Balance the following equation in acidic conditions. Phases are optional.
Solution
Cu in Cu has oxidation state of 0
Cu in Cu+2 has oxidation state of +2
So, Cu in Cu is oxidised to Cu+2
N in NO3- has oxidation state of +5
N in NO has oxidation state of +2
So, N in NO3- is reduced to NO
Reduction half cell:
NO3- + 3e- --> NO
Oxidation half cell:
Cu --> Cu+2 + 2e-
Balance number of electrons to be same in both half reactions
Reduction half cell:
2 NO3- + 6e- --> 2 NO
Oxidation half cell:
3 Cu --> 3 Cu+2 + 6e-
Lets combine both the reactions.

2 NO3- + 3 Cu --> 2 NO + 3 Cu+2
Balance Oxygen by adding water
2 NO3- + 3 Cu --> 2 NO + 3 Cu+2 + 4 H2O
Balance Hydrogen by adding H+
2 NO3- + 3 Cu + 8 H+ --> 2 NO + 3 Cu+2 + 4 H2O
This is balanced chemical equation in acidic medium
Answer:
2 NO 3
-
+ 3 Cu + 8 H +
--> 2 NO + 3 Cu 2+
+ 4 H 2 O