To Chemistry gurus, how to balance this redox reaction?

Question

Please explain step by step using the half reaction method to balance this redox reaction:

Cu + HNO3 --> Cu(NO3)2 + NO + H2O

When I separate into half reactions:

Cu --> Cu (NO3)2

HNO3 --> NO

For the top half reaction what do you add to balance both sides, and why?

I'm stuck on this step, so please walk me through this full problem step by step so I can understand how to balance this. Thank you.

Answer 1

Are you sure the equation is correct?

I thought the equation suppose to be:

Cu + 2 HNO3 -------> Cu( NO3)2 + H2

In this case, Cu ( 0 ) is oxidised to Cu ( +2 )

H ( +1 ) is reduced to H ( 0 )

* I'm not sure as well *

Answer 2

3(Cu --> Cu2+  +  2e-)

2(4H+ + NO3- + 3e- -->  NO  + 2H2O)

--------------------------------------...

8H+ + 3Cu + 2NO3- --> 3Cu2+ + 2NO + 4H2O

Balancing the copper oxidation reaction is straightforward.  The nitrogen in nitrate ion is reduced from +5 to +2 and therefore gains three electrons.  Balance up the oxygens by placing two water molecules on the right and 4 hydrogen ions on the left.  Multiply the oxidation half reaction by 3 and the reduction half reaction by 2 so that the number of electrons gained and lost is the same.  Then add the two half reactions together.