Edexcel A-Level Chemistry Paper 3, June 2018: Question 5

16 marks · Hard difficulty · Extended Writing

Analyse the properties of transition elements including vanadium oxidation states, reduction reactions with zinc, catalytic action in the contact process, and copper(II) complex ion reactions.

Practise this question

Question

A four-part A-Level Chemistry question about transition elements. Part (a) asks for the oxidation state of vanadium in NH4VO3. Part (b) asks to explain the sequence of reactions when excess zinc powder is added to acidified NH4VO3, providing a table of electrode potentials for various vanadium and zinc half-cells. Part (c) asks to explain how vanadium(V) oxide acts as a catalyst in the contact process. Part (d) asks to describe the reactions of aqueous copper(II) ions with sodium hydroxide, excess ammonia, and concentrated hydrochloric acid, linking observations with equations.
Question text

5 This question is about the properties of transition elements, their ions and their complexes.

(a) Give the oxidation state of vanadium in the compound NH4VO3.

(1)

(b) Excess zinc powder is added to an acidified solution of the compound NH4VO3.

Using the data in the table, explain the sequence of reactions that takes place.

In your answer, include a description of what you would see, and the relevant

ionic equations with their calculated Ecell values. State symbols are not required.

(7)

Electrode system E / V

V2+(aq) + 2e− V(s) −1.18

V3+(aq) + e− V2+(aq) −0.26

VO2+(aq) + 2H+(aq) + e− V3+(aq) + H O(l) +0.34

VO−(aq) + 4H+(aq) + e− VO2+(aq) + 2H O(l) +1.00

Zn2+(aq) + 2e− Zn(s) −0.76

… *P52304A01432*

(c) Explain how vanadium(V) oxide acts as a catalyst in one step of the contact

process. The equation for this step is

2SO2(g) + O2(g) 2SO3(g)

(2)

*(d) Describe the reactions of separate portions of aqueous copper(II) ions with

aqueous sodium hydroxide solution, with excess aqueous ammonia solution and

with concentrated hydrochloric acid. 15

In your answer you should link observations with equations which include the*P52304A01532*

formulae of any copper-containing complex ions. Include state symbols.

(6)

… *P52304A01632*

(Total for Question 5 = 16 marks)

Mark scheme

Show the mark scheme A comprehensive mark scheme providing acceptable answers, additional guidance, and mark allocations for all four parts of the question, including specific colour changes and ionic equations for vanadium reductions, catalytic steps for vanadium(V) oxide, and copper complex reactions.

Question

Acceptable Answers Additional Guidance Mark

Number

5(a) +5 Allow 5+ / +V / V+ / (V) / 5 (1)

Do not award V+

Question

Acceptable Answers Additional Guidance Mark

Number

5(b) A description that makes reference to the following points: (7)

M1 and M2 –colours

Yellow blue green violet / lavender / purple / mauve

2 or 3 colours linked to correct species / oxidation states / reactions (1)

4 colours linked to correct species / oxidation states / reactions (1)

M3 - statement

Statement that sequence is from +5 to +4 to +3 to +2 M3 can be implied from species in

or explanation or equations

(step-wise) reduction / zinc is a reducing agent (1)

M4, M5 and M6 - equations

These three equations, with appropriate Eo values Allow multiples

Zn + 2VΟ − + 8Η+ → Zn2+ + 2VΟ2+ + 4Η Ο and Eo = (+)1.76 (V) (1) Ignore state symbols even if

incorrect

Ζn + 2VΟ2+ + 4Η+ → Zn2+ + 2V3+ + 2Η Ο and Eo = (+)1.1(0) (V) (1) 3 correct equations with incorrect

Eo scores 2

Zn + 2V3+ → Ζn2+ + 2V2+ and Eo = (+)0.5(0) (V) (1) 2 correct equations with incorrect

Eo scores 1

3 correct Eo with incorrect

M7 – stops at V2+ equations scores 1

No (further) reduction (feasible) to V metal / V(0)

or

Zn + V2+ → Ζn2+ + V not feasible

or

Eo = −0.42 (V) (1)

Question

Acceptable Answers Additional Guidance Mark

Number

5(c) A explanation that makes reference to the following points: Ignore any references to heterogeneous (2)

catalysis

M1

V changes (its oxidation state / oxidation number) from +5 Allow Forms V2O4 / VO2 (as an intermediate)

to +4 (as it oxidises the sulfur dioxide)

OR Do not award VO2+ or VO – or VO +

The oxidation number of V decreases in the reaction

OR

Vanadium is reduced in the reaction with SO2

OR

V2O5 oxidises the SO2 / S

OR

V2O5 + SO2 V2O4 + SO3

(1)

M2

(Then) returns to +5 (oxidation state / oxidation number) by Allow (re-) forms V2O5

reacting with oxygen

OR

2 V2O4 + O2 2 V2O5 (1)

Question

Acceptable Answers Additional Guidance Mark

Number

*5(d) This question assesses a student’s ability to show a Guidance on how the mark scheme should be (6)

coherent and logically structured answer with linkages and applied:

fully-sustained reasoning. The mark for indicative content should be

added to the mark for lines of reasoning. For

Marks are awarded for indicative content and for how the example, an answer with five indicative

answer is structured and shows lines of reasoning. marking points that is partially structured with

some linkages and lines of reasoning scores 4

The following table shows how the marks should be marks (3 marks for indicative content and 1

awarded for indicative content. mark for partial structure and some linkages

and lines of reasoning).

If there are no linkages between points, the

Number of Number of same five indicative marking points would yield

indicative marks an overall score of 3 marks (3 marks for

marking awarded for indicative content and no marks for linkages).

points seen indicative

in answer marking In general it would be expected that 5 or 6

indicative points would score 2 reasoning

How to answer it

Transition Elements, Redox Potentials & Complex Ions

What this question tests

This comprehensive A-Level Chemistry question assesses your knowledge of transition metal chemistry, specifically: calculating oxidation states, predicting redox reactions and colour changes using standard electrode potentials (E-theta values), explaining heterogeneous catalysis mechanisms (Contact Process), and describing ligand exchange and precipitation reactions of aqueous copper(II) ions with complete balanced equations.

Part (a) — Oxidation State

Find the Oxidation State of Vanadium in NH₄VO₃

✅ Correct Answer

+5 (or 5 , +V , V )

💡 Key Knowledge

  • Ammonium ion ( NH₄⁺ ) has an oxidation state of +1.
  • Each oxygen atom is -2 (total for 3 oxygen atoms = -6).
  • Setting up the algebraic sum: +1 + V + 3(-2) = 0 , which gives V = +5 .
Mark: 1 mark
Part (b) — Redox Reactions & Electrode Potentials

Reduction of Vanadate(V) by Zinc Powder

✅ Correct Answers & Observations

  • Colour Sequence: Yellow to Blue to Green to Violet/Purple ( VO₃⁻ → VO²⁺ → V³⁺ → V²⁺ ).
  • Step 1 Equation: Zn + 2VO₃⁻ + 8H⁺ → Zn²⁺ + 2VO²⁺ + 4H₂O   (E-cell = +1.76 V)
  • Step 2 Equation: Zn + 2VO²⁺ + 4H⁺ → Zn²⁺ + 2V³⁺ + 2H₂O   (E-cell = +1.10 V)
  • Step 3 Equation: Zn + 2V³⁺ → Zn²⁺ + 2V²⁺   (E-cell = +0.50 V)
  • Limit Point: Reaction stops at V(II) because reduction to V metal is not feasible (E-cell is negative: -0.42 V).

🧠 Exam Technique & Calculations

  • To calculate E-cell, use: E-cell = E(reduction) - E(oxidation) or combine half-equations where the more positive system drives the reduction of the higher oxidation state.
  • Ensure all half-equations are correctly balanced for electrons before combining them into overall ionic equations.

❌ Common Errors

  • Failing to state all four distinct colours in the correct chronological order.
  • Continuing the sequence incorrectly to suggest vanadium metal forms (ignoring the negative E-cell value for the V²⁺/V system paired with Zn²⁺/Zn ).
Marks: 7 marks total
Part (c) — Catalysis

Vanadium(V) Oxide in the Contact Process

✅ Correct Explanation

  • Step 1: V₂O₅ oxidises SO₂ to SO₃ , and is itself reduced from +5 to +4 ( V₂O₅ + SO₂ → V₂O₄ + SO₃ ).
  • Step 2: The intermediate V₂O₄ is then re-oxidised back to V₂O₅ by reacting with oxygen ( 2V₂O₄ + O₂ → 2V₂O₅ ).

💡 Key Knowledge

  • Catalysts provide an alternative reaction pathway with a lower activation energy.
  • The catalyst is chemically changed during the intermediate step(s) but is regenerated completely by the end of the catalytic cycle.
Marks: 2 marks
Part (d) — Aqueous Copper(II) Reactions (*QWC Extended Response)

Reactions of Aqueous Copper(II) Ions

✅ Observations & Complex Equations

  • With NaOH(aq): Blue solution turns into a blue precipitate.
    [Cu(H₂O)₆]²⁺ + 2OH⁻ → [Cu(H₂O)₄(OH)²] + 2H₂O (or neutral copper hydroxide complex).
  • Dengan Excess NH₃(aq): Blue solution forms a blue precipitate initially, which dissolves in excess ammonia to form a deep blue solution.
    [Cu(H₂O)₆]²⁺ + 4NH₃ → [Cu(H₂O)₂(NH₃)₄]²⁺ + 4H₂O (ligand exchange).
  • With Concentrated HCl(aq): Blue solution turns yellow / green due to a ligand substitution reaction.
    [Cu(H₂O)₆]²⁺ + 4Cl⁻ → [CuCl₄]²⁻ + 6H₂O (octahedral to tetrahedral).

🧠 Extended Writing & Structure

  • This is a quality of written communication (*QWC) question. Structure your answer clearly with distinct paragraphs or headings for each reagent tested.
  • Explicitly link every visual observation (precipitate colours, solution changes) directly to the chemical equation and specific copper species responsible.
Marks: 6 marks total (including indicative content and structure)

Topics

Inorganic Chemistry · Physical Chemistry · Topic 15: Transition Metals · Topic 14: Redox II

Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 3, June 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.