Chapter 8, Problem 8.116QA

Interpretation Introduction

To assign:

Oxidation numbers to the elements in each compound and balance the redox reactions in basic solution

Answer to Problem 8.116QA

Solution:

a) Balanced redox reaction is:

$4Fe{O}_4^{2-}\left(aq\right)+6H_{2}O\left(l\right)\to 4FeOOH\left(s\right)+3O_2\left(g\right)+8O{H}^{-}\left(aq\right)$

Oxidation number of Fe in $Fe{O}_4^{2-}= +6$

Oxidation number of O in $Fe{O}_4^{2-}= -2$

Oxidation number of H in $H_{2}O= +1$

Oxidation number of O in $H_{2}O= -2$

Oxidation number of Fe in $FeOOH= +3$

Oxidation number of O in $FeOOH= -2$

Oxidation number of H in $FeOOH= +1$

Oxidation number of O in $O_2=0$

Oxidation number of O in $O{H}^{-}= -2$

Oxidation number of H in $O{H}^{-}= +1$

b) Balanced redox reaction is:

$4Fe{O}_4^{2-}\left(aq\right)+4H_{2}O\left(l\right)\to 2F{e}_2{O}_3\left(s\right)+3O_2\left(g\right)+8O{H}^{-}\left(aq\right)$

Oxidation number of Fe in $Fe{O}_4^{2-}= +6$

Oxidation number of O in $Fe{O}_4^{2-}= -2$

Oxidation number of H in $H_{2}O= +1$

Oxidation number of O in $H_{2}O= -2$

Oxidation number of Fe in $F{e}_2{O}_3= +3$

Oxidation number of O in $F{e}_2{O}_3= -2$

Oxidation number of O in $O_2=0$

Oxidation number of O in $O{H}^{-}= -2$

Oxidation number of H in $O{H}^{-}= +1$

Explanation of Solution

1) Concept:

For balancing redox reactions, first, the change in oxidation numbers ($∆O.N.$) is calculated. The $∆O.N.$ is adjusted by assigning coefficients to the molecules containing the unbalanced atoms. The ionic charges are adjusted by adding $H^{+}$ for acidic medium or $O{H}^{-}$ for basic medium. Finally, the H and O atoms are balanced by the addition of $H_{2}O$ to the appropriate sides.

2) Given:

The given redox reactions are

$Fe{O}_4^{2-}\left(aq\right)+H_{2}O\left(l\right)\to FeOOH\left(s\right)+O_2\left(g\right)+O{H}^{-}\left(aq\right)$

$Fe{O}_4^{2-}\left(aq\right)+H_{2}O\left(l\right)\to F{e}_2{O}_3\left(s\right)+O_2\left(g\right)+O{H}^{-}\left(aq\right)$

These reactions take place in the basic medium.

3) Calculations:

a)

Oxidation number of Fe in $Fe{O}_4^{2-}= +6$

Oxidation number of O in $Fe{O}_4^{2-}= -2$

Oxidation number of H in $H_{2}O= +1$

Oxidation number of O in $H_{2}O= -2$

Oxidation number of Fe in $FeOOH= +3$

Oxidation number of O in $FeOOH= -2$

Oxidation number of H in $FeOOH= +1$

Oxidation number of O in $O_2=0$

Oxidation number of O in $O{H}^{-}= -2$

Oxidation number of H in $O{H}^{-}= +1$

Consider the reaction

$Fe{O}_4^{2-}\left(aq\right)+H_{2}O\left(l\right)\to FeOOH\left(s\right)+O_2\left(g\right)+O{H}^{-}\left(aq\right)$

Let us remove the $O{H}^{-}$ and $H_{2}O$ species from this reaction.

$Fe{O}_4^{2-}\left(aq\right)\to FeOOH\left(s\right)+O_2\left(g\right)$

The oxidation number of $Fe$ on the left hand side is $+6$.

The oxidation number of $Fe$ on the right hand side is $+3$.

$\therefore ∆O.N.=\left(+3\right)-\left(+6\right)=-3$

The total oxidation number of O on the left hand side is $4\times \left(-2\right)=-8$.

 The total oxidation number of O on the right hand side is $\left(-2\right)+\left(-2\right)+2\times 0=-4$.

$\therefore ∆O.N.=\left(-4\right)-\left(-8\right)=+4$

Hence, we assign a coefficient of 4 to $Fe{O}_4^{2-}$ and $FeOOH\left(s\right)$, and we assign a coefficient of 3 to $O_2$.

$4Fe{O}_4^{2-}\left(aq\right)\to 4FeOOH\left(s\right)+3O_2\left(g\right)$

In order to adjust the ionic charges, we add eight $O{H}^{-}$ ions to the right hand side as this reaction takes place in a basic medium.

$4Fe{O}_4^{2-}\left(aq\right)\to 4FeOOH\left(s\right)+3O_2\left(g\right)+8O{H}^{-}\left(aq\right)$

On taking inventories of O atoms, we see that the equation is unbalanced. In order to balance the number of O atoms, we add six $H_{2}O$ to the left hand side.

$4Fe{O}_4^{2-}\left(aq\right)+6H_{2}O\left(l\right)\to 4FeOOH\left(s\right)+3O_2\left(g\right)+8O{H}^{-}\left(aq\right)$

On taking inventories of atoms on both sides, we see that the redox reaction is now completely balanced.

b)

Oxidation number of Fe in $Fe{O}_4^{2-}= +6$

Oxidation number of O in $Fe{O}_4^{2-}= -2$

Oxidation number of H in $H_{2}O= +1$

Oxidation number of O in $H_{2}O= -2$

Oxidation number of Fe in $F{e}_2{O}_3= +3$

Oxidation number of O in $F{e}_2{O}_3= -2$

Oxidation number of O in $O_2=0$

Oxidation number of O in $O{H}^{-}= -2$

Oxidation number of H in $O{H}^{-}= +1$

Consider the reaction

$Fe{O}_4^{2-}\left(aq\right)+H_{2}O\left(l\right)\to F{e}_2{O}_3\left(s\right)+O_2\left(g\right)+O{H}^{-}\left(aq\right)$

Let us remove the $O{H}^{-}$ and $H_{2}O$ species from this reaction.

$Fe{O}_4^{2-}\left(aq\right)\to F{e}_2{O}_3\left(s\right)+O_2\left(g\right)$

We assign a coefficient of 2 to $Fe{O}_4^{2-}$ in order to balance the $Fe$ atoms.

$2Fe{O}_4^{2-}\left(aq\right)\to F{e}_2{O}_3\left(s\right)+O_2\left(g\right)$

The oxidation number of $Fe$ on the left hand side is $+6$.

The oxidation number of $Fe$ on the right hand side is $+3$.

$\therefore ∆O.N.=\left(+3\right)-\left(+6\right)=-3$

The oxidation number of O on the left hand side is $-2$.

The oxidation number of O on the right hand in $O_2$ side is $0$.

$\therefore ∆O.N.=0-\left(-2\right)=+2$

Hence, we need a coefficient of 4 for $Fe{O}_4^{2-}$ and $2$ for $F{e}_2{O}_3$, and we assign a coefficient of 3 to $O_2$.

$4Fe{O}_4^{2-}\left(aq\right)\to 2F{e}_2{O}_3\left(s\right)+3O_2\left(g\right)$

In order to adjust the ionic charges, we add eight $O{H}^{-}$ ions to the right hand side as this reaction takes place in a basic medium.

$4Fe{O}_4^{2-}\left(aq\right)\to 2F{e}_2{O}_3\left(s\right)+3O_2\left(g\right)+8O{H}^{-}\left(aq\right)$

On taking inventories of O atoms, we see that the equation is unbalanced. In order to balance the number of O atoms, we add four $H_{2}O$ molecules to the left hand side.

$4Fe{O}_4^{2-}\left(aq\right)+4H_{2}O\left(l\right)\to 2F{e}_2{O}_3\left(s\right)+3O_2\left(g\right)+8O{H}^{-}\left(aq\right)$

On taking inventories of atoms on both sides, we see that the redox reaction is now balanced.

Conclusion:

The redox reactions are balanced using the rules for balancing redox reactions in basic medium.