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1.9.3States of Matter

Writing Half-Equations — Cambridge IGCSE Chemistry 0620 Revision Notes & Topical Questions

5 min read 5 parts Lesson 22 of 30 Unit quiz

📚 Cambridge IGCSE Chemistry 0620 Hub / Topical Revision
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Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
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Section 1 of 5Oxidation and Reduction in Electrolysis

In electrolysis, oxidation and reduction are defined by the transfer of electrons at the electrodes.

Oxidation
  • Oxidation Loss of electrons;
    • at the anode, negative ions lose electrons and are oxidised to form neutral atoms or molecules.
Reduction
  • Reduction Gain of electrons;
    • at the cathode, positive ions gain electrons and are reduced to form neutral atoms or molecules.
OIL RIG
  • OIL RIG A memory aid: Oxidation Is Loss, Reduction Is Gain, of electrons, which works for every electrode reaction.
Balancing
  • Balancing Both atoms and charges must balance, and the electrons gained at the cathode must equal the electrons lost at the anode.
Position of electrons
  • Position of electrons Cathode half-equations have electrons on the reactant side, while anode half-equations have electrons on the product side.

Section 2 of 5Common Half-Equations

Common Half-Equations
Acidic solutions
  • Acidic solutions Acidic solutions contain few hydroxide ions, so water molecules are discharged at the anode to give oxygen.
Sodium chloride
  • Sodium chloride Water is discharged at the cathode in sodium chloride solution, because sodium is too reactive to be discharged.
Copper sulfate
  • Copper sulfate Copper is less reactive than hydrogen, so copper ions are discharged at the cathode instead of hydrogen ions.
ElectrolyteCathode (reduction)Anode (oxidation)
Molten lead(II) bromidePb²⁺ + 2e⁻ → Pb2Br⁻ → Br₂ + 2e⁻
Concentrated aqueous sodium chloride2H₂O + 2e⁻ → H₂ + 2OH⁻2Cl⁻ → Cl₂ + 2e⁻
Dilute sulfuric acid2H⁺ + 2e⁻ → H₂2H₂O → O₂ + 4H⁺ + 4e⁻
Aqueous copper(II) sulfateCu²⁺ + 2e⁻ → Cu2H₂O → O₂ + 4H⁺ + 4e⁻

Section 3 of 5Method for Writing Half-Equations

1
Identify the ions in the electrolyte, including H⁺ and OH⁻ from water if the electrolyte is aqueous.
2
Decide which ion is discharged at each electrode, using the rules for metal ions and halide ions.
3
Write the species discharged at that electrode, showing its charge and the number of electrons transferred.
4
Balance the atoms first, then balance the charges by adding electrons to the side that needs them.
5
Make the electrons equal at both electrodes and add the two half-equations to give the overall equation.

Section 4 of 5Worked Example

Question Write the half-equations and the overall equation for the electrolysis of molten aluminium oxide, Al₂O₃.

  1. Al₂O₃ contains Al³⁺ and O²⁻ ions, so Al³⁺ moves to the cathode and O²⁻ moves to the anode.
  2. CathodeAl³⁺ + 3e⁻ → Al.
  3. Anode2O²⁻ → O₂ + 4e⁻.
  4. Multiply the cathode equation by 4 and the anode equation by 3, so that 12 electrons appear on each side.
  5. Add the equations4Al³⁺ + 6O²⁻ → 4Al + 3O₂.

Answer

  • Cathode: Al³⁺ + 3e⁻ → Al.
  • Anode: 2O²⁻ → O₂ + 4e⁻.
  • Overall: 4Al³⁺ + 6O²⁻ → 4Al + 3O₂.

Section 5 of 5Exam Checklist & Tips

1.9.3 – Key Points for the Exam

  • Half-equation An equation showing only the electrons and species involved at one electrode.
  • Oxidation The loss of electrons by a species.
  • Reduction The gain of electrons by a species.
  • Discharge The reaction of an ion at an electrode to form a neutral element or molecule.
  • Electron balance The equal number of electrons gained at the cathode and lost at the anode.

Common Mistakes

  • Using the oxygen-gain definition of oxidation for electrode reactions;
    • electrolysis is defined by electron transfer.
  • Putting the electrons on the wrong side of an electrode equation, such as 2Cl⁻ + 2e⁻ → Cl₂.
  • Using H⁺ in the cathode equation for sodium chloride solution;
    • there water is reduced to hydrogen and OH⁻ forms.
  • Balancing atoms but forgetting the charges on each side of a half-equation.

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Chemistry 0620 & Edexcel OL Product Prediction Rules

Aqueous Electrolysis Rules (Inert Electrodes)

  • At Cathode (-): Least reactive cation discharged. If metal is more reactive than hydrogen (Na+, K+, Mg2+), H2 gas is produced: 2H+ + 2e- → H2. If Cu2+ or Ag+, the metal coats the cathode.
  • At Anode (+): If concentrated halide present (Cl-, Br-, I-), halogen is formed (e.g. 2Cl- → Cl2 + 2e-). If sulfate (SO42-) or nitrate (NO3-), O2 gas is discharged: 4OH- → O2 + 2H2O + 4e-.
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