Chapter 20.2, Problem 14SP

a)

Interpretation Introduction

Interpretation : Oxidized and reduced atoms in the chemical reaction are to be identified.

Concept Introduction: The chemical reaction in which oxidation and reduction take place simultaneously is known as redox reaction.

Answer to Problem 14SP

The nitrogen atom of $N{O}_2^{-}$ is getting reduced and the nitrogen of $N{H}_4^{+}$ is getting oxidized.

Explanation of Solution

Oxidation is the removal of an electron and reduction is the addition of an electron or oxygen atom.

The given reaction,

  $N{H}_{4}N{O}_2(s)\to N_2(g)+2H_{2}O(g)$

The reaction can also be written as,

  $N{H}_4^{+}(aq)+N{O}_2^{-}(aq)\to N_2(g)+2H_{2}O(g)$

The oxidation number of Nitrogen is calculated as follows:

  $N{H}_{4}N{O}_2(s)\to N{H}_4^{+}(aq)+N{O}_2^{-}(aq)$

So, let the oxidation number of the nitrogen atom of $N{H}_4^{+}$ be $x$ and the charge of the hydrogen atom is always $\left(+1\right)$ . Hence,

  $\begin{array}{c}x+\left(4\times 1\right)=+1\\ x=1-3\\ x=-3\end{array}$

So, let the oxidation number of the nitrogen atom of $N{O}_2^{-}$ be $x$ and the charge of the oxygen atom is always $\left(-2\right)$ . Hence,

  $\begin{array}{c}x+2\times (-2)+(-1)=0\\ x-5=0\\ x=+5\end{array}$

The oxidation number of nitrogen in $N_2$

is calculated as follows:

Let the oxidation number of the nitrogen atom of $N_2$ be $x$ and the charge of the hydrogen atom is always $\left(+1\right)$ .

  $\begin{array}{l}2x=0\\ x=\frac{0}{2}\approx (notdefined)\\ x=0\end{array}$

In water molecules, the charge of the hydrogen atom is $\left(+1\right)$ and that of the oxygen atom is $\left(-2\right)$ .

  

Hence, the Nitrogen atom of $N{O}_2^{-}$ is getting reduced and the nitrogen of $N{H}_4^{+}$ is getting oxidized.

b)

Interpretation Introduction

Interpretation : Oxidized and reduced atoms in the chemical reaction are to be identified.

Concept Introduction: The chemical reaction in which oxidation and reduction take place simultaneously is known as a redox reaction.

Answer to Problem 14SP

The lead atom of $Pb{O}_2$ is getting reduced and the iodine is getting oxidized.

Explanation of Solution

Oxidation is the removal of an electron and reduction is the addition of an electron or oxygen atom.

The given reaction,

  $Pb{O}_2(aq)+4HI(aq)\to I_2(aq)+Pb{I}_2(s)+2H_{2}O(l)$

The reaction can also be written as,

  $Pb{O}_2(aq)+4HI(aq)\to I_2(aq)+Pb{I}_2(s)+2H_{2}O(l)$

The oxidation number of lead is calculated as follows:

  $Pb{O}_2(aq)+4HI(aq)\to I_2(aq)+Pb{I}_2(s)+2H_{2}O(l)$

So, let the oxidation number of the lead atom be $x$ and the charge of the oxygen atom is always $\left(-2\right)$ .

  $\begin{array}{c}x+\left(2\times \left(-2\right)\right)=0\\ x-4=0\\ x=+4\end{array}$

So, let the oxidation number of the iodine atom of $HI$ be $x$ and the charge of the hydrogen atom is always $\left(+1\right)$ .

  $\begin{array}{c}x+\left(1\times 1\right)=0\\ x+1=0\\ x=(-1)\end{array}$

The oxidation number of nitrogen in $I_2$

is calculated as follows:

Let the oxidation number of the nitrogen atom of $I_2$ be $x$ .

  $\begin{array}{c}2x=0\\ x=\frac{0}{2}\approx (notdefined)\\ x=0\end{array}$

The oxidation number of lead in $Pb{I}_2$

is calculated as follows:

Let the oxidation number of the nitrogen atom of $Pb$ be $x$ and the charge of the iodine atom is generally $\left(-1\right)$ .

  $\begin{array}{c}x+(2\times (-1))=0\\ x-2=0\\ x=+2\end{array}$

In water molecules, the charge of the hydrogen atom is $\left(+1\right)$ and that of the oxygen atom is $\left(-2\right)$ .

  

Hence, the lead atom of $Pb{O}_2$ is getting reduced and the iodine is getting oxidized.