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}

Step by step solution

01

(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 \(+1\).

02

(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 \(+3\).

03

(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 \(0\).

04

(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 \(-2\).

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