Edexcel A-Level Chemistry Paper 1, June 2017: Question 3

9 marks · Medium difficulty · Short Open Response

Explain trends in boiling temperatures of halogens, identify the strongest reducing agent among halides using oxidation numbers, and calculate cell potentials and ionic equations from standard electrode potentials.

Practise this question

Question

Question text

3 This question is about halogens and redox reactions.

(a) The boiling temperatures of three halogens are shown in the table.

Boiling temperature

Halogen

/ °C

chlorine –35

bromine 59

iodine 184

Explain why the boiling temperatures increase from chlorine to iodine.

(2)

(b) Potassium halides react with concentrated sulfuric acid to form

potassium hydrogensulfate and the different products shown in the table.

Potassium halide Products

potassium chloride hydrogen chloride

potassium bromide hydrogen bromide, bromine and sulfur dioxide

potassium iodide hydrogen iodide, iodine, hydrogen sulfide and sulfur

By referring to any changes in oxidation numbers when these halides react with

concentrated sulfuric acid, explain which halide is the strongest reducing agent.

(3)

… *P48058A0528*

Electrode reaction E / V

I (aq) + 2e– U 2I–(aq) +0.54

Fe3+(aq) + e– U Fe2+(aq) +0.77

Br (aq) + 2e– U 2Br–(aq) +1.09

MnO (s) + 4H+(aq) + 2e– U Mn2+(aq) + 2H O(l) +1.23

Cl (aq) + 2e– U 2Cl–(aq) +1.36

6 – + – 2+

MnO4(aq) + 8H*P48058A0628*(aq) + 5eUMn(aq) + 4H2O(l) +1.51

(i) Which species will oxidise Fe2+(aq) to Fe3+(aq)?

(1)

A Br2(aq)

B Cl–(aq)

C I2(aq)

D Mn2+(aq)

(ii) Write the ionic equation and calculate the Ecell value for the reaction between

MnO– ions and Br– ions in acidic solution.

State symbols are not required.

(3)

(Total for Question 3 = 9 marks)

Mark scheme

Show the mark scheme

Question

Answer Additional Guidance Mark

Number

3(a) An explanation that makes reference to the following An answer that states ‘covalent bonds break’ (2)

points: or ‘bonds between atoms break’ or refers to

‘ions’ scores (0) overall

Allow reverse argument for M1 and M2

from chlorine to iodine / down the group, the Allow iodine has more / most electron shells

number of electrons (in the molecule / atom) (than chlorine and/or bromine)

increases / changes from 34 to 106 / 17 to 53 (1)

Ignore ‘the size of the atoms /molecules

increases from chlorine to iodine’

Do not allow incorrect numbers of electrons

so the strength of the London / instantaneous Allow iodine has the strongest London force

dipole-(induced) dipole forces increases / there are and most energy is needed to separate the

more London / instantaneous dipole-(induced) dipole molecules

forces

and Allow more energy is need to overcome /

more energy is needed to separate the molecules break the London forces / bonds instead of

(1) separate the molecules

Allow dispersion forces / van der Waals forces

for London forces

Ignore higher temperature needed to

separate the molecules

Do not award dipole-dipole forces / just

‘intermolecular forces’

Question

Answer Additional Guidance Mark

Number

3(b) An explanation that makes reference to the following Allow the oxidation numbers written by the (3)

points: species in the table

(+)6 only needs to be mentioned once in M1

or M2

Allow references to potassium halides /

halogens / hydrogen halides instead of halide

ions

For full marks, the answer must identify

iodide as the strongest reducing agent

iodide ions are the strongest reducing agent because Only 1 oxidation number change is needed.

iodide ions / I− / (potassium) iodide reduces sulfur If both are given, both must be correct

(in sulfuric acid) from +6 to 0 in sulfur / −2 in H2S

(1)

(whereas) bromide ions / Br− / (potassium) bromide Allow bromide ions are stronger reducing

reduces sulfur (in sulfuric acid) from +6 to +4 (1) agents than chloride ions because they are

oxidised from −1 to 0

(whereas) chloride ions / Cl− / (potassium) chloride Allow just ‘it is not a redox reaction’

do not reduce sulfuric acid / sulfur / S (as there is

no change in oxidation number of Cl or S) (1)

Question

Answer Mark

Number

3(c)(i) The only correct answer is A (1)

B is not correct because Cl− is not an oxidising agent

C is not correct because I2 is not a powerful enough oxidising agent

D is not correct because Mn2+ is not an oxidising agent

Question

Answer Additional Guidance Mark

Number

3(c)(ii) Example of ionic equation (3)

2MnO − + 16H+ + 10Br−

→ 2Mn2+ + 8H O + 5Br

all species on correct sides of Allow ⇌

equation

Allow correct species if shown in working with half-equations but

and

slip made in final equation e.g. charge missing

no electrons / electrons

cancelled (1)

Ignore state symbols

Allow multiples

balancing correct species (1)

Allow M2 for almost correct species

Eo value (1)

cell Eo (= 1.51 − 1.09) = (+)0.42 (V)

cell

No TE on incorrect equation

(Total for Question 3 = 9 marks)

How to answer it

Halogens and Redox Reactions Study Guide

What this question tests

This exam question evaluates core inorganic and physical chemistry concepts from Edexcel A-Level Chemistry. It tests your understanding of intermolecular forces (London forces) in halogens, trends in reducing ability down Group 7 using oxidation numbers, interpreting electrochemical series data (E-cell values), and constructing complex ionic redox equations in acidic conditions.

Part (a) — Boiling Temperatures of Halogens

Explain why the boiling temperatures increase from chlorine to iodine. (2 marks)

✅ Correct Answer Structure

Point 1: Down the group (from chlorine to iodine), the number of electrons in the molecule/atom increases.
Point 2: Therefore, the strength of the London forces (instantaneous dipole-induced dipole forces) increases, requiring more energy to overcome/separate the molecules.

💡 Key Knowledge

Halogens exist as simple covalent diatomic molecules (Cl₂, Br₂, I₂). Their boiling points depend entirely on the strength of the intermolecular forces between molecules, not the covalent bonds within them.

❌ Common Errors

Saying that covalent bonds break when halogens boil (scores 0 overall).
Mentioning incorrect intermolecular forces like hydrogen bonding or permanent dipole-dipole forces.
Stating that the "size of the atoms increases" without linking it clearly to increased electron count.

Mark breakdown: 1 mark for stating electrons increase down the group; 1 mark for linking stronger London forces to more energy required to separate molecules.

Part (b) — Reducing Ability of Halide Ions

By referring to any changes in oxidation numbers when these halides react with concentrated sulfuric acid, explain which halide is the strongest reducing agent. (3 marks)

✅ Correct Answer

Iodide (I⁻) is the strongest reducing agent because iodide ions reduce sulfur in sulfuric acid from oxidation state +6 down to +4 (in SO₂), 0 (in S), or -2 (in H₂S), while being oxidized themselves.

🧠 Exam Technique & Marking Points

Point 1: Identify iodide as the strongest reducing agent and state its oxidation number change (sulfur goes from +6 to +4, 0, or -2).
Point 2: Mention bromide ions reducing sulfur from +6 to +4 (in SO₂).
Point 3: Explain that chloride ions do not reduce sulfuric acid because there is no change in oxidation number.

❌ Common Errors

Failing to explicitly name iodide as the strongest reducing agent at the start or end of the explanation.
Mixing up oxidation and reduction terms (e.g., stating the halogen is reduced instead of the sulfur).

Mark breakdown: 1 mark for iodide reducing sulfur +6 to lower states; 1 mark for bromide reducing sulfur +6 to +4; 1 mark for chloride not reducing sulfuric acid (no oxidation number change).

Part (c)(i) — Electrode Potentials (Multiple Choice)

Which species will oxidise Fe²⁺(aq) to Fe³⁺(aq)? (1 mark)

✅ Correct Answer: A — Br₂(aq)

For a species to oxidise Fe²⁺ to Fe³⁺, its standard electrode potential ( E° ) must be more positive than the Fe³⁺/Fe²⁺ system (+0.77 V).

💡 Why the others are incorrect

B (Cl⁻): Cannot act as an oxidising agent.
C (I₂): Has an E° of +0.54 V, which is less positive than +0.77 V, so it cannot oxidise Fe²⁺.
D (Mn²⁺): Is a reduction product, not an oxidising agent in this context.

Mark breakdown: 1 mark for selecting A.

Part (c)(ii) — Redox Equation and E-cell Calculation

Write the ionic equation and calculate the E°cell value for the reaction between MnO₄⁻ ions and Br⁻ ions in acidic solution. State symbols are not required. (3 marks)

📐 Step-by-Step Calculation & Equation

Step 1: Identify Half-Equations and E° values from the table:
Reduction: MnO₄⁻ + 8H⁺ + 5e⁻ ⇌ Mn²⁺ + 4H₂O ( E° = +1.51 V )
Oxidation: Br₂ + 2e⁻ ⇌ 2Br⁻ ( E° = +1.09 V )
Step 2: Balance electrons and construct the overall ionic equation:
Multiply the permanganate half-equation by 2 and the bromide half-equation by 5 to equalise electrons (10 electrons each):
2MnO₄⁻ + 16H⁺ + 10Br⁻ ➔ 2Mn²⁺ + 8H₂O + 5Br₂
Step 3: Calculate E°cell:
E°cell = E°(reduction) - E°(oxidation)
E°cell = 1.51 - 1.09 = +0.42 V (or 0.42 V )

❌ Common Calculation Traps

Multiplying the E° values by the balancing coefficients (2 and 5). Never multiply standard electrode potentials by stoichiometric multipliers!
Subtracting the values the wrong way round and forgetting the positive sign for E°cell .

Mark breakdown: 1 mark for correct species on correct sides with electrons cancelled; 1 mark for fully balanced species; 1 mark for correct E°cell value (+0.42 V).

Topics

Physical Chemistry · Inorganic Chemistry · Topic 2: Bonding and Structure · Topic 3: Redox I · Topic 14: Redox II · Topic 4: Inorganic Chemistry and the Periodic Table

Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 1, June 2017. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.