Sec 4 Combined Science (Chemistry) Redox Reactions notes

Oxidation originally meant gaining oxygen; reduction meant losing oxygen. When hydrogen reduces hot copper(II) oxide, both happen at once: CuO(s) + H₂(g) → Cu(s) + H₂O(g). The copper(II) oxide loses oxygen (it is reduced) while the hydrogen gains oxygen (it is oxidised).

The hydrogen view is the mirror: gaining hydrogen is reduction, losing hydrogen is oxidation. You will rarely need it, but it explains why "hydrogen removes oxygen from CuO" and "CuO is the oxidising agent" are the same fact stated two ways.

The oxygen view cannot explain reactions with no oxygen anywhere, so chemistry upgrades the definition: oxidation is the ==loss== of electrons; reduction is the ==gain== of electrons (memory aid: OIL RIG). Look at zinc dropping into copper(II) sulfate solution: Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s).

• Zinc atoms lose two electrons each: Zn → Zn²⁺ + 2e⁻ — zinc is oxidised (and dissolves away).
• Copper(II) ions gain those electrons: Cu²⁺ + 2e⁻ → Cu — copper(II) is reduced (and coats the zinc as pink-brown metal).
• The sulfate ions are spectators — unchanged on both sides.

Because the electrons lost and gained must match exactly, oxidation and reduction are inseparable: every redox equation can be split into two half-equations whose electrons cancel. Displacement reactions are the classic playground: a more reactive metal always oxidises itself and pushes the less reactive metal ion out of solution — Fe + CuSO₄ → FeSO₄ + Cu, but copper cannot displace iron.

An oxidation state is a number assigned to every atom in a species. The rules you need: an uncombined element is 0; a simple ion equals its charge; oxygen is −2 in compounds; hydrogen is +1; and the states add to 0 for a neutral compound (or to the charge for an ion).

The pay-off: an increase in oxidation state is oxidation; a decrease is reduction — exactly the electron transfer, counted. In Zn + CuSO₄ → ZnSO₄ + Cu: zinc goes 0 → +2 (up: oxidised); copper goes +2 → 0 (down: reduced); sulfur stays +6 and oxygen stays −2 (spectators).

Check the arithmetic: in Fe₂O₃ the three oxygens contribute 3 × (−2) = −6, so two irons must share +6 → each is +3. In CO, oxygen is −2 so carbon must be +2 to make zero. That subtraction is all the maths this topic asks for.

An oxidising agent oxidises something else while itself being reduced; a reducing agent reduces something else while itself being oxidised. Two standard solutions detect them by colour alone.

The chemistry under the KI test is the displacement you met above: an oxidising agent pulls electrons from iodide ions, 2I⁻ → I₂ + 2e⁻, and free iodine shows its red-brown colour (a starch indicator turns it dark blue for certainty). Under the manganate test, the reducing agent supplies electrons that turn purple MnO₄⁻ into nearly colourless Mn²⁺.

Watch the logic direction — it is the part students flip: the reagent (KI or KMnO₄) is not what you are testing. The unknown is. Brown iodine means the unknown took electrons, so the unknown is the oxidising agent; a fading purple means the unknown gave electrons, so it is the reducing agent.

Once you can see electron transfer, old chapters turn redox. Combustion is oxidation of the fuel: burning magnesium gives oxygen atoms to magnesium atoms (2Mg + O₂ → 2MgO — Mg 0 → +2, oxidised; O 0 → −2, reduced). Metals with dilute acids are displacement at the ionic level: Mg + 2HCl → MgCl₂ + H₂ has Mg → Mg²⁺ + 2e⁻ (oxidised) and 2H⁺ + 2e⁻ → H₂ (reduced) — the fizzing is the reduced hydrogen escaping.

This viewpoint also explains the reactivity series you will meet next chapter: a metal higher in the series is a stronger reducing agent — it loses electrons more willingly, so it can oxidise itself while reducing the ions of any metal below it. Redox is the engine under the whole topic.