Edexcel A-Level Chemistry Paper 3, November 2021: Question 1

9 marks · Medium difficulty · Short Open Response

Assess chlorine properties including subatomic particles of isotopes, mass spectrometry of molecular chlorine, nomenclature of chlorate(V), electron affinity equations, and standard electrode potentials.

Practise this question

Question

Exam question about chlorine consisting of six parts. Part (a)(i) provides a table to complete with protons, neutrons, and electrons for a 35Cl atom and a 37Cl- ion. Part (a)(ii) shows an m/z grid from 69 to 75 with a relative abundance axis from 0 to 100, asking to draw the mass spectrum for Cl2+ with 75% 35Cl and 25% 37Cl. Part (b) asks for the formula of potassium chlorate(V). Part (c) asks for the equation representing the first electron affinity of chlorine with state symbols. Part (d) gives the standard electrode potential for the chlorine/chloride half-cell and asks to identify an oxidising agent from the Data Booklet to convert chloride to chlorine and calculate the E_cell value.
Question text

1 This question is about chlorine.

(a) Chlorine has two isotopes with mass numbers 35 and 37.

(i) Complete the table to show the numbers of subatomic particles in a 35Cl atom

and a 37Cl− ion.

(2)

Particle Protons Neutrons Electrons

35Cl atom

37Cl− ion

(ii) A sample of chlorine contained 75 % of 35Cl and 25 % of 37Cl.

Complete the mass spectrum to show the peaks you would expect for the

molecular ion Cl+ from this sample of chlorine gas.

(2)

Relative 50

abundance

69 70 71 72 73 74 75

m/z

(b) Write the formula of potassium chlorate(V).

(1)

(c) Write the equation for the first electron affinity of chlorine. Include state symbols.

(2)

*P67806A0236*

(d) The standard electrode potential for the chlorine/chloride ion half-cell is

½C (aq) + e− ⇌ C −(aq) EO = +1.36 V

l2 l cell

(i) Identify an oxidising agent from the Data Booklet that will convert

chloride ions into chlorine under standard conditions.

(1)

(ii) Calculate the value of EO for the reaction in (d)(i).

cell

(1)

(Total for Question 1 = 9 marks)

Mark scheme

Show the mark scheme Mark scheme providing correct answers for all parts of Question 1. Part (a)(i) gives the numbers 17, 18, 17 for 35Cl and 17, 20, 18 for 37Cl-. Part (a)(ii) shows lines at m/z 70, 72, and 74 with a 9:6:1 relative abundance ratio. Part (b) gives KClO3. Part (c) gives Cl(g) + e- -> Cl-(g). Part (d)(i) gives acidified MnO4- or H2O2, and (d)(ii) gives +0.15 V or +0.41 V.

Question Answer Additional Guidance Mark

Number

1(a)(i) Example of table (2)

• all numbers for 35Cl correct (1)

Particle Protons Neutrons Electrons

• all numbers for 37Cl− correct (1) 35Cl atom 17 18 17

37Cl− ion 17 20 18

If no other mark is awarded, allow (1) for any four numbers correct

Number

1(a)(ii) Example of spectrum (2)

• lines at 70 and 72 and 74 (1)

• relative abundances 9:6:1 (1) Allow any abundances in an approximate 9:6:1 ratio e.g. 56:37-38:6

as %, or 75 : 50 : 8

Number

1(b) Allow (1)

• KClO K+ClO −

Number

1(c) Example of equation (2)

• equation (1) Cl(g) + e− → Cl−(g)

Allow just e for electron

• state symbols (1) Stand alone mark for species on both sides of equation

Ignore state symbol for electron

Number

1(d)(i) Either (1)

• identification of oxidising agent acidified (potassium) manganate(VII) / MnO − and H+

Or

acidified hydrogen peroxide / H O and H+

Allow H+ shown in equation in (i) or (ii)

If the acid is specified it must be sulfuric acid

Number

1(d)(ii) Either (1)

• value of Eo Eo = (+)0.15 (V) for

cell cell

acidified (potassium) manganate(VII)

Or

Eo = (+)0.41 (V) for

cell

acidified hydrogen peroxide

No TE on any other reagent in (i)

(Total for Question 1 = 9 marks)

How to answer it

Edexcel A-Level Chemistry: Chlorine Comprehensive Study Guide

What this question tests

This question assesses core physical and inorganic chemistry concepts relating to Group 7 (Halogens). Key skills tested include calculating subatomic particle numbers in isotopes and ions, predicting mass spectrometry fragmentation and relative abundance patterns for diatomic molecules, writing formulas using oxidation states, constructing electron affinity half-equations with state symbols, identifying oxidising agents from standard electrode potentials, and calculating standard cell potentials (E_cell).

Question 1 (a)(i) - Subatomic Particles

Atomic Structure of Isotopes and Ions

✅ Correct Answer

  • ³⁵Cl atom: Protons = 17, Neutrons = 18, Electrons = 17
  • ³⁷Cl⁻ ion: Protons = 17, Neutrons = 20, Electrons = 18

💡 Key Knowledge

  • Atomic number (Z) = number of protons = 17 for chlorine.
  • Mass number = protons + neutrons. Therefore, neutrons = mass number - atomic number.
  • Negative ions have gained electrons: a 1- charge means 1 extra electron compared to the neutral atom.
Mark breakdown: 1 mark for all correct values for the ³⁵Cl atom; 1 mark for all correct values for the ³⁷Cl⁻ ion.
Question 1 (a)(ii) - Mass Spectrometry

Predicting Molecular Ion Mass Spectra

✅ Correct Answer

  • Peaks required at: m/z = 70, 72, and 74.
  • Relative abundances ratio: 9 : 6 : 1 (or equivalent percentages: 56.25% : 37.5% : 6.25%, or using 75:50:8).

📐 Calculation & Probability Breakdown

Given: 75% (0.75) ³⁵Cl and 25% (0.25) ³⁷Cl.

  1. Peak at m/z 70 (³⁵Cl - ³⁵Cl): 0.75 × 0.75 = 0.5625 (relative proportion 9)
  2. Peak at m/z 72 (³⁵Cl - ³⁷Cl and ³⁷Cl - ³⁵Cl): 2 × (0.75 × 0.25) = 0.375 (relative proportion 6)
  3. Peak at m/z 74 (³⁷Cl - ³⁷Cl): 0.25 × 0.25 = 0.0625 (relative proportion 1)

❌ Common Errors

Students often forget that chlorine gas is diatomic (Cl₂). Failing to account for both isotopic combinations for the mixed peak (m/z 72) leads to an incorrect 3:1 ratio instead of the correct 9:6:1 molecular ion ratio.

Mark breakdown: 1 mark for identifying the correct m/z values (70, 72, 74); 1 mark for correct relative abundances in a 9:6:1 ratio.
Question 1 (b) - Nomenclature

Formula of Potassium Chlorate(V)

✅ Correct Answer

KClO₃ (also accept ionic representation K⁺ClO₃⁻ )

🧠 Exam Technique

Roman numerals in IUPAC nomenclature indicate the oxidation state of the central atom. Chlorine in chlorate(V) has an oxidation state of +5. Combine potassium (K⁺, Group 1) with the chlorate(V) ion (ClO₃⁻) to balance charges.

Mark breakdown: 1 mark for the correct formula.
Question 1 (c) - Electron Affinity

First Electron Affinity Equation

✅ Correct Answer

Cl(g) + e⁻ → Cl⁻(g)

💡 Key Knowledge

First electron affinity is defined as the enthalpy change when one mole of gaseous atoms gains one mole of electrons to form one mole of gaseous 1- ions. State symbols are strictly required for gaseous species.

❌ Common Errors

Omitting state symbols or using Cl₂ instead of atomic chlorine Cl(g) will lose you the equation mark.

Mark breakdown: 1 mark for chemical species and balancing (accept just e for electron); 1 mark for correct state symbols across all species.
Question 1 (d) - Electrode Potentials

Oxidising Agents and Cell Calculations

✅ Correct Answer (Part i)

Acidified potassium manganate(VII) ( MnO₄⁻ / H⁺ ) OR acidified hydrogen peroxide ( H₂O₂ / H⁺ ). If naming the acid explicitly, it must be sulfuric acid.

✅ Correct Answer (Part ii)

+0.15 V (for manganate(VII)) OR +0.41 V (for hydrogen peroxide).

📐 Calculation (Part ii)

Formula: E_cell = E_reduction - E_oxidation

  • Chlorine half-cell given: ½Cl₂(aq) + e⁻ ⇌ Cl⁻(aq) with E° = +1.36 V
  • For manganate(VII), E° = +1.51 V . Therefore, 1.51 - 1.36 = +0.15 V .
  • For hydrogen peroxide, E° = +1.77 V . Therefore, 1.77 - 1.36 = +0.41 V .

🧠 Exam Technique

To convert chloride ions ( Cl⁻ ) into chlorine ( Cl₂ ), the added oxidising agent must have a more positive standard electrode potential than +1.36 V so that the equilibrium position drives backwards to oxidize Cl⁻ .

Mark breakdown: Part (i) = 1 mark for valid oxidising agent with acid. Part (ii) = 1 mark for corresponding correct numerical value with sign and units.

Topics

Inorganic Chemistry · Physical Chemistry · Topic 4: Inorganic Chemistry and the Periodic Table · Topic 1: Atomic Structure and the Periodic Table · Topic 14: Redox II

Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 3, November 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.