Disproportionation and Redox Agents

Understanding Disproportionation

Disproportionation is a specific category of redox reaction. It occurs when a single species is simultaneously both oxidised and reduced during the chemical change. To identify such a reaction, you must calculate the oxidation states of the elements in the reactants and compare them to the oxidation states in the products.

Example: The Decomposition of Hydrogen Peroxide

A classic example of disproportionation is the decomposition of hydrogen peroxide:

$2\mathrm{H}_2\mathrm{O}_2 \rightarrow 2\mathrm{H}_2\mathrm{O} + \mathrm{O}_2$

In hydrogen peroxide ($\mathrm{H}_2\mathrm{O}_2$), the oxidation state of oxygen is $-1$. In the product water ($\mathrm{H}_2\mathrm{O}$), the oxidation state of oxygen is $-2$. In the product elemental oxygen ($\mathrm{O}_2$), the oxidation state is zero.

Looking at the stoichiometry, two of the four oxygen atoms in $2\mathrm{H}_2\mathrm{O}_2$ have decreased their oxidation state from $-1$ to $-2$ in water, which is reduction. The other two oxygen atoms have increased their oxidation state from $-1$ to zero in elemental oxygen, which is oxidation. Because oxygen has been both oxidised and reduced from the same starting species, it has undergone disproportionation.

Further Examples of Disproportionation

Chlorine and Sodium Hydroxide (Cold)

When chlorine gas reacts with a cold, dilute solution of sodium hydroxide, it disproportionates:

$2\mathrm{NaOH} + \mathrm{Cl}_2 \rightarrow \mathrm{NaCl} + \mathrm{NaClO} + \mathrm{H}_2\mathrm{O}$

1. In $\mathrm{Cl}_2$, the oxidation state of chlorine is zero.
2. In sodium chloride ($\mathrm{NaCl}$), sodium is $+1$, so chlorine must be $-1$ to ensure the sum of oxidation states is zero. This is reduction.
3. In sodium chlorate(I) ($\mathrm{NaClO}$), sodium is $+1$ and oxygen is $-2$. To ensure the sum is zero, chlorine must be $+1$. This is oxidation.

Since the chlorine in $\mathrm{Cl}_2$ is simultaneously oxidised and reduced, it undergoes disproportionation.

Copper(I) Ions in Aqueous Solution

Copper(I) ions ($\mathrm{Cu}^{+}$) are unstable in aqueous solution and undergo disproportionation:

$2\mathrm{Cu}^{+} \rightarrow \mathrm{Cu}^{2+} + \mathrm{Cu}$

The oxidation state of a simple ion is equal to its charge. Therefore, $\mathrm{Cu}^{+}$ is $+1$ and $\mathrm{Cu}^{2+}$ is $+2$. The oxidation state of elemental copper ($\mathrm{Cu}$) is zero. One $\mathrm{Cu}^{+}$ ion reduces its state from $+1$ to 0, while another $\mathrm{Cu}^{+}$ ion increases its state from $+1$ to $+2$.

Worked Example: Nitrogen Dioxide

Consider the reaction: $2\mathrm{NO}_2 + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{HNO}_3 + \mathrm{HNO}_2$

In $\mathrm{NO}_2$, oxygen is $-2$, so nitrogen is $+4$. In $\mathrm{HNO}_3$, hydrogen is $+1$ and three oxygens are $-6$, making nitrogen $+5$. In $\mathrm{HNO}_2$, hydrogen is $+1$ and two oxygens are $-4$, making nitrogen $+3$. The nitrogen atom in $\mathrm{NO}_2$ has been oxidised to $+5$ and reduced to $+3$ simultaneously.

Worked Example: Chlorine and Hot Alkali

When chlorine reacts with a hot aqueous solution of sodium hydroxide, the disproportionation follows a different path:

$3\mathrm{Cl}_2 + 6\mathrm{NaOH} \rightarrow 5\mathrm{NaCl} + \mathrm{NaClO}_3 + 3\mathrm{H}_2\mathrm{O}$

Chlorine starts at 0. In $\mathrm{NaCl}$, it is $-1$ (reduction). In $\mathrm{NaClO}_3$, sodium is $+1$ and three oxygens are $-6$, so chlorine must be $+5$ to make the compound neutral. This increase from 0 to $+5$ is oxidation. Since chlorine is both oxidised and reduced, this is a disproportionation reaction.

Oxidising and Reducing Agents

Definitions via Oxygen Transfer

1. Oxidising agents provide oxygen to another substance.
2. Reducing agents remove oxygen from another substance.

In the reaction $\mathrm{Fe_2O_3} + 3\mathrm{CO} \rightarrow 2\mathrm{Fe} + 3\mathrm{CO_2}$, the iron(III) oxide ($\mathrm{Fe_2O_3}$) is the oxidising agent because it supplies the oxygen. The carbon monoxide ($\mathrm{CO}$) is the reducing agent because it removes oxygen from the iron oxide.

Definitions via Electron Transfer

1. Oxidising agents oxidise other species. Because oxidation is the loss of electrons, the oxidising agent must accept those electrons. Therefore, the oxidising agent is itself reduced.
2. Reducing agents reduce other species. Because reduction is the gain of electrons, the reducing agent must provide those electrons. Therefore, the reducing agent is itself oxidised.

In the reaction $\mathrm{Zn} + \mathrm{Cu}^{2+} \rightarrow \mathrm{Zn}^{2+} + \mathrm{Cu}$:

• Zinc atoms lose electrons to become $\mathrm{Zn}^{2+}$, so Zn is the reducing agent.
• Copper(II) ions gain electrons to become Cu atoms, so $\mathrm{Cu}^{2+}$ is the oxidising agent.

Identifying Agents in Ionic Reactions

Consider the displacement: $\text{Cl}_2 + 2\text{I}^- \rightarrow 2\text{Cl}^- + \text{I}_2$

The iodide ions ($\text{I}^-$) lose electrons (oxidation) to form iodine molecules. Thus, $\text{I}^-$ is the reducing agent. The chlorine molecule ($\text{Cl}_2$) gains those electrons (reduction) to form chloride ions. Thus, $\text{Cl}_2$ is the oxidising agent.

Generally, Group 17 elements (halogens) are strong oxidising agents because they have a high tendency to gain electrons to form negative ions. Conversely, Group 2 elements are strong reducing agents because they easily lose electrons to form positive ions.