AQA A-Level Chemistry Paper 1, 2018: Question 9

7 marks · Medium difficulty · State/Explain/Describe

Use electrode potential data to identify a reducing agent for vanadium ions, determine an oxidation state, draw complex isomers, and write equations involving vanadium compounds and catalysis.

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Question

Question text

09 This question is about vanadium compounds and ions.

Use data from Table 4 to identify the species that can be used to reduce VO + ions to

09.1 2

VO2+ in aqueous solution and no further.

Explain your answer.

Table 4

Electrode half-equation EƟ / V

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

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

− –

Cl2(aq) + 2e → 2Cl (aq) +1.36

Fe3+(aq) + e− → Fe2+(aq) +0.77

Zn2+(aq) + 2e− → Zn(s) –0.76

[2 marks]

Reagent

Explanation

Give the oxidation state of vanadium in [VO(H O) ]2+

09.2 2 5

[1 mark]

The [V(H O) Cl ]+ ion exists as two isomers. One isomer is shown.

09.3 2 4 2

Draw the structure of the other isomer and state the type of isomerism.

[2 marks]

Type of isomerism

09.4 Heating NH4VO3 produces vanadium(V) oxide, water and one other product.

Give an equation for the reaction.

[1 mark]

09.5 Vanadium(V) oxide is the catalyst used in the manufacture of sulfur trioxide.

Give two equations to show how the catalyst is used and regenerated.

[1 mark]

Mark scheme

Show the mark scheme

Question Answers Additional Comments/Guidance Mark

Fe2+ Accept any Fe(II) compound – correct formula or 1

name

EƟ VO +(/ VO2+) > EƟ Fe3+(/Fe2+) > EƟ VO2+(/V3+)

If calculations of EMF are provided producing

09.1 EMFs = 0.23(V) and -0.43(V), with a comment,

allow M2

allow EƟ Fe3+ (/Fe2+) value of +0.77 is between the

EƟ values for the electrode half-equations

containing the V species or wtte

09.2 (+) 4 IV or four 1

+ Ignore absence of charge 1

H2O

H O Wedges, dotted lines and [ ] not required

Do not penalise bond from H to V (in water ligands)

09.3 Cl OH

H2O

Cis/trans allow E/Z, geometric and stereo(isomerism) 1

09.4 2 NH4VO3 V2O5 + H2O + 2NH3 Accept multiples 1

Ignore state symbols

09.5 V2O5 + SO2 V2O4 + SO3 Both equations needed for 1 mark in this order 1

V2O4 + ½ O2 V2O5 Allow multiples

Total 7

How to answer it

Vanadium Chemistry: Redox, Complex Isomerism & Catalysis

What this question tests

This question assesses your understanding of transition metal chemistry: selecting selective reducing agents using standard electrode potentials (E^⦵), determining oxidation states in oxovanadium ions, identifying and drawing geometric (cis/trans) isomers of octahedral complexes, balancing inorganic thermal decomposition equations, and detailing the two-step mechanism of vanadium(V) oxide in the Contact Process.

Question 09.1 • 2 Marks

Selective Reduction of Dioxovanadium(V) Ions

Using Electrode Potentials to Prevent Over-reduction

✅ Correct Answers

Reagent: Fe²⁺ (or any iron(II) compound, e.g., FeSO₄)

Explanation:

E^⦵(VO₂⁺/VO²⁺) > E^⦵(Fe³⁺/Fe²⁺) > E^⦵(VO²⁺/V³⁺)
Alternatively: The E^⦵ value for Fe³⁺/Fe²⁺ (+0.77 V) lies between +1.00 V and +0.34 V.
Or calculate EMFs: Fe²⁺ reduces VO₂⁺ (EMF = +1.00 − 0.77 = +0.23 V > 0, feasible), but cannot reduce VO²⁺ (EMF = +0.34 − 0.77 = −0.43 V < 0, not feasible).

💡 Key Knowledge

A reaction is feasible under standard conditions if E^⦵(reduction) > E^⦵(oxidation), meaning EMF > 0.
To reduce VO₂⁺ to VO²⁺ (+1.00 V), the reducing agent must have an E^⦵ < +1.00 V.
To stop at VO²⁺ and prevent further reduction to V³⁺ (+0.34 V), the reducing agent must have an E^⦵ > +0.34 V.

❌ Common Errors

Naming the oxidized form: Writing Fe³⁺ instead of the reducing agent Fe²⁺.
Choosing Zinc (Zn): Zn will reduce VO₂⁺ completely through VO²⁺ and V³⁺ down to V²⁺ because E^⦵(Zn²⁺/Zn) = −0.76 V is lower than all vanadium systems.
Choosing Cl⁻: E^⦵(Cl₂/2Cl⁻) = +1.36 V, so Cl⁻ cannot reduce VO₂⁺ (+1.00 V).

🧠 Exam Technique

Always state the species clearly as an ion or full chemical name. For the second mark, an inequality chain like +1.00 V > +0.77 V > +0.34 V clearly shows the examiner that you understand the "window" required for selective reduction.

Mark Breakdown: 1 mark for identifying Fe²⁺ (or named Fe(II) compound); 1 mark for a valid comparative explanation referencing E^⦵ values or calculated cell EMFs.

Question 09.2 • 1 Mark

Oxidation State in Oxovanadium Complexes

Deducing Vanadium's Oxidation State in [VO(H₂O)₅]²⁺

✅ Correct Answer

+4 (or 4 / IV )

📐 Step-by-Step Calculation

Overall charge on complex = +2
Water (H₂O) is a neutral ligand: charge = 0 × 5 = 0
The bonded oxygen is an oxo ligand (oxide ion, O²⁻): charge = −2
Algebraic sum: V + (−2) + 0 = +2 → V = +4

❌ Common Trap

Treating oxygen as neutral (like water) or mistaking the formula for an aqua complex with an OH⁻ ligand. Remember: "VO" contains an oxo group with a −2 charge.

🧠 Top Tip

Oxidation numbers must have the sign before or after (convention prefers +4), though the mark scheme generously accepts just '4' or 'IV'. Always write the '+' sign in inorganic chemistry to be safe.

Mark Breakdown: 1 mark for +4 / IV.

Question 09.3 • 2 Marks

Stereoisomerism in Octahedral Complexes

Drawing the trans-Isomer of [V(H₂O)₄Cl₂]⁺

✅ Correct Answer

Structure to draw:

An octahedral diagram showing the two Cl ligands directly opposite each other (180° apart):

One Cl at the top axial position and one Cl at the bottom axial position, with four equatorial H₂O ligands.
OR: Two Cl ligands opposite each other in the equatorial plane (e.g., front-left and back-right), with H₂O at the top, bottom, and remaining equatorial positions.

Type of isomerism: Cis/trans (also allow: geometric, stereoisomerism, or E/Z).

🧠 Diagram Drawing Requirements

The given diagram has Cl ligands at 90° to each other (cis). The other isomer must have Cl ligands at 180° (trans).
Square brackets, 3D wedges/dashed bonds, and the overall '+' charge are not strictly required to gain the mark, but it is best practice to include them.
Bonding directly from O to V in H₂O (e.g., OH₂) is good practice, though examiners do not penalize bonds drawn to H.

Mark Breakdown: 1 mark for correct trans structure with 180° Cl–V–Cl alignment; 1 mark for stating cis/trans (or geometric/stereoisomerism).

Question 09.4 • 1 Mark

Thermal Decomposition of Ammonium Metavanadate

Deducing Missing Reaction Products

✅ Balanced Equation

2 NH₄VO₃ → V₂O₅ + H₂O + 2 NH₃

Multiples are accepted. State symbols are not required.

📐 Balancing Strategy

The question specifies products: vanadium(V) oxide (V₂O₅), water (H₂O), and one other product.
Look at vanadium: V₂O₅ has two V atoms, so start with 2 NH₄VO₃ .
Count remaining atoms on LHS: 2 N, 8 H, 6 O.
Subtract V₂O₅ and H₂O from RHS: leaves 2 N and 6 H.
2 N + 6 H = 2 NH₃ (ammonia gas).

Mark Breakdown: 1 mark for the fully balanced equation.

Question 09.5 • 1 Mark

Vanadium(V) Oxide in Heterogeneous Catalysis

Two-Step Mechanism of the Contact Process

✅ Required Equations

Step 1 (Reduction of Catalyst):
V₂O₅ + SO₂ → V₂O₄ + SO₃
Step 2 (Regeneration of Catalyst):
V₂O₄ + ½ O₂ → V₂O₅
(or 2 V₂O₄ + O₂ → 2 V₂O₅)

💡 Key Knowledge

V₂O₅ acts as a catalyst by changing its oxidation state from +5 (in V₂O₅) down to +4 (in V₂O₄), then back to +5.
Overall reaction: SO₂ + ½ O₂ → SO₃ .
This is a classic example of variable oxidation states enabling transition metal catalysis.

❌ Common Errors

Reversed order: Writing the regeneration step first. The question asks how the catalyst is used then regenerated.
Writing the overall reaction: Providing only the net reaction ( 2SO₂ + O₂ → 2SO₃ ) scores 0 marks.
Partial credit trap: Both equations are required in the correct sequence to earn the single mark available.

Mark Breakdown: 1 mark for BOTH equations provided in the correct sequential order.

Topics

Physical Chemistry · Inorganic Chemistry · 3.1.7 Oxidation, Reduction and Redox Equations · 3.1.11 Electrode Potentials · 3.2.5 Transition Metals · 3.1.2 Amount of Substance

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 1, 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.