Chapter 19, Problem 61A

Interpretation Introduction

(a)

Interpretation:

The given redox reaction is to be balanced using oxidation number method.

Concept introduction:

The reaction can be balanced be oxidation number method.

Oxidation number method involve:-

• 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 61A

$2Al+3I_2\to 2A{l}^{3+}+6I^{-}$

Explanation of Solution

The given redox reaction is:

$Al+I_2\to A{l}^{3+}+I^{-}$

Assigning the oxidation number-

$\stackrel{0}{Al}+\stackrel{0}{I_2}\to \stackrel{+3}{A{l}^{3+}}+\stackrel{-1}{I^{-}}$

The oxidation numbers of atoms shows that

$Al$ is oxidized and $I_2$ is reduced.

Change in oxidation number of $Al=+3$

Change in oxidation number of $Iperatom=-2$

Thus adjusting the coefficients:

$2Al+3I_2\to A{l}^{3+}+I^{-}$

After readjustment:

$2Al+3I_2\to 2A{l}^{3+}+6I^{-}$

Interpretation Introduction

(b)

Interpretation:

The given redox reaction is to be balanced using oxidation number method.

Concept introduction:

Oxidation number method to balance ionic equation in acidic solution involve:-

• 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 add enough hydrogen ions and water molecules to the equation to balance hydrogen atoms on both sides.

Answer to Problem 61A

$Mn{O}_2+2B{r}^{-1}+4H^{+}\to M{n}^{+2}+B{r}_2+2H_{2}O$

Explanation of Solution

The given redox reaction is:

$Mn{O}_2+B{r}^{-1}\to M{n}^{+2}+B{r}_2$

Assigning the oxidation number-

$\stackrel{+4}{Mn}{O}_2+\stackrel{-1}{B{r}^{-1}}\to \stackrel{+2}{M{n}^{+2}}+\stackrel{0}{B{r}_2}$

The oxidation numbers of atoms shows that

$B{r}^{-1}$ is oxidized and $Mn$ is reduced.

Change in oxidation number of $B{r}^{-1}=+1$

Change in oxidation number of $Mn=-2$

Thus adjusting the coefficients:

$Mn{O}_2+2B{r}^{-1}\to M{n}^{+2}+B{r}_2$

Now adding hydrogen ions on left side of equation to balance negative charge

$Mn{O}_2+2B{r}^{-1}+4H^{+}\to M{n}^{+2}+B{r}_2$

Adding water to right side:

$Mn{O}_2+2B{r}^{-1}+4H^{+}\to M{n}^{+2}+B{r}_2+2H_{2}O$