Cu(s) + no3- (aq) > cu2+ (aq) + no
balance the following oxidation-reduction reaction that occurs in acidic solution using the half-reaction method.

Oxidation half reaction :- 

Cu > Cu2+ + 2e- ...(1) 

this is the balanced oxidation half reaction 

b) reduction half reaction :- 

NO3- > NO 

balance oxidation no. 

NO3- + 3e- > NO 

balance charge by adding H+ 

NO3- + 3e- + 4H+ > NO 

now balance O and H atoms by addin H2O... 

NO3- + 3e- + 4H+ > NO + 2H2O(2) 

this is balanced reduction half reaction 

multiply (1) by 3 and (2) by 2 and add 

3Cu > 3Cu2+ + 6e- 
2NO3- + 6e- + 8H+ > 2NO + 4H2O 
... 

6e- will cancel out fom both sides... 

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

Bis the answerhope it !