Edexcel International A Level Chemistry: Transition metals: redox, reactions and catalysis (YCH11) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

This practice set covers outcomes 17.18 to 17.33 of Topic 17: Transition Metals and their Chemistry, including Core Practical 14, from the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017. These outcomes are examined in the IA2 paper Unit 5: Transition Metals and Organic Nitrogen Chemistry. Attempt them only after the unit’s study guide and revision notes; the checklist and Edexcel chemistry hub help you see what is left.

Use these values throughout. E⦵ / V: VO₂⁺/VO²⁺ +1.00; VO²⁺/V³⁺ +0.34; V³⁺/V²⁺ −0.26; I₂/I⁻ +0.54; Zn²⁺/Zn −0.76; Cr₂O₇²⁻/Cr³⁺ +1.33; Cr³⁺/Cr²⁺ −0.41; O₂/H₂O +1.23; Fe³⁺/Fe²⁺ +0.77; S₂O₈²⁻/SO₄²⁻ +2.01. Ar: O 16.0, Ni 58.7, H 1.0, Cl 35.5, N 14.0. R = 8.31 J K⁻¹ mol⁻¹.

Questions

1. Zinc granules are added to acidified ammonium vanadate(V). The colours seen in order are yellow, green, blue, green, violet.

(a) Give the formula of the vanadium species responsible for the blue colour and for the violet colour. [2] (b) Explain why green is seen twice. [2]

2. This question uses the E⦵ values above.

(a) Show that iodide ions can reduce VO₂⁺ to VO²⁺ but cannot reduce VO²⁺ to V³⁺. [2] (b) Write the overall ionic equation for the reaction of I⁻ with VO₂⁺. [2] (c) A solution of V²⁺ is mixed with an equal amount of VO²⁺ in acid. Predict the vanadium product, justify it with E⦵cell, and write the equation. [3]

3. Orange potassium dichromate(VI) solution is acidified and left with excess zinc in a flask that lets hydrogen escape but keeps air out.

(a) Describe the colour changes and give the chromium species at the end. [2] (b) Calculate E⦵cell for the final reduction step and write its equation. [2] (c) Explain, using E⦵ values, why air must be kept out. [1]

4. A student converts Cr³⁺(aq) into dichromate(VI).

(a) State what is seen when excess NaOH(aq) is added to Cr³⁺(aq), and write the equation for the dissolving step. [2] (b) Hydrogen peroxide is added and the mixture warmed. State the colour change and write the equation. [2] (c) Explain why the excess H₂O₂ is removed before the solution is acidified, and state the final colour. [2]

5. Explain what is seen when dilute acid is added to yellow potassium chromate(VI) solution, and why the change is not a redox reaction. [3]

6. Solutions P, Q and R each hold a single cation from 17.22.

	P	Q	R
Solution colour	pink	colourless	green
Few drops NaOH(aq)	blue ppt	white ppt	green ppt
Excess NaOH(aq)	no change	ppt dissolves	no change
Excess NH₃(aq)	straw-coloured solution, darkens in air	ppt dissolves	blue solution

(a) Identify the cation in each of P, Q and R. [3] (b) Write the equation for Q’s precipitate dissolving in excess NaOH(aq), and name this behaviour. [2] (c) Write the equation for R’s precipitate dissolving in excess NH₃(aq). [1]

7. Hexaaquairon(III) ions react with aqueous ammonia.

(a) Write an equation for the reaction and name the type of reaction. [2] (b) Explain why the iron(III) precipitate does not dissolve in excess ammonia, although the copper(II) precipitate does. [1]

8. A pink solution of cobalt(II) chloride is treated with concentrated hydrochloric acid, then diluted with water.

(a) Write the equation for the change with concentrated acid and state the colour change. [2] (b) State the shape and coordination number of each complex. [2] (c) Explain the effect of then adding water. [2]

9. Ethane-1,2-diamine (en) replaces ammonia in a cobalt(II) complex:

[Co(NH₃)₆]²⁺ + 3en → [Co(en)₃]²⁺ + 6NH₃

Use these values at 298 K: ΔH = −7.0 kJ mol⁻¹, ΔSsystem = +88 J K⁻¹ mol⁻¹.

(a) Explain why ΔSsystem is positive. [1] (b) Calculate ΔStotal. [2] (c) Calculate K and explain what it shows about the stability of [Co(en)₃]²⁺. [3]

10. This question is about heterogeneous catalysis.

(a) Explain, with equations and oxidation numbers, how V₂O₅ catalyses the oxidation of sulfur dioxide. [3] (b) Describe how a catalytic converter removes CO and NO from exhaust gas. Include an equation. [3]

11. This question is about homogeneous catalysis.

(a) Explain why S₂O₈²⁻ reacts slowly with I⁻ without a catalyst, even though E⦵cell is large. [1] (b) Use E⦵ values to show that Fe²⁺ can catalyse this reaction. Write both steps. [4] (c) Acidified KMnO₄ is added to warm ethanedioate solution. Describe how [MnO₄⁻] changes with time, and explain the shape. [3]

12. Core Practical 14. A student prepares hexaamminenickel(II) chloride, [Ni(NH₃)₆]Cl₂, from 2.50 g of NiCl₂·6H₂O dissolved in the minimum of water. Excess concentrated ammonia is added in a fume cupboard, the mixture is cooled in ice and the crystals are filtered off.

(a) Calculate the Mr of NiCl₂·6H₂O and of the product. [2] (b) The dry product weighs 1.85 g. Calculate the percentage yield. [3] (c) Explain why the mixture is cooled in ice before filtering. [1] (d) Suggest two causes of the yield falling short of 100%. [2]

Answers

1. (a) Blue: VO²⁺ [1]; violet: V²⁺ [1] (b) First green: a mixture of yellow VO₂⁺ and blue VO²⁺ [1]; second green: V³⁺ [1] Examiner insight: The right colour with the wrong formula, such as VO₂⁺ for blue, does not gain credit; work out V’s oxidation number from the formula.

2. (a) VO₂⁺ → VO²⁺: E⦵cell = 1.00 − 0.54 = +0.46 V, positive so feasible [1]; VO²⁺ → V³⁺: 0.34 − 0.54 = −0.20 V, negative so not feasible [1] (b) Electrons balanced, two VO₂⁺ per I₂ [1]: 2VO₂⁺ + 4H⁺ + 2I⁻ → 2VO²⁺ + I₂ + 2H₂O [1] (c) E⦵cell = 0.34 − (−0.26) = +0.60 V, so VO²⁺ is reduced and V²⁺ oxidised [1]; both give V³⁺ [1]; VO²⁺ + V²⁺ + 2H⁺ → 2V³⁺ + H₂O [1] Examiner insight: “Show that” needs both E⦵cell values written out with signs and a conclusion for each; a bare “yes/no” is not enough.

3. (a) Orange to green (Cr³⁺) then blue [1]; final species Cr²⁺ [1] (b) E⦵cell = −0.41 − (−0.76) = +0.35 V [1]; 2Cr³⁺ + Zn → 2Cr²⁺ + Zn²⁺ [1] (c) Oxygen would oxidise Cr²⁺ back to Cr³⁺: E⦵cell = 1.23 − (−0.41) = +1.64 V, which is positive [1] Examiner insight: Give the colour sequence in order with the species; “turns blue” alone misses the green Cr³⁺ stage.

4. (a) Grey-green precipitate that dissolves to a dark green solution [1]; [Cr(OH)₃(H₂O)₃] + 3OH⁻ → [Cr(OH)₆]³⁻ + 3H₂O [1] (b) Dark green to yellow [1]; equation: 2[Cr(OH)₆]³⁻ + 3H₂O₂ → 2CrO₄²⁻ + 2OH⁻ + 8H₂O, balanced [1] (c) In acid, leftover H₂O₂ would reduce the dichromate(VI) back to Cr³⁺ [1]; final colour orange [1] Examiner insight: Balance the H₂O₂ equation by electrons first (each Cr loses 3, each H₂O₂ gains 2), then check charge.

5. Yellow turns orange [1]; H⁺ shifts 2CrO₄²⁻ + 2H⁺ ⇌ Cr₂O₇²⁻ + H₂O to the right [1]; chromium is +6 in both ions, so no oxidation number changes [1] Examiner insight: Quote the oxidation number of Cr in both species; “it is an equilibrium” does not explain why it is not redox.

6. (a) P Co²⁺ [1]; Q Zn²⁺ [1]; R Ni²⁺ [1] (b) [Zn(OH)₂(H₂O)₄] + 2OH⁻ → [Zn(OH)₄]²⁻ + 4H₂O [1]; amphoteric behaviour [1] (c) [Ni(OH)₂(H₂O)₄] + 6NH₃ → [Ni(NH₃)₆]²⁺ + 2OH⁻ + 4H₂O [1] Examiner insight: R’s green solution alone could suggest Fe²⁺ or Cr³⁺; use the excess-reagent results to justify the identity.

7. (a) [Fe(H₂O)₆]³⁺ + 3NH₃ → [Fe(OH)₃(H₂O)₃] + 3NH₄⁺ [1]; deprotonation (acid-base) [1] (b) NH₃ does not replace the ligands on iron(III) to form a soluble ammine complex, whereas with copper(II) ligand exchange gives [Cu(NH₃)₄(H₂O)₂]²⁺ [1] Examiner insight: With ammonia as the base, the products include NH₄⁺, not H₂O; check that charges balance.

8. (a) [Co(H₂O)₆]²⁺ + 4Cl⁻ ⇌ [CoCl₄]²⁻ + 6H₂O [1]; pink to blue [1] (b) [Co(H₂O)₆]²⁺: octahedral, 6 [1]; [CoCl₄]²⁻: tetrahedral, 4 [1] (c) Water increases the H₂O concentration and lowers [Cl⁻], so the equilibrium shifts left [1]; the solution turns pink again [1] Examiner insight: Explain the shape change by the larger size of Cl⁻; “the charge changes” does not account for four ligands.

9. (a) Four particles become seven, so disorder increases [1] (b) ΔSsurroundings = −ΔH/T = +7000/298 = +23.5 J K⁻¹ mol⁻¹ [1]; ΔStotal = 88 + 23.5 = +111.5 J K⁻¹ mol⁻¹ [1] (c) ln K = ΔStotal/R = 111.5/8.31 = 13.4 [1]; K = 7 × 10⁵ (about) [1]; K is very large, so the equilibrium lies far to the right and the chelate complex is much more stable [1] Examiner insight: Change ΔH from kJ to J (×1000) before you divide by T; skipping this makes ΔSsurroundings 1000 times too small.

10. (a) SO₂ + V₂O₅ → SO₃ + V₂O₄, V reduced from +5 to +4 [1]; V₂O₄ + ½O₂ → V₂O₅, V oxidised from +4 to +5 [1]; V₂O₅ is regenerated, so it is a catalyst [1] (b) CO and NO adsorb onto the metal surface, weakening their bonds [1]; they react: 2CO + 2NO → 2CO₂ + N₂ [1]; CO₂ and N₂ desorb, freeing the surface [1] Examiner insight: Name both adsorption and desorption; the specification describes both steps.

11. (a) S₂O₈²⁻ and I⁻ are both anions; their mutual repulsion gives a high activation energy [1] (b) Step 1: S₂O₈²⁻ + 2Fe²⁺ → 2SO₄²⁻ + 2Fe³⁺ [1]; E⦵cell = 2.01 − 0.77 = +1.24 V [1]; step 2: 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂ [1]; for step 2, 0.77 − 0.54 gives +0.23 V, so both steps are feasible and Fe²⁺ is regenerated [1] (c) Slow fall at first [1]; then a steep fall as Mn²⁺ (a product) catalyses the reaction [1]; then levels off once the reactants run out [1] Examiner insight: In part (b), each step needs its own E⦵cell; one overall value does not show that iron can both gain and lose electrons.

12. (a) NiCl₂·6H₂O = 237.7 [1]; [Ni(NH₃)₆]Cl₂ = 231.7 [1] (b) n(NiCl₂·6H₂O) = 2.50/237.7 = 0.01052 mol [1]; theoretical mass = 0.01052 × 231.7 = 2.437 g [1]; yield = 1.85/2.437 × 100 = 75.9% [1] (c) The complex is less soluble at low temperature, so more crystallises out [1] (d) Any two: product left in the filtrate [1]; losses on transfer or filtering, or product dissolved on washing [1] Examiner insight: Mole ratio is 1 : 1 Ni to complex; include all six waters of crystallisation in the starting Mr and round only at the end.

Where marks are usually lost

• Swapping VO₂⁺ and VO²⁺.
• Subtracting E⦵ values the wrong way, or writing E⦵cell without a sign.
• Stopping a reduction at the wrong oxidation state because the next step was not checked.
• Giving one observation when two (precipitate and excess) are needed.
• Writing precipitates as charged complexes.
• Missing NH₄⁺ in deprotonation by ammonia.
• Stating the chelate effect without counting particles.
• Leaving out desorption in heterogeneous catalysis.
• Describing autocatalysis as a steady decrease.
• Forgetting water of crystallisation in Mr values.

Next steps

• Transition metals revision notes
• Transition metals study guide
• Edexcel International A Level Chemistry hub
• Printable Edexcel Chemistry checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017, Pearson Education Limited – Unit 5, Topic 17: Transition Metals and their Chemistry, outcomes 17.18 to 17.33 (including Core Practical 14).