Chapter 16, Problem 1E

Identify each half-reaction below as either oxidation or reduction.

$\begin{array}{l}{\text{Fe}}^{3+}+3e^{-}\to \text{Fe}\\ \text{Cr}\to {\text{Cr}}^3{}^{+}+3e^{-}\\ {\text{MnO}}_4{}^{2-}\to {\text{MnO}}_4{}^{-}+e^{-}\\ {\text{Li}}^{+}+e^{-}\to \text{Li}\end{array}$

Interpretation Introduction

(a)

Interpretation:

The following half-reaction should be identified as either oxidation or reduction.

  ${\text{Fe}}^{3+}+3{\text{e}}^{-}\to \text{Fe}$

Concept Introduction:

The chemical reaction in which a substance loses an electron is known as oxidation reaction. The chemical reaction in which a substance gains an electron is known as reduction reaction.

Answer to Problem 1E

The given half-reaction is a reduction reaction.

Explanation of Solution

The oxidation state of a substance increases in oxidation reaction while the oxidation state decreases in reduction reaction.

In the given half-reaction, the oxidation state of Fe decreases due to addition of electron. Therefore, the reduction of Fe takes place in the given reaction.

Thus, the given half-reaction is a reduction reaction.

Interpretation Introduction

(b)

Interpretation:

The following half-reaction should be identified as either oxidation or reduction.

  $\text{Cr}\to {\text{Cr}}^{3+}+3{\text{e}}^{-}$

Concept Introduction:

The chemical reaction in which a substance loses an electron is known as oxidation reaction. The chemical reaction in which a substance gains an electron is known as reduction reaction.

Answer to Problem 1E

The given half-reaction is an oxidation reaction.

Explanation of Solution

The oxidation state of a substance increases in oxidation reaction while the oxidation state decreases in reduction reaction.

In the given half-reaction, the oxidation state of Cr increases due to loss of electron. Therefore, the oxidation of Cr takes place in the given reaction.

Thus, the given half-reaction is an oxidation reaction.

Interpretation Introduction

(c)

Interpretation:

The following half-reaction should be identified as either oxidation or reduction.

  ${\text{MnO}}_{\text{4}}{}^{2-}\to {\text{MnO}}_{\text{4}}{}^{-}+{\text{e}}^{-}$

Concept Introduction:

The chemical reaction in which a substance loses an electron is known as oxidation reaction. The chemical reaction in which a substance gains an electron is known as reduction reaction.

Answer to Problem 1E

The given half-reaction is an oxidation reaction.

Explanation of Solution

The oxidation state of a substance increases in oxidation reaction while the oxidation state decreases in reduction reaction.

In the given half-reaction, the oxidation state of Mn in ${\text{MnO}}_{\text{4}}{}^{2-}$ is +6 while in ${\text{MnO}}_{\text{4}}{}^{-}$ is +7. The oxidation state of Mn increases due to loss of electron. Therefore, the oxidation of Mn takes place in the given reaction.

Thus, the given half-reaction is an oxidation reaction.

Interpretation Introduction

(d)

Interpretation:

The following half-reaction should be identified as either oxidation or reduction.

  ${\text{Li}}^{+}+{\text{e}}^{-}\to \text{Li}$

Concept Introduction:

The chemical reaction in which a substance loses an electron is known as oxidation reaction. The chemical reaction in which a substance gains an electron is known as reduction reaction.

Answer to Problem 1E

The given half-reaction is a reduction reaction.

Explanation of Solution

The oxidation state of a substance increases in oxidation reaction while the oxidation state decreases in reduction reaction.

In the given half-reaction, the oxidation state of Li decreases due to addition of electron. Therefore, the reduction of Li takes place in the given reaction.

Thus, the given half-reaction is a reduction reaction.