AQA A-Level Chemistry Paper 3, 2019: Question 13

1 mark · Easy difficulty · Multiple Choice

Calculate the standard EMF of an electrochemical cell containing iron and copper electrodes using standard electrode potentials.

Practise this question

Question

Question 13 shows standard electrode potentials for two half-cells: Fe2+(aq) + 2e- to Fe(s) with E-standard = -0.44 V, and Cu2+(aq) + 2e- to Cu(s) with E-standard = +0.34 V. The question asks for the EMF of the electrochemical cell Fe(s)|Fe2+(aq)||Cu2+(aq)|Cu(s), with multiple choice options: A (+0.78 V), B (+0.10 V), C (-0.10 V), and D (-0.78 V).
Question text

The Eo values for two electrodes are shown.

Fe2+(aq) + 2 e– → Fe(s) Eo= –0.44 V

Cu2+(aq) + 2 e– → Cu(s) Eo= +0.34 V

What is the EMF of the cell Fe(s)|Fe2+(aq)||Cu2+(aq)|Cu(s)?

[1 mark]

A +0.78 V

B +0.10 V

C –0.10 V

D –0.78 V

Mark scheme

Show the mark scheme Mark scheme table showing that for Question 13, the correct answer is option A, worth 1 mark.

13 A 1

How to answer it

Calculating the EMF of an Electrochemical Cell

📋 What this question tests

This question assesses your ability to apply standard electrode potentials (E⦵) and conventional cell representations to:

  • Identify which electrode undergoes oxidation (anode / left-hand side) and which undergoes reduction (cathode / right-hand side) from IUPAC cell notation.
  • Calculate the standard cell potential (electromotive force, EMF) using the relationship: EMF = E°(RHS) − E°(LHS) .
  • Handle negative values correctly during algebraic subtraction.

Question 13

Electrochemical Cells & Standard Electrode Potentials

✅ Correct Answer

Option A: +0.78 V

Mark Scheme: 1 mark for selecting A.

💡 Key Knowledge

  • Cell Notation: In standard IUPAC cell representation:
    Left | LHS ion || RHS ion | Right
    The electrode on the Left undergoes oxidation; the electrode on the Right undergoes reduction.
  • EMF Formula:
    EMF = E°(Right) − E°(Left)
    or EMF = E°(reduction) − E°(oxidation)
  • A positive cell EMF indicates that the reaction is thermodynamically feasible in the direction written.

📐 Step-by-Step Calculation

  1. Identify Right-Hand Side (RHS) and Left-Hand Side (LHS) electrodes:
    From the cell notation: Fe(s) | Fe²⁺(aq) || Cu²⁺(aq) | Cu(s)
    • LHS: Fe²⁺(aq) / Fe(s) with E⦵ = −0.44 V
    • RHS: Cu²⁺(aq) / Cu(s) with E⦵ = +0.34 V
  2. Substitute values into the cell EMF formula:
    EMF = E°(RHS) − E°(LHS)
    EMF = (+0.34 V) − (−0.44 V)
  3. Evaluate the subtraction:
    EMF = +0.34 + 0.44 = +0.78 V

🧠 Exam Technique

  • RHS minus LHS: Always write down E°(cell) = E°(RHS) − E°(LHS) directly beneath the cell representation so you don't accidentally reverse them.
  • Feasible Reaction Check: The more positive electrode potential (+0.34 V for Cu²⁺/Cu) has a greater tendency to gain electrons (undergo reduction). Because Cu²⁺ is on the right and Fe is being oxidised on the left, the cell reaction is spontaneous, meaning the calculated EMF must be positive (+0.78 V). This immediately rules out C and D.

❌ Common Distractors & Traps

  • Option B (+0.10 V): Arises from adding the two values directly without subtracting: +0.34 + (−0.44) = −0.10 V , or subtracting without double-negative resolution: 0.44 − 0.34 = +0.10 V .
  • Option C (−0.10 V): Simple addition error: (+0.34) + (−0.44) = −0.10 V .
  • Option D (−0.78 V): Calculated E°(LHS) − E°(RHS) = −0.44 − 0.34 = −0.78 V by flipping the formula.

Topics

Physical Chemistry · 3.1.11 Electrode Potentials

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 3, 2019. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.