Chapter 14, Problem 59QP

Denitrification occurs when nitrogen in the soil is lost to the atmosphere. One way in which denitrification can occur in acidic soils rich in plant matter is described by the following equation:

${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\left(aq\right)+{\text{NO}}_3^{-}\left(aq\right)\to {\text{CO}}_{\text{2}}\left(g\right)+{\text{N}}_{\text{2}}\left(g\right)$

Balance this equation, adding ${\text{H}}^{+}\left(aq\right){\text{ and H}}_{\text{2}}\text{O}\left(l\right)$ as necessary.

Interpretation Introduction

Interpretation:

The given redox reaction is to be balanced.

Concept Introduction:

Redox reaction is a type of chemical reaction in which transfer of electrons takes place in between two species, one species loses electrons (undergo oxidation) and another species gains electrons (undergo reduction). Therefore, the oxidation and reduction processes occur simultaneously in a redox reaction.

Explanation of Solution

The half-cell reactions for the given chemical equations can be obtained by writing the oxidation numbers of each species present in the given reaction.

$\begin{array}{l}0+5-2+4-20\\ {\text{C}}_6{\text{H}}_{12}{\text{O}}_6\left(aq\right)+{\text{NO}}_3^{-}\left(aq\right)\to {\text{CO}}_2\left(g\right)+{\text{N}}_2\left(g\right)\end{array}$

According to the above equation, the oxidation number of nitrogen atom changes from $+5$ to $0$ and the oxidation number of carbon atom changes from $0$ to $+4$ , therefore, nitrogen is being reduced and carbon is being oxidized.

On the basis of this, the oxidation (anode) and reduction half-cell reactions (cathode) can be written as,

$\begin{array}{c}\text{Anode}:{\text{C}}_6{\text{H}}_{12}{\text{O}}_6\left(aq\right)\to {\text{CO}}_2\left(g\right)\\ \text{Cathode:}{\text{NO}}_3^{-}\left(aq\right)\to {\text{N}}_2\left(g\right)\end{array}$

Now, balance the reaction occurring at anode by using the stepwise approach as,

$\begin{array}{c}{\text{C}}_6{\text{H}}_{12}{\text{O}}_6\left(aq\right)\to 6{\text{CO}}_2\left(g\right)\\ {\text{C}}_6{\text{H}}_{12}{\text{O}}_6\left(aq\right)\to 6{\text{CO}}_2\left(g\right)+24{\text{e}}^{-}\\ {\text{C}}_6{\text{H}}_{12}{\text{O}}_6\left(aq\right)\to 6{\text{CO}}_2\left(g\right)+24{\text{H}}^{+}+24{\text{e}}^{-}\\ {\text{C}}_6{\text{H}}_{12}{\text{O}}_6\left(aq\right)+6{\text{H}}_2\text{O}\left(l\right)\to 6{\text{CO}}_2\left(g\right)+24{\text{H}}^{+}+24{\text{e}}^{-}\end{array}$

Now, balance the reaction occurring at cathode by using the stepwise approach as,

$\begin{array}{c}{\text{2NO}}_3^{-}\left(aq\right)\to {\text{N}}_2\left(g\right)\\ {\text{2NO}}_3^{-}\left(aq\right)+10{\text{e}}^{-}\to {\text{N}}_2\left(g\right)\\ {\text{2NO}}_3^{-}\left(aq\right)+12{\text{H}}^{+}\left(aq\right)+10{\text{e}}^{-}\to {\text{N}}_2\left(g\right)\\ {\text{2NO}}_3^{-}\left(aq\right)+12{\text{H}}^{+}\left(aq\right)+10{\text{e}}^{-}\to {\text{N}}_2\left(g\right)+6{\text{H}}_2\text{O}\left(l\right)\end{array}$

Multiply the reaction occurring at anode with $5$ and cathode with $12$ .

$\begin{array}{l}5\left({\text{C}}_6{\text{H}}_{12}{\text{O}}_6\left(aq\right)+6{\text{H}}_2\text{O}\left(l\right)\to 6{\text{CO}}_2\left(g\right)+24{\text{H}}^{+}+24{\text{e}}^{-}\right)\\ {\text{5C}}_6{\text{H}}_{12}{\text{O}}_6\left(aq\right)+30{\text{H}}_2\text{O}\left(l\right)\to 30{\text{CO}}_2\left(g\right)+120{\text{H}}^{+}+120{\text{e}}^{-}\\ 12\left({\text{2NO}}_3^{-}\left(aq\right)+12{\text{H}}^{+}\left(aq\right)+10{\text{e}}^{-}\to {\text{N}}_2\left(g\right)+6{\text{H}}_2\text{O}\left(l\right)\right)\\ {\text{24NO}}_3^{-}\left(aq\right)+144{\text{H}}^{+}\left(aq\right)+120{\text{e}}^{-}\to 12{\text{N}}_2\left(g\right)+72{\text{H}}_2\text{O}\left(l\right)\end{array}$

Add the reactions occurring at cathode and anode in order to obtain the overall reaction.

$\begin{array}{c}{\text{5C}}_6{\text{H}}_{12}{\text{O}}_6\left(aq\right)+30{\text{H}}_2\text{O}\left(l\right)\to 30{\text{CO}}_2\left(g\right)+120{\text{H}}^{+}\left(aq\right)+\cancel{120{\text{e}}^{-}}\\ {\text{24NO}}_3^{-}\left(aq\right)+144{\text{H}}^{+}\left(aq\right)+\cancel{120{\text{e}}^{-}}\to 12{\text{N}}_2\left(g\right)+72{\text{H}}_2\text{O}\left(l\right)\\ \underset{¯}{\overline{{\text{5C}}_6{\text{H}}_{12}{\text{O}}_6\left(aq\right)+30{\text{H}}_2\text{O}\left(l\right)+{\text{24NO}}_3^{-}\left(aq\right)+144{\text{H}}^{+}\left(aq\right)\to 30{\text{CO}}_2\left(g\right)+120{\text{H}}^{+}\left(aq\right)+12{\text{N}}_2\left(g\right)+72{\text{H}}_2\text{O}\left(l\right)}}\end{array}$

The overall reaction can be obtained by balancing the hydrogen ions and water molecules in the final equation as,

${\text{5C}}_6{\text{H}}_{12}{\text{O}}_6\left(aq\right)+{\text{24NO}}_3^{-}\left(aq\right)+24{\text{H}}^{+}\left(aq\right)\to 30{\text{CO}}_2\left(g\right)+12{\text{N}}_2\left(g\right)+42{\text{H}}_2\text{O}\left(l\right)$

Conclusion

The overall balanced chemical equation is,

${\text{5C}}_6{\text{H}}_{12}{\text{O}}_6\left(aq\right)+{\text{24NO}}_3^{-}\left(aq\right)+24{\text{H}}^{+}\left(aq\right)\to 30{\text{CO}}_2\left(g\right)+12{\text{N}}_2\left(g\right)+42{\text{H}}_2\text{O}\left(l\right)$