Edexcel A-Level Chemistry AS Paper 1, June 2023: Question 6

7 marks · Medium difficulty · Short Open Response

State the definitions of oxidation and reduction in terms of electron transfer, identify oxidising and reducing agents in a redox reaction, and complete and justify a disproportionation equation using oxidation numbers.

Practise this question

Question

Exam question 6 about redox reactions. Part (a) provides the equation 8KI + 5H2SO4 -> 4I2 + 4K2SO4 + H2S + 4H2O. Part (a)(i) asks to state oxidation and reduction in terms of electron transfer (1 mark). Part (a)(ii) asks to explain which species is the oxidising and reducing agent (2 marks). Part (b) states that bromine reacts with sodium hydroxide in a disproportionation reaction to form BrO3- and asks to complete the equation and justify balancing using oxidation numbers (4 marks). Total marks = 7.
Question text

6 Redox reactions can be understood in terms of electron transfer or in terms of

changes of oxidation number.

(a) The equation for a redox reaction is shown.

8KI + 5H2SO4 → 4I2 + 4K2SO4 + H2S + 4H2O

(i) State, in terms of electron transfer, what you understand by the terms

oxidation and reduction.

(1)

(ii) Explain, in terms of electron transfer, which of the species is the oxidising

agent and which is the reducing agent.

(2)

(b) Bromine reacts with sodium hydroxide solution in a disproportionation reaction

to form BrO–.

Complete the equation for this disproportionation.

Justify the balancing of the equation in terms of the changes in oxidation number.

(4)

Br + OH– → BrO– + +

… 2 … 3 …

… 14

… *P71926A01428*

(Total for Question 6 = 7 marks)

Mark scheme

Show the mark scheme Mark scheme for question 6. Part (a)(i) awards 1 mark for stating oxidation is loss of electrons and reduction is gain of electrons. Part (a)(ii) awards 2 marks for identifying iodide/potassium iodide as the reducing agent and sulfuric acid/sulfur in sulfate as the oxidising agent with electron transfer explanations. Part (b) awards 4 marks: 1 for correct species in the equation (3Br2 + 6OH- -> BrO3- + 5Br- + 3H2O), 1 for balancing, 1 for oxidation number changes for bromine from 0 to +5 and -1, and 1 for balancing the oxidation number changes.

Question Answer Additional Guidance Mark

Number

6(a)(i) An answer that makes reference to the following (1)

points:

• oxidation is loss of electrons and reduction

is gain of electrons

Number

6(a)(ii) An answer that makes reference to the following Allow a description of the transfer of one electron from iodide ion (2)

points: to the sulfur in the sulfate ion

• (Iodine / I in) potassium iodide / iodide ion Do not award just iodine / I / I2 is the reducing agent

/ KI / I‒ is the reducing agent because each Allow iodine is the reducing agent because iodide ion loses 1

iodide loses 1 electron (to form iodine) (1) electron (to form iodine)

• (sulfur in) sulfuric acid / sulfate ion / Allow sulfur (in sulfate) is the oxidising agent because each gains

H SO / SO2‒ is the oxidising agent (eight) electrons

24 4

because each sulfur gains (eight) electrons Do not award incorrect oxidation states of sulfur

(when forming H2S) (1)

If no other mark is awarded allow (1) for correct reducing agent

(from options in M1) and correct oxidising agent (from options in

M2)

If no other mark is scored allow 1 for iodide goes from –I/-1 to 0

and sulfur goes from +VI / +6 to –II / -2

Number

6(b) An answer that makes reference to the following (4)

points:

3Br + 6OH‒ → BrO‒ + 5Br‒ + 3H O

• substances in the equation (1) 2 3 2

Allow any bromide compound e.g. NaBr

Ignore state symbols even if incorrect

Do not award NaOH‒ and NaBrO‒

Do not award HBr

• balancing Ignore 1 in front of BrO‒

(1) 3

allow multiples

Allow TE on use of NaBr, but must balance

• oxidation number change for bromine from Allow incorrect descriptions of oxidation and reduction, e.g.

0 to +5 and ‒1 (1) bromine is reduced from 0 to +5 as long as oxidation states are

correct

NOTE: Redox was examined earlier

Ignore reference to balancing OH‒ and H O

• (oxidation number changes must be 2

balanced) so there must be five bromide

ion / Br‒ formed for each bromate ion /

BrO‒ (1)

(Total for Question 6 = 7 marks)

How to answer it

Redox Reactions & Disproportionation Study Guide

What this question tests

This question assesses your foundational and applied understanding of redox chemistry. You are tested on core definitions involving electron transfer, identifying oxidising and reducing agents within complex equations, completing and balancing disproportionation equations, and justifying stoichiometry using changes in oxidation numbers.

Question 6 (a)(i)

Definitions of Oxidation and Reduction

State, in terms of electron transfer, what you understand by the terms oxidation and reduction.

✅ Correct Answer

Oxidation: Loss of electrons.
Reduction: Gain of electrons.

Mark: (1) mark awarded for both definitions correctly stated.

💡 Key Knowledge

Remember the classic mnemonic: OIL RIG (Oxidation Is Loss, Reduction Is Gain of electrons). Always anchor these definitions explicitly to electrons when asked, rather than oxygen or hydrogen.

❌ Common Errors

Students often lose this mark by mixing up electron transfer with changes in oxidation numbers, or confusing the direction of transfer (stating oxidation is gain).

Question 6 (a)(ii)

Identifying Oxidising and Reducing Agents

Explain, in terms of electron transfer, which of the species is the oxidising agent and which is the reducing agent.

✅ Correct Answer

Reducing Agent: KI (or iodide ion, I⁻ ) because each iodide loses 1 electron to form iodine ( I₂ ).

Oxidising Agent: H₂SO₄ (or sulfate ion / sulfur in sulfate) because each sulfur gains 8 electrons (when forming H₂S ).

Mark: (2) marks total — (1) mark for identifying and explaining the reducing agent; (1) mark for identifying and explaining the oxidising agent.

🧠 Exam Technique

To secure both marks, you must link the identity of the agent to a particle/species and specify the exact number of electrons lost or gained. Vague answers like "iodine is the reducing agent" without referencing iodide ions will be heavily penalised.

Question 6 (b)

Disproportionation Equations and Oxidation Numbers

Complete the equation for this disproportionation and justify the balancing in terms of changes in oxidation numbers.

Equation frame: ...Br₂ + ...OH⁻ → ...BrO₃⁻ + ... + ...

✅ Correct Answer

Completed Balanced Equation:

3Br₂ + 6OH⁻ → BrO₃⁻ + 5Br⁻ + 3H₂O

Mark: (4) marks total breakdown:
• (1) mark for correct substances ( Br⁻ and H₂O alongside BrO₃⁻ and OH⁻ )
• (1) mark for overall equation balancing
• (1) mark for stating oxidation number changes (0 to +5 and 0 to -1)
• (1) mark for balancing oxidation number changes (5 bromide ions formed for every 1 bromate ion)

📐 Step-by-Step Justification & Balancing

  1. Identify oxidation states: Elemental bromine ( Br₂ ) has an oxidation number of 0 . In bromate ( BrO₃⁻ ), bromine is +5 . In bromide ( Br⁻ ), bromine is -1 .
  2. Determine changes: Bromine is simultaneously oxidised (0 to +5, increase of 5) and reduced (0 to -1, decrease of 1).
  3. Balance the electron transfer: Since 1 bromine atom gains 1 electron and another loses 5 electrons, you need a ratio of 5 Br⁻ produced for every 1 BrO₃⁻ to balance the electrons transferred.
  4. Balance mass/charge: Complete the balancing by placing coefficients: 3 in front of Br₂ , 6 in front of OH⁻ , and 3 in front of H₂O .

❌ Common Errors

Students frequently fail to balance the ratio of oxidation states properly—forgetting that the 5:1 ratio between Br⁻ and BrO₃⁻ dictates the stoichiometric coefficients for the bromine-containing products.

Topics

Physical Chemistry · Inorganic Chemistry · Topic 3: Redox I · Topic 4: Inorganic Chemistry and the Periodic Table

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