Chapter 4, Problem 4.126AP

(a)

Interpretation Introduction

Interpretation: The balanced equations for the given redox reactions are to be written. The oxidation numbers of the reactants and products are to be assigned.

Concept introduction: The reaction in which reduction and oxidation process occurs simultaneously is termed as redox reaction.

In oxidation there is loss of electrons occur which increases the oxidation state of an atom whereas in reduction process there is gaining of electrons occur which decreases the oxidation state of an atom.

Oxidation number represents the number electrons lost or gain by an atom.

To determine: The balanced equations for the given redox reaction and the oxidation numbers of the reactants and products.

Answer to Problem 4.126AP

Solution

The balanced equation of the given redox reaction is,

${\text{FeO}}_{\text{4}}{}^{\text{2-}}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\to \text{FeOOH}\left(s\right)+{\text{O}}_{\text{2}}\left(g\right)+{\text{OH}}^{-}\left(aq\right)$

In the reactant side oxidation number of $\text{Fe}$ in ${\text{FeO}}_{\text{4}}{}^{2-}$ ion is $\underset{\_}{+6}$ and the common oxidation number of oxygen and hydrogen are $\underset{\_}{-2}$ and $\underset{\_}{+1}$ respectively.

In the product side oxidation number of $\text{Fe}$ in $\text{FeOOH}$ is $\underset{\_}{+3}$, the common oxidation number of hydrogen is $\underset{\_}{+1}$, oxidation state of ${\text{O}}_{\text{2}}$ is $\underset{\_}{0}$. And lastly the oxidation number of ${\text{OH}}^{-}$ ion is $\underset{\_}{-2}$.

Explanation of Solution

Explanation

The given redox reaction is,

${\text{FeO}}_{\text{4}}{}^{\text{2-}}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\to \text{FeOOH}\left(s\right)+{\text{O}}_{\text{2}}\left(g\right)+{\text{OH}}^{-}\left(aq\right)$

The oxidation numbers of reactant and product will be assigned as follows:

The oxidation number of reactant side will be calculated as,

In the reactant side oxidation number of $\text{Fe}$ present in ${\text{FeO}}_{\text{4}}{}^{2-}$ is to be calculated. The oxidation state of oxygen is usually $-2$ as it requires two more electrons in order to achieve stable electronic configuration. The oxidation number of $\text{Fe}$ is assumed to be $x$ and is calculated by using the formula,

${\text{Charge on FeO}}_{\text{4}}{}^{2-}=\left[\begin{array}{c}\left(\text{Number of }\text{Fe}\text{atoms}\times \text{oxidation number of Fe}\right)+\\ \left(\text{Number of O atoms}\times \text{oxidation number of O}\right)\end{array}\right]$

Since, the charge on ${\text{FeO}}_{\text{4}}{}^{2-}$ is $-2$.

Substitute the number of atoms and their oxidation number in the above formula to calculate the oxidation number of $\text{Fe}$.

$\begin{array}{c}{\text{Charge on FeO}}_{\text{4}}{}^{2-}=\left[\left(x\right)+\left(4\times \left(-2\right)\right)\right]\\ -2=x+\left(-8\right)\\ x=-2+8\\ x=\underset{\_}{+6}\end{array}$

Hence, the oxidation number of $\text{Fe}$ in ${\text{FeO}}_{\text{4}}{}^{2-}$ is $\underset{\_}{+6}$.

Therefore the oxidation numbers on the reactant side of $\text{Fe}$ in ${\text{FeO}}_{\text{4}}{}^{2-}$ ion is $\underset{\_}{+6}$ and the common oxidation number of oxygen and hydrogen are $\underset{\_}{-2}$ and $\underset{\_}{+1}$ respectively.

In the product side oxidation number of $\text{Fe}$ present in $\text{FeOOH}$ is to be calculated. The oxidation state of oxygen is usually $-2$ as it requires two more electrons in order to achieve stable electronic configuration and the common oxidation state of hydrogen is $+1$. The oxidation number of $\text{Fe}$ is assumed to be $x$ and is calculated by using the formula,

$\text{Charge on FeOOH}=\left[\begin{array}{c}\left(\text{Number of }\text{Fe}\text{atoms}\times \text{oxidation number of Fe}\right)+\\ \left(\text{Number of O atoms}\times \text{oxidation number of O}\right)+\\ \left(\text{Number of H atoms}\times \text{oxidation number of H}\right)\end{array}\right]$

The charge on $\text{FeOOH}$ is $0$.

Substitute the number of atoms and their oxidation number in the above formula to calculate the oxidation number of $\text{Fe}$.

$\begin{array}{c}\text{Charge on FeOOH}=\left[\left(x\right)+\left(2\times \left(-2\right)\right)+1\left(+1\right)\right]\\ 0=x+\left(-4\right)+1\\ x=\underset{\_}{+3}\end{array}$

The common oxidation number of ${\text{OH}}^{-}$ ion is $\underset{\_}{-2}$.

Therefore, In the reactant side oxidation number of $\text{Fe}$ in ${\text{FeO}}_{\text{4}}{}^{2-}$ ion is $\underset{\_}{+6}$ and the common oxidation number of oxygen and hydrogen are $\underset{\_}{-2}$ and $\underset{\_}{+1}$ respectively.

In the product side oxidation number of $\text{Fe}$ in $\text{FeOOH}$ is $\underset{\_}{+3}$, the common oxidation number of hydrogen is $\underset{\_}{+1}$, oxidation state of ${\text{O}}_{\text{2}}$ is $\underset{\_}{0}$. And lastly the oxidation number of ${\text{OH}}^{-}$ ion is $\underset{\_}{-2}$.

Now balance the above given redox reaction as,

The given redox reaction is,

${\text{FeO}}_{\text{4}}{}^{\text{2-}}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\to \text{FeOOH}\left(s\right)+{\text{O}}_{\text{2}}\left(g\right)+{\text{OH}}^{-}\left(aq\right)$

In the given reaction, the number of $\text{Fe}$ $\text{O}$ and hydrogen atoms are present on the reactant side is equal to their number on the product side.

Therefore, the balanced equation is,

${\text{FeO}}_{\text{4}}{}^{\text{2-}}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\to \text{FeOOH}\left(s\right)+{\text{O}}_{\text{2}}\left(g\right)+{\text{OH}}^{-}\left(aq\right)$

(b)

Interpretation Introduction

To determine: The balanced equations for the given redox reaction and the oxidation numbers of the reactants and products.

Answer to Problem 4.126AP

Solution

The balanced equation of the given redox reaction is,

$4{\text{FeO}}_{\text{4}}{}^{\text{2-}}\left(aq\right)+4{\text{H}}_{\text{2}}\text{O}\left(l\right)\to 2{\text{Fe}}_2{\text{O}}_3\left(s\right)+3{\text{O}}_{\text{2}}\left(g\right)+8{\text{OH}}^{-}\left(aq\right)$

In the reactant side oxidation number of $\text{Fe}$ in ${\text{FeO}}_{\text{4}}{}^{2-}$ ion is $\underset{\_}{+6}$ and the common oxidation number of oxygen and hydrogen in ${\text{H}}_{\text{2}}\text{O}$ are $\underset{\_}{-2}$ and $\underset{\_}{+1}$ respectively.

In the product side oxidation number of $\text{Fe}$ in ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ is $\underset{\_}{+3}$, the common oxidation number of hydrogen is $\underset{\_}{+1}$ and oxidation state of ${\text{O}}_{\text{2}}$ is $\underset{\_}{0}$. And lastly the oxidation number of ${\text{OH}}^{-}$ ion is $\underset{\_}{-2}$.

Explanation of Solution

Explanation

The given unbalanced redox reaction is,

${\text{FeO}}_{\text{4}}{}^{\text{2-}}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\to {\text{Fe}}_2{\text{O}}_3\left(s\right)+{\text{O}}_{\text{2}}\left(g\right)+{\text{OH}}^{-}\left(aq\right)$

The oxidation numbers of reactant and product will be assigned as follows:

The oxidation number of reactant side will be calculated as,

In the reactant side oxidation number of $\text{Fe}$ present in ${\text{FeO}}_{\text{4}}{}^{2-}$ is to be calculated. The oxidation state of oxygen is usually $-2$ as it requires two more electrons in order to achieve stable electronic configuration. The oxidation number of $\text{Fe}$ is assumed to be $x$ and is calculated by using the formula,

${\text{Charge on FeO}}_{\text{4}}{}^{2-}=\left[\begin{array}{c}\left(\text{Number of }\text{Fe}\text{atoms}\times \text{oxidation number of Fe}\right)+\\ \left(\text{Number of O atoms}\times \text{oxidation number of O}\right)\end{array}\right]$

Since, the charge on ${\text{FeO}}_{\text{4}}{}^{2-}$ is $-2$.

Substitute the number of atoms and their oxidation number in the above formula to calculate the oxidation number of $\text{Fe}$.

$\begin{array}{c}{\text{Charge on FeO}}_{\text{4}}{}^{2-}=\left[\left(x\right)+\left(4\times \left(-2\right)\right)\right]\\ -2=x+\left(-8\right)\\ x=-2+8\\ x=\underset{\_}{+6}\end{array}$

Hence, the oxidation number of $\text{Fe}$ in ${\text{FeO}}_{\text{4}}{}^{2-}$ is $\underset{\_}{+6}$.

Therefore the oxidation numbers on the reactant side of $\text{Fe}$ in ${\text{FeO}}_{\text{4}}{}^{2-}$ ion is $\underset{\_}{+6}$ and the common oxidation number of oxygen and hydrogen are $\underset{\_}{-2}$ and $\underset{\_}{+1}$ respectively.

In the product side oxidation number of $\text{Fe}$ present in ${\text{Fe}}_2{\text{O}}_3$ is to be calculated. The oxidation state of oxygen is usually $-2$ as it requires two more electrons in order to achieve stable electronic configuration. The oxidation number of $\text{Fe}$ is assumed to be $x$ and is calculated by using the formula,

${\text{Charge on Fe}}_2{\text{O}}_3=\left[\begin{array}{c}\left(\text{Number of }\text{Fe}\text{atoms}\times \text{oxidation number of Fe}\right)+\\ \left(\text{Number of O atoms}\times \text{oxidation number of O}\right)\end{array}\right]$

Since, the charge on ${\text{Fe}}_2{\text{O}}_3$ is $0$ that is neutral.

Substitute the number of atoms and their oxidation number in the above formula to calculate the oxidation number of $\text{Fe}$.

$\begin{array}{c}{\text{Charge on Fe}}_2{\text{O}}_3=\left[\left(2x\right)+\left(2\times \left(-3\right)\right)\right]\\ 0=2x+\left(-6\right)\\ 2x=+6\\ x=\underset{\_}{+3}\end{array}$

The common oxidation number of ${\text{OH}}^{-}$ ion is $\underset{\_}{-2}$.

Therefore, In the reactant side oxidation number of $\text{Fe}$ in ${\text{FeO}}_{\text{4}}{}^{2-}$ ion is $\underset{\_}{+6}$ and for ${\text{H}}_{\text{2}}\text{O}$ molecule the common oxidation number of oxygen and hydrogen are $\underset{\_}{-2}$ and $\underset{\_}{+1}$ respectively.

In the product side oxidation number of $\text{Fe}$ in ${\text{Fe}}_2{\text{O}}_3$ is $\underset{\_}{+3}$, the common oxidation number of hydrogen is $\underset{\_}{+1}$, oxidation state of ${\text{O}}_{\text{2}}$ is $\underset{\_}{0}$. And lastly the oxidation number of ${\text{OH}}^{-}$ ion is $\underset{\_}{-2}$

Now balance the above given redox reaction as,

The given unbalanced redox reaction is,

${\text{FeO}}_{\text{4}}{}^{\text{2-}}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\to {\text{Fe}}_2{\text{O}}_3\left(s\right)+{\text{O}}_{\text{2}}\left(g\right)+{\text{OH}}^{-}\left(aq\right)$

In this reaction the number of iron atoms will be balanced by putting coefficient of $4$ in front of ${\text{FeO}}_{\text{4}}{}^{2-}$ in the reactant side and coefficient of $2$ in front of ${\text{Fe}}_2{\text{O}}_3$ in the product side then the reaction becomes as,

${\text{4FeO}}_{\text{4}}{}^{\text{2-}}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\to 2{\text{Fe}}_2{\text{O}}_3\left(s\right)+{\text{O}}_{\text{2}}\left(g\right)+{\text{OH}}^{-}\left(aq\right)$

The number of oxygen and hydrogen atoms is not equal in both the sides of the equation. Thus, the coefficient of $4$ is placed in front of water molecule and coefficient of $3$ before ${\text{O}}_{\text{2}}$ molecule in the reactant side and coefficient of $8$ is placed before ${\text{OH}}^{-}$ ions in the product side, then the reaction will become as,

$4{\text{FeO}}_{\text{4}}{}^{\text{2-}}\left(aq\right)+4{\text{H}}_{\text{2}}\text{O}\left(l\right)\to 2{\text{Fe}}_2{\text{O}}_3\left(s\right)+3{\text{O}}_{\text{2}}\left(g\right)+8{\text{OH}}^{-}\left(aq\right)$

Therefore the balanced equation is,

$4{\text{FeO}}_{\text{4}}{}^{\text{2-}}\left(aq\right)+4{\text{H}}_{\text{2}}\text{O}\left(l\right)\to 2{\text{Fe}}_2{\text{O}}_3\left(s\right)+3{\text{O}}_{\text{2}}\left(g\right)+8{\text{OH}}^{-}\left(aq\right)$

Conclusion

1. a) The balanced equation of the given redox reaction is,

   ${\text{FeO}}_{\text{4}}{}^{\text{2-}}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\to \text{FeOOH}\left(s\right)+{\text{O}}_{\text{2}}\left(g\right)+{\text{OH}}^{-}\left(aq\right)$

2. b) The balanced equation of the given redox reaction is,

   $4{\text{FeO}}_{\text{4}}{}^{\text{2-}}\left(aq\right)+4{\text{H}}_{\text{2}}\text{O}\left(l\right)\to 2{\text{Fe}}_2{\text{O}}_3\left(s\right)+3{\text{O}}_{\text{2}}\left(g\right)+8{\text{OH}}^{-}\left(aq\right)$