Chapter 20.2, Problem 20LC

(a)

Interpretation Introduction

Interpretation: The oxidizing and reducing agents present in the reaction, $2Na\left(s\right)\text{+}C{l}_2\left(l\right)\to 2NaCl\left(s\right)$ is to be determined.

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 20LC

Oxidizing agent is $C{l}_2$ and the reducing agent is $Na$ .

Explanation of Solution

In the reaction,

  $2Na\left(s\right)\text{+}C{l}_2\left(l\right)\to 2NaCl\left(s\right)$

The species that undergo oxidation is sodium, $Na$ by the loss of electrons. So it is the reducing agent in this reaction.

The species that undergo reduction is chlorine, $C{l}_2$ by the gain of electrons. So it is the oxidizing agent in this reaction.

(b)

Interpretation Introduction

Interpretation: The oxidizing and reducing agents present in the reaction, $2HN{O}_3\left(aq\right)+6HI\left(aq\right)\to 2NO\left(g\right)+3I_2\left(s\right)+4H_{2}O\left(l\right)$ is to be determined.

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 20LC

Oxidizing agent is $HN{O}_3$ and the reducing agent is $I^{-}$ .

Explanation of Solution

In the reaction,

  $2HN{O}_3\left(aq\right)+6HI\left(aq\right)\to 2NO\left(g\right)+3I_2\left(s\right)+4H_{2}O\left(l\right)$

The species that undergo oxidation is iodonium ion, $I^{-}$ by the loss of electrons. So it is the reducing agent in this reaction.

The species that undergo reduction is nitrogen in $HN{O}_3$ by the gain of electrons. So it is the oxidizing agent in this reaction.

(c)

Interpretation Introduction

Interpretation: The oxidizing and reducing agents present in the reaction, $2H_{2}S\left(g\right)+2HN{O}_3\left(aq\right)\to 3S\left(s\right)+2NO\left(g\right)+4H_{2}O\left(l\right)$ is to be determined.

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 20LC

Oxidizing agent is $HN{O}_3$ and the reducing agent is $H_{2}S$ .

Explanation of Solution

In the reaction,

  $2H_{2}S\left(g\right)+2HN{O}_3\left(aq\right)\to 3S\left(s\right)+2NO\left(g\right)+4H_{2}O\left(l\right)$

The species that undergo oxidation is sulfur present in $H_{2}S$ by the loss of electrons. So it is the reducing agent in this reaction.

The species that undergo reduction is nitrogen in $HN{O}_3$ by the gain of electrons. So it is the oxidizing agent in this reaction.

(d)

Interpretation Introduction

Interpretation: The oxidizing and reducing agents present in the reaction, $2PbS{O}_4\left(s\right)+2H_{2}O\left(l\right)\to 2Pb\left(s\right)+Pb{O}_2\left(s\right)+2H_{2}S{O}_4\left(aq\right)$ is to be determined.

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 20LC

Oxidizing agent is $PbS{O}_4$ and the reducing agent is $PbS{O}_4$ .

Explanation of Solution

In the reaction,

  $2PbS{O}_4\left(s\right)+2H_{2}O\left(l\right)\to 2Pb\left(s\right)+Pb{O}_2\left(s\right)+2H_{2}S{O}_4\left(aq\right)$

The species that undergo oxidation is lead, $P{b}^{2+}$ present in $PbS{O}_4$ by the loss of electrons and forming Pb. So it is the reducing agent in this reaction.

The species that undergo reduction is, lead, $P{b}^{2+}$ in $PbS{O}_4$ by the gain of electron and forming $Pb{O}_2$ . So it is the oxidizing agent in this reaction.