Question #755f2

1 Answer
Nov 4, 2015

Answer:

#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])^(-)#