Question

Iodine, used as an antiseptic agent, can be prepared in the laboratory by the following reaction: \(\begin{array}{l} 2 \mathrm{NaI}(s)+2 \mathrm{H}_{2} \mathrm{SO}_{4}(a q)+\mathrm{MnO}_{2}(s) \longrightarrow \\ \quad \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+\mathrm{MnSO}_{4}(a q)+\mathrm{I}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) \end{array}\) (a) Determine the oxidation number for the Mn and I on both sides of the equation. (b) Identify the oxidizing and reducing agents.

Short answer

(a) Mn changes from +4 to +2, I changes from -1 to 0. (b) MnO2 is the oxidizing agent, NaI is the reducing agent.

Step-by-step solution

Assign Oxidation Numbers to Reactants

First, we look at the Mn and I in the reactants: - For Mn in MnO2, since O has an oxidation number of -2 and there are two O atoms, Mn must be +4 (because 2*(-2) + Mn = 0). - For I in NaI, Na is +1 and since the compound is neutral, I must be -1.

Assign Oxidation Numbers to Products

Now, we evaluate the Mn and I in the products: - For Mn in MnSO4, sulfur has an oxidation number of +6 and oxygen -2 each, thus Mn must be +2 (total oxidation for SO4 is -2). This makes the sum of charges zero for the compound. - For I2, since it is a diatomic molecule, the oxidation number of I is 0.

Identify Change in Oxidation Numbers

Identify the changes: - Mn goes from +4 in MnO2 to +2 in MnSO4, which is a decrease of 2 units, indicating a reduction. - I goes from -1 in NaI to 0 in I2, which is an increase of 1 unit, indicating an oxidation.

Determine Oxidizing and Reducing Agents

Since Mn was reduced, MnO2 is the oxidizing agent as it accepts electrons. Since I was oxidized, NaI is the reducing agent as it provides electrons.

Key concepts

Oxidation Numbers

Understanding oxidation numbers is a fundamental step in analyzing redox reactions. Each element in a compound is assigned an oxidation number, which is a theoretical charge that an atom would have if all bonds were ionic. This helps identify which species undergo oxidation or reduction during the reaction.

To determine oxidation numbers, follow these basic rules:

• Pure elements have an oxidation number of zero, such as \( ext{I}_2\).
• The oxidation number of ions is the same as their charge, e.g., +1 for \( ext{Na}^+\).
• In compounds like \( ext{H}_2 ext{O}\) or acids like \( ext{H}_2 ext{SO}_4\), hydrogen is +1 and oxygen is -2, unless they are in specific exceptions like peroxides.
• The sum of oxidation numbers in a neutral compound is zero, and for polyatomic ions, it is equal to the ion's charge.

In our specific example, manganese ( ext{Mn}) in \( ext{MnO}_2\) starts at +4 and ends at +2 in \( ext{MnSO}_4\), while iodine ( ext{I}) changes from -1 in \( ext{NaI}\) to 0 in \( ext{I}_2\), showcasing the redox nature of the process.

Oxidizing and Reducing Agents

Oxidizing and reducing agents are agents of change in a redox reaction. They are critical players in the transfer of electrons. The oxidizing agent is the species that gets reduced and gains electrons, while the reducing agent is oxidized and loses electrons.

In the context of our reaction, manganese dioxide \(\text{MnO}_2\) acts as the oxidizing agent. This is because the manganese gains electrons as its oxidation number goes from +4 to +2, signifying reduction.

On the flip side, sodium iodide \(\text{NaI}\) functions as the reducing agent. Iodine, originally at an oxidation state of -1 in \(\text{NaI}\), donates electrons and is oxidized to a zero oxidation state in \(\text{I}_2\).

• Oxidizing Agent: Accepted electrons and was reduced (MnO ext{2}).
• Reducing Agent: Supplied electrons and was oxidized (NaI).

Observing how these agents are identified by changes in oxidation numbers can significantly help in visualizing the electron flow in redox reactions.

Chemical Equation Balancing

Balancing chemical equations is the process of ensuring that the number of atoms for each element is equal on both the reactant and product sides. It is a fundamental step in demonstrating the conservation of mass in a chemical reaction.

When dealing with redox reactions specifically, consider not just the number of atoms, but also the transfer of electrons. Balancing redox reactions often involves the use of the half-reaction method, where you separately balance the oxidation and reduction reactions before combining them.

In our iodination example, the reaction:
\[2 \text{NaI} + 2 \text{H}_2\text{SO}_4 + \text{MnO}_2 \rightarrow \text{Na}_2\text{SO}_4 + \text{MnSO}_4 + \text{I}_2 + 2\text{H}_2\text{O}\]
ensures equal quantities of each ion and molecule, matching the change in oxidation numbers for the elements involved. The equation, as presented, satisfies both mass and charge balance across the entire reaction.

• Always check each species atom balance and total charge.
• Consider starting with the more complex molecules first when balancing.

Mastering the art of balancing chemical equations enhances your understanding and ensures accuracy in chemistry problem-solving.