Chapter 20.1, Problem 9LC

(a)

Interpretation Introduction

Interpretation: The reducing and oxidizing agent in the given reaction, $2Na\left(s\right)\text{+}B{r}_2\left(l\right)\to 2NaBr\left(s\right)$ is to be found out.

Concept Introduction: In a redox reaction, the reagent that helps in the oxidation and loss of electrons of chemical species is referred to as a reducing agent. And the reagent that helps in the reduction of chemical species and gain of electrons is referred to as an oxidizing agent. So, oxidizing agents accept electrons while reducing agents lose electrons.

Answer to Problem 9LC

Reducing agent is $Na$ and oxidizing agent is $B{r}_2$ .

Explanation of Solution

By knowing the reduced and oxidized species in the reactions, the reducing and oxidizing agents can be easily found.

In the given reaction,

  $2Na\left(s\right)\text{+}B{r}_2\left(l\right)\to 2NaBr\left(s\right)$

Here, sodium, $Na$ undergoes oxidation by the loss of its electron. So, it is the reducing agent.

Bromine, $B{r}_2$ undergoes reduction by the gain of an electron. So, it is the oxidizing agent.

(b)

Interpretation Introduction

Interpretation: The reducing and oxidizing agent in the given reaction, $H_2\left(g\right)+C{l}_2\left(g\right)\to 2HCl\left(g\right)$ is to be found out.

Concept Introduction: In a redox reaction, the reagent that helps in the oxidation and loss of electrons of chemical species is referred to as a reducing agent. And the reagent that helps in the reduction of chemical species and gain of electrons is referred to as an oxidizing agent. So, oxidizing agents accept electrons while reducing agents lose electrons.

Answer to Problem 9LC

Reducing agent is $H_2$ and oxidizing agent is $C{l}_2$ .

Explanation of Solution

By knowing the reduced and oxidized species in the reactions the reducing and oxidizing agents can be easily found.

In the given reaction,

  $H_2\left(g\right)+C{l}_2\left(g\right)\to 2HCl\left(g\right)$

Here, $H_2$ undergoes oxidation by the loss of an electron. So, it is the reducing agent.

  $C{l}_2$ , chlorine undergoes reduction by the gain of electrons. So, it is the oxidizing agent.

(c)

Interpretation Introduction

Interpretation: The reducing and oxidizing agent in the given reaction, $2Li\left(s\right)+F_2\left(g\right)\to 2LiF\left(s\right)$ is to be found out.

Concept Introduction: In a redox reaction, the reagent that helps in the oxidation and loss of electrons of chemical species is referred to as a reducing agent. And the reagent that helps in the reduction of chemical species and gain of electrons is referred to as an oxidizing agent. So, oxidizing agents accept electrons while reducing agents lose electrons.

Answer to Problem 9LC

Reducing agent is $Li$ and oxidizing agent is $F_2$ .

Explanation of Solution

By knowing the reduced and oxidized species in the reactions the reducing and oxidizing agents can be easily found.

In the given reaction,

  $2Li\left(s\right)+F_2\left(g\right)\to 2LiF\left(s\right)$

Here, $Li$ undergoes oxidation by the loss of an electron. So, it is the reducing agent.

  $F_2$ , fluorine undergoes reduction by the gain of electrons. So, it is an oxidizing agent.

(d)

Interpretation Introduction

Interpretation: The reducing and oxidizing agent in the given reaction, $S\left(s\right)+C{l}_2\left(g\right)\to SC{l}_2\left(g\right)$ is to be found out.

Concept Introduction: In a redox reaction, the reagent that helps in the oxidation and loss of electrons of chemical species is referred to as a reducing agent. And the reagent that helps in the reduction of chemical species and gain of electrons is referred to as an oxidizing agent. So, oxidizing agents accept electrons while reducing agents lose electrons.

Answer to Problem 9LC

Reducing agent is $S$ and oxidizing agent is $C{l}_2$ .

Explanation of Solution

By knowing the reduced and oxidized species in the reactions the reducing and oxidizing agents can be easily found.

In the given reaction,

Here, sulphur, $S$ undergoes oxidation by the loss of electrons. So, it is the reducing agent.

  $C{l}_2$ , chlorine undergoes reduction by gaining electrons. So, it is the oxidizing agent.

(e)

Interpretation Introduction

Interpretation: The reducing and oxidizing agent in the given reaction, $N_2\left(g\right)+2O_2\left(g\right)\to 2N{O}_2\left(g\right)$ is to be found out.

Concept Introduction: In a redox reaction, the reagent that helps in the oxidation and loss of electrons of chemical species is referred to as a reducing agent. And the reagent that helps in the reduction of chemical species and gain of electrons is referred to as an oxidizing agent. So, oxidizing agents accept electrons while reducing agents lose electrons.

Answer to Problem 9LC

Reducing agent is $N_2$ and oxidizing agent is $O_2$ .

Explanation of Solution

By knowing the reduced and oxidized species in the reactions the reducing and oxidizing agents can be easily found.

In the given reaction,

  $N_2\left(g\right)+2O_2\left(g\right)\to 2N{O}_2\left(g\right)$

Here nitrogen, $N_2$ undergo oxidation by the loss of electrons. So, it is the reducing agent.

  $O_2$ , oxygen undergoes reduction by gaining electrons. So, it is the oxidizing agent.

(f)

Interpretation Introduction

Interpretation: The reducing and oxidizing agent in the given reaction, a $Mg\left(s\right)+Cu{\left(N{O}_3\right)}_2\left(aq\right)\to Mg{\left(N{O}_3\right)}_2\left(aq\right)+Cu\left(s\right)$ is to be found out.

Concept Introduction: In a redox reaction, the reagent that helps in the oxidation and loss of electrons of chemical species is referred to as a reducing agent. And the reagent that helps in the reduction of chemical species and gain of electrons is referred to as an oxidizing agent. So, oxidizing agents accept electrons while reducing agents lose electrons.

Answer to Problem 9LC

Reducing agent is $Mg$ and oxidizing agent is $Cu{\left(N{O}_3\right)}_2$ .

Explanation of Solution

By knowing the reduced and oxidized species in the reactions the reducing and oxidizing agents can be easily found.

In the given reaction,

  $Mg\left(s\right)+Cu{\left(N{O}_3\right)}_2\left(aq\right)\to Mg{\left(N{O}_3\right)}_2\left(aq\right)+Cu\left(s\right)$

Here $Mg$ undergoes es oxidation by the loss of electrons. So, it is the reducing agent.

  $C{u}^{2+}$ in the $Cu{\left(N{O}_3\right)}_2$ undergoes es reduction by gaining the electron. So, it is the oxidizing agent.