*** 4 ***
to balance the e's.. I'll multiply the reduction half by 2
6 Cu(+2) + 6 e- ---> 6 Cu(+1)
2 N(-3) ---> 2 N(0) + 6 e's
*** 5 ***
combining...
6 Cu(+2) + 6 e- + 2 N(-3) ---> 6 Cu(+1) + 2 N(0) + 6 e's
canceling e's
6 Cu(+2) + 2 N(-3) ---> 6 Cu(+1) + 2 N(0)
*** 6 ***
adding counter ions
6 CuF2 + 2 NH3 ---> 2 CuN + 1 N2.. .. notice the grouping?
*** 7 ***
adding additional chemical species
6 CuF2 + 2 NH3 + ___ NH3 ---> 2 Cu3N + __ NH4F + 1 N2
notice.. I added more NH3? that's the N that didn't "oxidize"
and balancing.. let's see 12 F's on the left.. I need 12 on the right
6 CuF2 + 2 NH3 + ___ NH3 ---> 2 Cu3N + 12 NH4F + 1 N2
then.. 16 N's on the right.. I need 14 more on the left
6 CuF2 + 2 NH3 + 14 NH3 ---> 2 Cu3N + 12 NH4F + 1 N2
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now. let's check all that
.. ..left.. . right...
Cu.. 6.. .. 6
F.. ..12.. .12
N.. ..16... 16.
H... .48... .48
and..
6 Cu's gained 1 electron each = 6 e's
2 N's gained 3 electrons each = 6 e's
and all is well..
**************
so.. all we have to do is combine the NH3's to make it neat and tidy and off we go
6 CuF2 + 16 NH3 ---> 2 Cu3N + 12 NH4F + 1 N2
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questions?