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

This guide teaches learning outcomes 17.18 to 17.33, with Core Practical 14, which make up the second half of Topic 17, Transition Metals and their Chemistry. The source is the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017. The topic sits in the IA2 (International A2) paper Unit 5: Transition Metals and Organic Nitrogen Chemistry. The specification states that Unit 6: Practical Skills in Chemistry II will predominantly assess knowledge and understanding related to Units 4 and 5. It builds on Topic 16 (electrode potentials) and Topic 12 (entropy).

Condensed recall is in the revision notes, and original exam-style questions are in the practice set. To plan, use the Edexcel chemistry hub, tick outcomes off on the checklist, and find weak spots with a 10-minute diagnostic.

Outcomes in this guide

Spec outcome	Skills you need
17.18-17.19	Give vanadium colours in the +5, +4, +3 and +2 states; explain interconversions with E⦵ values
17.20-17.21	Reduce dichromate(VI) with zinc in acid; make it from Cr³⁺ with H₂O₂ in alkali; the dichromate/chromate equilibrium
17.22-17.23	Observations and equations for eight aqua ions with NaOH(aq) and NH₃(aq), including excess
17.24-17.25	Ligand exchange with NH₃ and Cl⁻; why chelating ligands give more stable complexes
17.26-17.32	Heterogeneous and homogeneous catalysis: V₂O₅, catalytic converters, Fe²⁺, Mn²⁺
17.33	Core Practical 14: preparing a transition metal complex

Vanadium: four oxidation states (17.18-17.19)

Ion in acid	Name	Oxidation number of V	Colour
VO₂⁺	dioxovanadium(V)	+5	yellow
VO²⁺	oxovanadium(IV)	+4	blue
V³⁺	vanadium(III)	+3	green
V²⁺	vanadium(II)	+2	violet

Zinc in acid takes vanadium from +5 down to +2 one step at a time. While yellow VO₂⁺ and blue VO²⁺ are both present, the solution looks green, so green appears twice: once as a mixture, then as V³⁺.

Use these electrode potentials on this page:

Half-equation	E⦵ / V
VO₂⁺ + 2H⁺ + e⁻ ⇌ VO²⁺ + H₂O	+1.00
VO²⁺ + 2H⁺ + e⁻ ⇌ V³⁺ + H₂O	+0.34
V³⁺ + e⁻ ⇌ V²⁺	−0.26
Zn²⁺ + 2e⁻ ⇌ Zn	−0.76
Sn⁴⁺ + 2e⁻ ⇌ Sn²⁺	+0.15

A reducing agent can push vanadium down one step only if E⦵cell = E⦵(vanadium couple) − E⦵(reductant couple) is positive. Zinc gives +1.76 V, +1.10 V and +0.50 V for the three steps, so it reaches V²⁺.

Worked example 1: how far does tin(II) go?

Acidified ammonium vanadate(V) is shaken with excess Sn²⁺ ions.

1. +5 to +4: E⦵cell = 1.00 − 0.15 = +0.85 V. Feasible.
2. +4 to +3: E⦵cell = 0.34 − 0.15 = +0.19 V. Feasible.
3. +3 to +2: E⦵cell = −0.26 − 0.15 = −0.41 V. Not feasible.
4. The final species is V³⁺, so the solution ends green.
5. Equation for step 2, with electrons balanced (two VO²⁺ per Sn²⁺): 2VO²⁺ + 4H⁺ + Sn²⁺ → 2V³⁺ + 2H₂O + Sn⁴⁺

A positive E⦵cell shows feasibility only. It says nothing about rate.

Chromium: dichromate(VI), Cr³⁺ and Cr²⁺ (17.20-17.21)

Reduction with zinc in acid. Orange Cr₂O₇²⁻ is reduced first to green Cr³⁺, then to blue Cr²⁺. Data: Cr₂O₇²⁻/Cr³⁺ +1.33 V; Cr³⁺/Cr²⁺ −0.41 V.

Cr₂O₇²⁻ + 14H⁺ + 3Zn → 2Cr³⁺ + 7H₂O + 3Zn²⁺ E⦵cell = 1.33 − (−0.76) = +2.09 V

2Cr³⁺ + Zn → 2Cr²⁺ + Zn²⁺ E⦵cell = −0.41 − (−0.76) = +0.35 V

Keep Cr²⁺ away from air: with O₂/H₂O at +1.23 V, oxygen oxidises it back to Cr³⁺ (E⦵cell = +1.64 V).

Oxidation of Cr³⁺ with hydrogen peroxide. Add excess NaOH(aq) to Cr³⁺(aq): the grey-green precipitate redissolves to give dark green [Cr(OH)₆]³⁻. Warm with H₂O₂ and the solution turns yellow as chromate(VI) forms:

2[Cr(OH)₆]³⁻ + 3H₂O₂ → 2CrO₄²⁻ + 2OH⁻ + 8H₂O

Alkaline conditions matter. Use these values: CrO₄²⁻ + 4H₂O + 3e⁻ ⇌ [Cr(OH)₆]³⁻ + 2OH⁻, E⦵ = −0.13 V; H₂O₂ + 2e⁻ ⇌ 2OH⁻, E⦵ = +0.88 V. E⦵cell = 0.88 − (−0.13) = +1.01 V. In acid, the Cr(VI)/Cr(III) potential is far more positive (+1.33 V): chromium(III) is much harder to oxidise, and acidified dichromate(VI) oxidises hydrogen peroxide instead. So boil off leftover H₂O₂ first, then acidify: the yellow solution turns orange as dichromate(VI) forms.

The dichromate/chromate equilibrium (17.21).

Cr₂O₇²⁻ + H₂O ⇌ 2CrO₄²⁻ + 2H⁺ orange ⇌ yellow

Adding OH⁻ removes H⁺, so the position moves right and the colour goes yellow. Adding acid moves it left, back to orange. Chromium is +6 on both sides, so this is an acid-base equilibrium, not redox.

Aqua ions with NaOH(aq) and NH₃(aq) (17.22-17.23)

Both reagents first remove H⁺ from coordinated water molecules (deprotonation), giving a neutral hydroxide precipitate. What happens in excess is the part to learn.

Ion (solution colour)	Few drops NaOH or NH₃	Excess NaOH	Excess NH₃
Cr³⁺ (green)	grey-green ppt	dissolves: dark green [Cr(OH)₆]³⁻	dissolves: purple [Cr(NH₃)₆]³⁺
Mn²⁺ (very pale pink)	off-white ppt, darkens to brown in air	no change	no change
Fe²⁺ (pale green)	green ppt, turns brown at the surface in air	no change	no change
Fe³⁺ (yellow-brown)	red-brown ppt	no change	no change
Co²⁺ (pink)	blue ppt	no change	dissolves: straw-coloured [Co(NH₃)₆]²⁺, darkens in air
Ni²⁺ (green)	green ppt	no change	dissolves: blue [Ni(NH₃)₆]²⁺
Cu²⁺ (blue)	pale blue ppt	no change	dissolves: deep blue [Cu(NH₃)₄(H₂O)₂]²⁺
Zn²⁺ (colourless)	white ppt	dissolves: colourless [Zn(OH)₄]²⁻	dissolves: colourless [Zn(NH₃)₄]²⁺

The darkening of Mn(OH)₂, Fe(OH)₂ and the cobalt ammine is oxidation by air to a higher oxidation state.

Three kinds of equation (17.23)

Deprotonation (NaOH or a little NH₃, which acts as a base): [Mn(H₂O)₆]²⁺ + 2OH⁻ → [Mn(OH)₂(H₂O)₄] + 2H₂O [Fe(H₂O)₆]²⁺ + 2NH₃ → [Fe(OH)₂(H₂O)₄] + 2NH₄⁺

Amphoteric behaviour: a hydroxide that reacts with both acid and alkali. [Zn(OH)₂(H₂O)₄] + 2H⁺ → [Zn(H₂O)₆]²⁺ [Zn(OH)₂(H₂O)₄] + 2OH⁻ → [Zn(OH)₄]²⁻ + 4H₂O [Cr(OH)₃(H₂O)₃] + 3OH⁻ → [Cr(OH)₆]³⁻ + 3H₂O

Ligand exchange in excess ammonia (NH₃ now acts as a ligand): [Ni(OH)₂(H₂O)₄] + 6NH₃ → [Ni(NH₃)₆]²⁺ + 2OH⁻ + 4H₂O

A neutral precipitate has as many OH⁻ ligands as the metal’s charge.

Ligand exchange and colour change (17.24)

Copper with ammonia. Pale blue [Cu(H₂O)₆]²⁺ gives a pale blue precipitate, [Cu(OH)₂(H₂O)₄], then a deep blue solution:

[Cu(OH)₂(H₂O)₄] + 4NH₃ → [Cu(NH₃)₄(H₂O)₂]²⁺ + 2OH⁻ + 2H₂O

Overall, four NH₃ replace four of the six water ligands of [Cu(H₂O)₆]²⁺. Coordination number stays 6.

Chloride ligands. Concentrated hydrochloric acid supplies a high Cl⁻ concentration:

[Cu(H₂O)₆]²⁺ + 4Cl⁻ ⇌ [CuCl₄]²⁻ + 6H₂O (blue to yellow, often seen as green part-way) [Co(H₂O)₆]²⁺ + 4Cl⁻ ⇌ [CoCl₄]²⁻ + 6H₂O (pink to blue)

Cl⁻ is larger than H₂O, so only four fit: the shape changes from octahedral to tetrahedral and the coordination number from 6 to 4. Both colour changes come from a change of ligand and coordination number. Adding water reverses them.

Why chelating ligands win (17.25)

When a bidentate or hexadentate ligand replaces monodentate ones, the number of particles in solution rises. For example, [Mn(H₂O)₆]²⁺ + EDTA⁴⁻ → [Mn(EDTA)]²⁻ + 6H₂O goes from 2 particles to 7, so ΔSsystem is large and positive. The bonds broken and made are similar metal-ligand bonds, so ΔH is small and ΔSsurroundings (= −ΔH/T) is small too. ΔStotal is therefore positive, and since ΔStotal = R ln K, K is large: the chelated complex is more stable.

Worked example 2: an endothermic substitution that still happens

Use these values for the Mn²⁺/EDTA⁴⁻ reaction above at 298 K: ΔH = +4.0 kJ mol⁻¹, ΔSsystem = +190 J K⁻¹ mol⁻¹.

1. ΔSsurroundings = −ΔH/T = −4000/298 = −13.4 J K⁻¹ mol⁻¹.
2. ΔStotal = 190 + (−13.4) = +176.6 J K⁻¹ mol⁻¹.
3. Positive, so the reaction is feasible even though it is endothermic. The entropy gain of the system drives it.

Catalysis (17.26-17.32)

Transition metals and their compounds catalyse reactions because they have variable oxidation numbers and can form bonds to reactant molecules.

Heterogeneous catalysts (17.27) are not in the same phase as the reactants, and the chemistry takes place on the catalyst surface: reactants adsorb, their bonds weaken, they react, and products desorb.

V₂O₅ in the Contact process (17.28), for 2SO₂ + O₂ ⇌ 2SO₃:

SO₂ + V₂O₅ → SO₃ + V₂O₄ (V: +5 to +4) V₂O₄ + ½O₂ → V₂O₅ (V: +4 to +5)

V₂O₅ is regenerated each cycle.

Catalytic converters (17.29) remove CO and NO from engine exhaust:

2CO + 2NO → 2CO₂ + N₂

CO and NO adsorb onto the metal surface (platinum, rhodium or palladium on a ceramic honeycomb), which weakens their bonds so they react. CO₂ and N₂ then desorb, freeing the surface. The honeycomb gives a large surface area for a small mass of metal.

Homogeneous catalysts (17.30) are in the same phase as the reactants. The reaction goes via an intermediate species, giving two steps, each with a lower activation energy than the uncatalysed route.

Fe²⁺ with peroxodisulfate and iodide (17.31). S₂O₈²⁻ + 2I⁻ → 2SO₄²⁻ + I₂ is slow: both reactants carry negative charges and repel each other, which raises the activation energy. With Fe²⁺:

S₂O₈²⁻ + 2Fe²⁺ → 2SO₄²⁻ + 2Fe³⁺ E⦵cell = 2.01 − 0.77 = +1.24 V 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂ E⦵cell = 0.77 − 0.54 = +0.23 V

Each step involves oppositely charged ions. Fe³⁺ is the intermediate and Fe²⁺ is regenerated. Fe³⁺ works too, with the steps reversed.

Mn²⁺ as an autocatalyst (17.32). In 2MnO₄⁻ + 16H⁺ + 5C₂O₄²⁻ → 2Mn²⁺ + 10CO₂ + 8H₂O, the product Mn²⁺ catalyses the reaction. Mn²⁺ is oxidised to Mn³⁺ by MnO₄⁻, then Mn³⁺ oxidises C₂O₄²⁻ and returns to Mn²⁺. A plot of [MnO₄⁻] against time falls slowly at first, then steeply as Mn²⁺ builds up, then levels off as reactants run out.

Core Practical 14: preparing a complex (17.33)

The specification names it “the preparation of a transition metal complex”. A preparation that links to 17.20 is chromium(II) ethanoate:

1. Dissolve potassium dichromate(VI) in water and add zinc granules and concentrated hydrochloric acid.
2. Limit air entry. The colour goes orange, then green, then blue (Cr²⁺).
3. Push the blue solution into saturated sodium ethanoate solution. A red precipitate of chromium(II) ethanoate forms.
4. Filter quickly under reduced pressure and wash with water that has been boiled to remove oxygen, then ethanol.

Worked example 3: percentage yield

1.47 g of K₂Cr₂O₇ (Mr 294.2) gives 1.20 g of Cr₂(CH₃COO)₄(H₂O)₂ (Mr 376.0). One K₂Cr₂O₇ supplies two Cr, which is one formula unit of product.

1. n(K₂Cr₂O₇) = 1.47/294.2 = 4.997 × 10⁻³ mol.
2. Theoretical mass = 4.997 × 10⁻³ × 376.0 = 1.879 g.
3. Yield = 1.20/1.879 × 100 = 63.9%.

Losses come from air oxidation of Cr²⁺, product left in solution and transfer losses.

Common errors

• Writing VO²⁺ for +5 or VO₂⁺ for +4. Count: in VO₂⁺, V + 2(−2) = +1, so V is +5.
• Stopping zinc at Cr³⁺ or V³⁺ without checking the next E⦵ step.
• Calling Cr₂O₇²⁻ ⇌ CrO₄²⁻ a redox change.
• Saying excess NaOH dissolves Cu(OH)₂. Only the Cr and Zn hydroxides dissolve in excess NaOH.
• Saying [CuCl₄]²⁻ keeps coordination number 6.
• Explaining the chelate effect by “stronger bonds” rather than ΔSsystem.
• Describing a heterogeneous catalyst without both adsorption and desorption.

Next steps

Try the practice questions, then review with the revision notes. Outcomes 17.1-17.17 are in complexes and colour. Background: entropy and redox equilibria. Check gaps with the diagnostics.

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).