1. Variable Oxidation States
Transition metals can form ions with different oxidation states. The 3d and 4s electrons have similar energies, so different numbers of electrons can be lost. For example, iron forms Fe²⁺ and Fe³⁺, and manganese forms Mn²⁺ and MnO₄⁻ (+7).
Changing between oxidation states is a redox reaction. This page uses half-equations and oxidation states from Unit 1. Revise them on the Unit 1 redox page if you need to.
Variable oxidation states explain three things on this page:
- the colour changes of vanadium when it is reduced;
- redox titrations with manganate(VII);
- how transition metals act as catalysts.
Related practical: RP9, test-tube reactions of transition metal ions.
2. Vanadium: Four Oxidation States
Zinc and acid reduce vanadate(V) ions step by step. Each step lowers the oxidation state of vanadium by one, and each oxidation state has its own colour.
| Species | Oxidation state of V | Colour |
|---|---|---|
| VO₂⁺ | +5 | Yellow |
| VO²⁺ | +4 | Blue |
| V³⁺ | +3 | Green |
| V²⁺ | +2 | Violet |
Each arrow is a gain of one electron by vanadium. A green colour is also seen briefly at the start. This is not V³⁺. It is a mixture of yellow VO₂⁺ and blue VO²⁺.
Half-equations
- VO₂⁺ + 2H⁺ + e⁻ → VO²⁺ + H₂O (+5 to +4)
- VO²⁺ + 2H⁺ + e⁻ → V³⁺ + H₂O (+4 to +3)
- V³⁺ + e⁻ → V²⁺ (+3 to +2)
- Zinc is oxidised: Zn → Zn²⁺ + 2e⁻
For Zn²⁺ + 2e⁻ ⇌ Zn, E° = −0.76 V. This is more negative than the E° of each vanadium step. So zinc is a strong enough reducing agent to reduce vanadium(V) all the way to V²⁺. Use the rule from the electrode potentials page: the half-cell with the more negative E° supplies the electrons.
3. Redox Potentials Depend on pH and Ligand
The redox potential of a transition metal ion changing from a higher to a lower oxidation state is affected by the pH and by the ligand.
Effect of pH
In alkaline solution, metal ions are oxidised more easily than in acid. For example:
- Green Fe(OH)₂ darkens to brown Fe(OH)₃ when it stands in air.
- [Co(NH₃)₆]²⁺ is oxidised by air to [Co(NH₃)₆]³⁺.
Effect of the ligand: Tollens' reagent
Ligands change E°. In [Ag(NH₃)₂]⁺, the ammonia ligands make silver(I) a mild oxidising agent. This complex is Tollens' reagent.
- [Ag(NH₃)₂]⁺ + e⁻ → Ag + 2NH₃
- It oxidises aldehydes but not ketones, and it is reduced to silver (a silver mirror).
So Tollens' reagent can be used to tell aldehydes and ketones apart. See Aldehydes and Ketones in Unit 4.
4. Manganate(VII) Titrations
Potassium manganate(VII), KMnO₄, is a strong oxidising agent. It is used to find the amount of a reducing agent such as Fe²⁺ or ethanedioate (C₂O₄²⁻) ions.
The equations
- With iron(II): MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O
- With ethanedioate: 2MnO₄⁻ + 16H⁺ + 5C₂O₄²⁻ → 2Mn²⁺ + 10CO₂ + 8H₂O (warm the flask to about 60 °C)
The mole ratios you need are MnO₄⁻ : Fe²⁺ = 1 : 5 and MnO₄⁻ : C₂O₄²⁻ = 2 : 5.
The end point: self-indicating
MnO₄⁻ is purple and Mn²⁺ is very pale. While Fe²⁺ (or C₂O₄²⁻) is left in the flask, the purple colour disappears as each drop goes in. The end point is the first permanent pale pink colour, from the first drop of MnO₄⁻ in excess. No indicator is needed.
Which acid?
Acidify with dilute sulfuric acid. Other acids give wrong results:
- Not hydrochloric acid: MnO₄⁻ oxidises Cl⁻ to Cl₂. Extra MnO₄⁻ is used up, so the titre is too large.
- Not ethanoic acid: it is a weak acid, so there is not enough H⁺.
- Not nitric acid: it is an oxidising agent itself, so it oxidises some Fe²⁺ and the titre is too small.
Worked examples
Step 1: moles of MnO₄⁻ = 0.0200 × 22.50 / 1000 = 4.50 × 10⁻⁴ mol
Step 2: MnO₄⁻ : Fe²⁺ = 1 : 5, so moles of Fe²⁺ = 5 × 4.50 × 10⁻⁴ = 2.25 × 10⁻³ mol
Step 3: [Fe²⁺] = 2.25 × 10⁻³ / 0.0250 = 0.0900 mol dm⁻³
Step 1: moles of MnO₄⁻ = 0.00500 × 14.20 / 1000 = 7.10 × 10⁻⁵ mol
Step 2: moles of Fe²⁺ in 25.0 cm³ = 5 × 7.10 × 10⁻⁵ = 3.55 × 10⁻⁴ mol
Step 3: moles of Fe²⁺ in 250 cm³ = 10 × 3.55 × 10⁻⁴ = 3.55 × 10⁻³ mol
Step 4: mass of Fe = 3.55 × 10⁻³ × 55.8 = 0.198 g
Step 5: percentage of iron = 0.198 / 2.00 × 100 = 9.90%
Step 1: moles of MnO₄⁻ = 0.0200 × 24.00 / 1000 = 4.80 × 10⁻⁴ mol
Step 2: MnO₄⁻ : C₂O₄²⁻ = 2 : 5, so moles of C₂O₄²⁻ = 4.80 × 10⁻⁴ × 5/2 = 1.20 × 10⁻³ mol
Step 3: concentration = 1.20 × 10⁻³ / 0.0250 = 0.0480 mol dm⁻³
The titration technique (rinsing, reading the burette, concordant titres) is the same as in RP1.
5. Catalysts
Transition metals and their compounds are good catalysts. They can change oxidation state easily, and they can provide sites on their surface where reactions take place.
Heterogeneous catalysts
Usually the catalyst is a solid and the reactants are gases. Reactant molecules adsorb onto active sites on the surface. Bonds in the reactants weaken, the reaction happens, and the products desorb.
- Support medium: a ceramic honeycomb is coated with a thin layer of the catalyst. This gives a large surface area, so little of the expensive metal is needed.
- Poisoning: impurities bond to the active sites and block them. Sulfur poisons the iron catalyst in the Haber process. Lead poisons catalytic converters. A poisoned catalyst must be cleaned or replaced, which costs money.
The Contact process: vanadium(V) oxide
- Step 1: SO₂ + V₂O₅ → SO₃ + V₂O₄ (V reduced from +5 to +4)
- Step 2: V₂O₄ + ½O₂ → V₂O₅ (V oxidised back to +5)
- Overall: SO₂ + ½O₂ ⇌ SO₃
Vanadium changes between +5 and +4, and V₂O₅ is re-formed at the end. This shows why variable oxidation states matter in catalysis.
The Haber process: iron
Solid iron catalyses N₂ + 3H₂ ⇌ 2NH₃.
Homogeneous catalysts
Fe²⁺ catalysing the reaction of S₂O₈²⁻ with I⁻
S₂O₈²⁻ + 2I⁻ → 2SO₄²⁻ + I₂
The uncatalysed reaction is slow, because two negative ions repel each other. Fe²⁺ gives a route with two faster steps:
- Step 1: S₂O₈²⁻ + 2Fe²⁺ → 2SO₄²⁻ + 2Fe³⁺
- Step 2: 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂
Each step is between oppositely charged ions, so the ions attract. Fe²⁺ is re-formed in step 2. Fe³⁺ would also work as the catalyst, starting with step 2.
Autocatalysis by Mn²⁺
In the reaction of MnO₄⁻ with C₂O₄²⁻, a product (Mn²⁺) is the catalyst. This is called autocatalysis.
- 4Mn²⁺ + MnO₄⁻ + 8H⁺ → 5Mn³⁺ + 4H₂O
- 2Mn³⁺ + C₂O₄²⁻ → 2Mn²⁺ + 2CO₂
The rate starts slow, because there is no Mn²⁺ at first and the two negative ions repel. The rate speeds up as Mn²⁺ forms. Then it slows down as the reactants run out. This gives an S-shaped concentration-time graph.
Go deeper
These UK AQA pages cover the same ideas in more detail. They may include content that is not in 9620.
Exam-Style Practice Questions
Write your answer, then open the mark scheme. Each point marked (1) is worth one mark.
Question 1 (3 marks): Zinc and acid are added to a solution of VO₂⁺ ions. Give the colour and the oxidation state of vanadium in each species formed, in order, until the reaction stops.
Show Answer & Mark Scheme
- VO₂⁺ (+5) is yellow; it is reduced to blue VO²⁺ (+4) (1)
- then to green V³⁺ (+3) (1)
- then to violet V²⁺ (+2) (1)
A brief green colour at the very start is a mixture of yellow and blue, not V³⁺.
Question 2 (3 marks): 25.0 cm³ of a solution of Fe²⁺ ions reacts with 18.40 cm³ of 0.0150 mol dm⁻³ KMnO₄ in acid. Calculate the concentration of Fe²⁺ ions.
Show Answer & Mark Scheme
- Moles of MnO₄⁻ = 0.0150 × 18.40 / 1000 = 2.76 × 10⁻⁴ mol (1)
- Moles of Fe²⁺ = 5 × 2.76 × 10⁻⁴ = 1.38 × 10⁻³ mol (1)
- [Fe²⁺] = 1.38 × 10⁻³ / 0.0250 = 0.0552 mol dm⁻³ (1)
Question 3 (2 marks): Explain why hydrochloric acid is not used to acidify the solution in a manganate(VII) titration.
Show Answer & Mark Scheme
- MnO₄⁻ oxidises Cl⁻ to Cl₂ (1)
- So extra MnO₄⁻ is used and the titre is too large (1)
Question 4 (4 marks): Fe²⁺ ions catalyse the reaction between S₂O₈²⁻ and I⁻ ions. Explain why the uncatalysed reaction is slow, and write two equations to show how Fe²⁺ acts as a catalyst.
Show Answer & Mark Scheme
- The uncatalysed reaction is between two negative ions, which repel (1)
- S₂O₈²⁻ + 2Fe²⁺ → 2SO₄²⁻ + 2Fe³⁺ (1)
- 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂ (1)
- Each step is between oppositely charged ions, and Fe²⁺ is re-formed (1)
Question 5 (2 marks): Catalytic converters use a ceramic honeycomb coated with a thin layer of metal. Explain why a support medium is used, and why lead must be kept out of the exhaust gases.
Show Answer & Mark Scheme
- The support gives a large surface area, so little expensive metal is needed (1)
- Lead poisons the catalyst: it blocks the active sites, so the converter must be replaced, which costs money (1)