Electrolysis of aqueous copper (II) nitrate
There is a misunderstanding in the analysis of your first two half-reactions. Your first two half-reactions are fine. Remember that positive values of $E$ mean the reaction is spontaneous. Negative values of $E$ mean the reaction is nonspontaneous, since 

$$\Delta_\mathrm{r}G = -nFE_\mathrm{cell}.$$

Your first half-reaction is an oxidation (as written). You have the wrong sign on that $E$ value.
\begin{align}
  \ce{Cu (s) &amp;-&gt; Cu^2+ (aq) + 2e-}&amp;
  E^\circ_\mathrm{ox} &amp;= \pu{-0.34 V}
\end{align}

Your second half-reaction is a reduction (as written)
\begin{align}
  \ce{NO3- (aq) + 4H+ (aq) +3e- &amp;-&gt; NO (g) + 2H2O (l)}&amp;
  E^\circ_\mathrm{red} &amp;= \pu{+0.96 V}
\end{align}

The combination of these two half-reactions produces a positive $E^\circ_\mathrm{cell}$ which is a spontaneous reaction. You do not need to use the oxidation of water as your oxidation half-reaction. Copper is a much better reducing agent than water:

$$
\ce{3Cu (s) + 2NO3- (aq) +8H+ (aq) -&gt; 3Cu^2+ (aq) + 3NO (g) +4H2O (l)}\\ 
E^\circ_\mathrm{cell} = \pu{+0.96 V} +(\pu{-0.34 V}) = \pu{+0.62 V}
$$

How does this change your analysis?