Edexcel A-Level Chemistry AS Paper 1, November 2021: Question 6
13 marks · Medium difficulty · Short Open Response
Analyze reactions of halogens and halides involving redox properties, disproportionation, and stoichiometric calculations.
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Question text
6 This question is about the reactions of the halogens and their salts.
(a) The potassium halides react with concentrated sulfuric acid to form
hydrogen halides.
(i) The equation for this reaction for potassium chloride can be written
KCl + H2SO4 → HCl + KHSO4
The hydrogen chloride does not react further.
State why this reaction is not a redox reaction.
(1)
(ii) On descending Group 7, the hydrogen halides become better reducing agents.
Explain how the reactions of potassium chloride, potassium bromide and
potassium iodide with concentrated sulfuric acid provide evidence for
this statement.
No explanation of the trend is required.
(3)
… *P67083A01624*
(b) The reaction that occurs between chlorine and sodium hydroxide depends
on the temperature.
(i) At room temperature the reaction that occurs is
Cl2 + NaOH → NaClO + NaCl
Explain, with reference to oxidation numbers, why this is a
disproportionation reaction.
(2)
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(ii) With hot sodium hydroxide solution, a different disproportionation reaction
occurs. Sodium chlorate(V) is one of the products.
Complete the equation for this reaction. State symbols are not required.
(2)
… Cl2 + … NaOH →
(c) Chlorine is used as a bleach in the textiles industry. Any excess chlorine can be
removed by reduction to chloride ions.
The half-equation for the reaction of chlorine is
Cl + 2e– → 2Cl–
In one reaction, 768 cm3 of chlorine gas was reduced.
(i) Calculate the number of moles of electrons gained by chlorine molecules
during this reaction.
[Under these conditions one mole of gas occupies 24 dm3]
(2)
*P67083A01824*
(ii) The reducing agent was a solution containing thiosulfate ions, S O2–.
The chlorine reacted with 40 cm3 of a 0.20 mol dm–3 solution of these ions.
Deduce the number of moles of electrons lost by each atom of sulfur in the
thiosulfate ion, and hence the final oxidation state of the sulfur in the product.
(3)
(Total for Question 6 = 13 marks)
Mark scheme
Show the mark scheme
Question Answer Additional Guidance Mark
Number
6(a)(i) An answer that makes reference to the following point: (1)
• the oxidation number / state does not change for Accept there is no transfer of electrons
any element
Number
6(a)(ii) An explanation that makes reference to the following (3)
points: Mark independently
• (because) sulfur in sulfuric acid is reduced further (1) Allow potassium salt / halide ion for hydrogen halide
by hydrogen iodide than hydrogen bromide (and
hydrogen chloride)
• SO2 / S(IV) produced in the reaction with HBr (1) May be shown in an equation, but ignore incorrect
state symbols and/or balancing
• more negative oxidation states of sulfur / S / H2S
/ S2‒ are produced in the reaction with HI
(1) May be shown in an equation, but ignore incorrect
state symbols and/or balancing
Number
6(b)(i) An answer that makes reference to the following points: (2)
• chlorine / Cl2 is simultaneously oxidised and (1)
reduced
• the oxidation number of chlorine changes from 0 (1) Allow oxidation numbers underneath or above the
to –I and (+)I / 0 to –1 and (+)1 equation
/ increases by 1 and decreases by 1
Number
6(b)(ii) Example of equation (2)
• substances correct in equation (1) 3Cl2 + 6NaOH → NaClO3 + 5NaCl + 3H2O
• equation is balanced (1) Ignore state symbols even if incorrect
Number
6(c)(i) Example of calculation (2)
• calculation of the number of moles of chlorine (1) = 768 = 0.032 / 3.2 × 10-2 (mol) (answer 1)
gas reacting 24000
(1) = (answer 1) × 2 = 0.064 / 6.4 × 10-2 (mol)
• calculation of number of moles of electron gained
by chlorine
Allow TE on incorrect number of moles of chlorine
(M2 is for multiplying by 2)
Ignore SF except 1SF
Number
6(c)(ii) Example of calculation (3)
• calculation of number of moles of thiosulfate ions (1) = 40 × 0.20 = 0.008 / 8 × 10-3 (mol) (answer 1)
1000
• calculation of electrons lost per sulfur atom (1) = 6(c)(i) = 0.064 = 4 (electrons)
(answer 1) × 2 0.008 × 2
• deduction of the oxidation number of the sulfur (1) (In thiosulfate ion oxidation state of sulfur is (II)) and
and hence the product each loses four electrons so oxidation state is (+)(VI)
/ (VI)(+) / 6 (+) / (+)6
Allow TE throughout including on 6(c)(i)
Ignore SF
(Total for Question 6 = 13 marks)
How to answer it
Reactions of the Halogens and Their Salts
This question assesses your understanding of Group 7 redox chemistry, specifically the reducing ability of halide ions, disproportionation reactions of chlorine with sodium hydroxide under varying conditions, and mole calculations involving gas volumes, half-equations, and changes in oxidation numbers.
Part (a): Reactions of Halides with Concentrated Sulfuric Acid
✅ Correct Answer: (i)
The oxidation number / state does not change for any element (or: there is no transfer of electrons).
💡 Key Knowledge
- Chloride ions (Cl⁻) are weak reducing agents and cannot reduce sulfuric acid. Only an acid-base reaction occurs, producing HCl.
- Bromide and iodide ions are stronger reducing agents, reducing sulfur in H₂SO₄ down to lower oxidation states (like +4 in SO₂ or down to 0 and -2 in S and H₂S).
✅ Correct Answer: (ii)
Make three independent points:
- Sulfur in sulfuric acid is reduced further by hydrogen iodide than by hydrogen bromide.
- SO₂ (sulfur in oxidation state +IV) is produced in the reaction with HBr.
- More negative oxidation states of sulfur (S / H₂S / S²⁻) are produced in the reaction with HI.
❌ Common Errors
Students often lose marks by writing vague statements about "reactivity" instead of explicitly discussing the extent of reduction of sulfur and referencing specific sulfur-containing reduction products (SO₂, S, or H₂S).
Part (b): Chlorine and Sodium Hydroxide Reactions
✅ Correct Answer: (i)
Chlorine (Cl₂) is simultaneously oxidised and reduced. The oxidation number of chlorine changes from 0 in Cl₂ to -1 in NaCl and +1 in NaClO.
🧠 Exam Technique
When asked to explain a disproportionation reaction, always state both that the same element is oxidised and reduced, and explicitly quote the initial and final oxidation numbers to secure both marks.
✅ Correct Answer: (ii) - Equation
3Cl₂ + 6NaOH → NaClO₃ + 5NaCl + 3H₂O
❌ Common Errors
Forgetting that hot NaOH yields chlorate(V) ( NaClO₃ ) instead of chlorate(I) ( NaClO ), or failing to balance the resulting equation correctly.
Part (c): Calculations & Oxidation Numbers
📐 Calculation: Part (i)
Step 1: Calculate moles of chlorine gas
Moles = 768 cm³ / 24000 cm³ mol⁻¹ = 0.032 (or 3.2 × 10⁻² mol )
Step 2: Calculate moles of electrons gained
From half-equation: Cl₂ + 2e⁻ → 2Cl⁻ (1 mole of Cl₂ gains 2 moles of electrons).
0.032 mol × 2 = 0.064 (or 6.4 × 10⁻² mol )
📐 Calculation: Part (ii)
Step 1: Moles of thiosulfate ions (S₂O₃²⁻)
(40 / 1000) × 0.20 = 0.0080 (or 8.0 × 10⁻³ mol )
Step 2: Electrons lost per sulfur atom
Moles of e⁻ lost = Moles of e⁻ gained = 0.064 mol.
Since each thiosulfate ion has 2 sulfur atoms: Total sulfur atoms = 0.0080 × 2 = 0.016 mol.
Electrons lost per sulfur = 0.064 / 0.016 = 4 electrons
Step 3: Final oxidation state
Sulfur starts at oxidation state +II in S₂O₃²⁻. Losing 4 electrons increases its oxidation state by 4:
+2 + 4 = +VI (or +6)
❌ Calculation Traps
- Forgetting to multiply by 2 when converting moles of Cl₂ to moles of electrons.
- Failing to account for the two sulfur atoms present in each thiosulfate ion ( S₂O₃²⁻ ).
Topics
Inorganic Chemistry · Physical Chemistry · Topic 4: Inorganic Chemistry and the Periodic Table · Topic 3: Redox I · Topic 5: Formulae, Equations and Amounts of Substance
Question and mark scheme from the Edexcel A-Level Chemistry examination, AS Paper 1, November 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.