Question

Give balanced equations for the electrode reactions involved in the following conversions at the respective electrodes:
1. Aluminium oxide → Oxygen gas ← Copper [II] Sulphate
2. Copper metal → Copper ions → Copper metal
3. Lead [II] chloride → Chlorine gas ← Hydrochloric acid
4. Hydroxyl ions ← Acidified water → Oxygen gas
5. Potassium bromide → Bromine gas ← Lead bromide

  • Electrolysis
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Answer

  1. Aluminium oxide → Oxygen gas ← Copper [II] Sulphate

    Electrolysis of molten Al2O3:
    Al2O3 ⇌ 2Al3+ + 3O2-
    Reaction at anode:
    3O2- – 6e–⟶ 3[O]
    3[O] + 3[O] ⟶ 3O2 [product O2 gas]

    Electrolysis of copper (II) sulphate:
    CuSO4 ⇌ Cu2+ + SO42- [ions present]
    H2O ⇌ H1+ + OH1-
    Reaction at anode:
    OH1- – 1e–⟶ OH x 4
    4OH ⟶ 2H2O + O2 [product oxygen gas]

  2. Copper metal → Copper ions → Copper metal

    Reaction at anode:
    Cu – 2e– ⟶ Cu2+
    Reaction at cathode:
    Cu2+ + 2e– ⟶ Cu

  3. Lead [II] chloride → Chlorine gas ← Hydrochloric acid

    Electrolysis of molten PbCl2:
    PbCl2 ⇌ Pb2+ + 2Cl–
    Reaction at anode:
    2Cl– – 2e– ⟶ 2[Cl]
    Cl + Cl ⟶ Cl2

    Electrolysis of hydrochloric acid:
    HCl ⇌ H+ + Cl–
    Reaction at anode:
    2Cl– – 2e– ⟶ 2[Cl]
    Cl + Cl ⟶ Cl2

  4. Hydroxyl ions ← Acidified water → Oxygen gas

    Electrolysis of acidified water
    H2SO4 ⇌ 2H1+ + SO42-
    H2O ⇌ H1+ + OH1-
    Reaction at anode:
    OH1- – 1e– ⟶ OH x 4
    4OH ⟶ 2H2O + O2 [product oxygen gas]

  5. Potassium bromide → Bromine gas ← Lead bromide

    Electrolysis of molten KBr:
    KBr ⇌ K1+ + Br1-
    Reaction at anode:
    Br– – 1e– ⟶ Br
    Br + Br ⟶ Br2

    Electrolysis of molten PbBr2:
    PbBr2 ⇌ Pb2+ + 2Br–
    Reaction at anode:
    Br– – 1e– ⟶ Br
    Br + Br ⟶ Br2

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