Disproportionation and Redox Agents for the ESAT

What is Disproportionation?

Disproportionation is a specific type of redox reaction. In this process, a single species undergoes both oxidation and reduction at the same time. To identify such a reaction, you must calculate the oxidation states of the elements involved and see if a single element's oxidation state both increases and decreases across the reaction.

The Decomposition of Hydrogen Peroxide

A classic example of disproportionation is the decomposition of hydrogen peroxide into water and oxygen gas. The chemical equation is:

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

To see why this is disproportionation, we examine the oxidation states of oxygen:

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

In this reaction, two of the four oxygen atoms from the reactant have their oxidation state decreased from $-1$ to $-2$ in water, which is reduction. The other two oxygen atoms increase their oxidation state from $-1$ to 0 in oxygen gas, which is oxidation. Therefore, the oxygen in hydrogen peroxide has been simultaneously oxidised and reduced.

Chlorine in Sodium Hydroxide Solutions

The reaction of chlorine with sodium hydroxide is another fundamental example that varies based on temperature.

1. Cold Dilute Sodium Hydroxide

When chlorine reacts with cold dilute $\mathrm{NaOH}$, the following reaction occurs:

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

The oxidation state of chlorine in $\mathrm{Cl}_2$ is 0 because it is in its elemental state. In sodium chloride ($\mathrm{NaCl}$), the chlorine has an oxidation state of $-1$ (since $\mathrm{Na}$ is $+1$). In sodium chlorate(I) ($\mathrm{NaClO}$), the oxidation state of chlorine is $+1$, as the sum of $\mathrm{Na}$ ($+1$), $\mathrm{O}$ ($-2$), and $\mathrm{Cl}$ must equal zero. Because the chlorine has been reduced from 0 to $-1$ and oxidised from 0 to $+1$ simultaneously, it has undergone disproportionation.

2. Hot Aqueous Sodium Hydroxide

If the sodium hydroxide solution is hot, a different disproportionation occurs:

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

Here, the chlorine starts at 0. In $\mathrm{NaCl}$, it is $-1$. In $\mathrm{NaClO}_3$, the chlorine has an oxidation state of $+5$ (calculated as: $+1$ for $\mathrm{Na}$ plus $3 \times -2$ for $\mathrm{O}$, meaning $\mathrm{Cl}$ must be $+5$ to sum to zero). Five chlorine atoms are reduced, and one is oxidised, confirming disproportionation.

Disproportionation of Copper(I) Ions

In aqueous solution, $\mathrm{Cu}^{+}$ ions are unstable and undergo disproportionation:

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

The oxidation state of an ion is equal to its charge. Thus, $\mathrm{Cu}^{+}$ is $+1$ and $\mathrm{Cu}^{2+}$ is $+2$. Elemental copper ($\mathrm{Cu}$) is 0. One $\mathrm{Cu}^{+}$ ion reduces its state from $+1$ to 0, while the other 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 $\mathrm{NO}_2$, nitrogen is in the $+4$ state. In $\mathrm{HNO}_3$, nitrogen is $+5$, while in $\mathrm{HNO}_2$, nitrogen is $+3$. One nitrogen atom increases its oxidation state (oxidation) and the other decreases it (reduction), meaning $\mathrm{NO}_2$ undergoes disproportionation.

Oxidising and Reducing Agents

Understanding redox also requires identifying the agents that drive these changes. These can be defined by oxygen transfer or electron transfer.

1. Oxygen Transfer

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

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

2. Electron Transfer

• Oxidising agents oxidise something else by taking its electrons. Consequently, the oxidising agent gains electrons and is itself reduced.
• Reducing agents reduce something else by supplying electrons. Consequently, the reducing agent loses electrons and is itself oxidised.

In the reaction $\mathrm{Zn} + \mathrm{Cu}^{2+} \rightarrow \mathrm{Zn}^{2+} + \mathrm{Cu}$, the $\mathrm{Zn}$ atoms lose electrons (oxidation) and are the reducing agent. The $\mathrm{Cu}^{2+}$ ions gain electrons (reduction) and are the oxidising agent.

Periodic Trends in Redox Agents

The tendency of elements to act as oxidising or reducing agents is linked to their position in the Periodic Table:

• Group 2 elements lose electrons to form positive ions (oxidation). Because they are easily oxidised, they act as strong reducing agents.
• Group 17 elements (halogens) gain electrons to form negative ions (reduction). Because they are easily reduced, they act as strong oxidising agents.