How to balance this redox reaction
How to balance this redox reaction (using half reaction method):$$\ce{Cu + HNO3 -&gt; Cu(NO3)2 + NO + NO2}$$


You really can't balance this equation for two reasons. 

(1) There is hydrogen on the left hand side, but none on the right. 

(2) There are two competing redox reactions. One produces $\ce{NO}$ and the other produces $\ce{NO_2}$. The overall reactions are:$$\ce{Cu_{(s)}  +  4HNO3_{(aq)}  -&gt;  Cu(NO3)2_{(aq)}  +  2NO2_{(g)}  +  2H2O_{(l)}}$$  $$\ce{3Cu_{(s)}  +  8HNO3_{(aq)}  -&gt;  3Cu(NO3)2_{(aq)}  +  2NO_{(g)}  +  2H2O_{(l)}}$$

Thus the ration of $\ce{NO2_{(g)}}$ to $\ce{NO_{(g)}}$ is a kinetics problem. In concentrated acid the top reaction predominates and you get mostly the first reaction. In dilute acid the second reaction predominates and you get mostly the second reaction. 

Now as an example let's look at the redox reaction of the first reaction. 

Looking at the copper first we have: $$\ce{Cu_{(s)} -&gt; Cu^{2+}_{(aq)} + 2e^{-}}$$

Just looking at $\ce{NO3^{-} -&gt; NO2 + e^{-} }$ the charge balances, and the $\ce{N}$ atoms balance, but the $\ce{O}$ atoms do not balance and hence this is not a valid half cell reaction. The valid reaction is: $$\ce{e^{-} + 2HNO3_{(aq)}  -&gt; NO^{2-}_{3} + NO2_{(g)}  +  H2O_{(l)}}$$

Using "redox" I can simply balance the electrons. So I need the Cu reaction as is, but I need to multiple the nitrite acid reaction by two. $$\ce{Cu_{(s)} -&gt; Cu^{2+}_{(aq)} + 2e^{-}}$$ $$\ce{2e^{-} + 4HNO3_{(aq)}  -&gt; 2NO^{2-}_{3} + 2NO2_{(g)}  +  2H2O_{(l)}}$$

Now that the electrons match, I can add the two equations:
$$\ce{Cu_{(s)}  +  4HNO3_{(aq)}  -&gt;  Cu(NO3)2_{(aq)}  +  2NO2_{(g)}  +  2H2O_{(l)}}$$