A-Level Chemistry October/November 2024 Q3(c)(ii): Select two oxidising agents that can oxidise Fe2+ ions to Fe3+ ions under alkaline cond…

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  To find suitable oxidising agents, we must first identify the relevant half-equation for the oxidation of Fe²⁺ under alkaline conditions. In alkaline solution, Fe²⁺ exists as Fe(OH)₂ and Fe³⁺ exists as Fe(OH)₃. The half-equation from the Data Booklet is: Fe(OH)₃(s) + e⁻ ⇌ Fe(OH)₂(s) + OH⁻(aq) ; E^⦵ = -0.56 V

  For oxidation of Fe(OH)₂ to Fe(OH)₃, this half-reaction is reversed. A spontaneous reaction requires $E_{cell} > 0$. Since $E_{cell} = E_{red} - E_{ox}$, we need an oxidising agent (reduction half-cell) with $E_{red} > -0.56$ V.

  We select two suitable oxidising agents from the Data Booklet that function under neutral or alkaline conditions.

  oxidising agent 1: O₂ (dioxygen)

  • Reduction half-equation: O₂(g) + 2H₂O(l) + 4e⁻ ⇌ 4OH⁻(aq) ; E^⦵ = +0.40 V
  • Calculation of $E_{cell}$: $E_{cell} = E_{red} - E_{ox} = (+0.40) - (-0.56) = +0.96$ V
  • Overall equation:
    1. Oxidation: (Fe(OH)₂(s) + OH⁻(aq) → Fe(OH)₃(s) + e⁻) × 4
    2. Reduction: O₂(g) + 2H₂O(l) + 4e⁻ → 4OH⁻(aq)
    3. Combine and cancel 4OH⁻: 4Fe(OH)₂(s) + O₂(g) + 2H₂O(l) → 4Fe(OH)₃(s)

  equation: 4Fe(OH)₂(s) + O₂(g) + 2H₂O(l) → 4Fe(OH)₃(s) $E_{cell}$ = +0.96 V

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  oxidising agent 2: ClO⁻ (hypochlorite ion)

  • Reduction half-equation: ClO⁻(aq) + H₂O(l) + 2e⁻ ⇌ Cl⁻(aq) + 2OH⁻(aq) ; E^⦵ = +0.89 V
  • Calculation of $E_{cell}$: $E_{cell} = E_{red} - E_{ox} = (+0.89) - (-0.56) = +1.45$ V
  • Overall equation:
    1. Oxidation: (Fe(OH)₂(s) + OH⁻(aq) → Fe(OH)₃(s) + e⁻) × 2
    2. Reduction: ClO⁻(aq) + H₂O(l) + 2e⁻ → Cl⁻(aq) + 2OH⁻(aq)
    3. Combine and cancel 2OH⁻: 2Fe(OH)₂(s) + ClO⁻(aq) + H₂O(l) → 2Fe(OH)₃(s) + Cl⁻(aq)

  equation: 2Fe(OH)₂(s) + ClO⁻(aq) + H₂O(l) → 2Fe(OH)₃(s) + Cl⁻(aq) $E_{cell}$ = +1.45 V

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  How the marks are awarded

  • M1 — Awarded for correctly identifying one oxidising agent (O₂ or ClO⁻) and its corresponding correct Ecell value (+0.96 V for O₂ or +1.45 V for ClO⁻).
  • M2 — Awarded for correctly identifying the second oxidising agent and its corresponding correct Ecell value.
  • M3 — Awarded for the correct, fully balanced equation for the reaction with ClO⁻, including correct state symbols as shown in the half-equations used.
  • M4 — Awarded for the correct, fully balanced equation for the reaction with O₂, including correct state symbols as shown in the half-equations used.

  Common mistakes

  • Using the half-equations for acidic conditions (e.g., Fe³⁺/Fe²⁺) instead of alkaline conditions (Fe(OH)₃/Fe(OH)₂), leading to incorrect Ecell values and equations.
  • Incorrectly calculating Ecell, for example by subtracting in the wrong order (Eox - Ered) or adding the potentials when one has already been sign-flipped.
  • Failing to balance the overall equation correctly by not multiplying the half-equations to equalise the number of electrons transferred.
  • Forgetting to cancel spectator species (like OH⁻) from both sides of the combined equation, leading to an unsimplified final answer.

  Examiner tip: Always carefully check the reaction conditions (e.g., acidic or alkaline) and select the corresponding half-equations from the Data Booklet before calculating Ecell or constructing the overall equation.

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