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) \(\mathrm{PbS}(s)+4 \mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow \mathrm{PbSO}_{4}(s)+4 \mathrm{H}_{2} \mathrm{O}(l)\)

Short answer

(a) N is reduced, H is oxidized. (b) Al is oxidized, Fe is reduced. (c) Cl is reduced, I is oxidized. (d) S is oxidized, O is reduced.

Step-by-step solution

Identify oxidation states in reaction (a)

For the reaction \( \mathrm{N}_2(g) + 3\mathrm{H}_2(g) \rightarrow 2\mathrm{NH}_3(g) \), start by assigning oxidation states. In \( \mathrm{N}_2 \), nitrogen has an oxidation state of 0 because it is in its elemental form. In \( \mathrm{H}_2 \), hydrogen also has an oxidation state of 0. In \( \mathrm{NH}_3 \), nitrogen has an oxidation state of -3 and hydrogen has an oxidation state of +1. Thus, nitrogen is reduced as its oxidation state decreases, and hydrogen is oxidized as its oxidation state increases.

Determine oxidation and reduction in reaction (b)

In the reaction \( 3 \mathrm{Fe}\left(\mathrm{NO}_3\right)_2(aq) + 2 \mathrm{Al}(s) \rightarrow 3 \mathrm{Fe}(s) + 2 \mathrm{Al}\left(\mathrm{NO}_3\right)_3(aq) \), identify the changes in oxidation states. Aluminum in \( \mathrm{Al}(s) \) changes from 0 to +3 (oxidized), while iron in \( \mathrm{Fe}\left(\mathrm{NO}_3\right)_2 \) changes from +2 to 0 (reduced). Hence, aluminum is oxidized, and iron is reduced.

Analyze reaction (c) for redox changes

For \( \mathrm{Cl}_2(aq) + 2 \mathrm{NaI}(aq) \rightarrow \mathrm{I}_2(aq) + 2 \mathrm{NaCl}(aq) \), chlorine starts with an oxidation state of 0 and ends as -1 in \( \mathrm{NaCl} \) (gains electrons — reduced). Iodine starts with -1 in \( \mathrm{NaI} \) and becomes 0 in \( \mathrm{I}_2 \) (loses electrons — oxidized). Therefore, chlorine is reduced, and iodine is oxidized.

Identify oxidation and reduction for reaction (d)

In the reaction \( \mathrm{PbS}(s) + 4 \mathrm{H}_2 \mathrm{O}_2(aq) \rightarrow \mathrm{PbSO}_4(s) + 4 \mathrm{H}_2 \mathrm{O}(l) \), sulfur in \( \mathrm{PbS} \) changes from -2 to +6 in \( \mathrm{PbSO}_4 \) (oxidized), and oxygen in \( \mathrm{H}_2 \mathrm{O}_2 \) changes from -1 to -2 in \( \mathrm{H}_2 \mathrm{O} \) (reduced). Therefore, sulfur is oxidized, and oxygen is reduced.

Key concepts

Oxidation States

Oxidation states are vital in identifying redox reactions because they represent the imagined charges atoms would have in a compound. Every atom in a pure element has an oxidation state of 0. This means, for example, that in the molecule \( \mathrm{N}_2 \), nitrogen has an oxidation state of 0. In the compound \( \mathrm{NH}_3 \), nitrogen takes on an oxidation state of \(-3\), while hydrogen has an oxidation state of \(+1\). This difference suggests that nitrogen is gaining electrons (being reduced), and hydrogen is losing electrons (being oxidized).

• Oxidation state of an atom in its elemental form = 0
• Sum of oxidation states in a molecule = charge on the molecule

Understanding oxidation states allows us to track electron movements in reactions. By noting the change in oxidation states from reactants to products, we can determine which atoms are oxidized and which are reduced. This is pivotal for writing balanced chemical equations, particularly in redox reactions.

Oxidation

Oxidation is a process where an atom, ion, or molecule loses electrons, leading to an increase in oxidation state. A simple method to remember this could be using the mnemonic OIL (Oxidation Is Loss). Let's use the reaction \( \mathrm{Cl}_2(aq) + 2 \mathrm{NaI}(aq) \rightarrow \mathrm{I}_2(aq) + 2 \mathrm{NaCl}(aq) \) to explain this:

• In \( \mathrm{NaI} \), iodine has an oxidation state of \(-1\).
• In \( \mathrm{I}_2 \), iodine ends up with an oxidation state of 0.

This shift indicates that iodine loses electrons in this process, and thus, is oxidized. The electron loss is evident by the increase in oxidation state from \(-1\) to 0. Understanding oxidation is crucial as it helps in the identification of reducing agents, which are substances that lose electrons and are oxidized.

Reduction

Reduction is the gain of electrons, leading to a decrease in oxidation state. This can be recalled using the mnemonic RIG (Reduction Is Gain). Let’s analyze the reaction \( 3 \mathrm{Fe}(\mathrm{NO}_3)_2(aq) + 2 \mathrm{Al}(s) \rightarrow 3 \mathrm{Fe}(s) + 2 \mathrm{Al}(\mathrm{NO}_3)_3(aq) \) as an example:

• Iron in \( \mathrm{Fe}(\mathrm{NO}_3)_2 \) starts with an oxidation state of \(+2\).
• In \( \mathrm{Fe}(s)\), iron ends with an oxidation state of 0.

The decrease in oxidation state from \(+2\) to 0 signifies that iron gains electrons, marking it as reduced in the reaction. Recognizing which substances are undergoing reduction allows us to identify oxidizing agents, substances that accept electrons and are themselves reduced. This understanding is essential in predicting and manipulating chemical reactions.