Chapter 17, Problem 26E

Interpretation Introduction

(a)

Interpretation:

The balanced equation for the redox reaction, ${\text{MnO}}_4^{-}\left(aq\right)+{\text{S}}^{2-}\left(aq\right)\to {\text{MnO}}_2\left(aq\right)+\text{S}\left(s\right)$ in basic 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 26E

The balanced equation for the redox reaction, ${\text{MnO}}_4^{-}\left(aq\right)+{\text{S}}^{2-}\left(aq\right)\to {\text{MnO}}_2\left(aq\right)+\text{S}\left(s\right)$ in basic solution is shown below.

${\text{2MnO}}_4{}^{-}\left(aq\right)+4{\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{3S}}^{2-}\left(aq\right)\to 2{\text{MnO}}_2\left(aq\right)+8{\text{OH}}^{-}\left(aq\right)+3\text{S}\left(s\right)$

Explanation of Solution

The given redox reaction is shown below.

${\text{MnO}}_4^{-}\left(aq\right)+{\text{S}}^{2-}\left(aq\right)\to {\text{MnO}}_2\left(aq\right)+\text{S}\left(s\right)$

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 their number of atoms of an 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 manganese in ${\text{MnO}}_2$ is calculated below.

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

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

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

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

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

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

Calculate the value of n by simplifying the equation.

$\begin{array}{rll}\text{n}+\text{2}\left(-\text{2}\right)& =& 0\\ \text{n}+\text{(}-4\text{)}& =& 0\\ \text{n}& =& 0+4\\ \text{n}& =& +4\end{array}$

The oxidation state of manganese in ${\text{MnO}}_2$ is $+4$.

The oxidation number manganese decreases from ${\text{MnO}}_4{}^{-}\to {\text{MnO}}_2$ therefore, this step is reduction half step.

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

${\text{MnO}}_4{}^{-}\left(aq\right)\to {\text{MnO}}_2\left(aq\right)$

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 is balanced on both sides.

${\text{MnO}}_4{}^{-}\left(aq\right)\to {\text{MnO}}_2\left(aq\right)$

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

${\text{MnO}}_4{}^{-}\left(aq\right)\to {\text{MnO}}_2\left(aq\right)$

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.

${\text{MnO}}_4{}^{-}\left(aq\right)\to {\text{MnO}}_2\left(aq\right)+{\text{2H}}_{\text{2}}\text{O}\left(l\right)$

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.

${\text{MnO}}_4{}^{-}\left(aq\right)+{\text{4H}}^{+}\left(aq\right)\to {\text{MnO}}_2\left(aq\right)+{\text{2H}}_{\text{2}}\text{O}\left(l\right)$

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

${\text{MnO}}_4{}^{-}\left(aq\right)+{\text{4H}}^{+}\left(aq\right)+3{\text{e}}^{-}\to {\text{MnO}}_2\left(aq\right)+{\text{2H}}_{\text{2}}\text{O}\left(l\right)$

Step-6: Neutralize the all ${\text{H}}^{\text{+}}$ by adding hydroxide ions ${\text{OH}}^{-}$ on both sides.

Four hydroxide ions are added to both sides of the equation.

${\text{MnO}}_4{}^{-}\left(aq\right)+{\text{4H}}^{+}\left(aq\right)+4{\text{OH}}^{-}\left(aq\right)+3{\text{e}}^{-}\to {\text{MnO}}_2\left(aq\right)+{\text{2H}}_{\text{2}}\text{O}\left(l\right)+4{\text{OH}}^{-}\left(aq\right)$

Simplify the above equation by making the water of neutralized protons and balance out water molecules.

$\begin{array}{l}{\text{MnO}}_4{}^{-}\left(aq\right)+4{\text{H}}_{\text{2}}\text{O}\left(l\right)+3{\text{e}}^{-}\to {\text{MnO}}_2\left(aq\right)+{\text{2H}}_{\text{2}}\text{O}\left(l\right)+4{\text{OH}}^{-}\left(aq\right)\\ {\text{MnO}}_4{}^{-}\left(aq\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)+3{\text{e}}^{-}\to {\text{MnO}}_2\left(aq\right)+4{\text{OH}}^{-}\left(aq\right)\end{array}$

Step-7: Recheck the equation to be sure that it is perfectly balanced.

The equation is completely balanced and is shown below.

${\text{MnO}}_4{}^{-}\left(aq\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)+3{\text{e}}^{-}\to {\text{MnO}}_2\left(aq\right)+4{\text{OH}}^{-}\left(aq\right)$ …(1)

The oxidation state of sulfur in ${\text{S}}^{2-}$ is $-2$ comes from the charge on sulfur. The oxidation state of sulfur in $\text{S}$ is zero because it is elemental form.

The oxidation number of sulfur increases from ${\text{S}}^{2-}\to \text{S}$, therefore, it is an oxidation half-reaction.

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

${\text{S}}^{2-}\left(aq\right)\to \text{S}\left(s\right)$

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 sulfur is getting oxidized and their numbers of atoms are balanced on both sides.

${\text{S}}^{2-}\left(aq\right)\to \text{S}\left(s\right)$

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

${\text{S}}^{2-}\left(aq\right)\to \text{S}\left(s\right)$

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

${\text{S}}^{2-}\left(aq\right)\to \text{S}\left(s\right)$

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

${\text{S}}^{2-}\left(aq\right)\to \text{S}\left(s\right)$

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.

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

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

The equation is completely balanced and is shown below.

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

$\begin{array}{c}{\text{2MnO}}_4{}^{-}\left(aq\right)+4{\text{H}}_{\text{2}}\text{O}\left(l\right)+6{\text{e}}^{-}\to 2{\text{MnO}}_2\left(aq\right)+8{\text{OH}}^{-}\left(aq\right)\\ +\\ {\text{3S}}^{2-}\left(aq\right)\to 3\text{S}\left(s\right)+6{\text{e}}^{-}\end{array}$

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

${\text{2MnO}}_4{}^{-}\left(aq\right)+4{\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{3S}}^{2-}\left(aq\right)\to 2{\text{MnO}}_2\left(aq\right)+8{\text{OH}}^{-}\left(aq\right)+3\text{S}\left(s\right)$

Conclusion

The balanced equation of redox reaction is shown below.

${\text{2MnO}}_4{}^{-}\left(aq\right)+4{\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{3S}}^{2-}\left(aq\right)\to 2{\text{MnO}}_2\left(aq\right)+8{\text{OH}}^{-}\left(aq\right)+3\text{S}\left(s\right)$

Interpretation Introduction

(b)

Interpretation:

The balanced equation for the redox reaction, $\text{Cu}\left(s\right)+{\text{ClO}}^{-}\left(aq\right)\to {\text{Cu}}^{2+}\left(aq\right)+{\text{Cl}}^{-}\left(aq\right)$ in basic 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 26E

The balanced equation for the redox reaction, $\text{Cu}\left(s\right)+{\text{ClO}}^{-}\left(aq\right)\to {\text{Cu}}^{2+}\left(aq\right)+{\text{Cl}}^{-}\left(aq\right)$ in basic solution is shown below.

$\text{Cu}\left(s\right)+{\text{ClO}}^{-}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\to {\text{Cu}}^{2+}\left(aq\right)+{\text{Cl}}^{-}\left(aq\right)+2{\text{OH}}^{-}\left(aq\right)$

Explanation of Solution

The given redox reaction is shown below.

$\text{Cu}\left(s\right)+{\text{ClO}}^{-}\left(aq\right)\to {\text{Cu}}^{2+}\left(aq\right)+{\text{Cl}}^{-}\left(aq\right)$

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

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

The oxidation state of the chlorine in ${\text{ClO}}^{-}$ is calculated below.

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

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

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

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

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

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

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

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

The oxidation state of chlorine is $+3$ in ${\text{ClO}}_2{}^{-}$.

The oxidation number of chlorine is $-1$ in ${\text{Cl}}^{-}$ which comes from the charge on chlorine.

The oxidation number of chlorine decreases from ${\text{ClO}}^{-}\left(aq\right)\to {\text{Cl}}^{-}\left(aq\right)$ therefore, it is a reduction half-reaction.

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

${\text{ClO}}^{-}\left(aq\right)\to {\text{Cl}}^{-}\left(aq\right)$

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

${\text{ClO}}^{-}\left(aq\right)\to {\text{Cl}}^{-}\left(aq\right)$

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

${\text{ClO}}^{-}\left(aq\right)\to {\text{Cl}}^{-}\left(aq\right)$

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

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

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

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

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

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

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

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

Step-6: Neutralize the all ${\text{H}}^{\text{+}}$ by adding hydroxide ions ${\text{OH}}^{-}$ on both sides.

Two hydroxide ions are added to both sides of the equation.

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

Simplify the above equation by making water of neutralized protons and balance out water molecules.

$\begin{array}{c}{\text{ClO}}^{-}\left(aq\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)+2{\text{e}}^{-}\to {\text{Cl}}^{-}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)+2{\text{OH}}^{-}\left(aq\right)\\ {\text{ClO}}^{-}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)+2{\text{e}}^{-}\to {\text{Cl}}^{-}\left(aq\right)+2{\text{OH}}^{-}\left(aq\right)\end{array}$

Step-7: Recheck the equation to be sure that it is perfectly balanced.

The equation is completely balanced and is shown below.

${\text{ClO}}^{-}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)+2{\text{e}}^{-}\to {\text{Cl}}^{-}\left(aq\right)+2{\text{OH}}^{-}\left(aq\right)$ …(1)

The oxidation number of copper is zero in $\text{Cu}\left(s\right)$ because of elemental form.

The oxidation number of copper is $+2$ in ${\text{Cu}}^{2+}\left(aq\right)$ because of charge on the metal ion.

The oxidation of copper increases from $\text{Cu}\left(s\right)\to {\text{Cu}}^{2+}\left(aq\right)$, therefore, it is an oxidation half-reaction.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

The equation is completely balanced and is shown below.

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

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

Add equation (1) and equation (2).

$\begin{array}{c}\text{Cu}\left(s\right)\to {\text{Cu}}^{2+}\left(aq\right)+2{\text{e}}^{-}\\ +\\ {\text{ClO}}^{-}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)+2{\text{e}}^{-}\to {\text{Cl}}^{-}\left(aq\right)+2{\text{OH}}^{-}\left(aq\right)\end{array}$

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

$\text{Cu}\left(s\right)+{\text{ClO}}^{-}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\to {\text{Cu}}^{2+}\left(aq\right)+{\text{Cl}}^{-}\left(aq\right)+2{\text{OH}}^{-}\left(aq\right)$

Conclusion

The balanced equation of redox reaction is shown below.

$\text{Cu}\left(s\right)+{\text{ClO}}^{-}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\to {\text{Cu}}^{2+}\left(aq\right)+{\text{Cl}}^{-}\left(aq\right)+2{\text{OH}}^{-}\left(aq\right)$