Problem 51 - Chemistry: The Central Science

Question

Which element is oxidized and which is reduced in the following reactions? (a) \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\) (b) \(3 \mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{2}(a q)+2 \mathrm{Al}(s) \longrightarrow\) \(3 \mathrm{Fe}(s)+2 \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}(a q)\) (c) \(\mathrm{Cl}_{2}(a q)+2 \mathrm{NaI}(a q) \longrightarrow \mathrm{I}_{2}(a q)+2 \mathrm{NaCl}(a q)\) (d) \(\mathrm{PbS}(s)+4 \mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow \mathrm{PbSO}_{4}(s)+4 \mathrm{H}_{2} \mathrm{O}(l)\)

Step-by-Step Solution

Answer

(a) Nitrogen is reduced, and hydrogen is oxidized. (b) Iron is reduced, and aluminum is oxidized. (c) Chlorine is reduced, and iodine is oxidized. (d) Sulfur is oxidized, and no element is reduced.

Step 1: Identify Oxidation and Reduction

Oxidation is when an element loses electrons during a chemical reaction, which results in an increase of its oxidation state. Reduction is when an element gains electrons during a chemical reaction, resulting in a decrease of its oxidation state. We will now analyze each reaction to find which elements are undergoing oxidation and reduction. (a) \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\)

Step 2: Determine Oxidation State Changes in Reaction (a)

In this reaction, nitrogen starts in a molecule with oxidation state 0 and changes to -3 in the ammonia molecule, while hydrogen starts in a molecule with oxidation state 0 and changes to +1 in the ammonia molecule. Therefore, nitrogen is reduced by gaining electrons and hydrogen is oxidized by losing electrons. (b) \(3 \mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{2}(a q)+2 \mathrm{Al}(s) \longrightarrow 3 \mathrm{Fe}(s)+2 \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}(a q)\)

Step 3: Determine Oxidation State Changes in Reaction (b)

In this reaction, iron starts with an oxidation state of +2 in the ferrous nitrate and changes to an oxidation state of 0 as a solid, while aluminum starts with an oxidation state of 0 as a solid and changes to an oxidation state of +3 in the aluminum nitrate. Therefore, iron is reduced by gaining electrons and aluminum is oxidized by losing electrons. (c) \(\mathrm{Cl}_{2}(a q)+2 \mathrm{NaI}(a q) \longrightarrow \mathrm{I}_{2}(a q)+2 \mathrm{NaCl}(a q)\)

Step 4: Determine Oxidation State Changes in Reaction (c)

In this reaction, chlorine starts with an oxidation state of 0 in the chlorine molecules and changes to -1 in the sodium chloride molecules, while iodine starts with an oxidation state of -1 in the sodium iodide molecules and changes to 0 in the iodine molecules. Therefore, chlorine is reduced by gaining electrons and iodine is oxidized by losing electrons. (d) \(\mathrm{PbS}(s)+4 \mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow \mathrm{PbSO}_{4}(s)+4 \mathrm{H}_{2} \mathrm{O}(l)\)

Step 5: Determine Oxidation State Changes in Reaction (d)

In this reaction, lead starts with an oxidation state of +2 in the lead sulfide and maintains the same oxidation state in the lead sulfate, while sulfur starts with an oxidation state of -2 in the lead sulfide and changes to an oxidation state of +4 in the lead sulfate. Hence, sulfur is oxidized by losing electrons. However, no element is reduced in this reaction as hydrogen peroxide and water have the same oxidation state for oxygen.. In summary, the elements being oxidized and reduced in each reaction are as follows: (a) Nitrogen is reduced, and hydrogen is oxidized. (b) Iron is reduced, and aluminum is oxidized. (c) Chlorine is reduced, and iodine is oxidized. (d) Sulfur is oxidized, and no element is reduced.

Key Concepts

Oxidation State

Understanding the oxidation state of an element in a chemical compound is crucial for analyzing redox reactions. This state represents the theoretical charge that an atom would have if all its bonds to atoms of different elements were 100% ionic.

With simple guidelines, determining oxidation numbers becomes easier: elements in their elemental form have an oxidation state of 0; for a neutral compound, the sum of oxidation states equals zero; for ions, the sum equals the charge of the ion.

For example, in reaction (a), before the reaction occurs, both \( \mathrm{N}_{2}\) and \( \mathrm{H}_{2}\) are in their elemental forms, so their oxidation states are 0. Through the reaction, nitrogen and hydrogen change their oxidation states to -3 and +1, respectively, in the ammonia (\(\mathrm{NH}_{3}\)) molecule. These changes are essential for identifying which elements are oxidized or reduced.

Chemical Oxidation

Chemical oxidation involves the loss of electrons by an element, which leads to an increase in the oxidation state. It's half of the dual process occurring in any redox reaction. In our textbook examples, hydrogen in reaction (a) is oxidized as it goes from an oxidation state of 0 to +1.

Similarly, in reaction (b), aluminum is oxidized as its oxidation state rises from 0 to +3, and in reaction (c), iodine is oxidized from an oxidation state of -1 to 0.

Recognizing Oxidation

Look for elements that lose electrons and those whose oxidation state increases. The elemental identity of the oxidizing agent is also key here—it gains the electrons lost by the element being oxidized, undergoing a reduction process itself.

Chemical Reduction

The counterpart to oxidation is chemical reduction, and it involves the gain of electrons by an element, resulting in a decrease in oxidation state. This process signifies the acceptance of electrons that another element loses during oxidation.

In our reactions, nitrogen is reduced in (a), iron in (b), and chlorine in (c) as their oxidation states decrease from 0 to -3, +2 to 0, and 0 to -1, respectively.

Identifying Reduction

To pinpoint reduction, search for elements increasing their negative charge through electron gain, demonstrated by a decrease in oxidation state. In the broader context of the reaction, these elements act as oxidizing agents, driving the oxidation of other elements.