Chapter 18, Problem 25A

(a)

Interpretation Introduction

Interpretation: The given half-reaction which take place in acidic solution needs to be balanced.

  ${\text{O}}_{\text{2}}\left(\text{g}\right)\to {\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$

Concept Introduction: By a 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.

Answer to Problem 25A

Balanced half-reaction is-

${\text{O}}_{\text{2}}\left(\text{g}\right)+{\text{4H}}^{\text{+}}+4{\text{e}}^{-}\to 2{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$

Explanation of Solution

The given half-reaction is as follows:

${\text{O}}_{\text{2}}\left(\text{g}\right)\to {\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$

The reaction takes place in acidic medium. Balance oxygen by ${\text{H}}_{\text{2}}\text{O}$ and hydrogen by ${\text{H}}^{+}$ . To balance, O in reactant side give coefficient 2 to ${\text{H}}_{\text{2}}\text{O}$ . The equation becomes.

${\text{O}}_{\text{2}}\left(\text{g}\right)+{\text{4H}}^{\text{+}}\to 2{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$

Now, add electrons to balance charge. To balance charge on both side, add four electrons to the left. Hence, the balanced equation is-

${\text{O}}_{\text{2}}\left(\text{g}\right)+{\text{4H}}^{\text{+}}+4{\text{e}}^{-}\to 2{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$

(b)

Interpretation Introduction

Interpretation: The given half-reaction which take place in acidic solution needs to be balanced.

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.

Answer to Problem 25A

Balanced half-reaction is-

${\text{2IO}}_3^{-}\left(\text{aq}\right)+12{\text{H}}^{\text{+}}+12{\text{e}}^{-}\to {\text{I}}_{\text{2}}\left(\text{s}\right)+6{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$

Explanation of Solution

The given half-reaction is as follows:

${\text{IO}}_3^{-}\left(\text{aq}\right)\to {\text{I}}_{\text{2}}\left(\text{s}\right)$

First balance Iodine by multiplying reactant side ${\text{IO}}_3^{-}$ by 2.

${\text{2IO}}_3^{-}\left(\text{aq}\right)\to {\text{I}}_{\text{2}}\left(\text{s}\right)$

The reaction takes place in acidic medium. Balance oxygen by adding ${\text{H}}_{\text{2}}\text{O}$ and hydrogen by adding ${\text{H}}^{+}$ . To balance O in reactant side, multiply ${\text{H}}_{\text{2}}\text{O}$ by 6. The equation becomes.

${\text{2IO}}_3^{-}\left(\text{aq}\right)+12{\text{H}}^{\text{+}}\to {\text{I}}_{\text{2}}\left(\text{s}\right)+6{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$

Now, add electrons to balance charge. To balance charge on both sides, add 12 electrons to the left. Hence, the balanced equation is-

${\text{2IO}}_3^{-}\left(\text{aq}\right)+12{\text{H}}^{\text{+}}+12{\text{e}}^{-}\to {\text{I}}_{\text{2}}\left(\text{s}\right)+6{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$

(c)

Interpretation Introduction

Interpretation: The given half-reaction which take place in acidic solution needs to be balanced.

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.

Answer to Problem 25A

Balanced half-reaction is-

${\text{VO}}_2^{+}\left(\text{aq}\right)+2{\text{H}}^{\text{+}}+{\text{e}}^{-}\to {\text{V}}^{\text{3+}}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}$

Explanation of Solution

The given half-reaction is as follows:

${\text{VO}}_2^{+}\left(\text{aq}\right)\to {\text{V}}^{\text{3+}}\left(\text{aq}\right)$

The reaction takes place in acidic medium. Balance oxygen by ${\text{H}}_{\text{2}}\text{O}$ and hydrogen by ${\text{H}}^{+}$ . To balance, O in reactant side, multiply ${\text{H}}_{\text{2}}\text{O}$ by 1. The equation becomes.

${\text{VO}}_2^{+}\left(\text{aq}\right)+2{\text{H}}^{\text{+}}\to {\text{V}}^{\text{3+}}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}$

Now, add electrons to balance charge. To balance charge on both sides, add 1 electron to the left. Hence, the balanced equation is-

${\text{VO}}_2^{+}\left(\text{aq}\right)+2{\text{H}}^{\text{+}}+{\text{e}}^{-}\to {\text{V}}^{\text{3+}}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}$

(d)

Interpretation Introduction

Interpretation: The given half-reaction which take place in acidic solution needs to be balanced.

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.

Answer to Problem 25A

Balanced half-reaction is-

${\text{BiO}}^{+}\left(\text{aq}\right)+2{\text{H}}^{\text{+}}+2{\text{e}}^{-}\to \text{Bi}\left(\text{s}\right)+{\text{H}}_{\text{2}}\text{O}$

Explanation of Solution

The given half-reaction is as follows:

${\text{BiO}}^{+}\left(\text{aq}\right)\to \text{Bi}\left(\text{s}\right)$

The reaction takes place in acidic medium. Balance oxygen by ${\text{H}}_{\text{2}}\text{O}$ and hydrogen by ${\text{H}}^{+}$ . To balance, O in reactant side, multiply ${\text{H}}_{\text{2}}\text{O}$ , by 1. The equation becomes.

${\text{BiO}}^{+}\left(\text{aq}\right)+2{\text{H}}^{\text{+}}\to \text{Bi}\left(\text{s}\right)+{\text{H}}_{\text{2}}\text{O}$

Now, add electrons to balance charge. To balance charge on both sides, add two electrons to the left. Hence, the balanced equation is-

${\text{BiO}}^{+}\left(\text{aq}\right)+2{\text{H}}^{\text{+}}+2{\text{e}}^{-}\to \text{Bi}\left(\text{s}\right)+{\text{H}}_{\text{2}}\text{O}$