Question

14.135 Hydrogen peroxide can act as either an oxidizing agent or a reducing agent. (a) When \(\mathrm{H}_{2} \mathrm{O}_{2}\) is treated with aqueous \(\mathrm{KI}\), \(\mathrm{I}_{2}\) forms. In which role is \(\mathrm{H}_{2} \mathrm{O}_{2}\) acting? What oxygen-containing product is formed? (b) When \(\mathrm{H}_{2} \mathrm{O}_{2}\) is treated with aqueous \(\mathrm{KMnO}_{4}\), the purple color of \(\mathrm{MnO}_{4}^{-}\) disappears and a gas forms. In which role is \(\mathrm{H}_{2} \mathrm{O}_{2}\) acting? What is the oxygen-containing product formed?

Short answer

(a) Oxidizing agent; \(\mathrm{H_{2}O}\). (b) Reducing agent; \(\mathrm{O_{2}}\).

Step-by-step solution

Identify roles

Determine the role of \(\mathrm{H_{2}O_{2}}\) in each reaction. An oxidizing agent gains electrons and is reduced; a reducing agent loses electrons and is oxidized.

Reaction with KI (Part a)

Consider the reaction between \(\mathrm{H_{2}O_{2}}\) and \(\mathrm{KI}\). Write the balanced chemical equation: \[ \mathrm{H_{2}O_{2}} + 2 \mathrm{KI} + \mathrm{H_{2}SO_{4}} \rightarrow I_{2} + 2 \mathrm{H_{2}O} + \mathrm{K_{2}SO_{4}} \] Hydrogen peroxide is reduced, as evidenced by the formation of \(\mathrm{I_{2}}\) from \(\mathrm{I^{-}}\), hence \(\mathrm{H_{2}O_{2}}\) acts as the oxidizing agent.

Product of Part a

Identify the oxygen-containing product. From the balanced equation, \(\mathrm{H_{2}O}\) is the oxygen-containing product formed when \(\mathrm{H_{2}O_{2}}\) is reduced.

Reaction with KMnO4 (Part b)

Consider the reaction between \(\mathrm{H_{2}O_{2}}\) and \(\mathrm{KMnO_{4}}\). Write the balanced chemical equation: \[ 5 \mathrm{H_{2}O_{2}} + 2 \mathrm{KMnO_{4}} + 3 \mathrm{H_{2}SO_{4}} \rightarrow 2 \mathrm{MnSO_{4}} + 5 \mathrm{O_{2}} + 8 \mathrm{H_{2}O} + \mathrm{K_{2}SO_{4}} \] Hydrogen peroxide is oxidized, as evidenced by the disappearance of the purple color of \(\mathrm{MnO_{4}^{-}}\) and the formation of \(\mathrm{O_{2}}\) gas, hence \(\mathrm{H_{2}O_{2}}\) acts as the reducing agent.

Product of Part b

Identify the oxygen-containing product. From the balanced equation, \(\mathrm{O_{2}}\) is the oxygen-containing product formed when \(\mathrm{H_{2}O_{2}}\) is oxidized.

Key concepts

Oxidizing Agent

In any redox reaction, an oxidizing agent gains electrons from another substance. This means it gets reduced as it accepts electrons.

Take a look at Hydrogen peroxide’s reaction with aqueous \(KI\). The balanced chemical equation is: \[ \mathrm{H_{2}O_{2}} + 2 \mathrm{KI} + \mathrm{H_{2}SO_{4}} \rightarrow I_{2} + 2 \mathrm{H_{2}O} + \mathrm{K_{2}SO_{4}} \]

Here, \(H_{2}O_{2}\) is gaining electrons, which reduces it. That's why it's acting as an oxidizing agent and the key oxygen-containing product formed is \(H_{2}O\).

• An oxidizing agent gains electrons.
• It gets reduced.
• In the case of \(H_{2}O_{2} + KI\) reaction, hydrogen peroxide is the oxidizing agent.
• The formed oxygen-containing product is water ( \(H_{2}O\)).

Reducing Agent

Conversely, a reducing agent loses electrons, and thus it gets oxidized.

For instance, consider the reaction of Hydrogen peroxide with aqueous \(KMnO_{4}\). The balanced chemical equation is: \[ 5 \mathrm{H_{2}O_{2}} + 2 \mathrm{KMnO_{4}} + 3 \mathrm{H_{2}SO_{4}} \rightarrow 2 \mathrm{MnSO_{4}} + 5 \mathrm{O_{2}} + 8 \mathrm{H_{2}O} + \mathrm{K_{2}SO_{4}} \]

In this reaction, \(H_{2}O_{2}\) loses electrons, which means it is oxidized and thus acts as the reducing agent. The main oxygen-containing product here is the gas \(O_{2}\).

• A reducing agent loses electrons.
• It gets oxidized.
• In the reaction with \(KMnO_{4}\), \(H_{2}O_{2}\) is the reducing agent.
• The formed oxygen-containing product is \(O_{2}\) gas.

Balanced Chemical Equations

Balanced chemical equations are critical for understanding redox reactions. They ensure that the same number of each type of atom is present on both sides of the equation. This follows the Law of Conservation of Mass which states that matter can neither be created nor destroyed.

Let’s review the balancing process with our \(KI\) reaction example: \[ \mathrm{H_{2}O_{2}} + 2 \mathrm{KI} + \mathrm{H_{2}SO_{4}} \rightarrow I_{2} + 2 \mathrm{H_{2}O} + \mathrm{K_{2}SO_{4}} \]

• Inspect both sides of the equation to check if the counts for all atoms match.
• Hydrogen (H), Iodine (I), Sulfur (S), Oxygen (O), and Potassium (K) are all balanced here.
• This helps in clearly identifying reactants and products in redox reactions.

Reaction Identification

To identify what role \(H_{2}O_{2}\) is playing in a reaction, observe the changes in oxidation states of the involved elements. Determining whether the substance is gaining or losing electrons clarifies its role as either an oxidizing or reducing agent.

In the \(KI\) reaction:

• Look at change in oxidation states: Iodide ions (\(\text{I}^-\)) are oxidized to Iodine (\text{I}_2).
• This means \(H_{2}O_{2}\) must be reduced, thus acting as an oxidizing agent.

For the \(KMnO_{4}\) reaction:

• Manganese changes from \(+\text{7}\) in \(MnO_{4}^{-}\) to \(+2\) in \(MnSO_{4}\).
• This means \(H_{2}O_{2}\) is oxidized, acting as a reducing agent.