Question

In each of the following reactions, tell which substance is undergoing an oxidation and which a reduction, and identify the oxidizing and reducing agents. (a) \(\mathrm{SnO}_{2}(s)+2 \mathrm{C}(s) \longrightarrow \mathrm{Sn}(s)+2 \mathrm{CO}(g)\) (b) \(\mathrm{Sn}^{2+}(a q)+2 \mathrm{Fe}^{3+}(a q) \longrightarrow \mathrm{Sn}^{4+}(a q)+2 \mathrm{Fe}^{2+}(a q)\) (c) \(4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(l)\)

Short answer

(a) SnO₂ is oxidizing agent, C is reducing agent. (b) Fe³⁺ is oxidizing agent, Sn²⁺ is reducing agent. (c) O₂ is oxidizing agent, NH₃ is reducing agent.

Step-by-step solution

Determine Oxidation Numbers for Reaction (a)

Identify the oxidation states of each element in the reaction \( \mathrm{SnO}_{2}(s)+2 \mathrm{C}(s) \longrightarrow \mathrm{Sn}(s)+2 \mathrm{CO}(g) \).- \( \mathrm{Sn} \) in \( \mathrm{SnO}_{2} \): +4- \( \mathrm{C} \) in \( \mathrm{C} \): 0- \( \mathrm{Sn} \) in \( \mathrm{Sn} \): 0- \( \mathrm{C} \) in \( \mathrm{CO} \): +2Identify changes:- \( \mathrm{Sn} \) is reduced from +4 to 0.- \( \mathrm{C} \) is oxidized from 0 to +2.

Identify Oxidizing and Reducing Agents in Reaction (a)

In reaction (a):- \( \mathrm{SnO}_{2} \) undergoes reduction, so it is the oxidizing agent.- \( \mathrm{C} \) undergoes oxidation, so it is the reducing agent.

Determine Oxidation Numbers for Reaction (b)

Identify the oxidation states of elements in \( \mathrm{Sn}^{2+}(aq)+2 \mathrm{Fe}^{3+}(aq) \longrightarrow \mathrm{Sn}^{4+}(aq)+2 \mathrm{Fe}^{2+}(aq) \).- \( \mathrm{Sn} \): Changes from +2 to +4- \( \mathrm{Fe} \): Changes from +3 to +2Identify changes:- \( \mathrm{Sn} \) is oxidized from +2 to +4.- \( \mathrm{Fe} \) is reduced from +3 to +2.

Identify Oxidizing and Reducing Agents in Reaction (b)

In reaction (b):- \( \mathrm{Sn}^{2+} \) undergoes oxidation, so it is the reducing agent.- \( \mathrm{Fe}^{3+} \) undergoes reduction, so it is the oxidizing agent.

Determine Oxidation Numbers for Reaction (c)

Identify the oxidation states of elements in \( 4\mathrm{NH}_{3}(g)+5\mathrm{O}_{2}(g) \longrightarrow 4\mathrm{NO}(g)+6\mathrm{H}_{2}\mathrm{O}(l) \).- \( \mathrm{N} \) in \( \mathrm{NH}_3 \): -3- \( \mathrm{N} \) in \( \mathrm{NO} \): +2- \( \mathrm{O} \) in \( \mathrm{O}_2 \): 0- \( \mathrm{O} \) in \( \mathrm{H}_2\mathrm{O} \): -2Identify changes:- \( \mathrm{N} \) in \( \mathrm{NH}_3 \) is oxidized from -3 to +2.- \( \mathrm{O}_2 \) is reduced from 0 to -2.

Identify Oxidizing and Reducing Agents in Reaction (c)

In reaction (c):- \( \mathrm{NH}_3 \) undergoes oxidation, so it is the reducing agent.- \( \mathrm{O}_2 \) undergoes reduction, so it is the oxidizing agent.

Key concepts

Oxidation States

Oxidation states are a way to keep track of electron transfers in chemical reactions. They represent the degree of oxidation of an atom in a molecule or ion. You can think of the oxidation state as the hypothetical charge that an atom would have if all bonds to atoms of different elements were 100% ionic. This allows chemists to understand how electrons are distributed in a chemical compound.To determine the oxidation state of an element in a compound, we often use a set of rules, such as:

• The oxidation state of an element in its standard state (i.e., as an isolated atom or molecule such as \(O_2\), \(Sn\)) is 0.
• For monoatomic ions, the oxidation state is equal to the charge of the ion.
• Oxygen generally has an oxidation state of -2, while hydrogen is typically +1.

By applying these rules, we can find the oxidation states of elements in reactions. For example, in the reaction \[\mathrm{SnO}_2(s) + 2\mathrm{C}(s) \to \mathrm{Sn}(s) + 2\mathrm{CO}(g)\]Tin (\(\mathrm{Sn}\)) changes its oxidation state from +4 in \(\mathrm{SnO}_2\) to 0 in \(\mathrm{Sn}\), showing it is reduced. Carbon (\(\mathrm{C}\)) goes from 0 to +2, indicating it is oxidized.

Oxidizing Agents

An oxidizing agent is a substance that oxidizes another substance by accepting its electrons. Thus, oxidizing agents themselves are reduced in the process. Think of oxidizing agents as electron "thieves" in chemical reactions.In the first reaction from our example:\[\mathrm{SnO}_2(s) + 2\mathrm{C}(s) \to \mathrm{Sn}(s) + 2\mathrm{CO}(g)\]The oxidation state of tin (\(\mathrm{Sn}\)) decreases from +4 in \(\mathrm{SnO}_2\) to 0 in \(\mathrm{Sn}\). This decrease in oxidation state shows that tin gains electrons. This means \(\mathrm{SnO}_2\) acts as the oxidizing agent because it "takes" electrons from carbon.Identifying oxidizing agents is vital in predicting how chemical reactions occur, especially in assessing reactions involving energy or material transformations.

Reducing Agents

On the other hand, reducing agents are substances that donate electrons to another substance, causing the latter to be reduced. In the course of the reaction, reducing agents themselves become oxidized. Consider reducing agents as the generous "donors" in electron exchanges.Consider this reaction from the examples provided:\[4\mathrm{NH}_3(g) + 5\mathrm{O}_2(g) \to 4\mathrm{NO}(g) + 6\mathrm{H}_2\mathrm{O}(l)\]Here, ammonia (\(\mathrm{NH}_3\)) serves as the reducing agent. It undergoes a change in oxidation state of nitrogen from -3 in \(\mathrm{NH}_3\) to +2 in \(\mathrm{NO}\), indicating it has lost electrons. By losing these electrons, it helps in reducing \(\mathrm{O}_2\), which accepts them and gets reduced in the process.Understanding reducing agents is crucial in fields such as metallurgy, biology, and industrial chemistry, where specific products are often produced by carefully controlling redox balances.