Chapter 17, Problem 22E

Interpretation Introduction

(a)

Interpretation:

The balanced equation for the redox reaction, ${\text{Fe}}^{2+}\left(aq\right)+{\text{H}}_{\text{2}}{\text{O}}_2\left(aq\right)+{\text{H}}^{+}\left(aq\right)\stackrel{}{\to }{\text{Fe}}^{3+}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$, using oxidation number method is to be stated.

Concept introduction:

The oxidizer is the species whose oxidation state falls during the course of reaction and reducer is the species whose oxidation number increases. Oxidized product is the oxidation product of the reducer and reduced product is the reduction product of the oxidizer.

Answer to Problem 22E

The balanced equation for the redox reaction, ${\text{Fe}}^{2+}\left(aq\right)+{\text{H}}_{\text{2}}{\text{O}}_2\left(aq\right)+{\text{H}}^{+}\left(aq\right)\stackrel{}{\to }{\text{Fe}}^{3+}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$, using oxidation number method is shown below.

${\text{2Fe}}^{2+}\left(aq\right)+{\text{H}}_{\text{2}}{\text{O}}_2\left(aq\right)+2{\text{H}}^{+}\left(aq\right)\stackrel{}{\to }{\text{2Fe}}^{3+}\left(aq\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Explanation of Solution

The given redox reaction is shown below.

${\text{Fe}}^{2+}\left(aq\right)+{\text{H}}_{\text{2}}{\text{O}}_2\left(aq\right)+{\text{H}}^{+}\left(aq\right)\stackrel{}{\to }{\text{Fe}}^{3+}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The oxidation state of the central metal atom is calculated by knowing the standard oxidation states of few elements.

The oxidation state of oxygen in ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ is calculated below.

Step-1: Write down the oxidation number of every element and for unknown take “n”.

$\begin{array}{l}{\text{H}}_2{\text{ O}}_2\\ \text{+1 n}\end{array}$

Step-2: Multiply the oxidation state with their number of atoms of an element.

$\begin{array}{l}{\text{H}}_2{\text{ O}}_2\\ 2\left(\text{+1}\right)\text{ 2n}\end{array}$

Step-3: Add the oxidation numbers and set them equal to the charge of the species.

$\begin{array}{l}{\text{H}}_2{\text{ O}}_2\\ 2\left(\text{+1}\right)+\text{2n}=0\end{array}$

Calculate the value of n by simplifying the equation as shown below.

$\begin{array}{rll}2\left(\text{+1}\right)+\text{2n}& =& 0\\ \text{2n}+\text{(}+2\text{)}& =& 0\\ \text{2n}& =& 0-2\\ \text{2n}& =& -2\end{array}$

Divide the equation by two on both sides and simplify as shown below.

$\begin{array}{rll}\frac{\text{2n}}{\text{2}}& =& \frac{-2}{2}\\ \text{n}& =& -1\end{array}$

The oxidation state of oxygen in ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ is $-1$.

The oxidation state of oxygen in ${\text{H}}_{\text{2}}\text{O}$ is calculated below.

Step-1: Write down the oxidation number of every element and for unknown take “n”.

$\begin{array}{l}{\text{H}}_2\text{ O}\\ \text{+1 n}\end{array}$

Step-2: Multiply the oxidation state with their number of atoms.

$\begin{array}{l}{\text{H}}_2\text{ O}\\ 2\left(\text{+1}\right)\text{ n}\end{array}$

Step-3: Add the oxidation numbers and set them equal to the charge of the species.

$\begin{array}{l}{\text{H}}_2\text{ O}\\ 2\left(\text{+1}\right)+\text{n}=\text{0}\end{array}$

Calculate the value of n by simplifying the equation as shown below.

$\begin{array}{rll}2\left(\text{+1}\right)+\text{n}& =& \text{0}\\ \text{n}+\text{(}+2\text{)}& =& 0\\ \text{n}& =& 0-2\\ \text{n}& =& -2\end{array}$

The oxidation state of oxygen in ${\text{H}}_{\text{2}}\text{O}$ is $-2$.

The oxidation state of iron in ${\text{Fe}}^{2+}$ is $+2$ directly coming from the charge on iron. The oxidation state of iron in ${\text{Fe}}^{3+}$ is $+3$ directly coming from the charge on iron.

The oxidation number of oxygen is decreased from reactant to product and for iron, it increases from reactant to product.

To balance the redox reaction by oxidation number method the steps to be followed are shown below.

Step-1: First of all balance the number of atoms of elements getting oxidized and reduced.

The oxygen is getting reduced and iron is getting oxidized. The number of atoms of oxygen is not balanced on both sides of the equation. Balance them by multiplying the ${\text{H}}_{\text{2}}\text{O}$ by two on the left-hand side.

${\text{Fe}}^{2+}\left(aq\right)+{\text{H}}_{\text{2}}{\text{O}}_2\left(aq\right)+{\text{H}}^{+}\left(aq\right)\stackrel{}{\to }{\text{Fe}}^{3+}\left(aq\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Step-2: Determine the number of electrons lost and the number of electrons gained and balance them.

The iron is getting oxidized therefore, no of electrons lost by iron is one. The oxygen is getting reduced and no of electron gained by oxygen is two.

The number of electrons lost or gained comes from the difference in their oxidation state multiplied by its stoichiometry.

Balance the number of electrons lost or gained by multiplying ${\text{Fe}}^{2+}$ by two and then balance the iron atoms on both sides as shown below.

${\text{2Fe}}^{2+}\left(aq\right)+{\text{H}}_{\text{2}}{\text{O}}_2\left(aq\right)+{\text{H}}^{+}\left(aq\right)\stackrel{}{\to }{\text{2Fe}}^{3+}\left(aq\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Step-3: Balance the hydrogen and oxygen atoms on both sides of the equation.

Multiply the ${\text{H}}^{+}$ by two in order to balance hydrogen atoms on both sides of the equation as shown below.

${\text{2Fe}}^{2+}\left(aq\right)+{\text{H}}_{\text{2}}{\text{O}}_2\left(aq\right)+2{\text{H}}^{+}\left(aq\right)\stackrel{}{\to }{\text{2Fe}}^{3+}\left(aq\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The equation is now completely balanced and has equal charges on both sides.

Conclusion

The balanced equation for the redox reaction is shown below.

${\text{2Fe}}^{2+}\left(aq\right)+{\text{H}}_{\text{2}}{\text{O}}_2\left(aq\right)+2{\text{H}}^{+}\left(aq\right)\stackrel{}{\to }{\text{2Fe}}^{3+}\left(aq\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Interpretation Introduction

(b)

Interpretation:

The balanced equation for the redox reaction, ${\text{Cr}}_2{\text{O}}_7^{2-}\left(s\right)+{\text{Br}}^{-}\left(aq\right)+{\text{H}}^{+}\left(aq\right)\stackrel{}{\to }{\text{Cr}}^{3+}\left(aq\right)+{\text{Br}}_{\text{2}}\left(l\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$, using oxidation number method is to be stated.

Concept introduction:

The oxidizer is the species whose oxidation state falls during the course of reaction and reducer is the species whose oxidation number increases. Oxidized product is the oxidation product of the reducer and reduced product is the reduction product of the oxidizer.

Answer to Problem 22E

The balanced equation for the redox reaction, ${\text{Cr}}_2{\text{O}}_7^{2-}\left(s\right)+{\text{Br}}^{-}\left(aq\right)+{\text{H}}^{+}\left(aq\right)\stackrel{}{\to }{\text{Cr}}^{3+}\left(aq\right)+{\text{Br}}_{\text{2}}\left(l\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$, using oxidation number method is shown below.

${\text{Cr}}_2{\text{O}}_7^{2-}\left(s\right)+6{\text{Br}}^{-}\left(aq\right)+14{\text{H}}^{+}\left(aq\right)\stackrel{}{\to }{\text{2Cr}}^{3+}\left(aq\right)+3{\text{Br}}_{\text{2}}\left(l\right)+7{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Explanation of Solution

The given redox reaction is shown below.

${\text{Cr}}_2{\text{O}}_7^{2-}\left(s\right)+{\text{Br}}^{-}\left(aq\right)+{\text{H}}^{+}\left(aq\right)\stackrel{}{\to }{\text{Cr}}^{3+}\left(aq\right)+{\text{Br}}_{\text{2}}\left(l\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The oxidation state of the central metal atom is calculated by knowing the standard oxidation states of few elements.

The oxidation state of chromium in ${\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{}^{-}$ is calculated below.

Step-1: Write down the oxidation number of every element and for unknown take “n”.

$\begin{array}{l}{\text{Cr}}_2{\text{ O}}_7\\ \text{n }-\text{2}\end{array}$

Step-2: Multiply the oxidation state with their number of atoms of an element.

$\begin{array}{l}{\text{Cr}}_2{\text{ O}}_7\\ \text{2n 7}\left(-\text{2}\right)\end{array}$

Step-3: Add the oxidation numbers and set them equal to the charge of the species.

$\begin{array}{l}{\text{Cr}}_2{\text{ O}}_7\\ \text{2n}+\text{7}\left(-\text{2}\right)=-2\end{array}$

Calculate the value of n by simplifying the equation as shown below.

$\begin{array}{rll}\text{2n}+\text{7}\left(-\text{2}\right)& =& -2\\ \text{2n}+\text{(}-\text{14)}& =& -2\\ \text{2n}& =& -2+14\\ \text{2n}& =& +12\end{array}$

Divide by two on both sides and simplify as shown below.

$\begin{array}{rll}\frac{\text{2n}}{\text{2}}& =& \frac{+12}{2}\\ \text{n}& =& +6\end{array}$

The oxidation state of chromium in ${\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{}^{-}$ is $+6$.

The oxidation state of chromium in ${\text{Cr}}^{3+}$ is $+3$ directly coming from the charge on chromium.

The oxidation state of bromine is $-1$ in ${\text{Br}}^{-}$ directly coming from the charge on iodine. The oxidation state of bromine is zero in ${\text{Br}}_{\text{2}}$ because of the elemental form.

The oxidation number of chromium is decreased from reactant to product and for bromine, it is increased from reactant to product.

To balance the redox reaction by oxidation number method the steps to be followed are shown below.

Step-1: First of all balance the number of atoms of elements getting oxidized and reduced.

The chromium is getting reduced and bromine is getting oxidized. The number of atoms of chromium and bromine is balanced by multiplying ${\text{Cr}}^{3+}$ and ${\text{Br}}^{-}$ by two.

${\text{Cr}}_2{\text{O}}_7^{2-}\left(s\right)+2{\text{Br}}^{-}\left(aq\right)+{\text{H}}^{+}\left(aq\right)\stackrel{}{\to }{\text{2Cr}}^{3+}\left(aq\right)+{\text{Br}}_{\text{2}}\left(l\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Step-2: Determine the number of electrons lost and the number of electrons gained and balance them.

The bromine is getting oxidized therefore, no of electrons lost by bromine is two. The chromium is getting reduced and no of electron gained by chromium is six.

The number of electrons lost or gained comes from the difference in their oxidation state multiplied by its stoichiometry.

Balance the number of electrons lost or gained by multiply ${\text{Br}}^{-}$ by three then balance the chromium and bromine atoms on both sides as shown below.

${\text{Cr}}_2{\text{O}}_7^{2-}\left(s\right)+6{\text{Br}}^{-}\left(aq\right)+{\text{H}}^{+}\left(aq\right)\stackrel{}{\to }{\text{2Cr}}^{3+}\left(aq\right)+3{\text{Br}}_{\text{2}}\left(l\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Step-3: Balance the hydrogen and oxygen atoms on both sides of the equation.

Multiply the ${\text{H}}_{\text{2}}\text{O}$ by seven and ${\text{H}}^{+}$ by fourteen in order to balance hydrogen and oxygen elements as shown below.

${\text{Cr}}_2{\text{O}}_7^{2-}\left(s\right)+6{\text{Br}}^{-}\left(aq\right)+14{\text{H}}^{+}\left(aq\right)\stackrel{}{\to }{\text{2Cr}}^{3+}\left(aq\right)+3{\text{Br}}_{\text{2}}\left(l\right)+7{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The equation is now completely balanced and has equal charges on both sides.

Conclusion

The balanced equation for the redox reaction is shown below.

${\text{Cr}}_2{\text{O}}_7^{2-}\left(s\right)+6{\text{Br}}^{-}\left(aq\right)+14{\text{H}}^{+}\left(aq\right)\stackrel{}{\to }{\text{2Cr}}^{3+}\left(aq\right)+3{\text{Br}}_{\text{2}}\left(l\right)+7{\text{H}}_{\text{2}}\text{O}\left(l\right)$