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

This page condenses the study guide for last-minute review; go there for the full teaching and worked examples. It follows the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017, outcomes 17.18 to 17.33 of Topic 17, including Core Practical 14. All of it is examined in the IA2 paper Unit 5: Transition Metals and Organic Nitrogen Chemistry.

Test yourself afterwards with the practice questions, tick off outcomes on the printable checklist, find related pages on the course hub and spot weak areas with the free diagnostics.

Colours to know cold

Species	Oxidation state	Colour
VO₂⁺	V +5	yellow
VO²⁺	V +4	blue
V³⁺	V +3	green
V²⁺	V +2	violet
Cr₂O₇²⁻	Cr +6	orange
CrO₄²⁻	Cr +6	yellow
Cr³⁺(aq)	Cr +3	green
Cr²⁺(aq)	Cr +2	blue
[CuCl₄]²⁻	Cu +2	yellow
[CoCl₄]²⁻	Co +2	blue

Tip for vanadium formulae: in VO₂⁺, V = +1 + 4 = +5; in VO²⁺, V = +2 + 2 = +4.

Redox with E⦵ values (17.19-17.20)

Values used on this page (use the ones a question gives you):

Couple	E⦵ / V
VO₂⁺/VO²⁺	+1.00
VO²⁺/V³⁺	+0.34
V³⁺/V²⁺	−0.26
Zn²⁺/Zn	−0.76
Fe³⁺/Fe²⁺	+0.77
Cr₂O₇²⁻/Cr³⁺	+1.33
Cr³⁺/Cr²⁺	−0.41

Method: how far will a reductant take a metal?

1. List the metal’s couples from most positive to least positive.
2. For each step, E⦵cell = E⦵(couple being reduced) − E⦵(reductant couple).
3. Keep going while E⦵cell is positive. The first negative step is where it stops.
4. Balance electrons, then add the half-equations.

Worked reminder. Fe²⁺ with VO₂⁺: step 1 gives 1.00 − 0.77 = +0.23 V (goes); step 2 gives 0.34 − 0.77 = −0.43 V (does not). Fe²⁺ stops at blue VO²⁺:

VO₂⁺ + 2H⁺ + Fe²⁺ → VO²⁺ + H₂O + Fe³⁺

Zinc in acid reduces both metals to +2: V to violet V²⁺ and Cr to blue Cr²⁺. Exclude air, or O₂ reoxidises them.

Cr³⁺ to Cr(VI): excess NaOH, then warm with H₂O₂ (alkaline route), then boil off spare H₂O₂ and acidify.

green [Cr(OH)₆]³⁻ → yellow CrO₄²⁻ → orange Cr₂O₇²⁻

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

Equilibrium (17.21): Cr₂O₇²⁻ + H₂O ⇌ 2CrO₄²⁻ + 2H⁺. Alkali → yellow; acid → orange. Not redox (Cr stays +6).

Hydroxide and ammonia tests (17.22)

Ion	Precipitate	Excess NaOH	Excess NH₃
Cr³⁺	grey-green	dark green solution	purple solution
Mn²⁺	off-white, browns in air	stays	stays
Fe²⁺	green, browns in air	stays	stays
Fe³⁺	red-brown	stays	stays
Co²⁺	blue	stays	straw solution, darkens in air
Ni²⁺	green	stays	blue solution
Cu²⁺	pale blue	stays	deep blue solution
Zn²⁺	white	colourless solution	colourless solution

Memory hook: excess NaOH dissolves only Cr and Zn hydroxides (amphoteric). Excess NH₃ dissolves Cr, Co, Ni, Cu and Zn hydroxides (ligand exchange).

Equation patterns (17.23)

Type	Pattern	Example
Deprotonation	[M(H₂O)₆]ⁿ⁺ + nOH⁻ → [M(OH)ₙ(H₂O)₆₋ₙ] + nH₂O	[Fe(H₂O)₆]³⁺ + 3OH⁻ → [Fe(OH)₃(H₂O)₃] + 3H₂O
Deprotonation by NH₃	OH⁻ replaced by NH₃, product NH₄⁺	[Co(H₂O)₆]²⁺ + 2NH₃ → [Co(OH)₂(H₂O)₄] + 2NH₄⁺
Amphoteric (alkali)	hydroxide + OH⁻ → anion	[Zn(OH)₂(H₂O)₄] + 2OH⁻ → [Zn(OH)₄]²⁻ + 4H₂O
Amphoteric (acid)	hydroxide + H⁺ → aqua ion	[Cr(OH)₃(H₂O)₃] + 3H⁺ → [Cr(H₂O)₆]³⁺
Ligand exchange	hydroxide + NH₃ → ammine + OH⁻ + H₂O	[Zn(OH)₂(H₂O)₄] + 4NH₃ → [Zn(NH₃)₄]²⁺ + 2OH⁻ + 4H₂O

Check: atoms, charge, and that the precipitate is neutral.

Ligand exchange and stability (17.24-17.25)

• [Cu(H₂O)₆]²⁺ (pale blue) → [Cu(OH)₂(H₂O)₄] (pale blue ppt) → [Cu(NH₃)₄(H₂O)₂]²⁺ (deep blue). Four NH₃ replace four H₂O; coordination number stays 6.
• [Cu(H₂O)₆]²⁺ + 4Cl⁻ ⇌ [CuCl₄]²⁻ + 6H₂O: blue → yellow; 6 → 4; octahedral → tetrahedral.
• [Co(H₂O)₆]²⁺ + 4Cl⁻ ⇌ [CoCl₄]²⁻ + 6H₂O: pink → blue; same shape change. Water reverses both.
• Colour changes here come from a change of ligand and of coordination number.

Chelate stability in steps

1. Count particles on each side (EDTA⁴⁻ for six H₂O: 2 → 7; three en for six H₂O: 4 → 7).
2. More particles, so ΔSsystem is positive and large.
3. Similar bonds broken and made, so ΔH ≈ 0 and ΔSsurroundings = −ΔH/T is small.
4. ΔStotal = ΔSsystem + ΔSsurroundings is positive; ΔStotal = R ln K, so K is large.

Catalysis (17.26-17.32)

Catalyst	Type	Reaction	How
V₂O₅	heterogeneous	2SO₂ + O₂ ⇌ 2SO₃	V +5 → +4 oxidising SO₂; O₂ restores +5
Pt/Rh/Pd	heterogeneous	2CO + 2NO → 2CO₂ + N₂	adsorb, bonds weaken, react, desorb
Fe²⁺	homogeneous	S₂O₈²⁻ + 2I⁻ → 2SO₄²⁻ + I₂	Fe²⁺ → Fe³⁺ (intermediate) → Fe²⁺
Mn²⁺	homogeneous, autocatalyst	MnO₄⁻ + C₂O₄²⁻ in acid	Mn²⁺ → Mn³⁺ → Mn²⁺; Mn²⁺ is a product

• Heterogeneous: different phase; reaction at the surface.
• Homogeneous: same phase; goes via an intermediate species.
• V₂O₅ steps: SO₂ + V₂O₅ → SO₃ + V₂O₄, then V₂O₄ + ½O₂ → V₂O₅.
• Fe²⁺ steps: S₂O₈²⁻ + 2Fe²⁺ → 2SO₄²⁻ + 2Fe³⁺, then 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂. Uncatalysed reaction is slow because two anions repel.
• Autocatalysis curve of [MnO₄⁻] against time: slow fall, then steep fall, then levels off.

Worked reminder: checking a homogeneous catalyst with E⦵. A catalyst couple must lie between the two reacting couples. Fe³⁺/Fe²⁺ (+0.77 V) sits between S₂O₈²⁻/SO₄²⁻ (+2.01 V) and I₂/I⁻ (+0.54 V), so both steps have a positive E⦵cell: +1.24 V and +0.23 V.

Core Practical 14 in brief (17.33)

The specification asks for “the preparation of a transition metal complex”. Know a full method for one complex, for example chromium(II) ethanoate (zinc and acid reduce dichromate to blue Cr²⁺, which is added to sodium ethanoate to give a red precipitate), and how to work out percentage yield:

yield (%) = actual mass ÷ (moles of limiting reagent × ratio × Mr of product) × 100

Air-sensitive products need oxygen excluded during reaction, filtering and washing.

Must-know distinctions

• VO₂⁺ vs VO²⁺: +5 yellow vs +4 blue.
• Deprotonation vs ligand exchange: a little NH₃ acts as a base; excess NH₃ acts as a ligand.
• Amphoteric vs ligand exchange: Zn(OH)₂ dissolving in NaOH is amphoteric; dissolving in NH₃ is ligand exchange.
• Catalyst vs autocatalyst: an autocatalyst forms during the reaction and then catalyses it (Mn²⁺).
• Heterogeneous vs homogeneous: phase relative to the reactants, not “solid vs liquid”.

Quick self-test

1. Which vanadium ion is blue, and what is vanadium’s oxidation number in it?
2. Data as in the E⦵ table above. Can zinc reduce V³⁺ to V²⁺? Give E⦵cell.
3. How many moles of zinc reduce 0.0100 mol of VO₂⁺ to V²⁺?
4. What colour change shows Cr³⁺ being oxidised to chromate(VI)?
5. Why is Cr₂O₇²⁻ + H₂O ⇌ 2CrO₄²⁻ + 2H⁺ not a redox reaction?
6. Which two ions in 17.22 give precipitates that dissolve in excess NaOH(aq)?
7. Name the cobalt complex and its colour in excess NH₃(aq).
8. Write the equation for [Cu(H₂O)₆]²⁺ with concentrated hydrochloric acid.
9. A ligand swap has ΔH = −2.5 kJ mol⁻¹ and ΔSsystem = +62 J K⁻¹ mol⁻¹. Find ΔStotal at 298 K.
10. Write the two steps for V₂O₅ in the Contact process.
11. Why does Fe²⁺ speed up the reaction of S₂O₈²⁻ with I⁻?
12. Moles of zinc needed to reduce 0.00200 mol Cr₂O₇²⁻ all the way to Cr²⁺?

Answers

1. VO²⁺, +4.
2. Yes: −0.26 − (−0.76) = +0.50 V.
3. V goes +5 → +2: 3 e⁻ each, so 0.0300 mol e⁻; Zn gives 2 e⁻, so 0.0150 mol Zn.
4. Dark green to yellow.
5. Cr is +6 on both sides; only H⁺ is transferred.
6. Cr³⁺ and Zn²⁺.
7. [Co(NH₃)₆]²⁺, straw-coloured (darkens in air).
8. [Cu(H₂O)₆]²⁺ + 4Cl⁻ ⇌ [CuCl₄]²⁻ + 6H₂O.
9. ΔSsurroundings = +2500/298 = +8.4; ΔStotal = +70.4 J K⁻¹ mol⁻¹.
10. SO₂ + V₂O₅ → SO₃ + V₂O₄; V₂O₄ + ½O₂ → V₂O₅.
11. Each catalysed step is between oppositely charged ions, so the activation energy is lower than for two repelling anions.
12. Each Cr goes +6 → +2 (4 e⁻), so 8 e⁻ per Cr₂O₇²⁻ = 0.0160 mol e⁻; 0.00800 mol Zn.

Where marks are usually lost

• Swapping VO₂⁺ and VO²⁺, or giving the right colour with the wrong formula.
• Subtracting E⦵ values the wrong way round, so a feasible step looks infeasible.
• Forgetting that Cr²⁺ and V²⁺ are reoxidised by air.
• Giving only the precipitate colour when the question also asks about excess reagent.
• Writing NH₃ as OH⁻ in a deprotonation equation without forming NH₄⁺.
• Charged formulae for precipitates, such as [Cu(OH)₂(H₂O)₄]²⁺.
• Saying [CuCl₄]²⁻ is octahedral, or that its formation is redox.
• Explaining the chelate effect with bond strength instead of the increase in ΔSsystem.
• Leaving out desorption when describing a catalytic converter.
• Calling Mn²⁺ a reactant, or describing the autocatalysis curve as a steady fall.

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).