What you'll learn

• Why transition metals can exist in multiple stable oxidation states.
• The colours and formulas of the vanadium ions as they are sequentially reduced from $+5$ to $+2$.
• How pH and different ligands alter the redox potential of a transition metal ion.
• How to write equations and perform calculations for redox titrations involving manganate(VII).

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Why do transition metals have variable oxidation states?

One of the defining features of transition metals is their ability to form compounds with the metal in various oxidation states.

When a transition metal forms an ion, it loses its $4s$ electrons first, followed by one or more $3d$ electrons. Because the energy required to remove successive $3d$ electrons increases gradually rather than in massive jumps, transition metals can reach a variety of stable oxidation states.

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The colourful reduction of vanadium

A classic demonstration of variable oxidation states is the reduction of vanadium. In acidic solution, zinc metal is a sufficiently strong reducing agent to reduce vanadium sequentially from an oxidation state of $+5$ all the way down to $+2$.

As the oxidation state changes, the d-orbital electron configuration changes, which leads to a sequence of distinct colour changes.

1. Oxidation state +5: The reaction starts with the yellow dioxovanadium(V) ion, ${VO}_2^{+}$.
2. Oxidation state +4: Zinc reduces this to the blue oxovanadium(IV) ion, ${VO}^{2+}$.
3. Oxidation state +3: Further reduction produces the green vanadium(III) ion, $V^{3+}$.
4. Oxidation state +2: The final reduction step yields the violet vanadium(II) ion, $V^{2+}$.

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Factors influencing redox potentials

We measure the tendency of an ion to change from a higher to a lower oxidation state using its redox potential (standard electrode potential, $E^{\ominus }$). A more positive $E^{\ominus }$ value means the species is more easily reduced (it is a better oxidising agent).

For transition metals, this potential is not a single fixed number; it can be heavily influenced by the chemical environment.

1. The effect of pH

Many redox half-equations involving transition metals also feature hydrogen ions ($H^{+}$) or hydroxide ions (${OH}^{-}$). For example, the reduction of the yellow vanadium(V) ion:

$${VO}_2^{+}+2H^{+}+e^{-}\rightleftharpoons {VO}^{2+}+H_{2}O$$

According to Le Chatelier's principle, increasing the concentration of $H^{+}$ (lowering the pH) pushes the equilibrium to the right. Therefore, acidic conditions generally increase the redox potential, making it easier to reduce the transition metal from a higher to a lower oxidation state.

2. The effect of ligands

Transition metal ions form complex ions surrounded by ligands. Changing the ligands changes the stability of the complex. If a specific ligand stabilises the higher oxidation state more than the lower one, the redox potential will decrease (it becomes harder to reduce). If the ligand stabilises the lower oxidation state, the redox potential increases.

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Tollens' reagent: A transition metal test

You have likely used Tollens' reagent in organic chemistry to distinguish between aldehydes and ketones. This test is fundamentally a transition metal redox reaction!

Tollens' reagent contains the diamminesilver(I) complex, $[Ag({NH}_3{)}_2{]}^{+}$.

• Aldehydes are gentle reducing agents. They can reduce the silver(I) ion in the complex down to metallic silver ($Ag$), forming a silver mirror on the inside of the test tube.
• Ketones cannot be oxidised easily, so they do not reduce the silver complex, and no mirror forms.

The redox half-equation for the transition metal complex is:

$$[Ag({NH}_3{)}_2{]}^{+}+e^{-}\to Ag+2{NH}_3$$

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Redox titrations with manganate(VII)

Redox titrations are used to find the unknown concentration of a reducing agent by titrating it against a standard solution of an oxidising agent. The most common oxidising agent used at A-level is potassium manganate(VII), ${KMnO}_4$.

Manganate(VII) ions (${MnO}_4^{-}$) are deep purple. As they are reduced to manganese(II) ions (${Mn}^{2+}$), the solution becomes virtually colourless. This means the titration is self-indicating; the end-point is identified by the first permanent pale pink colour when a tiny excess of unreacted ${MnO}_4^{-}$ remains in the flask.

These titrations must be carried out in strongly acidic conditions (usually using dilute sulfuric acid) to ensure the ${MnO}_4^{-}$ is fully reduced to ${Mn}^{2+}$.

Iron(II) and Manganate(VII)

When titrating ${Fe}^{2+}$ ions (for example, to find the percentage of iron in an iron tablet or steel wire), the two half-equations combine to give an overall reacting ratio of $1:5$.

• Reduction: ${MnO}_4^{-}+8H^{+}+5e^{-}\to {Mn}^{2+}+4H_{2}O$
• Oxidation: ${Fe}^{2+}\to {Fe}^{3+}+e^{-}$
• Overall: ${MnO}_4^{-}+8H^{+}+5{Fe}^{2+}\to {Mn}^{2+}+4H_{2}O+5{Fe}^{3+}$

Ethanedioate and Manganate(VII)

Ethanedioate ions ($C_2{O}_4^{2-}$) can also be titrated with manganate(VII). This reaction requires heating at the start because the initial rate is very slow. The combined equation gives a reacting ratio of $2:5$.

• Reduction: ${MnO}_4^{-}+8H^{+}+5e^{-}\to {Mn}^{2+}+4H_{2}O$
• Oxidation: $C_2{O}_4^{2-}\to 2{CO}_2+2e^{-}$
• Overall: $2{MnO}_4^{-}+16H^{+}+5C_2{O}_4^{2-}\to 2{Mn}^{2+}+8H_{2}O+10{CO}_2$

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