OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2025: Question 20

Electrode Potentials, Standard Cells & Multi-Stage Redox Feasibility

WHAT THIS QUESTION TESTS

Core skills & knowledge assessed:

Electrochemical Cell Setup: Constructing and labelling standard cells, identifying when an inert platinum (Pt) electrode is required for ion-only half-cells.
Standard Conditions: Recalling standard concentration ( 1 mol dm⁻³ ) and temperature ( 298 K / 25 °C ).
Cell Potential Calculation: Calculating standard cell potential using E°_cell = E°(positive) - E°(negative) .
Predicting Redox Reactions: Using relative standard electrode potentials ( E° ) and Le Chatelier equilibrium shifts to explain feasibility and construct balanced redox equations across sequential steps.

Part (a)(i) — Standard Electrochemical Cell Diagram & Conditions

4 Marks • Redox System 2 [Zn²⁺/Zn] and System 4 [V³⁺/V²⁺]

📋 Diagram Layout Requirements:

Left Beaker (System 2): Zinc metal electrode labeled Zn(s) dipped in aqueous zinc ion solution labeled Zn²⁺(aq) (or 1 mol dm⁻³ Zn²⁺ ).
Right Beaker (System 4): Platinum wire or foil labeled Pt(s) immersed in an equimolar solution containing both V³⁺(aq) and V²⁺(aq) (each 1 mol dm⁻³ ).
External Circuit: Connecting wires from both electrodes via a high-resistance voltmeter labeled V (or voltmeter symbol).
Internal Circuit: An inverted U-tube or filter paper strip dipping into both solutions labeled salt bridge .

✅ Mark Breakdown

Mark 1: Complete circuit with voltmeter and labelled salt bridge physically dipping into both half-cells.
Mark 2: Half-cell 4 clearly shows an inert Pt electrode in contact with both V²⁺ and V³⁺ ions.
Mark 3: Half-cell 2 clearly shows Zn metal electrode in contact with Zn²⁺ ions.
Mark 4: Standard conditions clearly stated: 1 mol dm⁻³ (seen anywhere) and 298 K (or 25 °C ).

🧠 Exam Technique: Ion-Only Half-Cells

Whenever a half-equation has no solid metal present (e.g. V³⁺ + e⁻ ⇌ V²⁺ ), an inert conducting electrode is required. Always use platinum (Pt).

Ensure the liquid level in your drawing is high enough so that both the electrode and salt bridge end dip beneath the surface.

❌ Common Errors to Avoid

Wrong electrode in System 4: Drawing a solid vanadium metal electrode for System 4 instead of Pt. Vanadium is only present as aqueous ions in System 4!
Floating salt bridge: Drawing the salt bridge floating above the solutions. It must touch or dip into both solutions.
Imprecise concentration units: Writing "1 mole" instead of 1 mol dm⁻³ or 1 M . (Examiners do not allow "1 mol" for concentration).
Unnecessary gas pressure: Quoting 100 kPa when no gases are involved in the systems chosen. While not penalised, it wastes time.

Part (a)(ii) — Cell Potential Calculation

1 Mark • Determining E°_cell

📐 Calculation Steps

Identify half-cell potentials:
System 4: E° = -0.26 V (more positive / cathode)
System 2: E° = -0.76 V (more negative / anode)
Apply formula:
E°_cell = E°(positive electrode) - E°(negative electrode)
E°_cell = -0.26 V - (-0.76 V)
Solve:
E°_cell = +0.50 V (or +0.5 V )

❌ Calculation Trap

Watch out for double negatives! Subtracting a negative number means adding:

-0.26 - (-0.76) = -0.26 + 0.76 = +0.50 V

Never give a negative answer: A spontaneous standard cell must have a positive cell potential. -0.50 V scores zero.

Mark scheme guidance: Accept (+)0.5(0) V . DO NOT ALLOW -0.5V .

Part (b) — Two-Stage Reduction of Acidified VO²⁺ by Cobalt

4 Marks • Explaining Feasibility & Constructing Overall Equations

💡 Electrode Potential Comparison

Redox System 3: Co²⁺(aq) + 2e⁻ ⇌ Co(s) ( E° = -0.28 V )
Redox System 5: 2H⁺(aq) + VO²⁺(aq) + e⁻ ⇌ V³⁺(aq) + H₂O(l) ( E° = +0.34 V )
Redox System 4: V³⁺(aq) + e⁻ ⇌ V²⁺(aq) ( E° = -0.26 V )

✅ Model Answer

Stage 1 Equation:
Co + 2VO²⁺ + 4H⁺ → Co²⁺ + 2V³⁺ + 2H₂O

Stage 2 Equation:
Co + 2V³⁺ → Co²⁺ + 2V²⁺

Comparison of E° values:
The standard electrode potential of the Co²⁺/Co system ( -0.28 V ) is more negative (or less positive) than both:
• the VO²⁺/V³⁺ system ( +0.34 V )
• the V³⁺/V²⁺ system ( -0.26 V ).

Equilibria shifts:
• The more negative system (Co²⁺/Co) shifts to the left (oxidising Co to Co²⁺).
• The more positive systems (VO²⁺/V³⁺ in stage 1, and V³⁺/V²⁺ in stage 2) shift to the right (reducing vanadium species).

🧠 Examiner Guidance & Vocabulary Rules

Crucial Phrasing: Always state "more negative" or "less positive" when comparing values.
⚠️ Do NOT write "higher" or "lower" electrode potential (the mark scheme explicitly says: IGNORE higher / lower ).
Balancing Electrons: Co releases 2 electrons ( Co → Co²⁺ + 2e⁻ ), whereas each vanadium step only absorbs 1 electron. Multiply the vanadium half-equations by 2 before combining!
State symbols: State symbols are not required by the mark scheme (even if wrong, they are ignored), but formulas and balancing must be fully correct.

4 Marks awarded as:
[1 mark] Correct balanced equation for Stage 1 (VO²⁺ → V³⁺)
[1 mark] Correct balanced equation for Stage 2 (V³⁺ → V²⁺)
[1 mark] Comparing E° of Co to both VO²⁺ and V³⁺ using "more negative" or "less positive"
[1 mark] Linking equilibrium shift correctly to E° values (Co shifts left AND VO²⁺ / V³⁺ shifts right)