Edexcel A-Level Chemistry Paper 1, November 2020: Question 8
18 marks · Hard difficulty · Calculations
Perform calculations and write equations related to a redox titration of potassium manganate(VII) with sodium ethanedioate, including titration table completion, mass determination, and electrode potential reasoning.
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Question text
8 Tablets containing potassium manganate(VII), KMnO4, are dissolved in water forming
an antiseptic solution to treat skin conditions. The manufacturers claim that each
tablet contains 400mg of KMnO4.
To check the claim, the titration procedure outlined was carried out.
Five tablets were dissolved in distilled water to make 100.0 cm3 of solution.
Some of the KMnO4 solution was used to fill a burette.
25.0 cm3 of sodium ethanedioate solution, Na C O (aq), of concentration 0.200 mol dm−3,
22 4
was added to a conical flask and warmed.
Sulfuric acid, of concentration 2 mol dm−3, was also added to the conical flask.
The KMnO4 solution was added to the flask from the burette, until the end-point.
The equation for the reaction between MnO − ions from the KMnO and C O2− ions
44 2 4
from the sodium ethanedioate solution is shown.
16H+(aq) + 2MnO −(aq) + 5C O2−(aq) → 2Mn2+(aq) + 10CO (g) + 8H O(l)
42 4 2 2
(a) Give the colour change at the end-point of the titration.
(1)
(b) (i) Complete the dot-and-cross diagram for the ethanedioate ion.
Show the outer electrons only.
(2)
2−
O O
C C
O O
(ii) Determine the oxidation number of carbon in the ethanedioate ion, C O2−.
(1)
(c) Give the reason why sulfuric acid was also added to the conical flask.
(1)
(d) This redox reaction could be used in an electrochemical cell.
The cell half-equations are
2CO (g) + 2e− ⇌ C O2−(aq)
22 4
8H+(aq) + MnO −(aq) + 5e− ⇌ Mn2+(aq) + 4H O(l)
20 4 2
Write a cell diagram for this cell using the conventional representation.*P62668A02024*
(2)
(e) The results of the titration are shown.
Run Trial 1 2 3
Final volume / cm3 17.50 34.10 17.20 34.10
Initial volume / cm3 0.00 17.30 0.00 17.20
Titre / cm3 17.50 17.20
Concordant titres ( )
Mean titre / cm3
(i) Complete the table.
(2)
(ii) The equation for the reaction between MnO − ions from the KMnO and C O2− ions
44 2 4
from the sodium ethanedioate solution is shown.
16H+(aq) + 2MnO −(aq) + 5C O2−(aq) → 2Mn2+(aq) + 10CO (g) + 8H O(l)
42 4 2 2
Use this equation and your mean titre from (e)(i) to calculate the mass, in mg,
of KMnO4 in one tablet.
Give your answer to an appropriate number of significant figures.
(5)
(iii) A textbook suggested the conical flask should be heated during the titration,
as the reaction between the MnO − ions and the C O 2− ions is slow.
42 4
Use these electrode potentials and your knowledge of homogeneous catalysis
to deduce why the heating is very important at the start of the titration, but
less important as the titration proceeds. Justify your answer.
You may include equations in your justification.
O 21
Electrode system*P62668A02124*E/ V
2CO (g) + 2e− ⇌ C O 2−(aq) +0.64
22 4
Mn3+(aq) + e– ⇌ Mn2+(aq) +1.49
MnO −(aq) + 8H+(aq) + 5e− ⇌ Mn2+(aq) + 4H O(l) +1.51
(4)
… *P62668A02224*
(Total for Question 8 = 18 marks)
Mark scheme
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How to answer it
Redox Titration & Catalysis Study Guide
What this question tests
This comprehensive multi-part question assesses core physical and inorganic chemistry concepts: redox titrations involving manganate(VII), stoichiometry and mass calculations, bonding (dot-and-cross diagrams), oxidation numbers, electrochemical cell representation (conventional notation), and heterogeneous/homogeneous catalysis mechanisms.
Titration Colour Change
✅ Correct Answer
Colourless to (permanent pale/light) pink.
❌ Common Errors
Writing "purple for pink" or stating "pink to colourless" (forgetting that MnO₄⁻ is added from the burette into the conical flask).
Ethanedioate Ion Bonding & Oxidation Numbers
✅ Correct Answer
(i) Dot-and-cross: Double C=O bonds on left and right-hand sides with correct lone pairs and bonding pairs matching an ethanedioate structure (C-C single bond, 4 outer electrons on each oxygen).
(ii) Oxidation Number: (+)3
💡 Key Knowledge
For oxidation numbers, always include the sign before the number (+3, not just 3). Roman numerals (III) are also accepted by Edexcel.
Role of Sulfuric Acid
✅ Correct Answer
To provide the hydrogen ions (H⁺) needed as a reactant in the balanced equation, and to prevent the formation of brown manganese(IV) oxide (MnO₂).
❌ Common Errors
Vague statements like "acts as a catalyst" or "ensures the reaction goes to completion" do not score. You must link it directly to providing H⁺ ions required by the stoichiometry.
Electrochemical Cell Diagram
✅ Correct Answer
Pt(s) | C₂O₄²⁻(aq), 2CO₂(g) || MnO₄⁻(aq) + 8H⁺(aq), Mn²⁺(aq) + 4H₂O(l) | Pt(s)
A salt bridge (double vertical line) separates the two half-cells, platinum electrodes are included on both ends, and phases are correctly separated by single vertical lines or commas where appropriate.
🧠 Exam Technique
Remember standard cell convention: Reduced | Oxidised || Oxidised | Reduced. Platinum (Pt) inert electrodes are vital since all species in the half-cells are aqueous/gaseous.
Processing Titration Data & Tablet Purity
✅ Correct Answer
(i) Concordant titres & mean: Titres chosen are Run 2 (17.20) and Run 3 (17.20) — or Trials 1, 2, 3 depending on table structure. Mean titre = 16.85 cm³ (using accurate concordant values).
(ii) Final mass calculation: 380 mg or 375 mg (to 2 or 3 significant figures).
📐 Step-by-Step Calculation
- Moles of Na₂C₂O₄ added: (25.0 / 1000) × 0.200 = 5.00 × 10⁻³ mol
- Moles of KMnO₄ in mean titre: From equation, 2 MnO₄⁻ react with 5 C₂O₄²⁻. So, moles KMnO₄ = (5.00 × 10⁻³) × (2 / 5) = 2.00 × 10⁻³ mol in 16.85 cm³.
- Moles of KMnO₄ in total 100.0 cm³ solution: (2.00 × 10⁻³) × (100.0 / 16.85) = 0.01186 mol
- Mass of KMnO₄ in 5 tablets: Moles × M_r (158.0) = 0.01186 × 158 = 1.8754 g
- Mass per 1 tablet in mg: (1.8754 / 5) × 1000 = 375.08 mg = 380 mg (to 2 sf) or 375 mg (to 3 sf).
Homogeneous Catalysis & Kinetics
✅ Correct Answer
The reaction is slow initially because negatively charged ions (MnO₄⁻ and C₂O₄²⁻) repel each other, requiring a high activation energy. Once enough Mn²⁺ ions are formed, they autocatalyse the reaction:
- Mn²⁺ ions reduce MnO₄⁻ to form intermediate Mn³⁺ ions (E° is more negative/favourable).
- Mn³⁺ ions then oxidise C₂O₄²⁻ back into Mn²⁺ and CO₂.
💡 Key Knowledge
This is a classic example of homogeneous catalysis where the catalyst (Mn²⁺) is in the same physical state as the reactants. Heating at the start provides energy to overcome the initial electrostatic repulsion barrier.
Topics
Physical Chemistry · Inorganic Chemistry · Core Practicals · Topic 3: Redox I · Topic 5: Formulae, Equations and Amounts of Substance · Topic 14: Redox II · Topic 15: Transition Metals · Core Practical 3: Find the concentration of a solution of hydrochloric acid · Topic 2: Bonding and Structure
Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 1, November 2020. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.