Question

Which element is oxidized, and which is reduced in the following reactions? \begin{equation} \begin{array}{l}{\text { (a) } \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\longrightarrow} \\ {\text { (b) } 3 \mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{2}(a q)+2 \mathrm{Al}(s) \longrightarrow} \\\\\quad \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad{3 \mathrm{Fe}(s)+2 \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}(a q)}\\\\{\text { (c) } \mathrm{Cl}_{2}(a q)+2 \operatorname{Nal}(a q) \longrightarrow \mathrm{I}_{2}(a q)+2 \mathrm{NaCl}(a q)} \\ {\text { (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)}\end{array} \end{equation}

Short answer

In the given reactions: (a) N is reduced (from \(0\) to \(-3\)) and H is oxidized (from \(0\) to \(+1\)). (b) Fe is reduced (from \(+2\) to \(0\)) and Al is oxidized (from \(0\) to \(+3\)). (c) Cl is reduced (from \(0\) to \(-1\)) and I is oxidized (from \(-1\) to \(0\)). (d) S is oxidized (from \(-2\) to \(+6\)) and O is reduced (from \(-1\) to \(-2\)).

Step-by-step solution

(a) Reaction

The reaction is given as: \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\) Step 1: Identify the initial and final oxidation states For N: Initial \(N_{2}\) is \(0\) and final in \(NH_{3}\) is \(-3\) For H: Initial \(H_{2}\) is \(0\) and final in \(NH_{3}\) is \(+1\) Step 2: Identify oxidation and reduction N is reduced because it gains 3 electrons, changing its oxidation state from \(0\) to \(-3\). H is oxidized because it loses 1 electron, changing its oxidation state from \(0\) to \(+1\).

(b) Reaction

The reaction is given as: \(3\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{2}(a q)+2 \mathrm{Al}(s) \longrightarrow 3 \mathrm{Fe}(s)+2\mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}(a q)\) Step 1: Identify the initial and final oxidation states For Fe: Initial in \(Fe(NO_{3})_{2}\) is \(+2\) and final in \(Fe\) is \(0\) For Al: Initial \(Al\) is \(0\) and final in \(Al(NO_{3})_{3}\) is \(+3\) Step 2: Identify oxidation and reduction Fe is reduced because it gains 2 electrons, changing its oxidation state from \(+2\) to \(0\). Al is oxidized because it loses 3 electrons, changing its oxidation state from \(0\) to \(+3\).

(c) Reaction

The reaction is given as: \(\mathrm{Cl}_{2}(a q)+2 \operatorname{Nal}(a q) \longrightarrow \mathrm{I}_{2}(a q)+2 \mathrm{NaCl}(a q)\) Step 1: Identify the initial and final oxidation states For Cl: Initial \(Cl_{2}\) is \(0\) and final in \(NaCl\) is \(-1\) For I: Initial in \(NaI\) is \(-1\) and final in \(I_{2}\) is \(0\) Step 2: Identify oxidation and reduction Cl is reduced because it gains 1 electron, changing its oxidation state from \(0\) to \(-1\). I is oxidized because it loses 1 electron, changing its oxidation state from \(-1\) to \(0\).

(d) Reaction

The reaction is given as: \(\mathrm{PbS}(s)+4 \mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow \mathrm{PbSO}_{4}(s)+4 \mathrm{H}_{2} \mathrm{O}(l)\) Step 1: Identify the initial and final oxidation states For Pb: Initial in \(PbS\) is \(+2\) and final in \(PbSO_{4}\) is \(+2\) (No change) For S: Initial in \(PbS\) is \(-2\) and final in \(PbSO_{4}\) is \(+6\) For H: Initial in \(H_{2}O_{2}\) is \(+1\) and final in \(H_{2}O\) is \(+1\) (No change) For O: Initial in \(H_{2}O_{2}\) is \(-1\) and final in \(H_{2}O\) is \(-2\) Step 2: Identify oxidation and reduction S is oxidized because it loses 8 electrons, changing its oxidation state from \(-2\) to \(+6\). O is reduced because it gains 1 electron, changing its oxidation state from \(-1\) to \(-2\).

Key concepts

Oxidation State

The oxidation state is a concept that helps us determine the loss or gain of electrons in a chemical reaction. It assigns hypothetical charges to elements within a compound based on certain rules. This helps us track how electrons are transferred in reactions.

Some basic rules to find the oxidation state are:

• An element in its natural state (like O₂, N₂) has an oxidation state of 0.
• The sum of the oxidation states in a neutral compound is 0, while for an ion, it equals the ion's charge.
• In compounds, hydrogen usually has an oxidation state of +1 and oxygen -2.

These rules guide us in determining the oxidation states and help us identify which elements are interacting as oxidizers or reducers.

Oxidation

Oxidation refers to the process where an element loses electrons during a chemical reaction. When an element is oxidized, its oxidation state increases. This essentially means that the element is losing control over one or more electrons.

In the context of the exercise reactions:

• Hydrogen in the reaction (a), where its oxidation state changes from 0 in H₂ to +1 in NH₃, is oxidized as it loses electrons.
• Aluminum in reaction (b) is oxidized from an oxidation state of 0 in Al to +3 in Al(NO₃)₃.
• Iodine in reaction (c) sees an increase in oxidation state from -1 in NaI to 0 in I₂, indicating oxidation.
• Sulfur in reaction (d) is oxidized from -2 in PbS to +6 in PbSO₄.

This loss of electrons is crucial in redox reactions and identifies the substance that is oxidized.

Reduction

Reduction is the gain of electrons by an element in a chemical reaction. As an element gains electrons, its oxidation state decreases. This process is coupled with oxidation in what are known as redox reactions.

In the reactions provided:

• Nitrogen in reaction (a) is reduced as its oxidation state changes from 0 in N₂ to -3 in NH₃, indicating it has gained electrons.
• Iron in reaction (b) is reduced from an oxidation state of +2 in Fe(NO₃)₂ to 0 in Fe.
• Chlorine in reaction (c) has a reduction, with its oxidation state decreasing from 0 in Cl₂ to -1 in NaCl, showing electron gain.
• Oxygen in reaction (d) is reduced as its oxidation state shifts from -1 in H₂O₂ to -2 in H₂O.

Understanding reduction helps in grasping the complete picture of electron movement in reactions.