Balance the following redox equation:

CuF2 + NH3 -> CU3N + NH4F + N2

I'm not sure how to split this into half reactions because N is found in all the products.
Please show the steps you took to balancing this :) thanks!

use these steps..
(1).. identify the oxidation state of each and every atom
(2).. determine what is oxidized and what is reduced
(3).. write half reactions.. include electrons
(4).. balance half reactions.. the electrons that is!
(5).. combine half reactions and cancel electrons
(6).. add counter ions
(7).. add any remaining species.. THEN balance them.

here we go..
*** 1 ***
Cu in CuF2 is +2.. why? F is -1.. it has the higher electronegativity and gets the electron!
F in CuF2 is -1

N in NH3 is -3.. electronegavitiy of N = 3.04 and of H = 2.20.. N is higher so it gets the electrons
H in NH3 is +1

Cu in Cu3N is +1.. why? N is -3.. it has the higher electronegativity.. since there are 3 Cu's and since the overall molecule is neutral.. each Cu is +1
N in Cu3N is -3

N in NH4F is -3
H in NH4F is +1
F in NH4F is -1
N in N2 is 0.. why? because there is not difference in electronegativity between N and N.. and since the overall N2 molecule is neutral.. each N must be 0

*** 2 ***
ok.. now we know this
ALL Cu goes from +2 to +1.. and is reduced.. (reduction is defined as "reduction in charge")
SOME N goes from -3 to 0 and is oxidized.. (oxidation is an increase in charge)
SOME N goes from -3 to -3.. and remains unchanged.

all the H's and the F's remain unchanged.

*** 3 ***
here's the trick.. we're only going to consider the N that changes for now.. we'll add some more NH3 later.

reduction half..
3 Cu(+2) + 3 e- ---> 3 Cu(+1)

oxidation half
2 N(-3) ---> 2 N(0) + 6 e's

now.. notice that I wrote 3 Cu and 2 N instead of 1 Cu and 1 N? In a minute, I'm going to need 3 Cu's and 2 N's to form my molecules Cu3 and N2.. I might as well take care of that now.. ok with you?