Chapter 19, Problem 81A

Interpretation Introduction

(a)

Interpretation:

The given redox reaction is to be balanced.

Concept introduction:

Oxidation number method to balance ionic equation 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.

Answer to Problem 81A

$3Mg+2F{e}^{+3}\to 3M{g}^{+2}+2Fe$

Explanation of Solution

The given redox reaction is:

$Mg+F{e}^{+3}\to M{g}^{+2}+Fe$

Assigning the oxidation number-

$\stackrel{0}{Mg}+\stackrel{+3}{F{e}^{+3}}\to \stackrel{+2}{M{g}^{+2}}+\stackrel{0}{Fe}$

The oxidation numbers of atoms shows that

$Mg$ is oxidized and $Fe$ is reduced.

Change in oxidation number of $Mg=+2$

Change in oxidation number of $Fe=-3$

Thus adjusting the coefficients:

$3Mg+2F{e}^{+3}\to M{g}^{+2}+Fe$

Adjusting the rest atoms:

$3Mg+2F{e}^{+3}\to 3M{g}^{+2}+2Fe$

Interpretation Introduction

(b)

Interpretation:

The given redox reaction is to be balanced.

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 to balance the charge and water molecules to the equation to balance hydrogen atoms on both sides.

Answer to Problem 81A

$Cl{O}_3{}^{-1}+3S{O}_2+3H_{2}O\to C{l}^{-}+3S{O}_4{}^{-2}+6H^{+}$

Explanation of Solution

The given redox reaction is:

$Cl{O}_3{}^{-1}+S{O}_2\to C{l}^{-}+S{O}_4{}^{-2}$

Assigning the oxidation number-

$\stackrel{+5}{Cl}{O}_3{}^{-1}+\stackrel{+4}{S}{O}_2\to \stackrel{-1}{C{l}^{-}}+\stackrel{+6}{S}{O}_4{}^{-2}$

The oxidation numbers of atoms shows that

$S$ is oxidized and $Cl$ is reduced.

Change in oxidation number of $S=+2$

Change in oxidation number of $Cl=-6$

Thus adjusting the coefficients:

$Cl{O}_3{}^{-1}+3S{O}_2\to C{l}^{-}+S{O}_4{}^{-2}$

Balancing the atoms:

$Cl{O}_3{}^{-1}+3S{O}_2\to C{l}^{-}+3S{O}_4{}^{-2}$

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

$Cl{O}_3{}^{-1}+3S{O}_2\to C{l}^{-}+3S{O}_4{}^{-2}+6H^{+}$

Adding water to left side:

$Cl{O}_3{}^{-1}+3S{O}_2+3H_{2}O\to C{l}^{-}+3S{O}_4{}^{-2}+6H^{+}$