# Question 755f2

Nov 4, 2015

$6 {\text{H"_2"O"_text((aq]) +"N"_text(2(aq]) + 6"e"^(-) -> 2"NH"_text(3(aq]) + 6"OH}}_{\textrm{\left(a q\right]}}^{-}$

#### Explanation:

Start by assigning oxidation numbers to the atoms that take part in the half-reaction

${\stackrel{\textcolor{b l u e}{0}}{\text{N")_text(2(aq]) -> stackrel(color(blue)(-3))("N") stackrel(color(blue)(+1))("H}}}_{\textrm{3 \left(a q\right]}}$

Notice that the oxidation state of nitrogen changes from $\textcolor{b l u e}{0}$ on the reactants' side, to $\textcolor{b l u e}{- 3}$ on the products' side, which means that nitrogen is being reduced.

The first thing to focus on here is balancing the nitrogen atoms. To do that, multiply the ammonia moleucle by $2$

stackrel(color(blue)(0))("N")_text(2(aq]) -> 2stackrel(color(blue)(-3))("N")"H"_text(3(aq])

Now, each nitrogen atom will gain $3$ electrons, which means that a total of $6$ electrons must be gained by two nitrogen atoms.

stackrel(color(blue)(0))("N")_text(2(aq]) + 6"e"^(-) -> 2stackrel(color(blue)(-3))("N")"H"_text(3(aq])

In basic solution, any protons you may end up with after balancing the oxygen and hydrogen atoms must be neutralized by the addition of hydroxide anions, ${\text{OH}}^{-}$, to both sides of the equation.

So, notice that you have $6$ hydrogen atoms on the products' side, but none on the reactants' side, which means that you must add protons, ${\text{H}}^{+}$, to balance the hydrogen atoms.

$6 {\text{H"_text((aq])^(+) + stackrel(color(blue)(0))("N")_text(2(aq]) + 6"e"^(-) -> 2stackrel(color(blue)(-3))("N")"H}}_{\textrm{3 \left(a q\right]}}$

To neutralize these excess protons, add $6$ hydroxide anions to both sides of the equation

overbrace(6"OH"_text((aq])^(-) + 6"H"_text((aq])^(+))^(color(red)(6"H"_2"O"_text((l]))) + stackrel(color(blue)(0))("N")_text(2(aq]) + 6"e"^(-) -> 2stackrel(color(blue)(-3))("N")"H"_text(3(aq]) + 6"OH"_text((aq])^(-)#

The six hydroxide anions and the six protons will neutralize ach other to produce $6$ water molecules, which means that the balanced half-reaction will look like this

$6 {\text{H"_2"O"_text((aq]) + stackrel(color(blue)(0))("N")_text(2(aq]) + 6"e"^(-) -> 2stackrel(color(blue)(-3))("N")"H"_text(3(aq]) + 6"OH}}_{\textrm{\left(a q\right]}}^{-}$