Chapter 17, Problem 24E

Interpretation Introduction

(a)

Interpretation:

The balanced half-reaction in basic solution for the reaction, $\text{Ni}{\left(\text{OH}\right)}_{\text{2}}\left(s\right)\to {\text{NiO}}_{\text{2}}\left(s\right)$, 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 24E

The balanced half-reaction following in basic solution is shown below.

$\text{Ni}{\left(\text{OH}\right)}_{\text{2}}\left(s\right)+2{\text{OH}}^{-}\left(aq\right)\to {\text{NiO}}_{\text{2}}\left(s\right)+2{\text{H}}_2\text{O}\left(l\right)+2{\text{e}}^{-}$

Explanation of Solution

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

$\text{Ni}{\left(\text{OH}\right)}_{\text{2}}\left(s\right)\to {\text{NiO}}_{\text{2}}\left(s\right)$

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

The oxidation state of Nickel in $\text{Ni}{\left(\text{OH}\right)}_{\text{2}}$ is calculated below.

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

$\begin{array}{l}{\text{Ni O}}_2{\text{ H}}_2\\ \text{n }-2\text{ +1}\end{array}$

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

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

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

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

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

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

The oxidation state of nickel in $\text{Ni}{\left(\text{OH}\right)}_{\text{2}}$ is $+2$.

The oxidation state of oxygen in ${\text{NiO}}_{\text{2}}$ is calculated below.

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

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

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

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

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

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

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

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

The oxidation state of nickel in ${\text{NiO}}_{\text{2}}$ is $+4$.

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

$\text{Ni}{\left(\text{OH}\right)}_{\text{2}}\left(s\right)\to {\text{NiO}}_{\text{2}}\left(s\right)$

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

$\text{Ni}{\left(\text{OH}\right)}_{\text{2}}\left(s\right)\to {\text{NiO}}_{\text{2}}\left(s\right)$

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

Oxygen atoms are already balanced on both sides.

$\text{Ni}{\left(\text{OH}\right)}_{\text{2}}\left(s\right)\to {\text{NiO}}_{\text{2}}\left(s\right)$

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

Add two ${\text{H}}^{\text{+}}$ ions to the right-hand side of the equation.

$\text{Ni}{\left(\text{OH}\right)}_{\text{2}}\left(s\right)\to {\text{NiO}}_{\text{2}}\left(s\right)+2{\text{H}}^{+}\left(aq\right)$

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.

$\text{Ni}{\left(\text{OH}\right)}_{\text{2}}\left(s\right)\to {\text{NiO}}_{\text{2}}\left(s\right)+2{\text{H}}^{+}\left(aq\right)+2{\text{e}}^{-}$

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

Two hydroxide ions are added to both sides of equation.

$\begin{array}{l}\text{Ni}{\left(\text{OH}\right)}_{\text{2}}\left(s\right)+2{\text{OH}}^{-}\left(aq\right)\to {\text{NiO}}_{\text{2}}\left(s\right)+2{\text{H}}^{+}\left(aq\right)+2{\text{OH}}^{-}\left(aq\right)+2{\text{e}}^{-}\\ \text{Ni}{\left(\text{OH}\right)}_{\text{2}}\left(s\right)+2{\text{OH}}^{-}\left(aq\right)\to {\text{NiO}}_{\text{2}}\left(s\right)+2{\text{H}}_2\text{O}\left(l\right)+2{\text{e}}^{-}\end{array}$

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

The equation is completely balanced and is shown below.

$\text{Ni}{\left(\text{OH}\right)}_{\text{2}}\left(s\right)+2{\text{OH}}^{-}\left(aq\right)\to {\text{NiO}}_{\text{2}}\left(s\right)+2{\text{H}}_2\text{O}\left(l\right)+2{\text{e}}^{-}$

Conclusion

The balanced half-reaction is shown below.

$\text{Ni}{\left(\text{OH}\right)}_{\text{2}}\left(s\right)+2{\text{OH}}^{-}\left(aq\right)\to {\text{NiO}}_{\text{2}}\left(s\right)+2{\text{H}}_2\text{O}\left(l\right)+2{\text{e}}^{-}$

Interpretation Introduction

(b)

Interpretation:

The balanced half-reaction in basic solution for the reaction, ${\text{NO}}_2^{-}\left(aq\right)\to {\text{N}}_{\text{2}}\text{O}\left(g\right)$, 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 24E

The balanced half-reaction following in basic solution is shown below.

${\text{2NO}}_2^{-}\left(aq\right)+3{\text{H}}_{\text{2}}\text{O}\left(l\right)+4{\text{e}}^{-}\to {\text{N}}_{\text{2}}\text{O}\left(g\right)+6{\text{OH}}^{-}\left(aq\right)$

Explanation of Solution

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

${\text{NO}}_2^{-}\left(aq\right)\to {\text{N}}_{\text{2}}\text{O}\left(g\right)$

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

The oxidation state of nitrogen in ${\text{NO}}_2^{-}$ is calculated below.

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

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

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

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

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

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

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

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

The oxidation state of nitrogen in ${\text{NO}}_2^{-}$ is $+3$.

The oxidation state of nitrogen in ${\text{N}}_{\text{2}}\text{O}$ is calculated below.

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

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

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

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

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

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

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

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

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 nitrogen in ${\text{N}}_{\text{2}}\text{O}$ is $+1$.

The oxidation number of nitrogen 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 nitrogen is getting reduced and the number of atoms of that is not balanced on both sides. Balance them by multiplying ${\text{NO}}_2^{-}$ by two on the left-hand side of the equation.

${\text{2NO}}_2^{-}\left(aq\right)\to {\text{N}}_{\text{2}}\text{O}\left(g\right)$

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

${\text{2NO}}_2^{-}\left(aq\right)\to {\text{N}}_{\text{2}}\text{O}\left(g\right)$

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

Oxygen atoms are already balanced on both sides.

${\text{2NO}}_2^{-}\left(aq\right)\to {\text{N}}_{\text{2}}\text{O}\left(g\right)+3{\text{H}}_{\text{2}}\text{O}\left(l\right)$

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

Add six ${\text{H}}^{\text{+}}$ to the left-hand side of the equation.

${\text{2NO}}_2^{-}\left(aq\right)+6{\text{H}}^{+}\left(aq\right)\to {\text{N}}_{\text{2}}\text{O}\left(g\right)+3{\text{H}}_{\text{2}}\text{O}\left(l\right)$

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

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

${\text{2NO}}_2^{-}\left(aq\right)+6{\text{H}}^{+}\left(aq\right)+4{\text{e}}^{-}\to {\text{N}}_{\text{2}}\text{O}\left(g\right)+3{\text{H}}_{\text{2}}\text{O}\left(l\right)$

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

Two hydroxide ions are added to both sides of equation.

${\text{2NO}}_2^{-}\left(aq\right)+6{\text{H}}^{+}\left(aq\right)+6{\text{OH}}^{-}\left(aq\right)+4{\text{e}}^{-}\to {\text{N}}_{\text{2}}\text{O}\left(g\right)+3{\text{H}}_{\text{2}}\text{O}\left(l\right)+6{\text{OH}}^{-}\left(aq\right)$

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

$\begin{array}{l}{\text{2NO}}_2^{-}\left(aq\right)+6{\text{H}}_{\text{2}}\text{O}\left(l\right)+4{\text{e}}^{-}\to {\text{N}}_{\text{2}}\text{O}\left(g\right)+3{\text{H}}_{\text{2}}\text{O}\left(l\right)+6{\text{OH}}^{-}\left(aq\right)\\ {\text{2NO}}_2^{-}\left(aq\right)+3{\text{H}}_{\text{2}}\text{O}\left(l\right)+4{\text{e}}^{-}\to {\text{N}}_{\text{2}}\text{O}\left(g\right)+6{\text{OH}}^{-}\left(aq\right)\end{array}$

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

The equation is completely balanced and is shown below.

${\text{2NO}}_2^{-}\left(aq\right)+3{\text{H}}_{\text{2}}\text{O}\left(l\right)+4{\text{e}}^{-}\to {\text{N}}_{\text{2}}\text{O}\left(g\right)+6{\text{OH}}^{-}\left(aq\right)$

Conclusion

The balanced half-reaction is shown below.

${\text{2NO}}_2^{-}\left(aq\right)+3{\text{H}}_{\text{2}}\text{O}\left(l\right)+4{\text{e}}^{-}\to {\text{N}}_{\text{2}}\text{O}\left(g\right)+6{\text{OH}}^{-}\left(aq\right)$