                                   +1   (except metal hydrides -1)
6  Oxygen                                     -2   (except peroxides -1,
                                                    and OF2 where it is +2)
7  Sum in a neutral compound                    0
8  Sum in an ion                    = ion charge

The order matters: where rules conflict, the earlier one wins. In H₂O₂, oxygen must be −1 because hydrogen’s +1 takes priority.

Worked example — finding Mn in MnO₄⁻. Four oxygens contribute 4 × (−2) = −8. The ion’s overall charge is −1, so Mn’s oxidation number + (−8) = −1, giving Mn = +7. This is exactly why the ion is named manganate**(VII)** — the Roman numeral states the oxidation number directly, so naming and calculation should always agree.

Definitions

	Oxidation	Reduction
Electrons	Loss	Gain
Oxidation number	Increase	Decrease

OIL RIG. Oxidising agent is itself reduced; reducing agent is itself oxidised — the reversal that catches most candidates.

Balancing by half equations

1  Balance the atoms other than O and H
2  Balance O by adding H2O
3  Balance H by adding H+
4  Balance charge by adding electrons
5  Multiply the halves so electrons cancel, then add

Worked — manganate(VII) in acid:

MnO4-  ->  Mn2+
MnO4-  ->  Mn2+ + 4H2O           (balance O)
MnO4- + 8H+  ->  Mn2+ + 4H2O     (balance H)
MnO4- + 8H+ + 5e-  ->  Mn2+ + 4H2O   (balance charge)

Mn goes from +7 to +2 — a five-electron change, which is why the 5e⁻ appears in this half-equation. The overall reacting ratio between MnO₄⁻ and a given reducing agent is not always 1:5 — that specific ratio only holds for a reducing agent that supplies exactly one electron per species, such as Fe²⁺ → Fe³⁺ + e⁻ (giving the familiar MnO₄⁻ : Fe²⁺ = 1 : 5). For a reducing agent that supplies a different number of electrons per species (e.g. a typical two-electron reductant), combine the two balanced half-equations — matching electrons lost to electrons gained — to find the correct overall ratio.

Balancing by oxidation-number change (alternative method)

Instead of half equations, you can balance by tracking oxidation-number changes directly, since electrons lost must equal electrons gained.

Worked example. Balance MnO₄⁻ + Fe²⁺ + H⁺ → Mn²⁺ + Fe³⁺ + H₂O.

1  Mn: +7 -> +2  (down 5, gains 5e-)     Fe: +2 -> +3  (up 1, loses 1e-)
2  One Mn needs 5 Fe2+ to supply 5 electrons:
   MnO4- + 5Fe2+ + H+  ->  Mn2+ + 5Fe3+ + H2O
3  Balance O and H: 4 O in MnO4- need 4 H2O; those need 8 H+
   MnO4- + 5Fe2+ + 8H+  ->  Mn2+ + 5Fe3+ + 4H2O
4  Check charge: left (-1)+5(+2)+8(+1) = +17;  right (+2)+5(+3) = +17

Both atoms and overall charge must balance — equal electrons lost and gained is necessary but not sufficient on its own.

Titration essentials

Titration	Indicator	End point
Manganate(VII)	Self-indicating	First permanent pale pink
Thiosulfate/iodine	Starch, added near the end	Blue-black → colourless

Manganate(VII) titrations use sulfuric acid — not hydrochloric (chloride would be oxidised to chlorine) and not nitric (itself an oxidising agent).

Starch is added late because it forms a stable complex with iodine that releases slowly if added too early.

Disproportionation

The same element is simultaneously oxidised and reduced.

Cl2 + 2NaOH  ->  NaCl + NaClO + H2O
Cl:  0 -> -1  (reduced)  AND  0 -> +1 (oxidised)

To prove it, assign oxidation numbers to that element on both sides and show one rises and one falls.

Exam traps

• Calling the agent by what happens to it — the oxidising agent is reduced.
• Oxidation number is per atom, not for the whole formula unit.
• Forgetting the exceptions: H₂O₂, metal hydrides, OF₂.
• Using HCl in a manganate(VII) titration.
• Adding starch at the start of an iodine titration.
• Not checking that both atoms and charge balance in a half equation.
• Assuming a reaction is a simple redox without checking the oxidation number of the element in every product — a hidden disproportionation is easy to miss.

Self-test

1. Give the oxidation number of Mn in MnO₄⁻ and of Cr in Cr₂O₇²⁻.
2. Why is sulfuric acid used in manganate(VII) titrations?
3. Write the half equation for MnO₄⁻ → Mn²⁺ in acid.
4. When is starch added in an iodine–thiosulfate titration, and why?
5. Prove that Cl₂ + 2NaOH → NaCl + NaClO + H₂O is disproportionation.
6. Use oxidation-number changes to balance MnO₄⁻ + Fe²⁺ + H⁺ → Mn²⁺ + Fe³⁺ + H₂O.
7. In that reaction, identify the oxidising agent and the reducing agent, giving a reason for each.

Answers: 1. Mn = +7; Cr = +6. 2. Hydrochloric acid would be oxidised to chlorine and nitric acid is itself an oxidising agent, so both would give a false result. 3. MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. 4. Near the end point, when the solution is straw-coloured — added earlier, the starch–iodine complex releases iodine too slowly and the end point is indistinct. 5. Chlorine starts at 0; in NaCl it is −1 (reduced) and in NaClO it is +1 (oxidised) — the same element both oxidised and reduced. 6. MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O — Mn gains 5 electrons (+7→+2), so 5 Fe²⁺ are needed to lose 5 electrons (+2→+3) between them. 7. MnO₄⁻ is the oxidising agent — it causes Fe²⁺ to be oxidised while Mn itself is reduced (+7→+2); Fe²⁺ is the reducing agent — it causes Mn to be reduced while Fe itself is oxidised (+2→+3).