Chapter 17, Problem 19PE

(a)

Interpretation Introduction

Interpretation:

The below reaction has to be balanced.

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

Concept Introduction:

Rules to balance redox reactions by ion-electron method are as follows:

1. Two half-reactions with elements that got oxidized and reduced should be written.

2. Elements other than oxygen and hydrogen should be balanced.

3. Conditions to balance $\text{O}$ and $\text{H}$ are as follows:

In acidic solution, ${\text{H}}^{+}$ and ${\text{H}}_2\text{O}$ should be used to balance $\text{O}$ and $\text{H}$.

In basic solution, equal ${\text{OH}}^{-}$ ions to ${\text{H}}^{+}$ ions should be added on both sides of balanced reaction in acidic medium.

4. For acidic medium, charge is balanced by addition of ${\text{H}}^{+}$ ion. Then combine ${\text{H}}^{+}$ and ${\text{OH}}^{-}$ to form ${\text{H}}_2\text{O}$ and cancel common ${\text{H}}_2\text{O}$ in reaction.

5. Electrons should be added to balance electric charge.

6. Finally, two half-reactions should be added and any common substance including electrons must be cancelled.

(b)

Interpretation Introduction

Interpretation:

The below reaction has to be balanced.

  ${\text{Mn}}^{+2}\left(aq\right)+{\text{S}}_{\text{2}}{\text{O}}_8^{2-}\left(aq\right)\to {\text{MnO}}_4^{-}\left(aq\right)+{\text{SO}}_4^{2-}\left(aq\right)$

Concept Introduction:

Refer to part (a).

(c)

Interpretation Introduction

Interpretation:

The below reaction has to be balanced.

  $\text{Co}{\left({\text{NO}}_{\text{2}}\right)}_6^{3-}\left(aq\right)+{\text{MnO}}_4^{-}\left(aq\right)\to {\text{Co}}^{+2}\left(aq\right)+{\text{Mn}}^{+2}\left(aq\right)+{\text{NO}}_3^{-}\left(aq\right)$

Concept Introduction:

Refer to part (a).