AQA A-Level Chemistry Paper 1, 2018: Question 6

✅ Correct Answer

It contains mobile ions / ions can move through it / provides free ions.

❌ Common Errors

• Saying "electrons travel through the salt bridge" — this scores 0 marks instantly. Electrons only flow through the external wire.
• Vague answers like "it connects the solutions" or "it completes the circuit" without stating how (via ions moving).

🧠 Exam Technique

Always state the charge carrier involved. In external wiring: delocalised electrons flow. In solutions and salt bridges: mobile ions flow.

📐 Step-by-Step Calculation

1. Identify given data:
   Cell EMF = +0.16 V
   Right-hand electrode: standard conditions ( 1.0 mol dm⁻³ CuSO₄ ), so E_RHS = +0.34 V .
2. Apply EMF formula:
   EMF = E_RHS − E_LHS
3. Rearrange and solve:
   +0.16 = +0.34 − E_LHS
   E_LHS = +0.34 − 0.16 = +0.18 V

✅ Correct Answer

(+) 0.18 V

🧠 Top Tip

Double check with Le Chatelier's principle: The left-hand side has a lower [Cu²⁺] ( 0.15 mol dm⁻³ ) than standard. Since Cu²⁺(aq) + 2e⁻ ⇌ Cu(s) , lowering [Cu²⁺] shifts the equilibrium to the left, releasing electrons more readily and making the potential less positive ( +0.18 V vs +0.34 V ). The math agrees!

✅ Correct Answer

The concentration of Cu²⁺ is not 1.0 mol dm⁻³ (or is 0.15 mol dm⁻³ ).

❌ Common Errors

• Saying "conditions are not standard" without specifying that it is the concentration that differs (temperature is standard at 25 °C).
• Failing to specify the standard value ( 1.0 mol dm⁻³ ).

1 Mark: Fully correct sequence, phase boundaries ( | ), salt bridge ( || ), and state symbols included.

💡 IUPAC Convention Rules

• LHS = Oxidation (Anode): Reduced form on outer left → Oxidised form. Cu(s) | Cu²⁺(aq)
• Double vertical line: Represents the salt bridge ( || ).
• RHS = Reduction (Cathode): Oxidised form → Reduced form on outer right. Cu²⁺(aq) | Cu(s)
• Phase boundaries: Single line ( | ) separates different phases (solid and aqueous).

❌ Common Errors

• Missing state symbols: The question explicitly says "Include all state symbols". Omitting (s) or (aq) loses the mark.
• Inverting species order: writing Cu²⁺ | Cu || Cu | Cu²⁺ .

✅ Correct Answers

Change in [Cu²⁺] in LHS:
The concentration increases (or [Cu²⁺] ions increase).

Reason why EMF decreases to 0 V:
The [Cu²⁺] ions in the two solutions become equal / the same.

💡 Why does this happen?

• LHS (lower concentration): Acts as the negative electrode (anode). Oxidation occurs: Cu(s) → Cu²⁺(aq) + 2e⁻ . Hence, [Cu²⁺] increases.
• RHS (higher concentration): Acts as the positive electrode (cathode). Reduction occurs: Cu²⁺(aq) + 2e⁻ → Cu(s) . Hence, [Cu²⁺] decreases.
• Current flows until concentrations equalise. When [Cu²⁺]_LHS = [Cu²⁺]_RHS , there is no potential difference between the half-cells, so EMF = 0 V .

❌ Examiner Pitfall

For the reason EMF drops to 0 V, writing "the concentrations become constant" is NOT accepted.

In a standard chemical cell, dynamic equilibrium means concentrations become constant. However, for a concentration cell where both electrodes are identical, you must explicitly state that the concentrations become equal / the same as each other.