How do you balance the following redox reaction? I got #"Cu"(s) + 2"H"^(+)(aq) + "NO"_3^(-)(aq) -> "Cu"^(2+)(aq) + "NO"_2(g) + "H"_2"O"(l)# which is not right...

Well, you've got to start from the beginning for this one... Did you cancel the electrons? Surely the charges DO balance.

The core half reactions are:

#"NO"_3^(-)(aq) -> "NO"_2(g)#, reduction of #"N"#

#"Cu"(s) -> "Cu"^(2+)(aq)#, oxidation of #"Cu"#

Of course, these aren't balanced yet, but this is the main action. Now we balance in acid. Add water to balance the oxygens:

#"NO"_3^(-)(aq) -> "NO"_2(g) + "H"_2"O"(l)#

Add #"H"^(+)# to balance the hydrogens...

#2"H"^(+)(aq) + "NO"_3^(-)(aq) -> "NO"_2(g) + "H"_2"O"(l)#

Add electrons to balance the charge...

#e^(-) + 2"H"^(+)(aq) + "NO"_3^(-)(aq) -> "NO"_2(g) + "H"_2"O"(l)# #" "bb((1))#

Now for the oxidation half-reaction, which should be very simple.

#"Cu"(s) -> "Cu"^(2+)(aq) + 2e^(-)# #" "bb((2))#

There. Now we can add #(1)# and #(2)# together. The electrons are a fictitious construct of the method, so they must be eliminated.

#2(cancel(e^(-)) + 2"H"^(+)(aq) + "NO"_3^(-)(aq) -> "NO"_2(g) + "H"_2"O"(l))#
#ul("Cu"(s) -> "Cu"^(2+)(aq) + cancel(2e^(-))" "" "" "" "" "" "" "" "" "" ")#
#color(blue)("Cu"(s) + 4"H"^(+)(aq) + 2"NO"_3^(-)(aq) -> "Cu"^(2+)(aq) + 2"NO"_2(g) + 2"H"_2"O"(l))#

You should take it upon yourself to verify that the mass and charge are balanced.