# Question 0075f

Oct 15, 2015

${\text{H"_text((aq])^(+) + "HNO"_text(2(aq]) + 1e^(-) -> "NO"_text((g]) + "H"_2"O}}_{\textrm{\left(l\right]}}$

#### Explanation:

So, you knwo that nitrous acid, ${\text{HNO}}_{2}$, can be reduced to nitrogen gas, $\text{NO}$, in acidic solution.

${\text{HNO"_text(2(aq]) -> "NO}}_{\textrm{\left(g\right]}}$

Assign oxidation numbers to all the atoms that take part in the reaction

${\stackrel{\textcolor{b l u e}{+ 1}}{\text{H")stackrel(color(blue)(+3))("N")stackrel(color(blue)(-2))("O")_text(2(aq]) -> stackrel(color(blue)(+2))("N")stackrel(color(blue)(-2))("O}}}_{\textrm{\left(g\right]}}$

Nitrogen is being reduced from an oxidation state of $\textcolor{b l u e}{+ 3}$ on the reactants's side, to an oxidation state of $\textcolor{b l u e}{+ 2}$ on the products' side.

This means that you can write

stackrel(color(blue)(+3))("N")"O"_2^(-) + 1e^(-) -> stackrel(color(blue)(+2))("N")"O"

Each nitrogen atom gain one electron during the reduction half-reaction.

Notice that the oxygen atoms are not balanced. Since you're in acidic solution, you can use water to balance the oxygen atoms and protons to balance the hydrogen atoms.

This will give you

stackrel(color(blue)(+3))("N")"O"_2^(-) + 1e^(-) -> stackrel(color(blue)(+2))("N")"O" + "H"_2"O"#

Two oxygen atoms on the left-hand side, two on the right-hand side. Now balance the hydrogen atoms

$2 \text{H"^(+) + stackrel(color(blue)(+3))("N")"O"_2^(-) + 1e^(-) -> stackrel(color(blue)(+2))("N")"O" + "H"_2"O}$

And there you have it, the reduction half-reaction looks like this

${\text{H"_text((aq])^(+) + "HNO"_text(2(aq]) + 1e^(-) -> "NO"_text((g]) + "H"_2"O}}_{\textrm{\left(l\right]}}$