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)? $\endgroup$ – Satwik Pasani Oct 4 '13 at 3:45
  • $\begingroup$ @SatwikPasani The electrodes are inert, they are two solid slabs of copper just sitting in solution. $\endgroup$ – David Chen Oct 4 '13 at 3:49
  • $\begingroup$ Then the copper electrodes might also participate in the reaction at the anode. $$\ce{Cu_{(s)}->Cu^2+_{(aq)} + 2e^-}$$ $\endgroup$ – Satwik Pasani 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, $\ce{NO3^{-}}$? $\endgroup$ – David 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. $\endgroup$ – Satwik Pasani 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?


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