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5.1.1Unit 5. Transition Metals and Organic Nitrogen Chemistry

Reduction and Oxidation — IAL Edexcel Linear (Chemistry) Revision Notes & Topical Questions

5 min read 6 parts Lesson 7 of 43 Unit quiz

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Part 1 of 6Defining Redox

5.1.1 Reduction and Oxidation
Oxidation
  • Loss of electrons, or an increase in oxidation number, as in Fe²⁺ becoming Fe³⁺.
Reduction
  • Gain of electrons, or a decrease in oxidation number, as in Cu²⁺ becoming Cu.
Oxidising agent
  • This species accepts electrons from another and is itself reduced during the reaction.
Reducing agent
  • This species donates electrons to another and is itself oxidised during the reaction.
Memory aid
  • OIL RIGOxidation Is Loss, Reduction Is Gain, of electrons.

Part 2 of 6Oxidation Numbers

Oxidation Numbers
Elements
  • An uncombined element, such as O₂ or Fe, always has an oxidation number of zero.
Simple ions
  • The oxidation number equals the charge, so Al³⁺ is +3 and Cl⁻ is -1.
Oxygen and hydrogen
  • Oxygen is usually -2 and hydrogen usually +1, with exceptions such as peroxides (-1) and metal hydrides (-1 for H).
Totals
  • The oxidation numbers in a neutral compound add to zero and in an ion add to its overall charge.

Part 3 of 6Building Equations

1
Write each half-equation, balancing the atom being changed first.
2
Balance oxygen by adding H₂O to the side that needs it, then balance hydrogen by adding H⁺ in acidic solution.
3
Balance charge by adding electrons to the more positive side so that both sides carry the same total charge.
4
Multiply the half-equations so that the electrons gained equal the electrons lost, then add them and cancel the electrons.
5
Cancel any species that appear on both sides and check atoms and charges once more.

Part 4 of 6Disproportionation

Disproportionation
Meaning
  • One species is simultaneously oxidised and reduced, so the same element ends up in a higher and a lower oxidation state.
Example
  • Cl₂ + 2NaOH → NaCl + NaClO + H₂O, where chlorine goes from 0 to -1 and to +1.
Evidence
  • Show it by writing the oxidation number of the element before and after in every product.

Part 5 of 6Worked Example

Question Construct the ionic equation for acidified MnO₄⁻ oxidising Fe²⁺ to Fe³⁺.

  1. ReductionMnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O.
  2. OxidationFe²⁺ → Fe³⁺ + e⁻.
  3. Multiply the oxidation by 5 so five electrons are released, then add: MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺.
  4. Check charge
    • left -1 + 8 + 10 = +17;
      • right +2 + 15 = +17.

Answer MnO₄⁻(aq) + 8H⁺(aq) + 5Fe²⁺(aq) → Mn²⁺(aq) + 4H₂O(l) + 5Fe³⁺(aq)

Part 6 of 6Exam Checklist & Tips

5.1.1 – Key Points for the Exam

  • Oxidation number The charge an atom would have if all its bonds were ionic.
  • Half-equation An equation showing only the oxidation or the reduction with electrons.
  • Disproportionation Simultaneous oxidation and reduction of the same element.
  • Oxidising agent A species that removes electrons from another species.

Common Mistakes

  • Writing electrons on the wrong side of a half-equation.
  • Forgetting to multiply half-equations so electrons cancel.
  • Giving oxidation numbers as 2+ instead of +2; the sign comes first.
  • Calling a species the oxidising agent because it is oxidised; it is reduced.

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