Problem 44 - Chemistry The Molecular Science

Question

Metallic tin was obtained in ancient times by the reaction of the principal tin ore, cassiterite, \(\mathrm{SnO}_{2},\) with carbon from a charcoal fire: $$\mathrm{SnO}_{2}(\mathrm{~s})+2 \mathrm{C}(\mathrm{s}) \longrightarrow 2 \mathrm{CO(\mathrm{g})+\mathrm{Sn}(\mathrm{s})$$ Identify the oxidizing and reducing agents in this reaction.

Step-by-Step Solution

Answer

\( \mathrm{SnO}_2 \) is the oxidizing agent and \( \mathrm{C} \) is the reducing agent.

Step 1: Identify Oxidation States

Determine the oxidation states of the elements involved in the reaction. For \( \text{SnO}_2\), the oxidation state of Sn is +4 because oxygen is typically -2. For C (carbon), which is elemental at the outset, the oxidation state is 0.

Step 2: Analyze Changes in Oxidation States

During the reaction, \( \mathrm{Sn}^{4+} \) (from \( \mathrm{SnO}_2\)) is reduced to \( \mathrm{Sn}^0\). Carbon, initially 0 in \( \mathrm{C} \), is oxidized to \( \mathrm{CO} \) where its oxidation state is +2.

Step 3: Define the Oxidizing and Reducing Agents

The substance that gets reduced is the oxidizing agent, and the one that gets oxidized is the reducing agent. Here, Sn is reduced (from +4 to 0), so \( \mathrm{SnO}_2 \) is the oxidizing agent. Carbon is oxidized (from 0 to +2), so \( \mathrm{C} \) is the reducing agent.

Key Concepts

Oxidation States

Understanding oxidation states is crucial for identifying how electrons are transferred in a chemical reaction. Each element in a compound is assigned an oxidation state, reflecting its electron ownership compared to a neutral atom. For example, in the compound \( \text{SnO}_2 \), tin (Sn) has an oxidation state of +4 because each oxygen (O) typically has an oxidation state of -2. This convention helps us determine the electron gain or loss by elements.

During a reaction, changes in these oxidation states indicate which atoms are losing or gaining electrons. Using this analysis, we can understand redox reactions effectively, as the shift in oxidation states reveals the flow of electrons. Always begin by identifying these states before proceeding to determine which element is oxidized or reduced.

Oxidizing Agent

An oxidizing agent is a substance that gains electrons during a chemical reaction and is consequently reduced. It effectively "oxidizes" another substance by taking its electrons.

In our example, \( \text{SnO}_2 \) acts as the oxidizing agent. During the reaction, tin (Sn) in \( \text{SnO}_2 \) starts with an oxidation state of +4 and ends at 0 when it becomes metallic tin (Sn). By gaining electrons, \( \text{SnO}_2 \) is reduced, causing tin to transition from its initial oxidation to a more stable, neutral state.

This characteristic of electron gain is what makes \( \text{SnO}_2 \) the oxidizing agent, as it's responsible for oxidizing carbon, prompting the entire redox process.

Reducing Agent

A reducing agent is a substance that loses electrons, effectively "reducing" another substance by donating these electrons. In losing electrons, the reducing agent is itself oxidized.

In the reaction involving \( \text{SnO}_2 \) and carbon (C), carbon is the reducing agent. Initially, its oxidation state is 0 (since it is elemental), but it becomes +2 in carbon monoxide (CO). This change occurs because carbon loses electrons in the transformation, facilitating the reduction of tin.

Recognizing the reducing agent helps in identifying how substances interact at the molecular level, clarifying the roles of each reactant in electron exchange during redox reactions.