Electrolysis of aqueous copper (II) nitrate There are two copper blocks sitting in the $\ce{Cu(NO3)2 (aq)}$ solution, a battery is attached onto both of them, providing enough energy to start the reaction.

Since solid pieces of copper are involved, $\ce{Cu}$ must be considered in the reduction potential as well.

However, looking at the half reaction for copper:
\begin{align}
  \ce{Cu (s) &amp;-&gt; Cu^2+ (aq) + 2e-}&amp;
  (E^\circ &amp;= \pu{+0.34 V})
\end{align}

Compared to nitrate:
\begin{align}
  \ce{NO3- (aq) + 4H+ (aq) + 3e- &amp;-&gt; NO (g) + 2H2O (l)}&amp;
  (E^\circ &amp;= \pu{+0.96 V})
\end{align}

Since it produces more energy, I believe the second half reaction will occur instead.



I begin by finding all the species/ions I have:


$\ce{Cu^2+}$
$\ce{NO3^-}$
$\ce{H2O}$


Taking a look at the Standard Reduction Potentials at 25°C:

For my SOA (strongest oxidizing agent) half-reaction ($\ce{NO3^-}$):
\begin{align}
  \ce{NO3^- + 4H+ + 3e- &amp;-&gt; NO + 2H2O}&amp;
  (E^\circ = \pu{+0.96 V})
\end{align}

As for my SRA (strongest reducing agent) half-reaction ($\ce{H2O}$):
\begin{align}
  \ce{2H2O &amp;-&gt; O2 + 4H+ + 4e-}&amp;
  (E^\circ &amp;= \pu{-1.23 V})
\end{align}

For the balanced reaction I have:
\begin{align}
  \ce{4NO3- + 4H+ &amp;-&gt; 4NO + 2H2O + 3O2}&amp;
  (E^\circ &amp;= \pu{-0.27 V})
\end{align}

Some observations that can be made are that:


Nitric gas is forming at the anode
Oxygen gas is forming at the cathode
At least $\pu{0.27 V}$ must be put in


My question is:


Did I do all of this correctly? There's no answer key and I'm pretty sure that I might have made a mistake somewhere.
What would happen if you placed a necklace at the cathode?
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?