Chapter 18, Problem 48A

(a)

Interpretation Introduction

Interpretation: From the given reaction, the element that is oxidized and the one that is reduced are to be identified.

$3\text{Zn}\left(\text{s}\right)+{\text{N}}_{\text{2}}\left(\text{g}\right)\to {\text{Zn}}_{\text{3}}{\text{N}}_{\text{2}}\left(\text{s}\right)$

Concept Introduction: In oxidation-reduction reaction, transfer of electron/s takes place. The transfer of electron, results in the formation of ions. In oxidation, loss of electron/s takes place and in reduction, gain of electron/s takes place.

Answer to Problem 48A

Nitrogen is reduced and Zinc is oxidized in this reaction.

Explanation of Solution

In Zinc nitride ${\text{Zn}}_{\text{3}}{\text{N}}_{\text{2}}$ , oxidation state of Zn is $+2$ and N is $-3$ . As both Zn and ${\text{N}}_{\text{2}}$ are in a free state, their oxidation state is zero.

In this reaction, oxidation state of N decreases from $0$ to $-3$ . Hence, it is reduced.

Oxidation state of Zn increases from $0$ to $+2$ . Hence, it is oxidized.

(b)

Interpretation Introduction

Interpretation: From the given reaction, the element that is oxidized and the one that is reduced are to be identified.

$\text{Co}\left(\text{s}\right)+\text{S}\left(\text{s}\right)\to \text{CoS}\left(\text{s}\right)$

Concept Introduction: In oxidation-reduction reaction, transfer of electron/s takes place. The transfer of electron, results in the formation of ions. In oxidation, loss of electron/s takes place and in reduction, gain of electron/s takes place.

Answer to Problem 48A

Sulphur is reduced and Co is oxidized in this reaction.

Explanation of Solution

Oxidation state of Co is $+2$ and S is $-2$ in CoS. As both Co and S are in a free state so, there oxidation state is zero.

In this reaction, oxidation state of S decreases from $0$ to $-2$ . Hence, it is reduced.

Oxidation state of Co increases from $0$ to $+2$ . Hence, it is oxidized.

(c)

Interpretation Introduction

Interpretation: From the given reaction, the element that is oxidized and the one that is reduced are to be identified.

$4\text{K}\left(\text{s}\right)+{\text{O}}_{\text{2}}\left(\text{g}\right)\to 2{\text{K}}_{\text{2}}\text{O}\left(\text{s}\right)$

Concept Introduction: In oxidation-reduction reaction, transfer of electron/s takes place. The transfer of electron, results in the formation of ions. In oxidation, loss of electron/s takes place and in reduction, gain of electron/s takes place.

Answer to Problem 48A

Oxygen is reduced and potassium (K) is oxidized in this reaction.

Explanation of Solution

In ${\text{K}}_{\text{2}}\text{O}$ , oxidation state of K is $+1$ and O is $-2$ . In reactant side, oxidation state of K and ${\text{O}}_{\text{2}}$ is zero.

In this reaction oxidation state of O decreases from $0$ to $-2$ . Hence, it is reduced.

Oxidation state of K increases from $0$ to $+1$ . Hence, it is oxidized.

(d)

Interpretation Introduction

Interpretation: From the given reaction, the element that is oxidized and the one that is reduced are to be identified.

$4\text{Ag}\left(\text{s}\right)+{\text{O}}_{\text{2}}\left(\text{g}\right)\to 2{\text{Ag}}_{\text{2}}\text{O}$

Concept Introduction: In oxidation-reduction reaction, transfer of electron/s takes place. The transfer of electron, results in the formation of ions. In oxidation, loss of electron/s takes place and in reduction, gain of electron/s takes place.

Answer to Problem 48A

Oxygen is reduced and Silver (Ag) is oxidized in this reaction.

Explanation of Solution

In ${\text{Ag}}_{\text{2}}\text{O}$ , oxidation state of Ag is $+1$ and O is $-2$ . In reactant side, oxidation state of Ag and ${\text{O}}_{\text{2}}$ is zero.

In this reaction, oxidation state of O decreases from $0$ to $-2$ . Hence, it is reduced.

Oxidation state of Ag increases from $0$ to $+1$ . Hence, it is oxidized.