AQA A-Level Chemistry Paper 1, 2021: Question 9

✅ Correct Answer

Award 1 mark for each point:

• Mark 1: Lithium would react with water / react with the aqueous electrolyte (or water will oxidise Li to Li⁺ / Li will reduce water to H₂).
• Mark 2: The E° value for Li⁺/Li is more negative than that for water (E° Li⁺/Li < E° H₂O/H₂, OH⁻) OR the EMF for the reaction between Li and H₂O is positive (+2.21 V).

💡 Key Knowledge

• Species with more negative E° values are stronger reducing agents and release electrons more readily (shift left).
• Li⁺/Li is −3.04 V, whereas H₂O/H₂ is −0.83 V.
• Since −3.04 V is more negative than −0.83 V, solid Li readily reduces H₂O to H₂(g) and OH⁻(aq), making an aqueous cell chemically unstable and hazardous.

🧠 Exam Technique

Always structure "explain using data" questions in two steps:

1. Chemical statement: What actually happens chemically? (e.g., "Lithium reacts with water").
2. Data comparison: Explicitly quote or compare the numbers (e.g., "E°(Li⁺/Li) is more negative than E°(H₂O/H₂)").

❌ Common Errors

• Stating only that "lithium reacts violently with water" without referencing the E° data loses Mark 2.
• Writing "EMF is negative": the cell reaction between Li and water is spontaneous, meaning its EMF is +2.21 V (positive). The mark scheme specifically notes: "Ignore EMF is negative".

📐 Step-by-Step Calculation

1. Identify the two half-cells and their potentials:
   Positive electrode (reduction, more positive E°):
   ½ I₂(s) + e⁻ → I⁻(aq) | E° = +0.54 V
   Negative electrode (oxidation, more negative E°):
   Li⁺(aq) + e⁻ → Li(s) | E° = −3.04 V
2. Apply the EMF formula:
   EMF = E°(cathode / positive) − E°(anode / negative)
   EMF = +0.54 − (−3.04)
3. Evaluate:
   EMF = +3.58 V

✅ Correct Answer

+3.58 V (or 3.58)

✅ Correct Answer

• Non-standard conditions (Allow: non-aqueous conditions, different concentrations, or temperature not 298 K).

💡 Key Knowledge

Standard electrode potentials (E°) apply strictly under standard conditions:

• Concentration: 1.0 mol dm⁻³ of all ions
• Pressure: 100 kPa (for gases)
• Temperature: 298 K (25 °C)

Commercial pacemaker batteries use solid electrolytes or non-aqueous pastes with ion concentrations far from 1.0 mol dm⁻³, causing the operating cell voltage to deviate from E°.

📐 Step-by-Step Deduction

1. Rule 1: Group 1 metals always have an oxidation state of +1 → Li = +1 .
2. Rule 2: Oxygen in compounds is usually −2 (except peroxides and with F) → 4 oxygens = 4 × (−2) = −8 .
3. Rule 3: The overall compound is neutral (sum of oxidation states = 0):
   (+1) + Cl + (−8) = 0
   Cl − 7 = 0 ⇒ Cl = +7

✅ Correct Answer

+7 (Accept 7 or Roman numeral VII)

✅ 09.5: Positive Electrode Reaction (Discharge)

Cobalt is reduced from (+4) to (+3):

Li⁺ + CoO₂ + e⁻ → LiCoO₂

Also accepted: Li⁺ + CoO₂ + e⁻ → Li⁺CoO₂⁻

💡 Why is this the positive electrode? Reduction happens here (electrons are gained), which attracts electrons from the circuit.

✅ 09.6: Negative Electrode Reaction (Discharge)

Lithium is oxidised from (0) to (+1):

Li → Li⁺ + e⁻

💡 Why is this the negative electrode? Oxidation occurs here, supplying electrons to the external circuit.

❌ Common Errors in Lithium Battery Equations

• Reversing the directions: Writing charging equations instead of discharging equations. Unless specified as "charging", always provide the cell reaction during discharge (use).
• Swapping electrodes: Confusing which electrode is positive and which is negative in a galvanic/voltaic cell. Remember:
  • Negative electrode = Oxidation (Li → Li⁺ + e⁻)
  • Positive electrode = Reduction (Li⁺ + CoO₂ + e⁻ → LiCoO₂)
• Incorrect stoichiometry: Forgetting the incoming Li⁺ ion in the cobalt equation. Both mass and charge must balance.

💡 Specification Recall

You are expected to know the rechargeable lithium-ion cell reactions off by heart for AQA:

• Negative electrode: Li → Li⁺ + e⁻
• Positive electrode: Li⁺ + CoO₂ + e⁻ → LiCoO₂
• Overall cell reaction: Li + CoO₂ → LiCoO₂
• When recharging, reverse all arrows.