Disproportionation and Redox Agents

Understanding Disproportionation

Disproportionation is a specific category of redox reaction. In these reactions, a single species acts as both the oxidising agent and the reducing agent, meaning it is simultaneously oxidised and reduced. To identify disproportionation, you must track the oxidation states of a specific element across the reactants and products.

Example 1: Decomposition of Hydrogen Peroxide

A classic example is the decomposition of hydrogen peroxide into water and oxygen:

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

In hydrogen peroxide ($\mathrm{H}_2\mathrm{O}_2$), oxygen has an oxidation state of $-1$. In the products, the oxidation state changes as follows:

1. The oxygen in water ($\mathrm{H}_2\mathrm{O}$) has an oxidation state of $-2$. This is a decrease in oxidation state, which is reduction.
2. The oxygen in elemental oxygen ($\mathrm{O}_2$) has an oxidation state of zero. This is an increase in oxidation state, which is oxidation.

Because the oxygen in hydrogen peroxide has been both oxidised and reduced, it has undergone disproportionation.

Example 2: Chlorine in Sodium Hydroxide

When chlorine gas reacts with a cold, dilute solution of sodium hydroxide, the following reaction occurs:

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

The chlorine in $\mathrm{Cl}_2$ starts with an oxidation state of zero. In sodium chloride ($\mathrm{NaCl}$), the sodium is $+1$, so the chlorine must be $-1$ to ensure the compound is neutral. In sodium chlorate(I) ($\mathrm{NaClO}$), the sodium is $+1$ and oxygen is $-2$, so the chlorine must be $+1$ to balance the sum to zero. One chlorine atom is reduced (0 to $-1$) while another is oxidised (0 to $+1$). Therefore, $\mathrm{Cl}_2$ undergoes disproportionation.

Example 3: Copper(I) Ions

In aqueous solutions, copper(I) ions ($\mathrm{Cu}^{+}$) are unstable and undergo disproportionation:

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

The oxidation state of a monatomic ion is equal to its charge. Thus, $\mathrm{Cu}^{+}$ is $+1$ and $\mathrm{Cu}^{2+}$ is $+2$. Elemental copper ($\mathrm{Cu}$) is zero. One $\mathrm{Cu}^{+}$ ion is reduced to $\mathrm{Cu}$ (0), and another is oxidised to $\mathrm{Cu}^{2+}$ ($+2$).

Practice with Disproportionation Reactions

Exercise: Nitrogen Dioxide and Water

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

In $\mathrm{NO}_2$, nitrogen has an oxidation state of $+4$. In $\mathrm{HNO}_3$, using $H = +1$ and $O = -2$, nitrogen is calculated as $+5$. In $\mathrm{HNO}_2$, nitrogen is $+3$. One nitrogen atom increases from $+4$ to $+5$ (oxidation) and the other decreases from $+4$ to $+3$ (reduction). This is disproportionation.

Exercise: Chlorine and Hot Sodium Hydroxide

When the sodium hydroxide solution is hot, the reaction changes:

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

Chlorine in $\mathrm{Cl}_2$ is zero. In $5\mathrm{NaCl}$, it is $-1$. In $\mathrm{NaClO}_3$, the oxidation state of Cl is $+5$ (since $Na = +1$ and $3 \times O = -6$). Since chlorine is both reduced to $-1$ and oxidised to $+5$, this is a disproportionation reaction.

Oxidising and Reducing Agents

Redox reactions involve two specific roles: the oxidising agent and the reducing agent. These can be defined in two ways.

1. In terms of Oxygen Transfer

• Oxidising agents provide oxygen to another substance.
• Reducing agents remove oxygen from another substance.

For example, in the reaction $\mathrm{Fe_2O_3} + 3\mathrm{CO} \rightarrow 2\mathrm{Fe} + 3\mathrm{CO_2}$:

• $\mathrm{Fe_2O_3}$ is the oxidising agent because it provides the oxygen that the carbon monoxide gains.
• $\mathrm{CO}$ is the reducing agent because it removes the oxygen from the iron(III) oxide.

2. In terms of Electron Transfer

• Oxidising agents oxidise another species. Since oxidation is the loss of electrons, the oxidising agent must take those electrons. Therefore, the oxidising agent gains electrons and is itself reduced.
• Reducing agents reduce another species. Since reduction is the gain of electrons, the reducing agent must supply them. Therefore, the reducing agent loses electrons and is itself oxidised.

Consider the reaction between zinc and copper(II) ions: $\mathrm{Zn} + \mathrm{Cu}^{2+} \rightarrow \mathrm{Zn}^{2+} + \mathrm{Cu}$.

• $\mathrm{Zn}$ atoms lose electrons to become $\mathrm{Zn}^{2+}$. Thus, $\mathrm{Zn}$ is the reducing agent.
• $\mathrm{Cu}^{2+}$ ions gain electrons to become $\mathrm{Cu}$ atoms. Thus, $\mathrm{Cu}^{2+}$ is the oxidising agent.

Identifying Agents in Displacement Reactions

Look at the reaction: $\text{Cl}_2 + 2\text{I}^- \rightarrow 2\text{Cl}^- + \text{I}_2$.

Each iodide ion ($\text{I}^-$) loses an electron to form iodine ($\text{I}_2$). Their oxidation state increases from $-1$ to $0$. They are being oxidised, so $\text{I}^-$ is the reducing agent. The chlorine molecule ($\text{Cl}_2$) gains these electrons to form chloride ions ($\text{Cl}^-$). Its oxidation state decreases from $0$ to $-1$. It is being reduced, so $\text{Cl}_2$ is the oxidising agent.

Similarly, in the reaction $\text{Mg} + 2\text{H}^+ \rightarrow \text{Mg}^{2+} + \text{H}_2$:

• Magnesium ($\text{Mg}$) is oxidised (0 to $+2$), making it the reducing agent.
• Hydrogen ions ($\text{H}^+$) are reduced ($+1$ to 0), making them the oxidising agent.

Trends in the Periodic Table

Chemical behavior follows periodic trends:

1. Group 2 elements react by losing electrons to form positive ions. Because they lose electrons (oxidation), they are excellent reducing agents.
2. Group 17 elements react by gaining electrons to form negative ions. Because they gain electrons (reduction), they are excellent oxidising agents.