Chapter 20.3, Problem 22SP

a)

Interpretation Introduction

Interpretation : Whether the given chemical reaction is redox or not is to be identified and if it is redox then oxidized and reduced atoms in the chemical reaction are to be named.

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

Answer to Problem 22SP

The reaction $CaC{O}_3(s)+2HCl(aq)\to CaC{l}_2(aq)+H_{2}O(l)+C{O}_2(g)$ is not a redox reaction.

Explanation of Solution

The chemical reaction in which oxidation and reduction take place simultaneously is known as a redox reaction. Oxidation is the removal of an electron and reduction is the addition of an electron or oxygen atom.

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.

The given reaction,

  $CaC{O}_3(s)+2HCl(aq)\to CaC{l}_2(aq)+H_{2}O(l)+C{O}_2(g)$

The oxidation number of carbon in $CaC{O}_3$ is calculated as follows:

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

  $\begin{array}{c}x+\left(3\times \left(-2\right)\right)+\left(1\times 2\right)=0\\ x-6+2=0\\ x=+4\end{array}$

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

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

Also, from the rules, it is clear that the charge on calcium in both $CaC{O}_3$ and $CaC{l}_2$ is $\left(+2\right)$ . Similarly, in water molecules as, said the charge of the hydrogen atom is $\left(+1\right)$ and that of the oxygen atom is $\left(-2\right)$ .

Hence, the reaction is not a redox reaction.

b)

Interpretation Introduction

Interpretation : Whether the given chemical reaction is redox or not is to be identified and if it is redox then oxidized and reduced atoms in the chemical reaction are to be named.

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

Answer to Problem 22SP

The reaction $CuO(s)+H_2(g)\to Cu(s)+H_{2}O(l)$ is a redox reaction in which copper of $CuO$ is getting reduced and hydrogen of hydrogen gas is getting oxidized.

Explanation of Solution

The chemical reaction in which oxidation and reduction take place simultaneously is known as a redox reaction. Oxidation is the removal of an electron and reduction is the addition of an electron or oxygen atom.

The given reaction,

  $CuO(s)+H_2(g)\to Cu(s)+H_{2}O(l)$

The oxidation number of copper in $CuO$ is calculated as follows:

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

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

So, let the oxidation number of the carbon atom of $H_2$ be $x$ .

  $\begin{array}{c}2x=0\\ x=\frac{0}{2}\approx \text{ not defined}\\ \text{x=0}\end{array}$

In water molecules, the charge of the hydrogen atom is $\left(+1\right)$ and that of the oxygen atom is $\left(-2\right)$ . Also, copper in an elemental state has zero charges.

  

Thus, the reaction $CuO(s)+H_2(g)\to Cu(s)+H_{2}O(l)$ is a redox reaction in which copper of $CuO$ is getting reduced and hydrogen of hydrogen gas is getting oxidized.