Chapter 20, Problem 68A

(a)

Interpretation Introduction

Interpretation: The oxidizing agent among $S^{2-}$ and $S{O}_4{}^{2-}$ is to be identified.

Concept Introduction: The chemical species that undergo reduction by the gain of electrons are the oxidizing agent. When the oxidation number of species increases, the ability to accept electrons will also increase.

Answer to Problem 68A

  $S{O}_4{}^{2-}$ is the oxidizing agent.

Explanation of Solution

The oxidation number of $S$ in $S^{2-}$ is -2. While the oxidation number of $S$ in $S{O}_4{}^{2-}$ is +6, that is,

  $\begin{array}{c}x-8=-2\\ x=+6\end{array}$

As the oxidation number of $S$ in $S{O}_4{}^{2-}$ is high, it undergoes reduction and gain of electrons faster than $S^{2-}$ . So, $S^{2-}$ is the oxidizing agent.

(b)

Interpretation Introduction

Interpretation: The oxidizing agent among $H_{2}O$ and $H_2{O}_2$ is to be identified.

Concept Introduction: The chemical species that undergo reduction by the gain of electrons are the oxidizing agent. When the oxidation number of species increases, the ability to accept electrons will also increase.

Answer to Problem 68A

  $H_2{O}_2$ is the oxidizing agent.

Explanation of Solution

The oxidation number of H in $H_{2}O$ is +1. While the oxidation number of H in $H_2{O}_2$ is +2, that is,

  $\begin{array}{c}2x-4=0\\ x=+2\end{array}$

As the oxidation number of H in $H_2{O}_2$ is high, it undergoes reduction and gains electrons faster than $H_{2}O$ . So, $H_2{O}_2$ is the oxidizing agent.

(c)

Interpretation Introduction

Interpretation: The oxidizing agent among $N{O}_2{}^{-}$ and $N{O}_3{}^{-}$ is to be identified.

Concept Introduction: The chemical species that undergo reduction by the gain of electrons are the oxidizing agent. When the oxidation number of species increases, the ability to accept electrons will also increase.

Answer to Problem 68A

  $N{O}_3{}^{-}$ is the oxidizing agent.

Explanation of Solution

The oxidation number of N in $N{O}_2{}^{-}$ is,

  $\begin{array}{c}x-4=-1\\ x=+3\end{array}$

The oxidation number of N in $N{O}_3{}^{-}$ is,

  $\begin{array}{c}x-6=-1\\ x=+5\end{array}$

As the oxidation number of $N$ in $N{O}_3{}^{-}$ is high, it undergoes reduction and gain of electrons faster than $N{O}_2{}^{-}$ . So, $N{O}_3{}^{-}$ is the oxidizing agent.

(d)

Interpretation Introduction

Interpretation: To identify the oxidizing agent among $C{r}_2{O}_7{}^{2-}$ and $C{r}^{3+}$ .

Concept Introduction: The chemical species that undergo reduction by the gain of electrons are the oxidizing agent. When the oxidation number of species increases, the ability to accept electrons will also increase.

Answer to Problem 68A

  $C{r}_2{O}_7{}^{2-}$ is the oxidizing agent.

Explanation of Solution

The oxidation number of Cr in $C{r}^{3+}$ is +3. The oxidation number of Cr in $C{r}_2{O}_7{}^{2-}$ is,

  $\begin{array}{c}2x-14=-2\\ 2x=12\\ x=+6\end{array}$

The oxidation number of Cr in $C{r}_2{O}_7{}^{2-}$ is high, +6 is more likely undergo reduction.

So, $C{r}_2{O}_7{}^{2-}$ is the oxidizing agent.

(e)

Interpretation Introduction

Interpretation: The oxidizing agent among $H_2$ and $H_{2}O$ is to be identified.

Concept Introduction: The chemical species that undergo reduction by the gain of electrons are oxidizing agents. When the oxidation number of species increases, the ability to accept electrons will also increase.

Answer to Problem 68A

  $H_{2}O$ is the oxidizing agent.

Explanation of Solution

The oxidation number of H in $H_2$ is zero. While that in $H_{2}O$ is +1. So, $H_{2}O$ undergoes reduction more than that $H_2$ . So $H_{2}O$ is the oxidizing agent.