Chapter 17, Problem 25E

Interpretation Introduction

(a)

Interpretation:

The balanced equation for the redox reaction, $\text{Zn}\left(s\right)+{\text{NO}}_3^{-}\left(aq\right)\to {\text{Zn}}^{2+}\left(aq\right)+\text{NO}\left(g\right)$ in acidic solution 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 25E

The balanced equation for the redox reaction, $\text{Zn}\left(s\right)+{\text{NO}}_3^{-}\left(aq\right)\to {\text{Zn}}^{2+}\left(aq\right)+\text{NO}\left(g\right)$ in acidic solution is shown below.

$\text{3Zn}\left(s\right)+{\text{8H}}^{+}\left(aq\right)+{\text{2NO}}_3^{-}\left(aq\right)\to 3{\text{Zn}}^{2+}\left(aq\right)+2\text{NO}\left(g\right)+{\text{4H}}_{\text{2}}\text{O}\left(l\right)$

Explanation of Solution

The given redox reaction is shown below.

$\text{Zn}\left(s\right)+{\text{NO}}_3^{-}\left(aq\right)\to {\text{Zn}}^{2+}\left(aq\right)+\text{NO}\left(g\right)$

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

The oxidation state of the nitrogen in ${\text{NO}}_3^{-}$ is calculated below.

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

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

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

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

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

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

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 nitrogen is $+5$ in ${\text{NO}}_3^{-}$.

The oxidation number of nitrogen in $\text{NO}$ is calculated below.

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

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

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

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

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

$\begin{array}{l}\text{N O}\\ \text{n}+\left(-\text{2}\right)=0\end{array}$

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

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

The oxidation state of nitrogen in $\text{NO}$ is $+2$.

The oxidation number of nitrogen is decreased therefore, it is a reduction half step.

The reduction half-reaction for the above reaction is shown below.

${\text{NO}}_3^{-}\left(aq\right)\to \text{NO}\left(g\right)$

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 nitrogen is getting reduced and its number of atoms are balanced on both sides.

${\text{NO}}_3^{-}\left(aq\right)\to \text{NO}\left(g\right)$

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

${\text{NO}}_3^{-}\left(aq\right)\to \text{NO}\left(g\right)$

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

The number of oxygen atoms is balanced by adding two water molecules on the right-hand side of the equation.

${\text{NO}}_3^{-}\left(aq\right)\to \text{NO}\left(g\right)+{\text{2H}}_{\text{2}}\text{O}\left(l\right)$

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

The number of hydrogen atoms is balanced by adding four ${\text{H}}^{\text{+}}$ ions on the left-hand side of the equation.

${\text{NO}}_3^{-}\left(aq\right)+{\text{4H}}^{+}\left(aq\right)\to \text{NO}\left(g\right)+{\text{2H}}_{\text{2}}\text{O}\left(l\right)$

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

The charge is balanced by adding three electrons on the left-hand side of the equation.

${\text{NO}}_3^{-}\left(aq\right)+{\text{4H}}^{+}\left(aq\right)+3{\text{e}}^{-}\to \text{NO}\left(g\right)+{\text{2H}}_{\text{2}}\text{O}\left(l\right)$

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

The equation is completely balanced and is shown below.

${\text{NO}}_3^{-}\left(aq\right)+{\text{4H}}^{+}\left(aq\right)+3{\text{e}}^{-}\to \text{NO}\left(g\right)+{\text{2H}}_{\text{2}}\text{O}\left(l\right)$ …(1)

The oxidation state of zinc in $\text{Zn}$ is zero as it is the elemental form of zinc. The oxidation state of zinc in ${\text{Zn}}^{2+}$ is $+2$ coming from the charge on the zinc ion.

The oxidation number of zinc is increased, therefore, it is an oxidation half step.

The oxidation half-reaction for the above reaction is shown below.

$\text{Zn}\left(s\right)\to {\text{Zn}}^{2+}\left(aq\right)$

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 zinc is getting oxidized and its number of atoms are balanced on both sides.

$\text{Zn}\left(s\right)\to {\text{Zn}}^{2+}\left(aq\right)$

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

$\text{Zn}\left(s\right)\to {\text{Zn}}^{2+}\left(aq\right)$

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

$\text{Zn}\left(s\right)\to {\text{Zn}}^{2+}\left(aq\right)$

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

$\text{Zn}\left(s\right)\to {\text{Zn}}^{2+}\left(aq\right)$

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

The charge is balanced by adding two electrons on the left-hand side of the equation.

$\text{Zn}\left(s\right)\to {\text{Zn}}^{2+}\left(aq\right)+2{\text{e}}^{-}$

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

The equation is completely balanced and is shown below.

$\text{Zn}\left(s\right)\to {\text{Zn}}^{2+}\left(aq\right)+2{\text{e}}^{-}$ …(2)

The balanced redox equation is obtained by adding equation (1) and (2) in such a way that electrons are canceled out.

Multiply equation (1) by two and equation (2) by three and then add them.

$\begin{array}{c}\text{3Zn}\left(s\right)\to 3{\text{Zn}}^{2+}\left(aq\right)+6{\text{e}}^{-}\\ +\\ {\text{2NO}}_3^{-}\left(aq\right)+{\text{8H}}^{+}\left(aq\right)+6{\text{e}}^{-}\to 2\text{NO}\left(g\right)+{\text{4H}}_{\text{2}}\text{O}\left(l\right)\end{array}$

The balance redox equation after adding these equations is shown below.

$\text{3Zn}\left(s\right)+{\text{8H}}^{+}\left(aq\right)+{\text{2NO}}_3^{-}\left(aq\right)\to 3{\text{Zn}}^{2+}\left(aq\right)+2\text{NO}\left(g\right)+{\text{4H}}_{\text{2}}\text{O}\left(l\right)$

Conclusion

The balanced equation of redox reaction is shown below.

$\text{3Zn}\left(s\right)+{\text{8H}}^{+}\left(aq\right)+{\text{2NO}}_3^{-}\left(aq\right)\to 3{\text{Zn}}^{2+}\left(aq\right)+2\text{NO}\left(g\right)+{\text{4H}}_{\text{2}}\text{O}\left(l\right)$

Interpretation Introduction

(b)

Interpretation:

The balanced equation for the redox reaction, ${\text{Mn}}^{2+}\left(aq\right)+{\text{BiO}}_3^{-}\left(aq\right)\to {\text{MnO}}_4^{-}\left(aq\right)+{\text{Bi}}^{3+}\left(aq\right)$ in acidic solution 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 25E

The balanced equation for the redox reaction, ${\text{Mn}}^{2+}\left(aq\right)+{\text{BiO}}_3^{-}\left(aq\right)\to {\text{MnO}}_4^{-}\left(aq\right)+{\text{Bi}}^{3+}\left(aq\right)$ in acidic solution is shown below.

${\text{2Mn}}^{2+}\left(aq\right)+{\text{5BiO}}_3^{-}\left(aq\right)+{\text{14H}}^{+}\left(aq\right)\to 2{\text{MnO}}_4{}^{-}\left(aq\right)+5{\text{Bi}}^{3+}\left(aq\right)+{\text{7H}}_{\text{2}}\text{O}\left(l\right)$

Explanation of Solution

The given redox reaction is shown below.

${\text{Mn}}^{2+}\left(aq\right)+{\text{BiO}}_3^{-}\left(aq\right)\to {\text{MnO}}_4^{-}\left(aq\right)+{\text{Bi}}^{3+}\left(aq\right)$

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

The oxidation state of manganese in ${\text{MnO}}_4{}^{-}$ is calculated below.

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

$\begin{array}{l}{\text{Mn O}}_4\\ \text{n -2}\end{array}$

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

$\begin{array}{l}{\text{Mn O}}_4\\ \text{n 4}\left(\text{-2}\right)\end{array}$

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

$\begin{array}{l}{\text{Mn O}}_4\\ \text{n}+\text{4}\left(\text{-2}\right)=-1\end{array}$

Calculate the value of n by simplifying the equation.

$\begin{array}{rll}\text{n}+\text{4}\left(-\text{2}\right)& =& -1\\ \text{n}+\text{(}-\text{8)}& =& -1\\ \text{n}& =& -1+8\\ \text{n}& =& +7\end{array}$

The oxidation state of manganese in ${\text{MnO}}_4{}^{-}$ is $+7$.

The oxidation state of ${\text{Mn}}^{2+}$ is $+2$ which can be seen from the charge on the manganese.

The oxidation number of manganese is increased therefore, it is an oxidation half step.

The oxidation half-reaction for the above reaction is shown below.

${\text{Mn}}^{2+}\left(aq\right)\to {\text{MnO}}_4{}^{-}\left(aq\right)$

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

Step-1: Identify and balanced getting oxidized or reduced.

The manganese is getting reduced and its number of atoms are balanced on both sides.

${\text{Mn}}^{2+}\left(aq\right)\to {\text{MnO}}_4{}^{-}\left(aq\right)$

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

${\text{Mn}}^{2+}\left(aq\right)\to {\text{MnO}}_4{}^{-}\left(aq\right)$

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

The number of oxygen atoms is balanced by adding four water molecules on the left-hand side of the equation.

${\text{Mn}}^{2+}\left(aq\right)+{\text{4H}}_{\text{2}}\text{O}\left(l\right)\to {\text{MnO}}_4{}^{-}\left(aq\right)$

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

The number of hydrogen atoms is balanced by adding eight ${\text{H}}^{\text{+}}$ ions on the right-hand side of the equation.

${\text{Mn}}^{2+}\left(aq\right)+{\text{4H}}_{\text{2}}\text{O}\left(l\right)\to {\text{MnO}}_4{}^{-}\left(aq\right)+{\text{8H}}^{+}\left(aq\right)$

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

The charge is balanced by adding five electrons on the right-hand side

${\text{Mn}}^{2+}\left(aq\right)+{\text{4H}}_{\text{2}}\text{O}\left(l\right)\to {\text{MnO}}_4{}^{-}\left(aq\right)+{\text{8H}}^{+}\left(aq\right)+5{\text{e}}^{-}$

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

The equation is completely balanced and is shown below.

${\text{Mn}}^{2+}\left(aq\right)+{\text{4H}}_{\text{2}}\text{O}\left(l\right)\to {\text{MnO}}_4{}^{-}\left(aq\right)+{\text{8H}}^{+}\left(aq\right)+5{\text{e}}^{-}$ …(1)

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

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

$\begin{array}{l}{\text{Bi O}}_{\text{3}}\\ \text{n }-\text{2}\end{array}$

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

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

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

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

Calculate the value of n by simplifying the equation.

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

The oxidation of bismuth in ${\text{BiO}}_3^{-}$ is $+5$.

The oxidation of bismuth is $+3$ in ${\text{Bi}}^{3+}$ which comes from the charge on the bismuth ion.

The oxidation number of bismuth is reduced therefore, it is a reduction half step.

The reduction half-reaction for the above reaction is shown below.

${\text{BiO}}_3^{-}\left(aq\right)\to {\text{Bi}}^{3+}\left(aq\right)$

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 bismuth is getting reduced and its number of atoms are balanced on both sides.

${\text{BiO}}_3^{-}\left(aq\right)\to {\text{Bi}}^{3+}\left(aq\right)$

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

${\text{BiO}}_3^{-}\left(aq\right)\to {\text{Bi}}^{3+}\left(aq\right)$

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

The number of oxygen atoms is balanced by adding three water molecules on the right-hand side of the equation.

${\text{BiO}}_3^{-}\left(aq\right)\to {\text{Bi}}^{3+}\left(aq\right)+{\text{3H}}_{\text{2}}\text{O}\left(l\right)$

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

The number of hydrogen atoms is balanced by adding six ${\text{H}}^{\text{+}}$ ions on the left-hand side of the equation.

${\text{BiO}}_3^{-}\left(aq\right)+{\text{6H}}^{+}\left(aq\right)\to {\text{Bi}}^{3+}\left(aq\right)+{\text{3H}}_{\text{2}}\text{O}\left(l\right)$

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

The charge is balanced by adding two electrons on the left-hand side of the equation.

${\text{BiO}}_3^{-}\left(aq\right)+{\text{6H}}^{+}\left(aq\right)+2{\text{e}}^{-}\to {\text{Bi}}^{3+}\left(aq\right)+{\text{3H}}_{\text{2}}\text{O}\left(l\right)$

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

The equation is completely balanced and is shown below.

${\text{BiO}}_3^{-}\left(aq\right)+{\text{6H}}^{+}\left(aq\right)+2{\text{e}}^{-}\to {\text{Bi}}^{3+}\left(aq\right)+{\text{3H}}_{\text{2}}\text{O}\left(l\right)$ …(2)

The balanced redox equation is obtained by adding equation (1) and (2) in such a way that electrons are canceled out.

Multiply equation (1) by two and equation (2) by five and then add them.

$\begin{array}{c}{\text{2Mn}}^{2+}\left(aq\right)+{\text{8H}}_{\text{2}}\text{O}\left(l\right)\to 2{\text{MnO}}_4{}^{-}\left(aq\right)+{\text{16H}}^{+}\left(aq\right)+10{\text{e}}^{-}\\ +\\ {\text{5BiO}}_3^{-}\left(aq\right)+{\text{30H}}^{+}\left(aq\right)+10{\text{e}}^{-}\to 5{\text{Bi}}^{3+}\left(aq\right)+{\text{15H}}_{\text{2}}\text{O}\left(l\right)\end{array}$

The balance redox equation after adding these equations is shown below.

${\text{2Mn}}^{2+}\left(aq\right)+{\text{5BiO}}_3^{-}\left(aq\right)+{\text{14H}}^{+}\left(aq\right)\to 2{\text{MnO}}_4{}^{-}\left(aq\right)+5{\text{Bi}}^{3+}\left(aq\right)+{\text{7H}}_{\text{2}}\text{O}\left(l\right)$

Conclusion

The balanced equation of redox reaction is shown below.

${\text{2Mn}}^{2+}\left(aq\right)+{\text{5BiO}}_3^{-}\left(aq\right)+{\text{14H}}^{+}\left(aq\right)\to 2{\text{MnO}}_4{}^{-}\left(aq\right)+5{\text{Bi}}^{3+}\left(aq\right)+{\text{7H}}_{\text{2}}\text{O}\left(l\right)$