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) &-> Cu^2+ (aq) + 2e-}& (E^\circ &= \pu{+0.34 V}) \end{align}

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

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

For the balanced reaction I have: \begin{align} \ce{4NO3- + 4H+ &-> 4NO + 2H2O + 3O2}& (E^\circ &= \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?
  • $\begingroup$ Are the electrodes inert or made of any reactive metal, (say for example copper)? $\endgroup$ – stochastic13 Oct 4 '13 at 3:45
  • $\begingroup$ @SatwikPasani The electrodes are inert, they are two solid slabs of copper just sitting in solution. $\endgroup$ – Dave Chen Oct 4 '13 at 3:49
  • $\begingroup$ Then the copper electrodes might also participate in the reaction at the anode. $$\ce{Cu_{(s)}->Cu^2+_{(aq)} + 2e^-}$$ $\endgroup$ – stochastic13 Oct 4 '13 at 3:55
  • $\begingroup$ @SatwikPasani $\ce{Cu_{(s)}->Cu^2+_{(aq)} + 2e^-}$ has a potential of +0.34V while $\ce{NO3^{-}_{(aq)} + 4H^{+}_{(aq)} + 3e^{-}->NO_{(g)} + 2H2O}$ has +0.96V. Since it produces more energy, wouldn't the stronger oxidizing agent be used instead, i.e, $\ce{NO3^{-}}$? $\endgroup$ – Dave Chen Oct 4 '13 at 4:09
  • 1
    $\begingroup$ Have you considered the overpotential of nitrates on copper surface. Although I am not sure, considering the kinetics (overpotential), cooper is oxidised in preference to Nitrate at the anode. $\endgroup$ – stochastic13 Oct 4 '13 at 4:16

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) &-> Cu^2+ (aq) + 2e-}& E^\circ_\mathrm{ox} &= \pu{-0.34 V} \end{align}

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

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

How does this change your analysis?

| improve this answer | |

This seems to be a thought experiment, and many factors have been considered, but not enough to correspond to reality.

One item to consider is that copper is electrolytically purified by psssing current between a thin cathode of pure copper and a thick cathode of impure copper. An electrical power supply drives the copper (in solution) from the anode (+) to the cathode (-), and less reactive metals precipitate while more reactive metals remain in solution. Sulfate is the most usual anion.

But nitrate could be used as well. Oh, but nitrate is an oxidizer, so it complicates the issue. Perhaps. At high concentrations (say, >25%), nitric acid is such an oxidizer that it will passivate iron, and not dissolve it. Reduce the acid concentration to 10% or less, and iron will dissolve in it, because the acidity function outweighs the oxidizing function. So let's consider our copper nitrate to be at a moderate concentration (~10%), and without added acid ($H^+$), which is an important reactant in all the oxidation reactions.

With pH near neutral (pH = 7), the potential of the nitrate reduction half cell drops from +0.96 volts to +0.408, so then the combined cell voltage is only +0.068. What this means is that a copper block in a solution of copper sulfate will corrode slightly, consuming hydrogen ion. The reaction will not proceed as far as developing pH = 8 before stopping. The copper blocks would show a bit of staining or etching.

When an external potential is applied, one reaction is that $Cu^{+2}$ is forced off the anode into solution and some other $Cu^{+2}$ in the solution is driven onto the cathode to maintain neutrality everywhere. But nitrate ion is a a bit more active than a mere spectator ion and some will be reduced at the anode to $NO$. The pH will increase because $H^+$ is consumed.

At the cathode, copper will plate out, not oxygen. I can imagine copper hydroxide being precipitated at the anode and $NO$ gas being liberated there as well, but it could get more complicated, as $NO$ reacts with $OH^-$.

| improve this answer | |

Not the answer you're looking for? Browse other questions tagged or ask your own question.