Chapter 18, Problem 24A

(a)

Interpretation Introduction

Interpretation: The given half-reaction needs to be balanced.

  $\text{Cu}\left(\text{s}\right)\to {\text{Cu}}^{\text{2+}}\left(\text{aq}\right)$

Concept Introduction: By transfer of electrons, oxidation-reduction reactions are characterized. The transfer occurs to form ions. Oxidation reaction is caused by losing electrons and reduction caused by gaining electrons. In a reduction half-reaction, electrons are placed on the reactant side and in oxidation half-reaction; the electrons are shown in the product side.

Answer to Problem 24A

Balanced half-reaction is-

$\text{Cu}\left(\text{s}\right)\to {\text{Cu}}^{\text{2+}}\left(\text{aq}\right)+2{\text{e}}^{-}$

Explanation of Solution

The given half-reaction is as follows:

$\text{Cu}\left(\text{s}\right)\to {\text{Cu}}^{\text{2+}}\left(\text{aq}\right)$

The charge on the left is zero and on the right side is $2+$ . Overall charge becomes zero if $2{\text{e}}^{-}$ are added on the right. Hence, the balanced reaction is-

$\text{Cu}\left(\text{s}\right)\to {\text{Cu}}^{\text{2+}}\left(\text{aq}\right)+2{\text{e}}^{-}$

(b)

Interpretation Introduction

Interpretation: The given half-reaction needs to be balanced.

  ${\text{Fe}}^{\text{3+}}\left(\text{aq}\right)\to {\text{Fe}}^{\text{2+}}\left(\text{aq}\right)$

Concept Introduction: By transfer of electrons, oxidation-reduction reactions are characterized. The transfer occurs to form ions. Oxidation reaction is caused by losing electrons and reduction caused by gaining electrons. In a reduction half-reaction, electrons are placed on the reactant side and in oxidation half-reaction; the electrons are shown in the product side.

Answer to Problem 24A

Balanced half-reaction is-

${\text{Fe}}^{\text{3+}}\left(\text{aq}\right)+{\text{e}}^{-}\to {\text{Fe}}^{\text{2+}}\left(\text{aq}\right)$

Explanation of Solution

The given half-reaction is as follows:

${\text{Fe}}^{\text{3+}}\left(\text{aq}\right)\to {\text{Fe}}^{\text{2+}}\left(\text{aq}\right)$

The charge on the left is $3+$ and on the right side is $2+$ . Overall charge becomes equal if one ${\text{e}}^{-}$ is added on the left. Hence, the balanced reaction is-

${\text{Fe}}^{\text{3+}}\left(\text{aq}\right)+{\text{e}}^{-}\to {\text{Fe}}^{\text{2+}}\left(\text{aq}\right)$

(c)

Interpretation Introduction

Interpretation: The given half-reaction needs to be balanced.

  ${\text{Br}}^{-}\left(\text{aq}\right)\to {\text{Br}}_{\text{2}}\left(\text{l}\right)$

Concept Introduction: By transfer of electrons, oxidation-reduction reactions are characterized. The transfer occurs to form ions. Oxidation reaction is caused by losing electrons and reduction caused by gaining electrons. In a reduction half-reaction, electrons are placed on the reactant side and in oxidation half-reaction; the electrons are shown in the product side.

Answer to Problem 24A

Balanced half-reaction is-

${\text{2Br}}^{-}\left(\text{aq}\right)\to {\text{Br}}_{\text{2}}\left(\text{l}\right)+2{\text{e}}^{-}$

Explanation of Solution

The given half-reaction is as follows:

${\text{Br}}^{-}\left(\text{aq}\right)\to {\text{Br}}_{\text{2}}\left(\text{l}\right)$

The charge on the left is $-2$ and on the right side is zero. Overall charge becomes zero if ${\text{2e}}^{-}$ are added on the right. Hence, the balanced reaction is-

${\text{2Br}}^{-}\left(\text{aq}\right)\to {\text{Br}}_{\text{2}}\left(\text{l}\right)+2{\text{e}}^{-}$

(d)

Interpretation Introduction

Interpretation: The given half-reaction needs to be balanced.

  ${\text{Fe}}^{\text{2+}}\left(\text{aq}\right)\to \text{Fe}\left(\text{s}\right)$

Concept Introduction: By transfer of electrons, oxidation-reduction reactions are characterized. The transfer occurs to form ions. Oxidation reaction is caused by losing electrons and reduction caused by gaining electrons. In a reduction half-reaction, electrons are placed on the reactant side and in oxidation half-reaction; the electrons are shown in the product side.

Answer to Problem 24A

Balanced half-reaction is-

${\text{Fe}}^{\text{2+}}\left(\text{aq}\right)+2{\text{e}}^{-}\to \text{Fe}\left(\text{s}\right)$

Explanation of Solution

The given half-reaction is as follows:

${\text{Fe}}^{\text{2+}}\left(\text{aq}\right)\to \text{Fe}\left(\text{s}\right)$

The charge on the left is $2+$ and on the right side is zero. Overall charge becomes equal if two ${\text{e}}^{-}$ are added on the right. Hence, the balanced reaction is-

${\text{Fe}}^{\text{2+}}\left(\text{aq}\right)+2{\text{e}}^{-}\to \text{Fe}\left(\text{s}\right)$