Chapter 17, Problem 23E

Interpretation Introduction

(a)

Interpretation:

The balanced half-reaction in acidic solution for the reaction, ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(aq\right)\to {\text{H}}_{\text{2}}\text{O}\left(l\right)$, 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 23E

The balanced half-reaction in acidic solution is shown below.

${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(aq\right)+2{\text{H}}^{+}\left(aq\right)+2{\text{e}}^{-}\to 2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Explanation of Solution

The given half-reaction equation to be balanced is shown below.

${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(aq\right)\to {\text{H}}_{\text{2}}\text{O}\left(l\right)$

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 the number of atoms of the 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 of an element.

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

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 number of oxygen is decreased. Therefore, it is an reduction half-reaction.

The balancing of the half-reactions is done by the following the steps shown below.

Step-1: Identify and balance the element getting oxidized or reduced.

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

${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(aq\right)\to 2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Step-2: Balance elements other than oxygen and hydrogen if any.

${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(aq\right)\to 2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Step-3: Balance oxygen atoms by adding water on the appropriate side.

Oxygen atoms are already balanced on both sides.

${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(aq\right)\to 2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Step-4: Balance the hydrogen atoms by adding ${\text{H}}^{\text{+}}$ to the relevant side when necessary.

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

Step-5: Balance the charge by adding electrons to the appropriate side.

Two electrons are added to the left-hand side in order to balance the charge.

${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(aq\right)+2{\text{H}}^{+}\left(aq\right)+2{\text{e}}^{-}\to 2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Step-6: Recheck the equation to be sure that it is perfectly balanced.

The equation is completely balanced and is shown below.

${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(aq\right)+2{\text{H}}^{+}\left(aq\right)+2{\text{e}}^{-}\to 2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Conclusion

The balanced half-reaction is shown below.

${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(aq\right)+2{\text{H}}^{+}\left(aq\right)+2{\text{e}}^{-}\to 2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Interpretation Introduction

(b)

Interpretation:

The balanced half-reaction in acidic solution for the reaction, ${\text{AsO}}_3{}^{3-}\left(aq\right)\to {\text{AsO}}_3{}^{-}\left(aq\right)$, 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 23E

The balanced half-reaction in acidic solution is shown below.

${\text{AsO}}_3{}^{3-}\left(aq\right)\to {\text{AsO}}_3{}^{-}\left(aq\right)+2{\text{e}}^{-}$

Explanation of Solution

The given half-reaction equation to be balanced is shown below.

${\text{AsO}}_3{}^{3-}\left(aq\right)\to {\text{AsO}}_3{}^{-}\left(aq\right)$

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

The oxidation state of arsenic in ${\text{AsO}}_3{}^{3-}$ is calculated below.

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

$\begin{array}{l}{\text{As O}}_3\\ \text{n }-\text{2}\end{array}$

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

$\begin{array}{l}{\text{As O}}_3\\ \text{n 3}\left(-\text{2}\right)\end{array}$

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

$\begin{array}{l}{\text{As O}}_3\\ \text{n}+\text{3}\left(-\text{2}\right)=-3\end{array}$

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

$\begin{array}{rll}\text{n}+\text{3}\left(-\text{2}\right)& =& -3\\ \text{n}+\text{(}-\text{6)}& =& -3\\ \text{n}& =& -3+6\\ \text{n}& =& +3\end{array}$

The oxidation state of arsenic in ${\text{AsO}}_3{}^{3-}$ is $+3$.

The oxidation state of arsenic in ${\text{AsO}}_3{}^{-}$ is calculated below.

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

$\begin{array}{l}{\text{As O}}_3\\ \text{n }-\text{2}\end{array}$

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

$\begin{array}{l}{\text{As O}}_3\\ \text{n 3}\left(-\text{2}\right)\end{array}$

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

$\begin{array}{l}{\text{As O}}_3\\ \text{n}+\text{3}\left(-\text{2}\right)=-1\end{array}$

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

$\begin{array}{rll}\text{n}+\text{3}\left(-\text{2}\right)& =& -1\\ \text{n}+\text{(}-\text{6)}& =& -1\\ \text{n}& =& -1+6\\ \text{n}& =& +5\end{array}$

The oxidation state of arsenic in ${\text{AsO}}_3{}^{-}$ is $+5$.

The oxidation number of arsenic is increased. Therefore, it is an oxidation half-reaction.

The balancing of the half-reactions is done by the following the steps shown below.

Step-1: Identify and balance the element getting oxidized or reduced.

The arsenic is getting oxidized and the number of atoms of that is balanced on both sides.

${\text{AsO}}_3{}^{3-}\left(aq\right)\to {\text{AsO}}_3{}^{-}\left(aq\right)$

Step-2: Balance elements other than oxygen and hydrogen if any.

${\text{AsO}}_3{}^{3-}\left(aq\right)\to {\text{AsO}}_3{}^{-}\left(aq\right)$

Step-3: Balance oxygen atoms by adding water on the appropriate side.

Oxygen atoms are already balanced on both sides.

${\text{AsO}}_3{}^{3-}\left(aq\right)\to {\text{AsO}}_3{}^{-}\left(aq\right)$

Step-4: Balance the hydrogen atoms by adding ${\text{H}}^{\text{+}}$ to the relevant side when necessary.

${\text{AsO}}_3{}^{3-}\left(aq\right)\to {\text{AsO}}_3{}^{-}\left(aq\right)$

Step-5: Balance the charge by adding electrons to the appropriate side.

Two electrons are added to the right-hand side in order to balance the charge.

${\text{AsO}}_3{}^{3-}\left(aq\right)\to {\text{AsO}}_3{}^{-}\left(aq\right)+2{\text{e}}^{-}$

Step-6: Recheck the equation to be sure that it is perfectly balanced.

The equation is completely balanced and is shown below.

${\text{AsO}}_3{}^{3-}\left(aq\right)\to {\text{AsO}}_3{}^{-}\left(aq\right)+2{\text{e}}^{-}$

Conclusion

The balanced half-reaction is shown below.

${\text{AsO}}_3{}^{3-}\left(aq\right)\to {\text{AsO}}_3{}^{-}\left(aq\right)+2{\text{e}}^{-}$