Disproportionation and Redox Agents for the ESAT

What is Disproportionation

Disproportionation is a unique category of redox reaction. It occurs when a single chemical species undergoes both oxidation and reduction at the same time to form two different products. To identify a disproportionation reaction, you must calculate the oxidation states of the atoms before and after the reaction to see if atoms of the same element have moved in both positive and negative directions.

The Decomposition of Hydrogen Peroxide

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

$2H_2O_2 \rightarrow 2H_2O + O_2$

In hydrogen peroxide ($H_2O_2$), oxygen has an oxidation state of -1. In the product water ($H_2O$), the oxidation state of oxygen is -2. In the product elemental oxygen ($O_2$), the oxidation state is zero. Therefore, some oxygen atoms have their oxidation state decreased (reduction), while others have it increased (oxidation). Because the oxygen in $H_2O_2$ is both oxidised and reduced, it has undergone disproportionation.

Chlorine in Alkaline Solutions

The reaction of chlorine with cold, dilute sodium hydroxide is another important example:

$2NaOH + Cl_2 \rightarrow NaCl + NaClO + H_2O$

The oxidation state of chlorine in $Cl_2$ is 0. In sodium chloride ($NaCl$), the chlorine has an oxidation state of -1. In sodium chlorate(I) ($NaClO$), the sodium is +1 and oxygen is -2, meaning the chlorine must be +1 to ensure the compound is neutral. Since the chlorine atoms from the same $Cl_2$ molecule are both reduced to -1 and oxidised to +1, the $Cl_2$ has undergone disproportionation.

Disproportionation of Copper(I) Ions

In aqueous solution, $Cu^{+}$ ions are unstable and will disproportionate:

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

The oxidation state of an ion is equal to its charge. Here, one $Cu^{+}$ ion (oxidation state +1) is reduced to elemental copper (oxidation state 0), while another $Cu^{+}$ ion is oxidised to $Cu^{2+}$ (oxidation state +2).

Advanced Worked Examples

Consider the reaction between nitrogen dioxide and water:

$2NO_2 + H_2O \rightarrow HNO_3 + HNO_2$

In $NO_2$, nitrogen has an oxidation state of +4. In $HNO_3$, the oxidation state of nitrogen is +5. In $HNO_2$, the oxidation state of nitrogen is +3. One nitrogen atom has been oxidised, and the other has been reduced, confirming this is a disproportionation reaction.

Next, observe the reaction of chlorine with hot aqueous sodium hydroxide:

$3Cl_2 + 6NaOH \rightarrow 5NaCl + NaClO_3 + 3H_2O$

The chlorine atoms in $Cl_2$ start at an oxidation state of 0. In $NaCl$, the state is -1. In $NaClO_3$, sodium is +1 and the three oxygens total -6, so the chlorine must have an oxidation state of +5. Five chlorine atoms are reduced, and one is oxidised. This remains a disproportionation reaction because the same species ($Cl_2$) is both oxidised and reduced.

Oxidising and Reducing Agents

In any redox reaction, we can identify specific reactants as either oxidising agents or reducing agents based on their behaviour during the process.

Definitions via Oxygen Transfer

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

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

Definitions via Electron Transfer

1. Oxidising agents oxidise something else. 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 something else. Because reduction is the gain of electrons, the reducing agent must provide those electrons. Therefore, the reducing agent is itself oxidised.

Identifying Agents in Ionic Reactions

Look at the reaction between zinc and copper(II) ions: $Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu$. The $Zn$ atoms are oxidised (loss of electrons) to $Zn^{2+}$. The $Cu^{2+}$ ions are reduced (gain of electrons) to $Cu$. The oxidising agent is $Cu^{2+}$ because it gains electrons, and the reducing agent is $Zn$ because it loses electrons.

In the displacement reaction $Cl_2 + 2I^{-} \rightarrow 2Cl^{-} + I_2$, the iodide ions ($I^{-}$) lose electrons to form $I_2$, meaning they are oxidised and act as the reducing agent. The chlorine molecule ($Cl_2$) gains electrons to form $Cl^{-}$, meaning it is reduced and acts as the oxidising agent.

Periodic Trends in Redox Agents

Group 2 elements (metals) react by losing electrons to form positive ions. Since they lose electrons, they are oxidised and therefore act as strong reducing agents. Group 17 elements (non-metals) react by gaining electrons to form negative ions. Since they gain electrons, they are reduced and therefore act as strong oxidising agents.