Transition metals have variable oxidation states

Transition metals can exist in various positive oxidation states in compounds and complexes. This occurs due to the similar energies of the outer d orbital electrons, allowing multiple oxidation states to be stable.

For example, vanadium exhibits four oxidation states from +2 to +5, each with different coloured ions:

Vanadium(V) ions exhibit a stepwise colour change when reduced by zinc metal in acidic solution:

1. Yellow VO₂⁺ solution changes to blue VO²⁺. The reduction half-equation is: VO₂⁺ + 2H⁺ + e^- ➔ VO²⁺ + H₂O
2. Blue VO²⁺ solution changes to green V³⁺. The reduction half-equation is: VO²⁺ + 2H⁺ + e^- ➔ V³⁺ + H₂O
3. Green V³⁺ solution changes to violet V²⁺. The reduction half-equation is: V³⁺ + e^- ➔ V²⁺

Redox potentials indicate ease of reduction

The redox potential, E^⦵, indicates how easily an ion is reduced. Ions with higher, more positive E^⦵ values are more easily reduced to lower oxidation states.

For example, Cu²⁺ has a higher redox potential than Cr³⁺ so is more easily reduced.

However, E^⦵ values are measured under standard conditions, so the actual redox potential can vary depending on:

• Ligands - E^⦵ assumes metal ions are surrounded by water ligands in aqueous solution. Other ligands that form stronger bonds with the metal ion can raise or lower the potential by stabilising a particular oxidation state.
• pH - Acidic conditions provide excess H⁺ ions needed for reduction of some metal ions. For example: VO₂⁺_(aq) + 2H⁺_(aq) + e^- ➔ VO²⁺_(aq) + H₂O_(l)

Alkaline conditions favour oxidation reactions that consume OH^- ions instead. For example:

Cr(OH)_3(s) + 5OH^-_(aq) ➔ CrO₄^2-_(aq) + 4H₂O_(l) + 3e^-

In general, more acidic solutions give higher, more positive redox potentials.

Titrations using transition metals are redox titrations

The variable oxidation states of transition metals allows them to act as oxidising or reducing agents in titrations. Their characteristic colour changes clearly indicate the endpoint.

A common oxidising agent used in titrations is manganate(VII), MnO₄^-.

During the titration purple MnO₄^- ions are reduced to colourless Mn²⁺ ions by a reducing agent such as Fe²⁺ or C₂O₄^2-.

The reduction half-equation is:

MnO₄^- + 8H⁺ + 5e^- ➔ Mn²⁺ + 4H₂O

Titration procedure:

1. Use a pipette to transfer a measured volume of reducing agent into a conical flask.
2. Add dilute sulfuric acid to the conical flask to provide excess acidic conditions.
3. Fill a burette with MnO₄^-_(aq) ions and slowly add it to the conical flask while swirling continuously. The purple MnO₄^- ions react to produce a colourless solution of Mn²⁺ ions.
4. Record the volume of MnO₄^-_(aq) needed to reach the end point. The end point is indicated by the sudden appearance of a purple solution, indicating an excess of MnO₄^- ions.
5. Repeat the titrations until concordant titres are obtained (within 0.10 cm³ of each other).

If the concentration and volume of MnO₄^- is known, the moles and concentration of the reducing agent can be calculated.

Worked example 1 - Titration calculation of Fe2+ with MnO4-

25.0 cm³ of 0.0200 mol dm^-3 aqueous potassium manganate(VII) reacted with 50.0 cm³ of acidified iron(II) sulfate solution.

Calculate the concentration of Fe²⁺ ions in the solution. Give your answer to 3 significant figures.

2MnO₄^-_(aq) + 6H⁺_(aq) + 5Fe²⁺_(aq) ➔ 2Mn²⁺_(aq) + 3H₂O_(l) + 5Fe³⁺_(aq)

Step 1: Conversion of cm³ into dm³

To convert from cm³ into dm³, divide by 1,000

25.0 cm³ = 0.0250 dm³

50.0 cm³ = 0.0500 dm³

Step 2: Calculate number of moles of MnO₄^-

$\text{n }=\text{ c}\times \text{ V }=0.0200\times 0.0250=5.00\times {10}^{-4}\text{ mol}$

Step 3: Calculate number of moles of Fe²⁺

Fe²⁺ : MnO₄^- mole ratio = 5:2

Moles of Fe²⁺ = $(5.00\times {10}^{-4})\times \frac{5}{2}=1.25\times {10}^{-3}\text{ mol}$

Step 4: Calculate concentration of Fe²⁺

$\text{c }=\frac{\text{n}}{\text{V}}=\frac{1.25\times {10}^{-3}}{0.0500}=0.0250{\text{ mol dm}}^{-3}$

Worked example 2 - Titration calculation of C2O42- with MnO4-

30.0 cm³ of 0.0100 mol dm^-3 aqueous potassium manganate(VII) reacted with 25.0 cm³ of oxalate ion (C₂O₄^2-) solution.

Calculate the concentration of C₂O₄^2- ions in the solution. Give your answer to 3 significant figures.

2MnO₄^-_(aq) + 5C₂O₄^2-_(aq) + 16H⁺_(aq) ➔ 2Mn²⁺_(aq) + 10CO_2(g) + 8H₂O_(l)

Step 1: Conversion of cm³ into dm³

To convert from cm³ into dm³, divide by 1,000

30.0 cm³ = 0.0300 dm³

25.0 cm³ = 0.0250 dm³

Step 2: Calculate number of moles of MnO₄^-

$\text{n }=\text{ c}\times \text{ V }=0.0100\times 0.0300=3.00\times {10}^{-4}\text{ mol}$

Step 3: Calculate number of moles of C₂O₄^2-

C₂O₄^2- : MnO₄^- mole ratio = 5:2

Moles of C₂O₄^2-= $(3.00\times {10}^{-4})\times \frac{5}{2}=7.50\times {10}^{-4}\text{ mol}$

Step 4: Calculate concentration of C₂O₄^2-

$\text{c }=\frac{\text{n}}{\text{V}}=\frac{7.50\times {10}^{-4}}{0.0250}=0.0300{\text{ mol dm}}^{-3}$

Tollens' reagent application

Tollens' reagent contains the silver(I) complex [Ag(NH₃)₂]⁺ which is easily reduced to silver metal by certain organic molecules.

It is prepared by adding just enough ammonia to silver nitrate solution to give a clear, colourless solution.

Tollen's reagent oxidises aldehydes (RCHO) to carboxylic acids (RCOO^-), producing a silver mirror on the inside of the test tube:

RCHO_(aq) + 2[Ag(NH₃)₂]⁺_(aq) + 3OH^-_(aq) ➔ RCOO^-_(aq) + 2Ag_(s) + 4NH_3(aq) + 2H₂O_(l)

Unlike aldehydes, Tollens’ reagent cannot oxidise ketones so no silver mirror forms.