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:

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.

Manganate(VII) titrations

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.

Copper(II) titrations

Copper(II), Cu²⁺, can act as an oxidising agent in iodine/thiosulfate titrations.

Titration procedure:

1. Add a measured volume of colourless I^-_(aq) ions to a solution of blue Cu²⁺_(aq) ions. A white copper(I) iodide precipitate and a brown I₂ solution form: 2Cu²⁺_(aq) + 4I^-_(aq) ➔ 2CuI_(s) + I_2(aq)

2. Titrate this brown iodine solution against a standard sodium thiosulfate (Na₂S₂O₃^2-) solution in a burette. The thiosulfate reduces the iodine back to iodide ions, causing the brown colour to fade. I_2(aq) + 2S₂O₃^2-_(aq) ➔ 2I^-_(aq) + S₄O₆^2-_(aq)

3. Add a starch indicator near the endpoint to sharpen the colour change observed at the end point. In the presence of starch, iodine is blue-black.

4. Record the volume of S₂O₃^2-_(aq) needed to reach the end point. The end point is indicated by the sudden appearance of a blue-black solution, indicating an excess of I₂.

5. Repeat until concordant titres are achieved.

If the concentration and volume of S₂O₃^2- is known, the moles and concentration of Cu²⁺ 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-

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

Worked example 3 - Titration calculation of Cu2+ with I-

In a titration experiment to determine the concentration of copper(II) ions, a solution of sodium thiosulfate with a concentration of 0.0500 mol dm^-3 was used. A volume of 20.0 cm³ of this thiosulfate solution was required to react completely with the iodine formed in the reaction between copper(II) ions and excess iodide ions.

The relevant reactions are as follows:

2Cu²⁺_(aq) + 4I^-_(aq) ➔ 2CuI_(s) + I_2(aq)

I_2(aq) + 2S₂O₃^2-_(aq) ➔ 2I^-_(aq) + S₄O₆^2-_(aq)

Calculate the concentration of copper(II) ions in the solution, assuming the volume of the copper(II) solution is 25.0 cm³. Give your answer to 3 significant figures.

Step 1: Conversion of cm³ into dm³

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

20.0 cm³ = 0.0200 dm³

25.0 cm³ = 0.0250 dm³

Step 2: Calculate number of moles of S₂O₃^2-

$\text{n }=\text{ c }\times \text{ V}=0.0500\times 0.0200=1.00\times {10}^{-3}\text{ mol}$

Step 3: Calculate number of moles of I₂

I₂ : S₂O₃^2- mole ratio = 1:2

Moles of I₂ = $\frac{1.00\times {10}^{-3}}{2}=5.00\times {10}^{-4}\text{ mol}$

Step 4: Calculate number of moles of Cu²⁺

Cu²⁺ : I₂ mole ratio = 2:1

Moles of Cu²⁺ = (5.00 x 10^-4) x 2 = 1.00 x 10^-3 mol

Step 5: Calculate concentration of Cu²⁺

$\text{c }=\frac{\text{n}}{\text{V}}=\frac{1.00\times {10}^{-3}}{0.0250}=0.0400{\text{ mol dm}}^{-3}$