4.9 Redox Reactions

Oxidation-Reduction (Redox) Reactions

Electrons flow along a curved arrow from reducing to oxidizing agent. Oxidation numbers, half-reactions, and the OIL RIG mnemonic update for each of 5 reactions.

Electron TransferOIL RIGCombustionHalf-Reactions
Topic 4.9

Oxidation–Reduction (Redox) Reactions

Represent a balanced redox reaction equation using half-reactions.

Balanced chemical equations for redox reactions can be constructed from half-reactions. A half-reaction shows just the oxidation or just the reduction, with the electrons written explicitly.

The method (acidic solution):

  1. Split the reaction into an oxidation half-reaction and a reduction half-reaction.
  2. Balance all atoms except O and H.
  3. Balance O by adding H₂O.
  4. Balance H by adding H⁺.
  5. Balance charge by adding electrons — to the right for oxidation, to the left for reduction.
  6. Multiply each half-reaction so the electron counts match, then add them and cancel anything appearing on both sides.

For basic solution, balance as if acidic, then add enough OH⁻ to both sides to neutralize every H⁺, converting H⁺ + OH⁻ into H₂O, and cancel duplicate waters.

Why this matters beyond Unit 4: half-reactions are the language of electrochemistry. In Unit 9 the oxidation half-reaction happens at the anode, the reduction half-reaction at the cathode, and the number of electrons transferred is the n in ΔG° = −nFE°.

The essential check: a balanced redox equation must balance atoms and total charge. Summing the charges on each side is the fastest way to catch an error.

Key points

  • Electrons lost must equal electrons gained — that constraint is what balances the equation.
  • Balance O with water, H with H⁺, and charge with electrons, in that order.
  • For basic solution, balance as acidic first, then neutralize the H⁺ with OH⁻.
  • Always verify total charge on both sides, not just atom counts.

Common mistakes

  • Adding electrons to the wrong side. Oxidation loses electrons, so they appear as a product.
  • Forgetting to scale both halves. The electron counts must match exactly before adding.
  • Leaving H⁺ in a basic-solution answer.
  • Checking only atoms. Charge balance catches errors that atom counting misses.

Worked example

Balance in acidic solution: Cr₂O₇²⁻(aq) + Fe²⁺(aq) → Cr³⁺(aq) + Fe³⁺(aq)

Oxidation half-reaction:
Fe²⁺ → Fe³⁺ + e⁻ (atoms already balanced; one electron balances the charge)

Reduction half-reaction:
Cr₂O₇²⁻ → 2 Cr³⁺ (balance Cr)
Cr₂O₇²⁻ → 2 Cr³⁺ + 7 H₂O (balance O with water)
14 H⁺ + Cr₂O₇²⁻ → 2 Cr³⁺ + 7 H₂O (balance H with H⁺)
Charge: left = 14(+1) + (−2) = +12; right = 2(+3) = +6. Add 6 e⁻ to the left:
14 H⁺ + Cr₂O₇²⁻ + 6 e⁻ → 2 Cr³⁺ + 7 H₂O

Scale and add: multiply the oxidation half by 6 so both involve 6 e⁻.
6 Fe²⁺ → 6 Fe³⁺ + 6 e⁻

Overall:
Cr₂O₇²⁻(aq) + 6 Fe²⁺(aq) + 14 H⁺(aq) → 2 Cr³⁺(aq) + 6 Fe³⁺(aq) + 7 H₂O(l)

Charge check: left = −2 + 12 + 14 = +24; right = +6 + 18 = +24 ✓

Full notes for topic 4.9 →