Chapter 19, Problem 60A

(a)

Interpretation Introduction

Interpretation: The given redox reaction is to be balanced.

Concept Introduction :

The reaction can be balanced by the oxidation number method.

The oxidation number method involves:-

• Assign the oxidation number to each element.
• Check which element is oxidized and which element is reduced.
• Check the change in oxidation number for the elements which are oxidized and reduced.
• Balance the change in oxidation number by adjusting the coefficients in the equation.
• Then balance the other elements.

Answer to Problem 60A

  $2\text{A}{\text{u}}^{3+}+6{\text{I}}^{-}\to 2\text{A}\text{u}+3{\text{I}}_2$

Explanation of Solution

The given redox reaction is:

  $\text{A}{\text{u}}^{3+}+{\text{I}}^{-}\to \text{A}\text{u}+{\text{I}}_2$

Assigning the oxidation number-

  $\stackrel{+3}{\text{A}{\text{u}}^{3+}}+\stackrel{-1}{{\text{I}}^{-}}\to \stackrel{0}{\text{A}\text{u}}+\stackrel{0}{{\text{I}}_2}$

The oxidation numbers of atoms show that

${\text{I}}^{-}$ is oxidized and $\text{A}{\text{u}}^{3+}$ is reduced.

Change in oxidation number of $\text{I}=+1$

Change in oxidation number of $\text{A}\text{u}=-3$

Thus, adjusting the coefficients:

  $\text{A}{\text{u}}^{3+}+3{\text{I}}^{-}\to \text{A}\text{u}+{\text{I}}_2$

After readjustment:

  $2\text{A}{\text{u}}^{3+}+6{\text{I}}^{-}\to 2\text{A}\text{u}+3{\text{I}}_2$

(b)

Interpretation Introduction

Interpretation: The given redox reaction is to be balanced.

Concept Introduction :

The reaction can be balanced by the oxidation number method.

The oxidation number method involves:-

• Assign the oxidation number to each element.
• Check which element is oxidized and which element is reduced.
• Check the change in oxidation number for the elements which are oxidized and reduced.
• Balance the change in oxidation number by adjusting the coefficients in the equation.
• Then balance the other elements.

Answer to Problem 60A

  $2\text{C}{\text{e}}^{4+}+\text{S}{\text{n}}^{2+}\to 2\text{C}{\text{e}}^{3+}+\text{S}{\text{n}}^{4+}$

Explanation of Solution

The given redox reaction is:

  $\text{C}{\text{e}}^{4+}+\text{S}{\text{n}}^{2+}\to \text{C}{\text{e}}^{3+}+\text{S}{\text{n}}^{4+}$

Assigning the oxidation number-

  $\stackrel{+4}{\text{C}{\text{e}}^{4+}}+\stackrel{+2}{\text{S}{\text{n}}^{2+}}\to \stackrel{+3}{\text{C}{\text{e}}^{3+}}+\stackrel{+4}{\text{S}{\text{n}}^{4+}}$

The oxidation numbers of atoms show that

$\text{S}{\text{n}}^{2+}$ is oxidized and $\text{C}{\text{e}}^{4+}$ is reduced.

Change in oxidation number of $\text{S}\text{n}=+2$

Change in oxidation number of $\text{C}\text{e}=-1$

Thus, adjusting the coefficients:

  $2\text{C}{\text{e}}^{4+}+\text{S}{\text{n}}^{2+}\to 2\text{C}{\text{e}}^{3+}+\text{S}{\text{n}}^{4+}$