I have read that $\ce{Cu}$ is deposited at the cathode when copper (II) nitrate is electrolyzed. However, when I looked at reduction potentials, the reduction of nitrate to $\ce{NO}$ and $\ce{H2O}$ $(0.96\rm~V)$ is more spontaneous than the reduction of $\ce{Cu^2+}$ to $\ce{Cu}$ $(0.34\rm~ V)$. So why is Cu formed rather than NO?

What I read doesn't mention what material the electrodes are, would it make a difference if they were inert or made of Cu? I supposed if they were made of Cu, then the Cu will be oxidized and the $\ce{NO3-}$ reduced:

$$\ce{Cu + NO3- -> Cu^2+ + NO + H2O} {\rm \qquad \qquad(unbalanced)}$$

However this has a positive voltage of $1.30$, so why would electrolysis even be necessary?

If the electrodes were inert, then there were be no Cu to oxidize, so what would happen? Would the water be oxidized? $$\ce{2H2O -> 4H+ + 4e- + O2} \\\ \rm V=-1.23$$

  • $\begingroup$ Copper is a cation which goes to the cathode where reduction takes place. Nitrate is an anion which goes to the anode where oxidation takes place. $\endgroup$ – aventurin Apr 6 '16 at 15:38
  • $\begingroup$ But not according to the reduction potentials..nitrate is more likely to be reduced to NO and H2O than Cu2+ to Cu is $\endgroup$ – user28625 Apr 6 '16 at 15:40
  • $\begingroup$ @user28625 Reduction potentials have nothing to do with the fact that copper is a cation, and cations tend to cathode where the reduction happens, while the anions tend away from it. Maybe nitrate would reduce quite willingly; but it is simply not there. $\endgroup$ – Ivan Neretin Apr 6 '16 at 16:26
  • $\begingroup$ If it would reduce quite willingly, why wouldn't it go to the cathode, a source of electrons, to get reduced? $\endgroup$ – user28625 Apr 6 '16 at 16:27
  • $\begingroup$ Because anions must go to the anode (see, they aren't called "anions" for nothing), and breaking the law would cause a hefty penalty. That being said, I guess some tiny portion of $\ce{NO3-}$ eventually gets reduced. $\endgroup$ – Ivan Neretin Apr 6 '16 at 17:56

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