Oxidising Agents Reducing Agents and Disproportionation

Understanding Disproportionation

Disproportionation is a specific category of redox reaction. It occurs when a single species is simultaneously both oxidised and reduced. To identify these reactions, you must track the oxidation states of a specific element as it moves from the reactants to the products.

A classic example 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 ($\mathrm{H}_2\mathrm{O}_2$), oxygen has an oxidation state of $-1$. In the products, the oxygen in water has an oxidation state of $-2$, while the oxygen in elemental oxygen ($\mathrm{O}_2$) has an oxidation state of $0$. Consequently, two of the four oxygen atoms in the reactants have seen their oxidation state decrease from $-1$ to $-2$ (reduction), while the other two have seen their oxidation state increase from $-1$ to $0$ (oxidation). Because the oxygen has been simultaneously oxidised and reduced, it has undergone disproportionation.

Further Examples of Disproportionation

1. The reaction of chlorine with cold, dilute sodium hydroxide solution:

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

The chlorine atoms in $\mathrm{Cl}_2$ start with an oxidation state of $0$. In sodium chloride ($\mathrm{NaCl}$), the chlorine has an oxidation state of $-1$ (reduction). In sodium chlorate(I) ($\mathrm{NaClO}$), where sodium is $+1$ and oxygen is $-2$, the chlorine must have an oxidation state of $+1$ to ensure the sum of oxidation states is zero (oxidation). Thus, chlorine has undergone disproportionation.

1. The disproportionation of $\mathrm{Cu}^{+}$ ions in aqueous solution:

One $\mathrm{Cu}^{+}$ ion (oxidation state $+1$) reduces to elemental copper (oxidation state $0$), while another $\mathrm{Cu}^{+}$ ion oxidises to $\mathrm{Cu}^{2+}$ (oxidation state $+2$).

Worked Exercises: Identifying Disproportionation

Exercise 56: 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 is in the $+4$ oxidation state. In nitric acid ($\mathrm{HNO}_3$), the nitrogen is $+5$ (oxidation). In nitrous acid ($\mathrm{HNO}_2$), the nitrogen is $+3$ (reduction). This confirms the nitrogen in $2\mathrm{NO}_2$ has undergone disproportionation.

Exercise 57: Chlorine and Hot Sodium Hydroxide

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

Elemental chlorine ($\mathrm{Cl}_2$) has an oxidation state of $0$. In $\mathrm{NaCl}$, it is $-1$. In $\mathrm{NaClO}_3$, the oxidation state of chlorine is $+5$ because the sodium is $+1$ and the three oxygens total $-6$. Five chlorine atoms are reduced, and one is oxidised. This is a disproportionation reaction.

Oxidising and Reducing Agents: Oxygen Transfer

In reactions involving oxygen transfer, we use the following definitions:

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

In the reaction between iron(III) oxide and carbon monoxide: $\mathrm{Fe_2O_3} + 3\mathrm{CO} \rightarrow 2\mathrm{Fe} + 3\mathrm{CO_2}$. The iron(III) oxide is reduced to iron by losing oxygen, making it the oxidising agent. The carbon monoxide is oxidised to carbon dioxide by gaining oxygen, making it the reducing agent.

Oxidising and Reducing Agents: Electron Transfer

When reactions are viewed through electron movement:

• An oxidising agent oxidises another species by taking its electrons. Therefore, the oxidising agent gains electrons and is itself reduced.
• A reducing agent reduces another species by supplying electrons. Therefore, the reducing agent loses electrons and is itself oxidised.

In the reaction between zinc and copper(II) ions: $\mathrm{Zn} + \mathrm{Cu}^{2+} \rightarrow \mathrm{Zn}^{2+} + \mathrm{Cu}$. Zinc atoms lose electrons to become $\mathrm{Zn}^{2+}$ (oxidation), so zinc is the reducing agent. Copper(II) ions gain electrons to become copper atoms (reduction), so $\mathrm{Cu}^{2+}$ is the oxidising agent.

Systematic Identification Method

To identify the agents in a reaction such as $\mathrm{Cl}_2 + 2\mathrm{I}^- \rightarrow 2\mathrm{Cl}^- + \mathrm{I}_2$, follow these steps:

1. Determine what has been oxidised and reduced. Here, iodide ions ($\mathrm{I}^-$) lose electrons to form $\mathrm{I}_2$ (oxidation), while chlorine ($\mathrm{Cl}_2$) gains electrons to form $\mathrm{Cl}^-$ (reduction).
2. Apply the definitions. The species that is reduced is the oxidising agent ($\mathrm{Cl}_2$). The species that is oxidised is the reducing agent ($\mathrm{I}^-$).

In the case of Group 2 and Group 17 elements: Group 2 metals lose electrons to form positive ions (oxidation), making them effective reducing agents. Group 17 non-metals gain electrons to form negative ions (reduction), making them effective oxidising agents.