Edexcel A-Level Chemistry Paper 1, June 2023: Question 10

15 marks · Hard difficulty · Synoptic Questions

Calculate the concentration of dissolved oxygen in river water using titration data, explain its solubility and the instability of Mn3+ ions using electrode potentials.

Practise this question

Question

A multi-part chemistry question about manganese compounds used to determine dissolved oxygen in water samples. Part (a) asks to draw a dot-and-cross diagram for an oxygen molecule. Part (b) asks to comment on the solubility of oxygen in water based on intermolecular forces. Part (c) outlines a Winkler titration method involving several equations and requires calculating the concentration of dissolved oxygen in g dm^-3 and in ppm. Part (d) provides a table of half-cell electrode systems and asks to explain why Mn3+ ions are unstable in aqueous solution.
Question text

10 Manganese compounds can be used to determine the amounts of dissolved

molecular oxygen in water samples.

(a) Draw the dot‑and‑cross diagram for an oxygen molecule, O2 .

Show outer shell electrons only.

(1)

(b) The solubility of oxygen in water under standard conditions

is 1.22 × 10–3 mol dm–3.

Comment on this value by considering the type and strength of the

intermolecular forces in

• pure water

• pure oxygen

• a mixture of water and oxygen.

Detailed descriptions of the forces involved are not required.

(4)

(c) The amount of dissolved oxygen in a sample of river water was found using the

process outlined.

• excess alkaline manganese(II) sulfate, MnSO*P71912A02328*, was added to a 150 cm3 sample

of river water

• the Mn2+ ions reacted with the dissolved oxygen forming a precipitate of

manganese(IV) oxide hydroxide

2Mn2+(aq) + O (aq) + 4OH–(aq) → 2MnO(OH) (s)

• the precipitate was then dissolved using excess sulfuric acid, forming

Mn4+(aq) ions

MnO(OH) (s) + 4H+(aq) → Mn4+(aq) + 3H O(l)

• excess potassium iodide solution was then added, forming iodine

Mn4+(aq) + 2I–(aq) → Mn2+(aq) + I (aq)

• the liberated iodine was then titrated with sodium thiosulfate solution,

Na S O (aq), of concentration 0.00518 mol dm–3

22 3

I (aq) + 2S O2–(aq) → 2I–(aq) + S O2–(aq)

22 3 4 6

• the mean volume of the titre of Na S O (aq) was 34.20 cm3.

22 3

(i) Calculate the concentration of dissolved oxygen in the sample of river water,

in g dm–3.

(5)

You may use this space to continue your answer to 10(c)(i).

*P71912A02428*

(ii) The concentration of oxygen in water is often expressed in parts per million

(ppm), where 1 ppm equals 1 g of solute dissolved in 1 × 106 g of solvent.

Calculate the concentration of the oxygen in the sample of river water in ppm.

Assume the density of the river water is 1.00 g cm–3.

(1)

(d) Some data is shown for electrode systems involving the Mn3+(aq) ion.

Half‑cell Electrode system E d / V

A MnO (s) + 4H+(aq) + e– Mn3+(aq) + 2H O(l) +0.95

B Mn3+(aq) + e– Mn2+(aq) +1.51

Explain why Mn3+ ions are unstable in aqueous solution.

Include an equation and the type of reaction that occurs.

(4)

*P71912A02528*

(Total for Question 10 = 15 marks)

Mark scheme

Show the mark scheme The mark scheme provides detailed guidance and marking points for all parts of Question 10. Part (a) shows the expected dot-and-cross diagram for O2. Part (b) lists marking points for discussing London forces and hydrogen bonds. Parts (c)(i) and (c)(ii) provide step-by-step example calculations for determining oxygen concentration in g dm^-3 and ppm. Part (d) details the required points regarding disproportionation, cell potentials, and the overall equation for the instability of Mn3+ ions.

(Total for Question 9 = 13 marks)

Question

Answer Additional guidance Mark

number

10(a) An answer that makes reference to the following points (1)

• dot-and-cross diagram for oxygen

Allow diagram without circles

Allow all dots or all crosses

Allow bond pairs shown axially X X

O O

If circles are drawn, allow electrons anywhere in the overlap

region / on the lines in the overlap region

Ignore inner shells

Question

Answer Additional Guidance Mark

Number

10(b) An answer that makes reference to the following points Allow abbreviations such as LDF / VDW as (4)

this has been penalised in q4

Allow H-bonds for hydrogen bonds

• the oxygen is only sparingly soluble (in water) (1) Allow not very soluble / slightly soluble /

low solubility / doesn’t dissolve very well

• oxygen has (only weak) London forces (between molecules) (1) Do not award any additional forces / bonds

between O2 molecules

• water has (London forces, dipole- dipole attractions and strong)

Allow this shown on a diagram

hydrogen bonds (between molecules) (1)

• the (intermolecular) forces between oxygen and water would be Allow the resultant forces between them

London forces would be London forces, which are not

and strong enough to overcome the hydrogen

which are weaker than the hydrogen bonds (in water) (1) bonds (in water)

Allow oxygen molecules are not able to

disrupt the hydrogen bonds between water

molecules

Allow dispersion / temporary dipole-induced

dipole / instantaneous dipole / van der Waals

as alternatives to London forces throughout

Ignore just ‘results in fewer hydrogen bonds

between water molecules’

Ignore hydrogen bonds between oxygen and

water

Question

Answer Additional Guidance Mark

Number

10(c)(i) Example of calculation (5)

(34.2 ÷ 1000) × 0.00518 = 1.7716 × 10−4 (mol)

• calculation of amount of Na2S2O3(aq) (1)

• calculation of amount of I (aq) (1) = (1.7716 × 10−4) ÷ 2 = 8.8578 × 10−5 (mol)

• calculation of amount of O2(aq) (1) = (8.8578 × 10−5) ÷ 2 = 4.4289 × 10−5 (mol)

• calculation of concentration of O in mol dm−3 (aq) (1)

2 = 4.4289 × 10−5 × (1000/150) = 2.9526 × 10−4 (mol dm−3)

• calculation of concentration of O in g dm−3 (aq) (1) = 2.9526 10−4 × 32 = 9.4483 × 10−3 / 0.0094483 (g dm−3)

Allow steps in different orders

Allow TE throughout

Allow intermediate values quoted as fractions

Ignore SF except 1 SF

Ignore units even if incorrect

Correct answer, with or without working, scores (5)

Penalise incorrect rounding / truncation once only

Question

Answer Additional Guidance Mark

Number

10(c)(ii) Example of calculation (1)

• calculation of concentration of O in ppm 9.4483 × 10−3 × (1000000 ÷ 1000)

= 9.4483 (ppm)

Allow TE from (i)

Ignore SF except 1 SF

Question

Answer Additional Guidance Mark

Number

10(d) An answer that makes reference to the following points Allow use of oxidation numbers instead of (4)

formulae of ions / molecules

Allow Mn4+ to represent manganese(IV) in

MnO2

• (some) Mn3+ (ions from half-cell B) will oxidise (other) Mn3+ (ions Allow Mn3+ is oxidised in one (half) equation

from half-cell A) and reduced in the other

Allow Mn3+ is an oxidising agent in one (half)

and

(some) Mn3+ (ions) will reduce (other) Mn3+ ions (1) equation and a reducing agent in the other

• as Eo for half-cell B is more positive / higher than Eo for half-cell A M2 can be shown by anti-clockwise rule, eg

in table

or Ignore just Eo is positive so reaction is

cell

as Eo = (+)0.56 V (1)

cell feasible

• 2Mn3+(aq) + 2H O(l) → Mn2+(aq) + MnO (s) + 4H+(aq) (1) Allow reversible arrow

Ignore state symbols

Ignore just redox

• disproportionation (reaction) (1)

(Total for Question 10 = 15 marks)

TOTAL FOR PAPER = 90 MARKS

How to answer it

Manganese Compounds & Dissolved Oxygen Analysis

What this question tests

This multi-topic Edexcel A-Level Chemistry question assesses bonding (dot-and-cross diagrams), intermolecular forces and solubility, multi-step redox titrations with concentration conversions (g dm⁻³ and ppm), and electrode potentials applied to disproportionation reactions. Mastering stoichiometry and mole ratios is essential for full marks.

Part (a): Dot-and-Cross Diagram for O₂

(1 Mark)

✅ Correct Answer

A double covalent bond between two oxygen atoms showing 4 shared electrons in the overlap region (2 pairs) and 4 non-bonding (lone pair) electrons on the outer shell of each oxygen atom.

💡 Key Knowledge

Oxygen is in Group 6 and needs 2 electrons to complete its octet. It shares two pairs of electrons to form a double covalent bond ( O=O ).

Mark scheme guidance: Allow dot-and-cross diagrams using all dots or all crosses, or mixed. Ignore inner shells.

Part (b): Solubility of Oxygen & Intermolecular Forces

(4 Marks)

✅ Correct Answer

  • Oxygen is only sparingly / slightly soluble in water.
  • Pure oxygen has only weak London forces between molecules.
  • Pure water has London forces, dipole-dipole attractions, and strong hydrogen bonds between molecules.
  • The intermolecular forces between oxygen and water would be London forces, which are weaker than the hydrogen bonds in water (so oxygen cannot disrupt water's hydrogen-bonding network effectively).

🧠 Exam Technique

To score all 4 marks, you must explicitly describe the forces present in both individual substances (pure oxygen and pure water) AND describe the forces in the mixture, explaining why dissolution is unfavourable.

Mark scheme guidance: 1 mark per distinct marking point. Accept abbreviations such as LDF or VDW for London forces.

Part (c)(i): Titration Calculation for Dissolved Oxygen

(5 Marks)

📐 Step-by-Step Calculation

  1. Find moles of thiosulfate ( Na₂S₂O₃ ):
    Moles = Volume (dm³) × Concentration
    = (34.20 ÷ 1000) × 0.00518 = 1.7716 × 10⁻⁴ mol
  2. Find moles of iodine ( I₂ ):
    From equation 4: 2 mol S₂O₃²⁻ reacts with 1 mol I₂.
    Moles of I₂ = (1.7716 × 10⁻⁴) ÷ 2 = 8.8578 × 10⁻⁵ mol
  3. Find moles of oxygen ( O₂ ) in 150 cm³:
    Tracing back through the equations:
    1 mol O₂ forms 1 mol Mn(IV), which forms 1 mol I₂.
    Therefore, moles of O₂ = 8.8578 × 10⁻⁵ mol
  4. Calculate concentration of O₂ in mol dm⁻³:
    Conc = (Moles ÷ Volume in cm³) × 1000
    = (8.8578 × 10⁻⁵ ÷ 150) × 1000 = 2.9526 × 10⁻⁴ mol dm⁻³
  5. Convert concentration to g dm⁻³:
    Molar mass of O₂ = 16.0 × 2 = 32.0 g mol⁻¹.
    Conc in g dm⁻³ = 2.9526 × 10⁻⁴ × 32.0 = 9.448 × 10⁻³ g dm⁻³ (or 0.00945 g dm⁻³)

❌ Common Calculation Traps

Forgetting stoichiometry ratios (dividing by 2 at the wrong step) or failing to scale the 150 cm³ river water sample up to 1 dm³ (multiplying by 1000/150) are the most frequent causes of lost marks.

🧠 Significant Figures

Data is provided to 3 significant figures. Give your final answer to 2, 3, or 4 significant figures. Avoid rounding intermediate steps.

Mark scheme guidance: Correct answer with or without working scores full marks (5 marks). Penalise incorrect rounding only once.

Part (c)(ii): Converting to Parts Per Million (ppm)

(1 Mark)

✅ Correct Answer

9.45 ppm (derived from 9.4483 × 10⁻³ g dm⁻³ multiplied by 1000 to convert mg L⁻¹ / ppm, given density of water is 1.00 g cm⁻³).

💡 Key Knowledge

1 ppm is defined as 1 mg of solute per 1 kg (10⁶ mg) of solvent. For water (density 1.00 g cm⁻³), 1 mg dm⁻³ = 1 ppm . Multiply your g dm⁻³ answer by 1000.

Mark scheme guidance: Allow Error Carried Forward (TE) from part (c)(i).

Part (d): Electrode Potentials & Instability of Mn³⁺

(4 Marks)

✅ Correct Answer

  • Explanation: Mn³⁺ ions undergo disproportionation because the E° value for half-cell B (+1.51 V) is more positive than half-cell A (+0.95 V). Therefore, one Mn³⁺ is oxidised to MnO₂ while another is reduced to Mn²⁺.
  • Equation: 2Mn³⁺(aq) + 2H₂O(l) ⇌ Mn²⁺(aq) + MnO₂(s) + 4H⁺(aq) (allow reversible or irreversible arrow).
  • Type of reaction: Disproportionation.

❌ Common Errors

Students often lose marks by stating the reaction is "just redox" instead of specifically identifying disproportionation (where the same element is both simultaneously oxidised and reduced).

Mark scheme guidance: 1 mark for relative E° values / thermodynamic feasibility, 1 mark for stating Mn³⁺ is both oxidised and reduced, 1 mark for correct chemical equation, 1 mark for identifying disproportionation.

Topics

Physical Chemistry · Inorganic Chemistry · Topic 2: Bonding and Structure · Topic 3: Redox I · Topic 5: Formulae, Equations and Amounts of Substance · Topic 14: Redox II

Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 1, June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.