Chapter 17, Problem 21E

Interpretation Introduction

(a)

Interpretation:

The balanced equation for the redox reaction,

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

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 21E

The balanced equation for the redox reaction,

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

using oxidation number method is shown below.

${\text{2MnO}}_4^{-}\left(aq\right)+10{\text{I}}^{-}\left(aq\right)+16{\text{H}}^{+}\left(aq\right)\stackrel{}{\to }{\text{2Mn}}^{2+}\left(aq\right)+5{\text{I}}_{\text{2}}\left(s\right)+8{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Explanation of Solution

The given redox reaction is shown below.

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

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

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

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

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

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

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

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 ${\text{Mn}}^{2+}$ is $+2$ directly coming from the charge on manganese.

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

The oxidation number of manganese is decreased from reactant to product and for iodine, 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 manganese is getting reduced and iodine is getting oxidized. The number of atoms of iodine is not balanced on both sides of the equation. Balance them by multiplying the ${\text{I}}^{-}$ by two on the left-hand side.

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

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

The iodine is getting oxidized therefore, number of electrons lost by iodide ion is two. The manganese is getting reduced and number of electron gained by manganese is five.

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 ${\text{I}}^{-}$ by five and ${\text{MnO}}_4{}^{-}$ by two and then balance the manganese and iodine atoms on both sides as shown below.

${\text{2MnO}}_4^{-}\left(aq\right)+10{\text{I}}^{-}\left(aq\right)+{\text{H}}^{+}\left(aq\right)\stackrel{}{\to }{\text{2Mn}}^{2+}\left(aq\right)+5{\text{I}}_{\text{2}}\left(s\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

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

Multiply the ${\text{H}}_{\text{2}}\text{O}$ by eight and ${\text{H}}^{+}$ by sixteen in order to balance hydrogen and oxygen elements.

${\text{2MnO}}_4^{-}\left(aq\right)+10{\text{I}}^{-}\left(aq\right)+16{\text{H}}^{+}\left(aq\right)\stackrel{}{\to }{\text{2Mn}}^{2+}\left(aq\right)+5{\text{I}}_{\text{2}}\left(s\right)+8{\text{H}}_{\text{2}}\text{O}\left(l\right)$

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

Conclusion

The balanced equation for the redox reaction is shown below.

${\text{2MnO}}_4^{-}\left(aq\right)+10{\text{I}}^{-}\left(aq\right)+16{\text{H}}^{+}\left(aq\right)\stackrel{}{\to }{\text{2Mn}}^{2+}\left(aq\right)+5{\text{I}}_{\text{2}}\left(s\right)+8{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Interpretation Introduction

(b)

Interpretation:

The balanced equation for the redox reaction, $\text{Cu}\left(s\right)+{\text{H}}^{+}\left(aq\right)+{\text{SO}}_4^{2-}\left(aq\right)\stackrel{}{\to }{\text{Cu}}^{2+}\left(aq\right)+{\text{SO}}_2\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$, 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 21E

The balanced equation for the redox reaction, $\text{Cu}\left(s\right)+{\text{H}}^{+}\left(aq\right)+{\text{SO}}_4^{2-}\left(aq\right)\stackrel{}{\to }{\text{Cu}}^{2+}\left(aq\right)+{\text{SO}}_2\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$, using oxidation number method is shown below.

$\text{Cu}\left(s\right)+4{\text{H}}^{+}\left(aq\right)+{\text{SO}}_4^{2-}\left(aq\right)\stackrel{}{\to }{\text{Cu}}^{2+}\left(aq\right)+{\text{SO}}_2\left(g\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Explanation of Solution

The given redox reaction is shown below.

$\text{Cu}\left(s\right)+{\text{H}}^{+}\left(aq\right)+{\text{SO}}_4^{2-}\left(aq\right)\stackrel{}{\to }{\text{Cu}}^{2+}\left(aq\right)+{\text{SO}}_2\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

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

The oxidation state of manganese in ${\text{SO}}_4^{2-}$ is calculated below.

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

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

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

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

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

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

Calculate the value of n by simplifying the equation.

$\begin{array}{rll}\text{n}+\text{4}\left(-\text{2}\right)& =& -2\\ \text{n}+\text{(}-\text{8)}& =& -2\\ \text{n}& =& -2+8\\ \text{n}& =& +6\end{array}$

The oxidation state of sulfur in ${\text{SO}}_4^{2-}$ is $+6$.

The oxidation state of manganese in ${\text{SO}}_2$ is calculated below.

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

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

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

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

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

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

Calculate the value of n by simplifying the equation.

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

The oxidation state of sulfur in ${\text{SO}}_2$ is $+4$.

The oxidation state of copper in ${\text{Cu}}^{2+}$ is $+2$ directly coming from the charge on copper. The oxidation state of copper is zero in $\text{Cu}$ because of the elemental form.

The oxidation number of sulfur is decreased from reactant to product and for copper, 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 sulfur is getting reduced and copper is getting oxidized. The number of atoms of both copper and sulfur is balanced on both sides as shown below.

$\text{Cu}\left(s\right)+{\text{H}}^{+}\left(aq\right)+{\text{SO}}_4^{2-}\left(aq\right)\stackrel{}{\to }{\text{Cu}}^{2+}\left(aq\right)+{\text{SO}}_2\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

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

The copper is getting oxidized and number of electrons lost by copper is two. The sulfur is getting reduced and number of electron gained by sulfur is two.

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

The number of electrons lost or gained are balanced as equal number of electrons are lost and gained as shown below.

$\text{Cu}\left(s\right)+{\text{H}}^{+}\left(aq\right)+{\text{SO}}_4^{2-}\left(aq\right)\stackrel{}{\to }{\text{Cu}}^{2+}\left(aq\right)+{\text{SO}}_2\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

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

Multiply the ${\text{H}}_{\text{2}}\text{O}$ by two and ${\text{H}}^{+}$ by four in order to balance hydrogen and oxygen elements.

$\text{Cu}\left(s\right)+4{\text{H}}^{+}\left(aq\right)+{\text{SO}}_4^{2-}\left(aq\right)\stackrel{}{\to }{\text{Cu}}^{2+}\left(aq\right)+{\text{SO}}_2\left(g\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

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

Conclusion

The balanced equation for the redox reaction is shown below.

$\text{Cu}\left(s\right)+4{\text{H}}^{+}\left(aq\right)+{\text{SO}}_4^{2-}\left(aq\right)\stackrel{}{\to }{\text{Cu}}^{2+}\left(aq\right)+{\text{SO}}_2\left(g\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$