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

In the given reactions: (a) N is reduced and H is oxidized. (b) Fe is reduced and Al is oxidized. (c) Cl is reduced and I is oxidized. (d) S is oxidized and O is reduced.

Step-by-step solution

(a) Assigning Oxidation Numbers

For the reaction $\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g):$ Oxidation numbers: \(N_{2}\): \(0 \longrightarrow -3\) in \(NH_{3}\), \(H_{2}\): \(0 \longrightarrow +1\) in \(NH_{3}\)

(a) Identifying Oxidizing and Reducing Elements

\(N_{2}\) goes from 0 to -3, meaning it is reduced (gain in electrons). \(H_{2}\) goes from 0 to +1, meaning it is oxidized (loss of electrons).

(b) Assigning Oxidation Numbers

For the reaction \(3 \mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{2}(a q)+2 \mathrm{Al}(s) \longrightarrow 3 \mathrm{Fe}(s)+2 \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}(a q):\) Oxidation numbers: \(Fe\) in \(\mathrm{Fe(NO_{3})_{2}}\): \(+2 \longrightarrow 0\) in \(Fe\), \(Al\): \(0 \longrightarrow +3\) in \(\mathrm{Al(NO_{3})_{3}}\)

(b) Identifying Oxidizing and Reducing Elements

\(Fe\) goes from +2 to 0, meaning it is reduced (gain in electrons). \(Al\) goes from 0 to +3, meaning it is oxidized (loss of electrons).

(c) Assigning Oxidation Numbers

For the reaction \(\mathrm{Cl}_{2}(a q)+2 \mathrm{NaI}(a q) \longrightarrow \mathrm{I}_{2}(a q)+2 \mathrm{NaCl}(a q):\) Oxidation numbers: \(Cl_{2}\): \(0 \longrightarrow -1\) in \(\mathrm{NaCl}\), \(I\) in \(\mathrm{NaI}\): \(-1 \longrightarrow 0\) in \(I_{2}\)

(c) Identifying Oxidizing and Reducing Elements

\(Cl_{2}\) goes from 0 to -1, meaning it is reduced (gain in electrons). \(I\) goes from -1 to 0, meaning it is oxidized (loss of electrons).

(d) Assigning Oxidation Numbers

For the reaction \(\mathrm{PbS}(s)+4 \mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow \mathrm{PbSO}_{4}(s)+4 \mathrm{H}_{2} \mathrm{O}(l)\): Oxidation numbers: \(S\) in \(\mathrm{PbS}\): \(-2 \longrightarrow +6\) in \(\mathrm{PbSO_{4}}\), \(O\) in \(\mathrm{H_{2}O_{2}}\): \(-1 \longrightarrow -2\) in \(\mathrm{H_{2}O}\)

(d) Identifying Oxidizing and Reducing Elements

\(S\) goes from -2 to +6, meaning it is oxidized (loss of electrons). \(O\) goes from -1 to -2, meaning it is reduced (gain in electrons).