4
$\begingroup$

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?
$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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