Chapter 19.1, Problem 2PP

Interpretation Introduction

(a)

Interpretation:

The given reaction is to be classified as oxidation or reduction reaction.

Concept introduction:

To classify a change as oxidation and reduction, we should know what is oxidation and reduction.

Oxidation involves:

• Loss of electrons.
• Addition of oxygen
• Removal of hydrogen
• Addition of electronegative element
• Removal of electropositive element
• Increase in oxidation number of the element

Reduction involves:

• Gain of electrons.
• Addition of hydrogen
• Removal of oxygen
• Addition of electropositive element
• Removal of electronegative element
• Decrease in oxidation number of the element

The substance which undergoes oxidation is it self oxidized and which undergoes reduction is it self reduced.

Answer to Problem 2PP

$B{r}^{-}$is oxidized and $C{l}_2$ is reduced

Explanation of Solution

The given reaction is:

$C{l}_2+2B{r}^{-}\to 2C{l}^{-}+B{r}_2$

Oxidation half reaction:

$2B{r}^{-}\to B{r}_2+2e^{-}$

Reduction half reaction:

$C{l}_2+2e^{-}\to 2C{l}^{-}$

Thus chlorine is reduced and bromine is oxidized.

Interpretation Introduction

(b)

Interpretation:

The given reaction is to be classified as oxidation or reduction reaction.

Concept introduction:

To classify a change as oxidation and reduction, we should know what is oxidation and reduction.

Oxidation involves:

• Loss of electrons.
• Addition of oxygen
• Removal of hydrogen
• Addition of electronegative element
• Removal of electropositive element
• Increase in oxidation number of the element

Reduction involves:

• Gain of electrons.
• Addition of hydrogen
• Removal of oxygen
• Addition of electropositive element
• Removal of electronegative element
• Decrease in oxidation number of the element

The substance which undergoes oxidation is it self oxidized and which undergoes reduction is it self reduced.

Answer to Problem 2PP

$Ce$is oxidized and $C{u}^{+2}$ is reduced

Explanation of Solution

The given reaction is:

$2Ce+3C{u}^{2+}\to 2C{e}^{+3}+3Cu$

Oxidation half reaction:

$Ce\to C{e}^{+3}+3e^{-}$

Reduction half reaction:

$C{u}^{+2}+2e^{-}\to Cu$

Thus copper is reduced and cerium is oxidized.

Interpretation Introduction

(c)

Interpretation:

The given reaction is to be classified as oxidation or reduction reaction.

Concept introduction:

To classify a change as oxidation and reduction, we should know what is oxidation and reduction.

Oxidation involves:

• Loss of electrons.
• Addition of oxygen
• Removal of hydrogen
• Addition of electronegative element
• Removal of electropositive element
• Increase in oxidation number of the element

Reduction involves:

• Gain of electrons.
• Addition of hydrogen
• Removal of oxygen
• Addition of electropositive element
• Removal of electronegative element
• Decrease in oxidation number of the element

The substance which undergoes oxidation is it self oxidized and which undergoes reduction is it self reduced.

Answer to Problem 2PP

$Zn$is oxidized and $O_2$ is reduced

Explanation of Solution

The given reaction is:

$O_2+2Zn\to 2ZnO$

Zinc involves addition of oxygen so it is oxidized and Oxygen involves addition of electropositive element so it is reduced.

Interpretation Introduction

(d)

Interpretation:

The given reaction is to be classified as oxidation or reduction reaction.

Concept introduction:

To classify a change as oxidation and reduction, we should know what is oxidation and reduction.

Oxidation involves:

• Loss of electrons.
• Addition of oxygen
• Removal of hydrogen
• Addition of electronegative element
• Removal of electropositive element
• Increase in oxidation number of the element

Reduction involves:

• Gain of electrons.
• Addition of hydrogen
• Removal of oxygen
• Addition of electropositive element
• Removal of electronegative element
• Decrease in oxidation number of the element

The substance which undergoes oxidation is it self oxidized and which undergoes reduction is it self reduced.

Answer to Problem 2PP

$Na$is oxidized and $H^{+}$ is reduced

Explanation of Solution

The given reaction is:

$2Na+2H^{+}\to 2N{a}^{+}+H_2$

Oxidation half reaction:

$2Na\to 2N{a}^{+}+2e^{-}$

Reduction half reaction:

$2H^{+}+2e^{-}\to H_2$

Thus hydrogen is reduced and sodium is oxidized.