Chapter 19, Problem 41A

Interpretation Introduction

(a)

Interpretation:

The species oxidized and reduced in given reaction is to be determined.

Concept introduction:

The species which undergo oxidation is said to be oxidized and which undergo reduction is said to be reduced.

Oxidation number helps to know the oxidized and reduced species. The substance which undergoes decrease in oxidation number is reduced and which undergoes increase in oxidation number is oxidized.

Answer to Problem 41A

$Ga$ is oxidized to $GaB{r}_3$ and $B{r}_2$ is reduced to $GaB{r}_3$

Explanation of Solution

The given equation is:

$3B{r}_2+2Ga\to 2GaB{r}_3$

Oxidation number of various atoms is:

$3{\stackrel{0}{Br}}_2+2\stackrel{0}{Ga}\to 2\stackrel{+3}{Ga}{\stackrel{-1}{Br}}_3$

Oxidation no. of bromine is decreasing hence it is reduced.

Oxidation no. of galium is increasing hence it is oxidized.

Interpretation Introduction

(b)

Interpretation:

The species oxidized and reduced in given reaction is to be determined.

Concept introduction:

The species which undergo oxidation is said to be oxidized and which undergo reduction is said to be reduced.

Oxidation number helps to know the oxidized and reduced species. The substance which undergoes decrease in oxidation number is reduced and which undergoes increase in oxidation number is oxidized.

Answer to Problem 41A

$Zn$ is oxidized to $ZnC{l}_2$ and $HCl$ is reduced to $H_2$

Explanation of Solution

The given equation is:

$HCl+Zn\to ZnC{l}_2+H_2$

Oxidation number of various atoms is:

$\stackrel{+1}{H}\stackrel{-1}{Cl}+\stackrel{0}{Zn}\to \stackrel{+2}{Zn}{\stackrel{-1}{Cl}}_2+{\stackrel{0}{H}}_2$

Oxidation no. of hydrogen is decreasing hence it is reduced.

Oxidation no. of zinc is increasing hence it is oxidized.

Interpretation Introduction

(c)

Interpretation:

Determine the species oxidized and reduced in given reaction.

Concept introduction:

The species which undergo oxidation is said to be oxidized and which undergo reduction is said to be reduced.

Oxidation number helps to know the oxidized and reduced species. The substance which undergoes decrease in oxidation number is reduced and which undergoes increase in oxidation number is oxidized.

Answer to Problem 41A

$Mg$ is oxidized to $M{g}_3{N}_2$ and $N_2$ is reduced to $M{g}_3{N}_2$

Explanation of Solution

The given equation is:

$Mg+N_2\to M{g}_3{N}_2$

Oxidation number of various atoms is:

$\stackrel{0}{Mg}+{\stackrel{0}{N}}_2\to {\stackrel{+2}{Mg}}_3{\stackrel{-3}{N}}_2$

Oxidation no. of nitrogen is decreasing hence it is reduced.

Oxidation no. of magnesium is increasing hence it is oxidized.