AQA A-Level Chemistry Paper 1, 2021: Question 8

✅ Acceptable Answers

A list of electrode potentials (or E°) in numerical order.

OR

A list of half-cells (or equations) in numerical order of electrode potential (or E°).

❌ Common Errors

• Writing "list of EMF values in order" — EMF is strictly rejected because EMF refers to the potential difference of a complete cell, not standard half-cell potentials.
• Forgetting to specify "in order" (e.g. just writing "a list of electrode potentials").

✅ Acceptable Answers (Any TWO from)

• Temperature: 298 K (or 25 °C)
• Concentration: [H⁺] = 1.00 mol dm⁻³ (or 1 mol dm⁻³)
• Pressure: 100 kPa

🧠 Exam Technique

• Both conditions are required to score the single mark.
• Always include correct units ( K , mol dm⁻³ , kPa ).
• AQA prefers 100 kPa (IUPAC standard state). The mark scheme notes to ignore 1 atm, meaning it won't earn credit as a standard value.

✅ Correct Answer

[Co(H₂O)₆]²⁺

(Co²⁺ or [Co(H₂O)₆]²⁺(aq) accepted; absence of square brackets is not penalised).

💡 Key Knowledge

• Reducing agents lose electrons (they get oxidised) and are found on the right-hand side of reduction half-equations.
• The strongest reducing agent is oxidised most readily → associated with the most negative E° (here, Fe).
• The weakest reducing agent is least readily oxidised → associated with the most positive E° (+1.81 V, which strongly favours the forward reduction). Hence, the species on the right is [Co(H₂O)₆]²⁺.

❌ Common Errors

• Selecting [Co(H₂O)₆]³⁺: This is the strongest oxidising agent, not the weakest reducing agent! Oxidising agents are on the left; reducing agents are on the right.
• Selecting Fe(s): Iron is the strongest reducing agent in this table (−0.44 V).

📐 Step-by-Step Deduction

1. Step 1: Check Fe to Fe²⁺
   VO₂⁺/VO²⁺ (+1.00 V) is more positive than Fe²⁺/Fe (−0.44 V).
   Reaction is feasible: Fe is oxidised to Fe²⁺.
2. Step 2: Check Fe²⁺ to Fe³⁺ (The Examiner Trap!)
   Can VO₂⁺ oxidise Fe²⁺ further?
   E°(VO₂⁺/VO²⁺) = +1.00 V is still more positive than E°([Fe(H₂O)₆]³⁺/[Fe(H₂O)₆]²⁺) = +0.77 V!
   Therefore, VO₂⁺ oxidises Fe²⁺ to Fe³⁺ as well. The final iron product must be Fe³⁺!
3. Step 3: Combine half-equations
   Oxidation: Fe(s) + 6 H₂O(l) → [Fe(H₂O)₆]³⁺(aq) + 3e⁻
   Reduction: 3 × [VO₂⁺(aq) + 2 H⁺(aq) + e⁻ → VO²⁺(aq) + H₂O(l)]

✅ Correct Overall Equations (2 Marks)

3 VO₂⁺ + 6 H⁺ + Fe + 3 H₂O → 3 VO²⁺ + [Fe(H₂O)₆]³⁺

OR (uncomplexed form):

3 VO₂⁺ + 6 H⁺ + Fe → 3 VO²⁺ + 3 H₂O + Fe³⁺

State symbols are not required.

💡 Explanation (Mark 1)

To justify feasibility, compare the standard electrode potentials explicitly:

• E°([Co(H₂O)₆]³⁺/[Co(H₂O)₆]²⁺) > E°([Fe(H₂O)₆]³⁺/[Fe(H₂O)₆]²⁺)
• OR quote values: +1.81 V > +0.77 V
• OR calculate cell EMF: E°_cell = +1.81 − (+0.77) = +1.04 V (> 0)

Examiner Note: You must explicitly refer to E° or electrode potential. Stating only "cobalt is more positive than iron" without mentioning E° scores 0.

✅ Equation (Mark 2)

Both half-equations involve 1 electron transfer:

[Co(H₂O)₆]³⁺ + [Fe(H₂O)₆]²⁺ → [Co(H₂O)₆]²⁺ + [Fe(H₂O)₆]³⁺

Since both complexes exchange 1 mole of electrons, their stoichiometry is 1:1.

✅ Correct Answer

They have different ligands (NH₃ vs H₂O).

💡 Chemical Insight

Ligands affect the energy levels of d-orbitals and differently stabilise the +2 and +3 oxidation states:

• NH₃ ligands bind more strongly to Co³⁺ than Co²⁺, stabilising the +3 state and making reduction harder (E° drops from +1.81 V down to +0.11 V).
• Simply identifying different ligands is all AQA requires for this 1-mark question.

❌ Do NOT Write

• "Different oxidation states" — Both cobalt ions undergoing reduction start in the +3 state! Writing this is explicitly penalised.
• "Different concentrations" — E° values are measured under standard conditions (1.00 mol dm⁻³), so concentrations are identical.