Chapter 20.3, Problem 23SP

a)

Interpretation Introduction

Interpretation : Using the oxidation number change method, the chemical reaction is to be balanced.

Concept Introduction : The special type of chemical reaction in which one reacting species undergo oxidation while other reacting species undergo reduction is known as a redox reaction.

Answer to Problem 23SP

The required balance reaction is as follows:

  $\text{ 2K}Cl{O}_3(s)\to \text{ 2K}Cl\text{(s)}+{\text{ 3O}}_2(g)$

Explanation of Solution

Rules to calculate the oxidation number are as follows:

• All elements of the alkali group in a combined form with other atoms always show $\left(+1\right)$ oxidation state.
• All elements of alkaline earth metal groups in a combined form with other atoms always show $\left(+2\right)$ oxidation state.
• The charge of the oxygen atom is generally $\left(-2\right)$ .
• All elements of halogen groups in a combined form with other atoms always show $\left(-1\right)$ oxidation state.
• The charge of a hydrogen atom is always $\left(+1\right)$ .
• Atoms in their elemental state have zero oxidation state.

Given,

  $\text{ K}Cl{O}_3(s)\to \text{ K}Cl\text{(s)}+{\text{ O}}_2(g)$ .

The steps of the method are as follows:

Step 1: Write the oxidation number of atoms in the unbalanced chemical equation.

  $\begin{array}{l}+1\text{ +5 -2 +1 -1 0}\\ \text{K}Cl{O}_3(s)\to \text{ K}Cl\text{(s)}+{\text{ O}}_2(g)\end{array}$

Step 2: Calculation of the oxidation number of atoms is as follows:

The oxidation number of $Cl$ in $Cl{O}_{{}_3}^{-}$ is calculated as follows:

Let the oxidation number of the chlorine atom be x and the charge of oxygen atoms is always $\left(-2\right)$ .

  $\begin{array}{l}x+\left(3\times \left(-2\right)\right)+1=0\\ x-6+1=0\\ x=+5\end{array}$

Here, $\left(+1\right)$ is the negative charge.

So, let the oxidation number of the iodine atom of $O_2$ be x.

  $\begin{array}{l}2x=0\\ x=\frac{0}{2}\approx notdefiend\\ x=0\end{array}$

Step 3: The reaction is a redox reaction so, the oxidized and reduced atom is to be identified.

Now, balance the charge by adding electrons

  

The reaction after balancing the charge is as follows:

  $\text{K}Cl{O}_3(s)\to \text{ K}Cl\text{(s)}+{\text{ O}}_2(g)$

This can be understood as follows:

  $\begin{array}{l}+5\left(-2\times 3\right)-1\\ Cl{O}_3^{-}(aq)\to C{l}^{-}\text{(aq)}\end{array}$

Step 3: Balance the atoms other and oxygen on both sides of the reaction.

  $\text{2K}Cl{O}_3(s)\to \text{ 2K}Cl\text{(s)}+{\text{ 3O}}_2(g)$

The required balance reaction is as follows:

  $\text{2K}Cl{O}_3(s)\to \text{ 2K}Cl\text{(s)}+{\text{ 3O}}_2(g)$

b)

Interpretation Introduction

Interpretation : Using the oxidation number change method, the chemical reaction is to be balanced.

Concept Introduction : Oxidation reaction is a part of redox reaction in which the removal of electrons or addition of oxygen takes place.

Answer to Problem 23SP

The required balance reaction is as follows:

  $2HN{O}_2(aq)+2HI\text{(aq)}\to 2NO\text{(g)}+I_2(s)+2H_{2}O(l)$

Explanation of Solution

Rules to calculate the oxidation number are as follows:

• All elements of the alkali group in a combined form with other atoms always show $\left(+1\right)$ oxidation state.
• All elements of alkaline earth metal groups in a combined form with other atoms always show $\left(+2\right)$ oxidation state.
• The charge of the oxygen atom is generally $\left(-2\right)$ .
• All elements of halogen groups in a combined form with other atoms always show $\left(-1\right)$ oxidation state.
• The charge of the hydrogen atom is always $\left(+1\right)$ .
• Atoms in their elemental state have zero oxidation state.

Given,

  $HN{O}_2(aq)+HI\text{(aq)}\to NO\text{(g)}+I_2(s)+H_{2}O(l)$ .

The steps of the method are as follows:

Step 1: Write the oxidation number of atoms in the unbalanced chemical equation.

  $\begin{array}{l}+1\text{ -1 +1 -1 +2 -2 0 }\left(\text{ +1}\times \text{2}\right)\text{ -2}\\ HN{O}_2(aq)+HI\text{(aq)}\to NO\text{(g)}+I_2(s)+H_{2}O(l)\end{array}$

Step 2: The reaction is a redox reaction so, the oxidized and reduced atom is to be identified.

Now, balance the charge by adding electrons

  

The reaction after balancing the charge is as follows:

  

  $HN{O}_2(aq)+2HI\text{(aq)}\to NO\text{(g)}+I_2(s)+H_{2}O(l)$

Step 3: Balance all the atoms on both sides of the reaction.

  $2HN{O}_2(aq)+2HI\text{(aq)}\to 2NO\text{(g)}+I_2(s)+2H_{2}O(l)$

So, the required balance reaction is as follows:

  $2HN{O}_2(aq)+2HI\text{(aq)}\to 2NO\text{(g)}+I_2(s)+2H_{2}O(l)$

The required balance reaction is as follows:

  $2HN{O}_2(aq)+2HI\text{(aq)}\to 2NO\text{(g)}+I_2(s)+2H_{2}O(l)$