Edexcel A-Level Chemistry Paper 1, June 2018: Question 5

7 marks · Medium difficulty · Short Open Response

Construct an electrochemical cell diagram, state salt bridge composition, write a redox equation, and calculate standard cell potential.

Practise this question

Question

An exam question with parts (a) to (c) regarding an electrochemical cell made from copper and manganese half-equations. Part (a) requires labeling a diagram of the electrochemical cell apparatus showing two beakers, a salt bridge, electrodes, and a voltmeter. Parts (b)(i) and (b)(ii) ask about the chemical in the salt bridge and why a metal wire cannot replace it. Parts (c)(i) and (c)(ii) ask for the overall ionic equation and the calculation of the standard electrode potential for the Mn3+/Mn2+ half-cell.
Question text

5 An electrochemical cell is made from the electrode systems represented by these

half-equations.

Cu2+(aq) + 2e− Cu(s)

Mn3+(aq) + e− Mn2+(aq)

The Ecell value is measured using the apparatus shown.

copper

1 mol dm−3

Cu2+(aq)

salt …

bridge

(a) Complete the diagram by adding labels on the dotted lines provided.

(3)

(b) A salt bridge is used to connect the two half-cells.

(i) State what chemical is contained in the salt bridge.

(1)

(ii) Give a possible reason why the salt bridge cannot be replaced by an

unreactive metal wire.

(1)

(c) In this cell, the copper is oxidised and Ecell = +1.15V.

Cu2+(aq) + 2e− Cu(s) E = +0.34 V

Mn3+(aq) + e− Mn2+(aq)

(i) Write the overall ionic equation for the reaction taking place.

8 State symbols are not required.

*P52302RA0828* (1)

(ii) Calculate the value of the standard electrode potential for the

Mn3+(aq) Ι Mn2+(aq) half-cell.

(1)

(Total for Question 5 = 7 marks)

Mark scheme

Show the mark scheme The mark scheme for question 5 detailing acceptable answers for labeling the diagram, identifying salt bridge components (such as potassium nitrate), explaining the role of ions versus electrons, writing the correct redox equation, and calculating the standard electrode potential as +1.49 V.

Question

Acceptable Answer Additional Guidance Mark

Number

5(a) Allow potentiometer / Wheatstone (3)

(high resistance) voltmeter (1) bridge / just ‘V’

Ignore high voltage

Do not award voltameter

platinum /Pt (electrode) (1) Ignore just ‘inert metal’

Do not award manganese / Mn

manganese(II) and manganese(III) ions / Allow any named manganese(II) salt

Mn2+ and Mn3+ (1) and manganese(III) salt

Ignore concentration and units

Question

Acceptable Answer Additional Guidance Mark

Number

5(b)(i) If name and formula are given, both must be correct (1)

potassium nitrate / KNO3

If more than one substance given, all must be

correct

Allow

potassium chloride / KCl

sodium nitrate / NaNO3

sodium chloride / NaCl

ammonium nitrate / NH4NO3

ammonium chloride / NH4Cl

Ignore concentration

Question

Acceptable Answer Additional Guidance Mark

Number

5(b)(ii) Allow any indication of movement for flow in all (1)

wire does not allow the flow of ions points

or

wire (only) allows flow of electrons Allow the salt bridge donates / removes ions (to

or balance the charges in the solution and the wire

salt bridge allows flow of ions does not do this)

or

salt bridge does not allow the flow of electrons Ignore just ‘the circuit is not complete’

or

a flow of ions is needed to complete the circuit Ignore references to changes in potential difference

/ Eo / Eo

or cell

ions (need to) flow between the half-cells /

between the solutions

Question

Acceptable Answer Additional Guidance Mark

Number

5(c)(i) Example of equation (1)

correct equation 2Mn3+ + Cu → 2Mn2+ + Cu2+

Allow multiples

Allow ⇌ provided equation is written in

the direction shown

Ignore state symbols, even if incorrect

Ignore cancelled electrons e.g.

2Mn3++ Cu + 2e → 2Mn2+ + Cu2+ + 2e

Do not award equation with uncancelled

electrons

Question

Acceptable Answer Additional Guidance Mark

Number

5(c)(ii) Stand alone mark (1)

Eo = 1.15 – (−0.34) = (+)1.49 (V)

Correct answer with no working scores

the mark

(Total for Question 5 = 7 marks)

How to answer it

Electrochemical Cells and Standard Electrode Potentials

Edexcel A-Level Chemistry — Exam Mastery Guide

What this question tests

This question assesses your understanding of electrochemical cell setup, the function of salt bridges versus external wires, redox equation balancing, and applying the standard cell potential equation ( E(cell) = E(right) - E(left) ).

Part (a): Completing the Electrochemical Cell Diagram

3 Marks

✅ Correct Answers

  • Top box: (high resistance) voltmeter
  • Middle label (electrode): platinum / Pt (electrode)
  • Bottom labels (solution): manganese(II) and manganese(III) ions (or Mn²⁺ and Mn³⁺ )

💡 Key Knowledge

An ion-ion half-cell requires an inert conductor (platinum) because neither species is a solid metal that can act as an electrode. Both oxidation states of the transition metal must be present in solution.

❌ Common Errors

  • Writing "voltameter" instead of voltmeter.
  • Suggesting solid manganese metal as the electrode instead of platinum.
  • Omitting one of the two manganese ion oxidation states from the solution label.
Mark breakdown: 1 mark for voltmeter, 1 mark for Pt electrode, 1 mark for both Mn²⁺ and Mn³⁺ ions.

Part (b): The Salt Bridge

2 Marks total (1 + 1)

(i) Chemical in the salt bridge

✅ Correct Answer

potassium nitrate / KNO₃ (or other unreactive ionic substances like KCl , NH₄NO₃ )

🧠 Exam Technique

Always state both the name and correct chemical formula if unsure, but ensure they match. If multiple substances are written, all must be valid.

(ii) Why a metal wire cannot replace the salt bridge

✅ Correct Answers

  • A wire only allows the flow of electrons, not ions.
  • A salt bridge allows the flow of ions to complete the circuit between solutions.

❌ Common Errors

Stating vaguely that "the circuit is not complete" without explaining why (i.e., failure to mention the movement of ions vs. electrons).

Mark breakdown: 1 mark for (b)(i), 1 mark for (b)(ii).

Part (c): Calculations and Overall Equations

2 Marks total (1 + 1)

(i) Overall Ionic Equation

✅ Correct Answer

2Mn³⁺ + Cu → 2Mn²⁺ + Cu²⁺

(Multiples allowed; state symbols not required)

🧠 Exam Technique

The question states copper is oxidised ( Cu → Cu²⁺ + 2e⁻ ), meaning manganese(III) must be reduced ( Mn³⁺ + e⁻ → Mn²⁺ ). Multiply the manganese half-equation by 2 to balance electrons before combining.

❌ Common Errors

Leaving uncancelled electrons in the final equation or failing to balance the moles of electrons between the two half-cells.

(ii) Calculating Standard Electrode Potential

📐 Step-by-Step Calculation

  1. Recall the equation: E(cell) = E(right) - E(left)
  2. Identify components: Copper is oxidised, meaning it acts as the negative (left) electrode. Manganese is the positive (right) electrode.
  3. Rearrange for E(right): E(right) = E(cell) + E(left)
  4. Substitute values: E = 1.15 + (-0.34) = +1.49 V (or 1.15 - (-0.34) depending on how half-cells are subtracted).

✅ Final Answer

E = +1.49 V (accept 1.49 )

Mark breakdown: 1 mark for the correct numerical calculation and sign. Stand-alone mark.

Topics

Physical Chemistry · Core Practicals · Topic 14: Redox II · Core Practical 10: Construct electrochemical cells and measure electrode potentials

Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 1, June 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.