WJEC A-Level Chemistry Unit 3, June 2025: Question 6

✅ Correct Answer

Initially forms a grey-green precipitate which dissolves in excess (sodium hydroxide) to form a green solution.

💡 Key Knowledge

Chromium(III) hydroxide is amphoteric:

• Initial precipitation:
  [Cr(H₂O)₆]³⁺(aq) + 3OH⁻(aq) → [Cr(OH)₃(H₂O)₃](s) + 3H₂O(l)
  (Grey-green solid forms)
• Excess OH⁻:
  [Cr(OH)₃(H₂O)₃](s) + 3OH⁻(aq) → [Cr(OH)₆]³⁻(aq) + 3H₂O(l)
  (Precipitate acts as an acid, dissolves to form a deep green solution)

🧠 Exam Technique

The prompt asks for observation(s) when reagent is added gradually until in excess. Whenever you see this phrasing for transition metal ions, always describe a two-stage process:

1. What happens with a few drops (colour + state: "precipitate").
2. What happens with excess (whether it dissolves + resulting colour + state: "solution").

❌ Common Errors

• Writing "a green precipitate" without stating that it dissolves in excess.
• Stating that it forms a "blue precipitate" (confusing chromium with copper).
• Failing to mention physical states (e.g. saying just "turns green" rather than stating "precipitate" and "solution").

✅ Correct Answer

• A: Salt bridge
• B: Platinum (electrode) / Pt

💡 Component Functions

• A (Salt Bridge): Completes the circuit and allows the movement of ions between half-cells to balance charge without mixing the solutions (commonly filter paper soaked in KNO₃).
• B (Platinum Electrode): Provides an unreactive, conductive surface for electron transfer. Platinum is required because neither half-cell contains a solid metal participating in the redox equilibrium.

❌ Common Errors

• Naming B as "graphite" or "metal electrode" (platinum is the standard requirement in standard potential determinations).
• Naming A as "delivery tube" or "glass tube".
• Getting one component right and losing the mark because both parts are required for 1 mark.

🧠 Exam Technique

Whenever both species in a half-reaction are in solution (e.g., Cr₂O₇²⁻ and Cr³⁺) or one is a gas (H₂ and H⁺), an inert platinum electrode must be specified.

✅ Correct Answer

Direction of electron flow: from left to right (from the hydrogen half-cell to the chromium half-cell along the connecting wire).

📐 Step-by-Step Reasoning

1. Compare electrode potentials:
   • Left (SHE): 2H⁺ + 2e⁻ ⇌ H₂    Eᶿ = 0.00 V
   • Right: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ ⇌ 2Cr³⁺ + 7H₂O    Eᶿ = +1.33 V
2. Identify oxidation vs. reduction:
   • The more positive system (+1.33 V) undergoes reduction (accepts electrons).
   • The more negative system (0.00 V) undergoes oxidation (releases electrons):
     H₂ → 2H⁺ + 2e⁻
3. Trace the flow: Electrons flow externally through the wire from the negative electrode (site of oxidation: SHE) to the positive electrode (site of reduction: Cr half-cell).

🧠 Diagram Annotation Tip

Draw a clear arrow directly above or on the connecting wire labeled with e⁻ → pointing to the right. Avoid drawing arrows through the salt bridge; electrons only travel through the wire, never through solution or the salt bridge!

❌ Common Errors

• Drawing arrows pointing through the salt bridge (only ions move in the salt bridge).
• Confusing the flow of conventional current with electron flow.
• Assuming electrons flow towards the 0.00 V half-cell because 0 is "lower" in magnitude without considering the positive sign of +1.33 V.