Edexcel A-Level Chemistry Paper 1, June 2024: Question 7

✅ Correct Answer: Option D

Expression:

Units: atm^-½

📐 Unit Derivation

Substitute pressure units (atm) directly into the K_p expression:

Units = atm / (atm × atm^½)

Units = 1 / atm^½ = atm^-½

Alternatively, calculate overall reaction order: 1 - (1 + 0.5) = -0.5. Hence, (atm)^-0.5 = atm^-½ .

❌ Common Errors

• Inverting the fraction (Options B and C): Writing reactants over products instead of products over reactants.
• Power arithmetic mistakes (Option A): Incorrectly cancelling atm / (atm × atm^½) to obtain atm² .

🧠 Exam Technique

Always write partial pressures with standard lower-case p notation (e.g. p(SO₃)), never square brackets [SO₃] , which denote aqueous concentrations.

✅ Correct Answer: Option B

B: temperature

Temperature is the only factor that alters the numerical value of an equilibrium constant (K_c or K_p).

💡 Key Knowledge

• Temperature: Changes K_p. For an exothermic forward reaction (ΔH = -196 kJ mol⁻¹), increasing temperature shifts equilibrium to the left and decreases K_p.
• Pressure & Concentration: Shift equilibrium positions, but the value of K_p remains completely constant at a given temperature.
• Catalyst: Increases rates of both forward and reverse reactions equally. Affects rate, not yield or K_p.

✅ Correct Equations

Step 1: Oxidation of SO₂ (Reduction of catalyst)

SO₂ + V₂O₅ → SO₃ + V₂O₄

Vanadium is reduced from +5 in V₂O₅ to +4 in V₂O₄.

Step 2: Regeneration of Catalyst (Oxidation by O₂)

V₂O₄ + ½O₂ → V₂O₅

(or 2V₂O₄ + O₂ → 2V₂O₅)

💡 Transition Metal Catalysis

Transition metals make effective catalysts because they possess variable oxidation states and can easily transfer electrons to and from reactants.

Summing the two catalytic steps restores the overall stoichiometry:

SO₂ + ½O₂ → SO₃

Notice that V₂O₅ is reformed unchanged at the end of the cycle.

❌ Common Errors

• Including electrons: Writing half-equations involving e⁻ instead of balanced chemical equations with oxygen species.
• Incorrect vanadium oxides: Confusing V₂O₄ with VO (vanadium(II) oxide) or writing unbalanced oxygen coefficients.

🧠 Examiner Guidance

• State symbols are not required.
• Reversible arrows are permitted by the mark scheme.
• Using 2VO₂ instead of V₂O₄ is fully acceptable (e.g. SO₂ + V₂O₅ → SO₃ + 2VO₂ ).

✅ Correct Answer: Option C

Aqueous Ion: VO₂⁺ (dioxovanadium(V))

Colour of solution: Yellow

Oxidation number: +5

💡 Vanadium Oxidation States Summary

❌ Why Other Options Are Wrong

• A: VO₂⁺ is yellow, but its oxidation state is +5, not +3: V + 2(-2) = +1 ⇒ V = +5.
• B: VO²⁺ has oxidation state +4, but its colour is blue, not green.
• D: VO²⁺ is blue, but its oxidation state is +4: V + (-2) = +2 ⇒ V = +4, not +5.

🧠 Memory Aid

Remember: You Better Get Vanadium:

+5 (Yellow) → +4 (Blue) → +3 (Green) → +2 (Violet)

✅ Correct Answer: Option A

E°_cell = +1.76 V

📐 Step-by-Step Calculation

1. Look up standard electrode potentials from Data Booklet:
   Reduction: VO₂⁺ + 2H⁺ + e⁻ ⇌ VO²⁺ + H₂O   (E° = +1.00 V)
   Oxidation: Zn²⁺ + 2e⁻ ⇌ Zn   (E° = -0.76 V)
2. Identify Reduction and Oxidation half-cells:
   • VO₂⁺ is reduced (+5 to +4) → acts as cathode (positive electrode)
   • Zn is oxidised (0 to +2) → acts as anode (negative electrode)
3. Apply the cell potential formula:
   E°_cell = E°(reduction) - E°(oxidation)
   E°_cell = (+1.00 V) - (-0.76 V) = +1.76 V

❌ Common Distractors Explained

• Option B (+0.24 V): Arises from adding values incorrectly: +1.00 + (-0.76) = +0.24 V.
• Option C (-0.24 V): Arises from adding and reversing signs incorrectly.
• Option D (-1.76 V): The potential for the non-spontaneous reverse reaction (-0.76 - 1.00). A feasible cell reaction as written must have a positive E°_cell.

🧠 Exam Tip: Never Multiply E° by Stoichiometry!

Even though the overall equation multiplies the vanadium half-reaction by 2 to balance electrons, E° is an intensive property and must never be multiplied by stoichiometric coefficients.