How do I balance the electrons and equation in this redox?

Br2 + NH3 %26gt;%26gt;%26gt;%26gt;%26gt; NH4Br + N2





Maybe im over complicating it but im unsure of how to balance the electrons in this equation the Br changes from 0 to +1 and the N changes from +3 to +5 and also 0 so what do i do about both the Nitrogens? And how do i balance it|||this is a redox reaction.. let's balance it like one...





here's the steps to balancing redox reactions....


1) identify all oxidation states


2) determine which species was oxidized and which reduced


3) write half reactions. include electrons


4) balance electrons in half reactions


5) combine half reactions then cancel electrons


6) add counter ions


7) add other species and THEN balance them...





see if you can follow along...





*** 1 ***


Br in Br2 is 0


N in NH3 is -3


H in NH3 is +1





N in NH4Br is -3


H in NH4Br is +1


Br in NH4Br is -1


N in N2 is 0





*** 2 ***


Br goes from 0 to -1 and is reduced


SOME N goes from -3 to 0 and is oxidized





*** 3 ***


Br2 + 2 e's ----%26gt; 2 Br(-1)


2 N(-3) ----%26gt; N2 + 6 e's





*** 4 ***


3 Br2 + 6 e's ----%26gt; 6 Br(-1)


2 N(-3) ----%26gt; N2 + 6 e's





*** 5 ***


3 Br2 + 6 e's + 2 N(-3) ----%26gt; 6 Br(-1) + N2 + 6 e's





3 Br2 + 2 N(-3) ----%26gt; 6 Br(-1) + N2





*** 6 ***


3 Br2 + 2 NH3 ----%26gt; 6 NH4Br + N2





*** 7 ***


3 Br2 + 2 NH3 + __ NH3 ----%26gt; 6 NH4Br + N2





(notice that I added more NH3? this is because the 2 NH3 only accounts for the NH3 that was oxidized. Some wasn't right? we know that from steps 1 and 2)





3 Br2 + 2 NH3 + 6 NH3 ----%26gt; 6 NH4Br + N2








**************


and finally we simplify...





3 Br2 + 8 NH3 ----%26gt; 6 NH4Br + N2





***********************************


fyi.. you assumed the oxidation state of N was +3 and H was -1... H- is a "hydride ion" and can exist if the other ion has lower electronegativity. LiH is an example. Li+ and H-...But nitrogen has a higher electronegativity than H and therefore withdraws electron density from the H. making it N-3 and H+1.. not the other way around.|||Balance it by inspection:





(3/2)Br2 + 4NH3 %26gt;%26gt;%26gt;%26gt;%26gt; 3NH4Br + (1/2)N2





whole number:





3Br2 + 8NH3 %26gt;%26gt;%26gt;%26gt;%26gt; 6NH4Br + N2