Chapter 17, Problem 18PE

(a)

Interpretation Introduction

Interpretation:

The below redox reaction in basic solution has to be balanced.

  ${\text{MnO}}_4^{-}+{\text{SO}}_{{}^3}^{2-}\to {\text{MnO}}_2+{\text{SO}}_4^{2-}$

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 redox reaction in basic solution has to be balanced.

  ${\text{ClO}}_2+{\text{SbO}}_2^{-}\to {\text{ClO}}_2^{-}+\text{Sb}{\left(\text{OH}\right)}_6^{-}$

Concept Introduction:

Refer to part (a).

(c)

Interpretation Introduction

Interpretation:

The below redox reaction in basic solution has to be balanced.

  $\text{Al}+{\text{NO}}_{{}^3}^{-}\to {\text{NH}}_{\text{3}}+\text{Al}{\left(\text{OH}\right)}_4^{-}$

Concept Introduction:

Refer to part (a).

(d)

Interpretation Introduction

Interpretation:

The below redox reaction in basic solution has to be balanced.

  ${\text{P}}_{\text{4}}\to {\text{HPO}}_3^{2-}+{\text{PH}}_{\text{3}}$

Concept Introduction:

Refer to part (a).

(e)

Interpretation Introduction

Interpretation:

The below redox reaction in basic solution has to be balanced.

  $\text{Al}+{\text{OH}}^{-}\to \text{Al}{\left(\text{OH}\right)}_4^{-}+{\text{H}}_2$

Concept Introduction:

Refer to part (a).