Question

Why and how do we get ${\text{N}}{{\text{H}}_{\text{4}}}{\text{N}}{{\text{O}}_{\text{3}}}$ when reacting very dilute ${\text{HN}}{{\text{O}}_{\text{3}}}$ with ${\text{Zn}}$?

Answer 1

Hint: We know that zinc belongs to the d-block of the periodic table. Zinc exists in $ + 2$ oxidation state as a ${\text{Z}}{{\text{n}}^{2 + }}$ion. ${\text{HN}}{{\text{O}}_{\text{3}}}$ i.e. nitric acid is a strong oxidising agent. Depending on its concentration, nitric acid is dilute or concentrated. In reaction with zinc, nitric acid gives nitrogenous compounds of ammonia.

Complete answer:
We know that zinc belongs to the d-block of the periodic table. Zinc exists in $ + 2$ oxidation state as a ${\text{Z}}{{\text{n}}^{2 + }}$ion. ${\text{HN}}{{\text{O}}_{\text{3}}}$ i.e. nitric acid is a strong oxidising agent. Depending on its concentration, nitric acid is dilute or concentrated. In reaction with zinc, nitric acid gives nitrogenous compounds of ammonia.
When zinc metal reacts with nitric acid, the reaction produces hydrogen gas. This hydrogen gas is immediately oxidised to water. The reactions are as follows:
${\text{Z}}{{\text{n}}_{\left( {\text{s}} \right)}} + {\text{2HN}}{{\text{O}}_{{\text{3}}\left( {{\text{aq}}} \right)}} \to {\text{Zn}}{\left( {{\text{N}}{{\text{O}}_{\text{3}}}} \right)_{{\text{2}}\left( {{\text{aq}}} \right)}} + {{\text{H}}_{{\text{2}}\left( {\text{g}} \right)}}$
${{\text{H}}_{{\text{2}}\left( {\text{g}} \right)}} \to {\text{2H}}_{\left( {{\text{aq}}} \right)}^ + + {\text{2}}{{\text{e}}^ - }$
Simultaneously nitric acid is reduced to ammonia. More dilute the nitric acid more is its reduction. The reaction is as follows:
${\text{HN}}{{\text{O}}_{{\text{3}}\left( {{\text{aq}}} \right)}} + {{4 \times }}\left( {{\text{2H}}_{\left( {{\text{aq}}} \right)}^ + + {\text{2}}{{\text{e}}^ - }} \right) \to {\text{N}}{{\text{H}}_{{\text{3}}\left( {{\text{aq}}} \right)}} + {\text{3}}{{\text{H}}_{\text{2}}}{{\text{O}}_{\left( {\text{l}} \right)}}$
Nitrogen is thus reduced from $ + 5$ oxidation state to $ - 3$ oxidation state.
The acid then neutralizes ammonia to ammonium nitrate. The reaction is as follows:
${\text{N}}{{\text{H}}_{{\text{3}}\left( {{\text{aq}}} \right)}} + {\text{HN}}{{\text{O}}_{{\text{3}}\left( {{\text{aq}}} \right)}} \to {\text{N}}{{\text{H}}_{\text{4}}}{\text{N}}{{\text{O}}_{{\text{3}}\left( {{\text{aq}}} \right)}}$
Putting all the reactions together we get the reaction as follows:
${\text{4Z}}{{\text{n}}_{\left( {\text{s}} \right)}} + {\text{10HN}}{{\text{O}}_{{\text{3}}\left( {{\text{aq}}} \right)}} \to {\text{4Zn}}{\left( {{\text{N}}{{\text{O}}_{\text{3}}}} \right)_{{\text{2}}\left( {{\text{aq}}} \right)}} + {\text{N}}{{\text{H}}_{\text{4}}}{\text{N}}{{\text{O}}_{{\text{3}}\left( {{\text{aq}}} \right)}} + {\text{3}}{{\text{H}}_{\text{2}}}{{\text{O}}_{\left( {\text{l}} \right)}}$
Thus, we get ${\text{N}}{{\text{H}}_{\text{4}}}{\text{N}}{{\text{O}}_{\text{3}}}$ when reacting very dilute ${\text{HN}}{{\text{O}}_{\text{3}}}$ with ${\text{Zn}}$.

Note: The reaction of nitric acid with zinc is different depending on the reaction conditions:
Warm and dilute ${\text{HN}}{{\text{O}}_{\text{3}}}$ with ${\text{Zn}}$ gives zinc nitrate and nitrous oxide.
Cold and very dilute ${\text{HN}}{{\text{O}}_{\text{3}}}$ with ${\text{Zn}}$ gives zinc nitrate and ammonium nitrate.
Hot and concentrated ${\text{HN}}{{\text{O}}_{\text{3}}}$ with ${\text{Zn}}$ gives zinc nitrate and nitric oxide.