Chapter 20, Problem 37A

(a)

Interpretation Introduction

Interpretation: The reactant that oxidized and reduced in the reaction, $H_2\left(g\right)+S\left(s\right)\to H_{2}S\left(g\right)$ is to be identified.

Concept Introduction: The species with high electronegativity was found to accept electrons toward it. The gaining of electrons is the reduction and the loss of electrons is oxidation.

Answer to Problem 37A

The reactant that is oxidized is $H_2$ while the reduced is S.

Explanation of Solution

In the given reaction the electronegativity of sulfur is high. So, it gains or accepts electrons present in $H_2$ . While S is a less electronegative species so it loses its electrons. The reactions are,

  $H_2\to 2H^{+}+2e^{-}$

  $S+2e^{-}\to S^{2-}$

So, $H_2$ is said to be oxidized and S is found to be reduced.

(b)

Interpretation Introduction

Interpretation: The reactant that oxidized and reduced in the reaction, $N_2\left(g\right)+3H_2\left(g\right)\to 2N{H}_3\left(g\right)$ is to be identified.

Concept Introduction: The species with high electronegativity was found to accept electrons toward it. The gaining of electrons is a reduction and the loss of electrons is oxidation.

Answer to Problem 37A

The reactant that is oxidized is $H_2$ while the reduced is $N_2$ .

Explanation of Solution

In the given reaction the electronegativity of $N_2$ , and nitrogen is high. So it gains or accepts electrons present in $H_2$ . While $H_2$ is a less electronegative species so it loses its electrons. The reactions are,

  $3H_2\to 6H^{+}+6e^{-}$

  $N_2+6e^{-}\to 2N^{3-}$

So, $H_2$ is said to be oxidized and $N_2$ is found to be reduced.

(c)

Interpretation Introduction

Interpretation: The reactant that oxidized and reduced in the reaction, $O_2\left(g\right)+S\left(s\right)\to S{O}_2\left(g\right)$ is to be identified.

Concept Introduction: The species with high electronegativity was found to accept electrons toward it. The gaining of electrons is a reduction and the loss of electrons is oxidation.

Answer to Problem 37A

The reactant that is oxidized is S while the reduced is $O_2$ .

Explanation of Solution

In the given reaction the electronegativity of oxygen is high. So, it gains or accepts electrons present in S, sulfur. While S is a less electronegative species so it loses its electrons. The reactions are,

  $S\to S^{4+}+4e^{-}$

  $O_2+4e^{-}\to O_2{}^{4-}$

So, S is said to be oxidized and $O_2$ is found to be reduced.

(d)

Interpretation Introduction

Interpretation: The reactant that oxidized and reduced in the reaction, $H_2\left(g\right)+O_2\left(g\right)\to H_{2}O\left(l\right)$ is to be identified.

Concept Introduction: The species with high electronegativity was found to accept electrons toward it. The gaining of electrons is a reduction and the loss of electrons is oxidation.

Answer to Problem 37A

The reactant that is oxidized is $H_2$ while the reduced is $O_2$ .

Explanation of Solution

In the given reaction the electronegativity of oxygen is high. So, it gains or accepts electrons present in $H_2$ . While $H_2$ is a less electronegative species so it loses its electrons. The reactions are,

  $2H_2\to 4H^{+}+4e^{-}$

  $O_2+4e^{-}\to 2O^{2-}$

So, $H_2$ is said to be oxidized and $O_2$ is found to be reduced.