# Question #33342

Jan 19, 2015

$N {H}_{3} + {O}_{2} \to N {O}_{2} + {H}_{2} O$

Now, let's focus on assigning oxidation numbers for all atoms involved in the reaction

${N}^{- 3} {H}_{3}^{+ 1} + {O}_{2}^{0} \to {N}^{+ 4} {O}_{2}^{- 2} + {H}_{2}^{+ 1} {O}^{- 2}$

Notice that nitrogen went from an oxidation number (ON) of -3 on the reactants' side, to an ON of +4 on the products' side, while oxygen went from an ON of 0 on the reactants' side, to an ON of -2 on the products' side. The half-reactions look like this

${N}^{- 3} \to {N}^{+ 4} + 7 {e}^{-}$
${O}_{2}^{0} + 4 {e}^{-} \to {O}_{2}^{2 -}$

Since there are two oxygen atoms on the reactants' side, and each oxygen atom gained $2 {e}^{-}$, oxygen gained a total number of $4 {e}^{-}$. The total number of electrons transferred in this reaction is $4 \cdot 7 = 28 {e}^{-}$, which means that the oxidation half-reaction must be multiplied by 4, and the reduction half-reaction must be multiplied by 7.

This will produce these coefficients for the oxidized and reduced reactants

$4 N {H}_{3} + 7 {O}_{2} \to N {O}_{2} + {H}_{2} O$

Now, balance nitrogen like you would normally, and you'll get

$4 N {H}_{3} + 7 {O}_{2} \to 4 N {O}_{2} + {H}_{2} O$

The hydrogen and oxygen are not balanced, since we have 12 H on the left side and only 2 on the right side, and 14 O on the left side and only 9 on the right. Let's balance hydrogen first.

In basic solution, for every H atom you need you must add ${H}_{2} O$ on the side that needs it, and $O {H}^{-}$ on the opposite side. We need 10 H atoms on the right side, so

$10 O {H}^{-} + 4 N {H}_{3} + 7 {O}_{2} \to 4 N {O}_{2} + {H}_{2} O + 10 {H}_{2} O$

Now the hydrogen is balanced. For every oxygen atom we need, $2 O {H}^{-}$ must be added to that side, and ${H}_{2} O$ to the opposite side. We need 5 oxygen atoms on the right side, so

$5 {H}_{2} O + 10 O {H}^{-} + 4 N {H}_{3} + 7 {O}_{2} \to 4 N {O}_{2} + {H}_{2} O + 10 {H}_{2} O + 10 O {H}^{-}$

The balanced equation thus is

$4 N {H}_{3} + 7 {O}_{2} \to 4 N {O}_{2} + 6 {H}_{2} O$

The $10 O {H}^{-}$ cancelled out, and we were left with 6 ${H}_{2} O$ molecules on the right side.