Question

Question #755f2

Answer 1

`6"H"_2"O"_text((aq]) +"N"_text(2(aq]) + 6"e"^(-) -> 2"NH"_text(3(aq]) + 6"OH"_text((aq])^(-)`

Explanation:

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

`stackrel(color(blue)(0))("N")_text(2(aq]) -> stackrel(color(blue)(-3))("N") stackrel(color(blue)(+1))("H")_text(3(aq])`

Notice that the oxidation state of nitrogen changes from `color(blue)(0)` on the reactants' side, to `color(blue)(-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, `"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, `"H"^(+)`, to balance the hydrogen atoms.

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

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"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"_text((aq])^(-)`