Define oxidation in terms of: (a) loss of electrons (b) an increase in oxidation number

Cambridge IGCSE Chemistry 0620 – Redox: Oxidation

Core definition (what the exam expects)

1. Quick summary of oxidation & reduction

• Oxidation: loss of e⁻ ⇔ oxidation number becomes more positive (or less negative).
• Reduction: gain of e⁻ ⇔ oxidation number becomes more negative (or less positive).
• A redox reaction always contains one oxidation and one reduction occurring simultaneously.

2. Assigning oxidation numbers (Roman numerals)

In the IGCSE syllabus oxidation numbers are written with Roman numerals (Fe II, Fe III, etc.). Follow these steps for every element in a formula:

1. Assign the known oxidation numbers:
   • Elements in their elemental form: 0
   • Group 1 metals: +I
   • Group 2 metals: +II
   • Fluorine: –I (always)
   • Oxygen: –II (except in peroxides = –I, super‑oxides = –½, and when bonded to fluorine)
   • Hydrogen: +I (except when bonded to metals, then –I)
2. For a polyatomic ion, the sum of the oxidation numbers must equal the ion’s overall charge.
3. For a neutral compound, the sum of the oxidation numbers must be zero.
4. Solve for the unknown oxidation numbers using the above rules.
5. Write the final oxidation numbers with Roman numerals (e.g. +II → II, +III → III).

Example – Assigning oxidation numbers in FeSO₄

1. Known: O = –II, S in sulphate = +VI (because 4 × –II = –VIII, so S + (–VIII) = –2 → S = +VI).
2. Let Fe = x. The compound is neutral: x + (+VI) + 4(–II) = 0 → x + 6 – 8 = 0 → x = +II.
3. Write: Fe II, S VI, O II.

3. How to recognise a redox reaction

Check any ONE of the following:

1. Is there a change in oxidation number for any element?
2. Can the reaction be written as a pair of half‑reactions showing loss/gain of electrons?
3. Does the wording “gain of oxygen” or “loss of oxygen” appear (historical definition)?

Worked example – 2 Fe + 3 Cl₂ → 2 FeCl₃

• Assign oxidation numbers: Fe 0 → Fe III, Cl 0 → Cl I.
• Fe: 0 → +III (increase) → oxidation.
• Cl: 0 → –I (decrease) → reduction.
• Both processes occur → redox reaction.

4. Oxidation expressed as electron loss & oxidation‑number increase

When an atom, ion or molecule loses electrons, its charge becomes more positive. This is recorded as an increase in its oxidation number.

Half‑reaction example – Magnesium burning in oxygen

Overall reaction

• Oxidation half‑reaction (loss of electrons)
  Mg → Mg²⁺ + 2 e⁻
  Oxidation number: 0 → II (increase of +2).
• Reduction half‑reaction (gain of electrons)
  O₂ + 4 e⁻ → 2 O²⁻
  Oxidation number: 0 → II (decrease of 2).

5. Typical oxidation‑number changes (exam‑friendly table)

6. Extended worked example – Acidic medium

Reaction

• Assign oxidation numbers (key changes):
  • Fe: II → III (oxidation, loss of 1 e⁻ per Fe)
  • Mn: VII → II (reduction, gain of 5 e⁻ per Mn)
• Electron balance: 5 Fe × 1 e⁻ = 5 e⁻ lost = 1 Mn × 5 e⁻ = 5 e⁻ gained.
• Thus the equation is correctly balanced as a redox process.

7. Checklist for identifying oxidation & reduction in any equation

1. Assign oxidation numbers using the step‑by‑step guide (Roman numerals).
2. Compare each element’s number on the reactant and product sides.
   • Number becomes **more positive** (or less negative) → oxidation.
   • Number becomes **more negative** (or less positive) → reduction.
3. If required, write the half‑reactions to confirm that total electrons lost = total electrons gained.
4. Remember the historical wording: “gain of oxygen” = oxidation, “loss of oxygen” = reduction.

8. Key points to remember (exam quick‑ref)

• Oxidation = **gain of oxygen** (historical) **or** loss of electrons → increase in oxidation number.
• Reduction = loss of oxygen **or** gain of electrons → decrease in oxidation number.
• Oxidation numbers must be written with Roman numerals in the IGCSE syllabus.
• Every oxidation is paired with a reduction; together they constitute a redox reaction.
• Use the three‑check method (oxidation‑number change, electron transfer, oxygen‑transfer wording) to spot redox reactions quickly.