Chapter 19, Problem 45E

Interpretation Introduction

Interpretation:

The oxidizer and reducer with oxidized and reduced products are to be identified. The oxidation and reduction half-reaction equations are to be stated assuming reaction takes place in acidic solution. The balanced redox equation 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 45E

The oxidizer is ${\text{NO}}_3^{-}$ and reducer is $\text{Sn}$ and their corresponding reduced and oxidized product are ${\text{NO}}_2$ and ${\text{H}}_{\text{2}}{\text{SnO}}_{\text{3}}$.

The oxidation half-reaction equation is shown below.

$\text{Sn}+3{\text{H}}_{\text{2}}\text{O}\to {\text{H}}_{\text{2}}{\text{SnO}}_{\text{3}}+4{\text{H}}^{\text{+}}+4{\text{e}}^{-}$

The reduction half-reaction equation is shown below.

${\text{NO}}_3^{-}+2{\text{H}}^{+}+{\text{e}}^{-}\to {\text{NO}}_2+{\text{H}}_{\text{2}}\text{O}$

The balanced redox equation is shown below.

$4{\text{NO}}_3^{-}+4{\text{H}}^{+}+\text{Sn}\to 4{\text{NO}}_2+{\text{H}}_{\text{2}}{\text{O+H}}_{\text{2}}{\text{SnO}}_{\text{3}}$

Explanation of Solution

The given redox reaction equation to be balanced is shown below.

$\text{Sn}+{\text{NO}}_3^{-}\to {\text{H}}_{\text{2}}{\text{SnO}}_{\text{3}}+{\text{NO}}_2$

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 their number of atoms of an 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 ${\text{NO}}_2$ 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 their number of atoms of an element.

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

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

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

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

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

The oxidation state of nitrogen in ${\text{NO}}_2$ is $+4$.

The oxidation state of tin in ${\text{H}}_{\text{2}}{\text{SnO}}_{\text{3}}$ is calculated below.

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

$\begin{array}{l}\text{H}\text{Sn }\text{O}\\ \text{+1}\text{n }-\text{2}\end{array}$

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

$\begin{array}{l}{\text{H}}_2\text{Sn }{\text{O}}_3\\ 2\left(\text{+1}\right)\text{n 3}\left(-\text{2}\right)\end{array}$

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

$\begin{array}{l}{\text{H}}_2\text{Sn }{\text{O}}_3\\ 2\left(\text{+1}\right)\text{n 3}\left(-\text{2}\right)=0\end{array}$

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

$\begin{array}{rll}2\left(\text{+1}\right)+\text{n }+\text{ 3}\left(-\text{2}\right)& =& 0\\ \text{2+n-6}& =& 0\\ \text{n}& =& +4\end{array}$

The oxidation state of tin in ${\text{H}}_{\text{2}}{\text{SnO}}_{\text{3}}$ is $+4$.

The oxidation state of tin in $\text{Sn}$ is $0$ which comes from the zero charge on the tin.

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

$\text{Sn}\to {\text{H}}_{\text{2}}{\text{SnO}}_{\text{3}}$

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

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

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

$\text{Sn}\to {\text{H}}_{\text{2}}{\text{SnO}}_{\text{3}}$

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

$\text{Sn}\to {\text{H}}_{\text{2}}{\text{SnO}}_{\text{3}}$

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

Oxygen atoms are balanced by adding three water molecules to the left-hand side of the equation.

$\text{Sn}+3{\text{H}}_{\text{2}}\text{O}\to {\text{H}}_{\text{2}}{\text{SnO}}_{\text{3}}$

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 the four ${\text{H}}^{\text{+}}$ on the right-hand side of the equation.

$\text{Sn}+3{\text{H}}_{\text{2}}\text{O}\to {\text{H}}_{\text{2}}{\text{SnO}}_{\text{3}}+4{\text{H}}^{\text{+}}$

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

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

$\text{Sn}+3{\text{H}}_{\text{2}}\text{O}\to {\text{H}}_{\text{2}}{\text{SnO}}_{\text{3}}+4{\text{H}}^{\text{+}}+4{\text{e}}^{-}$

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

The equation is completely balanced and is shown below.

$\text{Sn}+3{\text{H}}_{\text{2}}\text{O}\to {\text{H}}_{\text{2}}{\text{SnO}}_{\text{3}}+4{\text{H}}^{\text{+}}+4{\text{e}}^{-}$…(1)

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

${\text{NO}}_3^{-}\to {\text{NO}}_2$

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 is balanced on both sides.

${\text{NO}}_3^{-}\to {\text{NO}}_2$

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

${\text{NO}}_3^{-}\to {\text{NO}}_2$

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

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

${\text{NO}}_3^{-}\to {\text{NO}}_2+{\text{H}}_{\text{2}}\text{O}$

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.

${\text{NO}}_3^{-}+2{\text{H}}^{+}\to {\text{NO}}_2+{\text{H}}_{\text{2}}\text{O}$

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

The charge is balanced by adding one electron on the left-hand side of the equation.

${\text{NO}}_3^{-}+2{\text{H}}^{+}+{\text{e}}^{-}\to {\text{NO}}_2+{\text{H}}_{\text{2}}\text{O}$

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

The equation is completely balanced and is shown below.

${\text{NO}}_3^{-}+2{\text{H}}^{+}+{\text{e}}^{-}\to {\text{NO}}_2+{\text{H}}_{\text{2}}\text{O}$…(2)

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

Multiply equation (2) by four in order to cancel out the number of electrons.

$\begin{array}{l}4\times \left({\text{NO}}_3^{-}+2{\text{H}}^{+}+{\text{e}}^{-}\to {\text{NO}}_2+{\text{H}}_{\text{2}}\text{O}\right)\\ {\text{4NO}}_3^{-}+8{\text{H}}^{+}+4{\text{e}}^{-}\to 4{\text{NO}}_2+4{\text{H}}_{\text{2}}\text{O}\end{array}$…(3)

Add equation (3) and (1) to get a balanced redox equation as shown below.

$\begin{array}{l}\text{Reduction}{\text{4NO}}_3^{-}+8{\text{H}}^{+}+4{\text{e}}^{-}\to 4{\text{NO}}_2+4{\text{H}}_{\text{2}}\text{O}\\ \underset{\_}{\text{Oxidation}\text{Sn}+3{\text{H}}_{\text{2}}\text{O}\to {\text{H}}_{\text{2}}{\text{SnO}}_{\text{3}}+4{\text{H}}^{\text{+}}+4{\text{e}}^{-}}\\ \text{Redox}4{\text{NO}}_3^{-}+4{\text{H}}^{+}+\text{Sn}\to 4{\text{NO}}_2+{\text{H}}_{\text{2}}{\text{O+H}}_{\text{2}}{\text{SnO}}_{\text{3}}\end{array}$

The common things on both sides of the equation canceled out to give the balanced redox equation.

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

$4{\text{NO}}_3^{-}+4{\text{H}}^{+}+\text{Sn}\to 4{\text{NO}}_2+{\text{H}}_{\text{2}}{\text{O+H}}_{\text{2}}{\text{SnO}}_{\text{3}}$

Conclusion

The oxidizer and reducer with oxidized and reduced products, oxidation and reduction half-reaction equations, and balanced redox equation are rightfully stated above.