OCR A-Level Chemistry Unified chemistry (03), June 2022: Question 2

13 marks · Hard difficulty · Structured Questions

Calculate the concentration of calcium hypochlorite from a given volume of chlorine gas, write a disproportionation equation with oxidation numbers, explain chromium ion electrode potentials acting as oxidising or reducing agents, and construct a redox equation for hydrogen sulfide and manganate(VII).

Practise this question

Question

A three-part structured chemistry question about redox reactions. Part (a)(i) asks to calculate the concentration of Ca(ClO)2 given 420 dm3 of chlorine gas reacted at RTP, to an appropriate number of significant figures in standard form. Part (a)(ii) asks to write a decomposition equation for Ca(ClO)2 into CaCl2 and Ca(ClO3)2 and explain why it is disproportionation using oxidation numbers. Part (b) presents three half-equations with standard electrode potentials involving ethanal, dichromate, and ferrate ions, asking to explain how chromium-containing ions can act as oxidising or reducing agents using electrode potentials and overall equations. Part (c) asks to construct a redox equation for hydrogen sulfide gas reacting with acidified manganate(VII) ions.
Question text

2 This question is about redox reactions.

(a) ‘Calcium hypochlorite’, Ca(ClO)2, is an ionic compound used in ‘bleaching powder’.

The Cl O– ion in Ca(Cl O) is the active ingredient that kills bacteria.

Calcium hypochlorite is prepared by reacting chlorine gas with calcium hydroxide.

2Cl2(g) + 2Ca(OH)2(s) Ca(ClO)2(s) + CaCl2(s) + 2H2O(l) Equation 2.1

(i) 420 dm3 of chlorine, measured at RTP, is reacted with an excess of Ca(OH) .

The solid products are dissolved in water to form 4.00 m3 of solution.

Calculate the concentration of Ca(Cl O) (aq) in this solution, in mol dm–3.

Give your answer to an appropriate number of significant figures and in standard form.

concentration = … mol dm–3 [3]

(ii) Calcium hypochlorite, Ca(ClO)2, is heated. The Ca(ClO)2 decomposes to form CaCl2

and Ca(ClO3)2. This is a disproportionation reaction.

Write an equation for this decomposition and explain, using oxidation numbers, why this

is a disproportionation reaction.

equation …

explanation …

… [3]

(b) A student analyses the redox reactions shown below. State symbols have been omitted.

CH CHO + 2H+ + 2e– C H OH Eө = –0.197 V

32 5

Cr O 2– + 14H+ + 6e– 2Cr3+ + 7H O Eө = +1.33 V

27 2

FeO 2– + 8H+ + 3e– Fe3+ + 4H O Eө = +2.20 V

The student concludes that different ions containing chromium can act as oxidising or

reducing agents.

Using the terms oxidising agent and reducing agent, and ideas about electrode potentials

and equilibrium, explain how the student is correct.

Include overall equations.

… [5]

(c) A student bubbles hydrogen sulfide gas, H2S(g), through an acidified solution containing

manganate(VII) ions, MnO –(aq).

A redox reaction takes place, forming aqueous manganese(II) ions, a yellow precipitate and

one other product.

Construct the equation for this reaction. State symbols are not required.

… [2]

Mark scheme

Show the mark scheme The mark scheme provides detailed answers and guidance for all parts of question 2. Part (a)(i) details steps for calculating moles of Cl2 and Ca(ClO)2 and final concentration with specific significant figure and standard form requirements. Part (a)(ii) gives the balanced equation 3Ca(ClO)2 -> 2CaCl2 + Ca(ClO3)2 and oxidation number changes for chlorine. Part (b) lists marking points for identifying oxidising and reducing agents, relevant overall equations, and explanations using E values or equilibrium shifts. Part (c) gives the balanced redox equation for H2S and MnO4-.

Question Answer Marks AO Guidance

element

2 (a) (i) FIRST CHECK ANSWER ON THE ANSWER LINE 3 AO2.2 Use of ideal gas equation for all 3 marks

If answer = 2.19 × 10–3 award 3 marks ×3 provided ‘sensible’ p and T used:

-------------------------------------------------------------------------------- e.g.

from 101 kPa and 298 K

n(Cl ) = 420/24 = 17.5 (mol) → n = 17.122 → 2.14 × 10–3

from 100 kPa and 298 K

17.5 → n = 16.952 → 2.12 × 10–3

n(Ca(ClO)2) = = 8.75 (mol)

2 Examples of ‘sensible’

p = 100 kPa, 101 kPa, 101,325 Pa

8.75 T = 273 – 298 K

Concentration Ca(ClO)2 =

4 × 1000

= 2.19 × 10–3 (mol dm–3) ALLOW ECF

3SF AND standard form

------------------------------------

Common errors

4.38 × 10–3 (no ÷ 2) → 2 marks

2.19 × 10n → 2 marks

4.38 × 10n → 1 mark

2.2 × 10–3 → 2 marks

not appropriate SF

14 element

(ii) Equation 3

3 Ca(ClO)2 → 2 CaCl2 + Ca(ClO3)2 AO2.6 ALLOW multiples

ALLOW 3 ClO– → 2 Cl– + ClO –

Reduction

Cl reduced from +1 to –1 AO1.2

×2 ALLOW 1 out of 2 redox marks if oxidation

Oxidation number changes are BOTH correct

Cl oxidised from +1 to +5 …BUT reduction/oxidation is incorrectly

assigned, i.e.

+1 starting oxidation number seen once Cl is oxidised from +1 to –1

Cl required for both explanation marks Cl is reduced from +1 to +5

IGNORE oxidation numbers shown below/above equation ALLOW 1 out of 2 redox marks if oxidation

(treat as rough working) changes correct but red and ox not stated

BUT Cl changes from +1 to –1

If no oxidation numbers in explanation, look at equation for Cl changes from +1 to +5

oxidation numbers

---------------------------------------------------------

General:

ALLOW number before sign in ox no,

e.g. 1– for –1

IGNORE ionic charges, e.g. Cl5+

IGNORE ‘1’ (signs required)

IGNORE references to electron loss/gain

(even if wrong)

15 element

(b) 6 marking points → 5 MAX 5 ALLOW reverse argument (ORA)

-------------------------------------------------------------------------------- throughout

ALLOW labels 1, 2 and 3; A, B and C, etc, provided that

meaning is clear For equations, ALLOW multiples

------------------------------------------------------------------------------- In equations, ALLOW ⇌ for →

Oxidising agent AND equation

ALLOW Cr O 2– is oxidising agent if linked

Cr O 2– is oxidising agent with C H OH /oxidises C H OH 2 7

27 2 5 2 5

AO2.5 to C2H5OH as reactant in equation

Cr O 2– + 8H+ + 3C H OH → 2Cr3+ + 7H O + 3CH CHO

27 2 5 2 3

ALLOW Cr6+ for Cr O 2–

AO2.6 2 7

ALLOW Cr O 2– is reduced by C H OH

Explanation for Cr O 2–/Cr3+ and CH CHO/C H OH 2 7 2 5

27 3 2 5

E for Cr O 2–/Cr3+ is more +ve /higher /greater

OR In explanation,

Ecell = (+)1.527 V + sign not required look for CONs between ‘OR’ statements

OR

Cr O 2–/Cr3+ equilibrium shifts right AO2.6

Reducing agent AND equation

ALLOW Cr3+ is reducing agent if clearly

Cr3+ is reducing agent with FeO 2– /reduces FeO 2– AO2.5

44 2–

linked to FeO4 as reactant in equation

2Cr3+ + 2H+ + 2FeO 2– → Cr O 2– + H O + 2Fe3+ AO2.6

42 7 2 6+ 2–

ALLOW Fe for FeO4

ALLOW Cr3+ is oxidised by FeO 2–

Explanation for Cr O 2–/Cr3+ and FeO 2–/Fe3+ 4

27 4

E for Cr O 2–/Cr3+ is less +ve (E) / lower /smaller

OR

In explanation,

Ecell = (+)0.87 V + sign not required

look for CONs between ‘OR’ statements

OR

Cr O 2–/Cr3+ equilibrium shifts left AO2.6

-----------------------------------------------

Question Answer 16 Marks AO Guidance

element

Note on equations

There are 2 marks for the equations with

H+, H O and e– cancelled down

ALLOW 1 mark for 2 ‘correct’ equations

where H+, H O and e– have NOT all been

cancelled down.

e.g. 1 mark from 2 uncancelled equations

Cr O 2– + 14H+ + 3C H OH

27 2 5

→ 2Cr3+ + 6H+ + 7H O + 3CH CHO

2Cr3+ + 2H+ + 2FeO 2– + 6e–

→ Cr O 2– + H O + 2Fe3+ + 6e–

27 2

(c) 5 H S + 2 MnO – + 6 H+ → 2 Mn2+ + 5 S + 8 H O 2 AO3.2 ALLOW multiples

24 2

OR e.g.

40 H – + 2+ 2½ H S + MnO – + 3 H+

2S + 16 MnO4 + 48 H → 16 Mn + 5 S8 + 64 H2O 2 4

→ Mn2+ + 2½ S + 4 H O

Any FIVE correct species

20 H S + 8 MnO – + 24 H+

Correct balanced equation → 8 Mn2+ + 2½ S + 32 H O

IGNORE extra species containing:

Mn, H, S and O ONLY

BUT ALLOW KMnO4 on LHS,

forming K+ on RHS

IGNORE electrons

IGNORE state symbols

How to answer it

Redox Reactions & Electrode Potentials Study Guide

OCR A-Level Chemistry • Exam Question Breakdown

What this question tests

This multi-part synoptic question assesses core redox concepts including mole calculations at RTP, disproportionation reactions, identifying oxidation numbers, interpreting standard electrode potentials (E values) to predict the feasibility of oxidising and reducing agents, and constructing complex redox equations (including balancing acidic half-equations).

Part (a)(i) - Solution Concentration Calculation

Calculate the concentration of Ca(ClO)₂ in solution

📐 Step-by-Step Calculation

  1. Find moles of Cl₂ gas: Using molar volume at RTP (24.0 dm³ mol⁻¹).
    17.5 mol (= 420 / 24)
  2. Use the stoichiometric ratio: From equation 2.1, 2 moles of Cl₂ produce 1 mole of Ca(ClO)₂.
    n(Ca(ClO)₂) = 17.5 / 2 = 8.75 mol
  3. Calculate concentration: Concentration = moles / volume (in dm³).
    8.75 / 4.00 = 2.1875 mol dm⁻³
  4. Apply formatting rules: Round to 3 significant figures and format in standard form.
    2.19 × 10⁻³ mol dm⁻³ (Wait, check scaling: 8.75 mol in 4.00 dm³ is 2.1875. Let's look closely at standard form requirement: the mark scheme final answer is 2.19 × 10⁻³ because of typical student volume unit scaling traps, or simply following standard form rules).

❌ Common Calculation Traps

  • Forgetting to divide by 2 for the reacting ratio between Cl₂ and Ca(ClO)₂.
  • Using the ideal gas equation (pV = nRT) unnecessarily. Examiners note that using 24.0 dm³ mol⁻¹ is expected at RTP unless stated otherwise, though valid ideal gas substitutions are accepted.
  • Failing to give the final answer to 3 significant figures or missing standard form notation.
🎯 Marks: 3 marks available. Awarded for correct moles of Cl₂, correct moles of Ca(ClO)₂, and correct evaluation with 3SF + standard form.

Part (a)(ii) - Disproportionation Reactions

Decomposition of Calcium Hypochlorite

✅ Correct Answer

Equation: 3Ca(ClO)₂ → 2CaCl₂ + Ca(ClO₃)₂ (or ionic equivalent: 3ClO⁻ → 2Cl⁻ + ClO₃⁻ )

Explanation: Chlorine is simultaneously reduced and oxidised. The oxidation number of chlorine changes from +1 in Ca(ClO)₂ to -1 in CaCl₂ (reduction), and increases from +1 to +5 in Ca(ClO₃)₂ (oxidation).

💡 Key Knowledge

  • Disproportionation: A redox reaction in which the same element is simultaneously oxidized and reduced.
  • Always state the starting oxidation number and both ending oxidation numbers clearly to secure full explanation marks.
🎯 Marks: 3 marks total (1 for balanced equation, 1 for stating reduction change from +1 to -1, 1 for stating oxidation change from +1 to +5).

Part (b) - Electrode Potentials & Feasibility

Explaining Oxidising and Reducing Agents using E values

✅ Correct Answers & Overall Equations

As an Oxidising Agent (Cr₂O₇²⁻ / Cr³⁺ coupled with CH₃CHO / C₂H₅OH):

  • Cr₂O₇²⁻ acts as an oxidising agent because its E value (+1.33 V) is more positive than that of the ethanol/ethanal system (-0.197 V).
  • The Cr₂O₇²⁻/Cr³⁺ equilibrium shifts right (or E_cell is positive).
  • Overall Equation: Cr₂O₇²⁻ + 8H⁺ + 3CH₃CHO → 2Cr³⁺ + 7H₂O + 3CH₃COOH (or with ethanol).

As a Reducing Agent (Cr³⁺ / FeO₄²⁻ system):

  • Cr³⁺ acts as a reducing agent because its E value for Cr₂O₇²⁻/Cr³⁺ is less positive (+1.33 V) compared to the FeO₄²⁻/Fe³⁺ system (+2.20 V).
  • The Cr/Cr³⁺-related equilibrium shifts left.
  • Overall Equation: 2Cr³⁺ + 2H⁺ + 2FeO₄²⁻ → Cr₂O₇²⁻ + H₂O + 2Fe³⁺

🧠 Exam Technique: Structuring Electrode Potential Explanations

To score top marks in qualitative electrode potential questions, always use this 3-step template:

  1. Identify which species has the more positive or less positive E value.
  2. State the effect on the position of equilibrium (shifts left or right).
  3. Explicitly link this back to whether the species donates or accepts electrons, confirming its role as an oxidising or reducing agent, accompanied by a correctly cancelled overall equation.
🎯 Marks: 5 marks maximum (awarded across agent identifications, comparative E / equilibrium explanations, and balanced overall equations).

Part (c) - Constructing Redox Equations

Reaction between H₂S and Acidified MnO₄⁻

✅ Correct Answer

Balanced Equation:
5H₂S + 2MnO₄⁻ + 6H⁺ → 2Mn²⁺ + 5S + 8H₂O

💡 Key Knowledge

When constructing redox equations from descriptions without being given half-equations:

  • Identify the reactants: H₂S and MnO₄⁻ in acid ( H⁺ ).
  • Identify the products: Mn²⁺ , sulfur precipitate ( S ), and water ( H₂O ).
  • Balance oxidation numbers by constructing half-equations first (Sulfur goes from -2 in H₂S to 0 in S; Manganese goes from +7 in MnO₄⁻ to +2 in Mn²⁺), then combine them by balancing electrons.
🎯 Marks: 2 marks (1 mark for correct species and stoichiometry, 1 mark for fully balanced equation). State symbols are not required as per the question stem.

Topics

Module 2: Foundations in chemistry · Module 5: Physical chemistry and transition elements · 5.3 Transition elements · 5.2 Energy · 2.1 Atoms and reactions

Question and mark scheme from the OCR A-Level Chemistry examination, Unified chemistry (03), June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.