Since we know that copper anode will release $\ce{Cu^2+}$ ion, so I was wondering whether the cathode will thicken as copper has a lower position on the electrochemical series.

Based on my research, I found that some said that cathode will only thicken if the solution is stirred, some said that hydrogen gas will form.

Hence, are there any difference if electrolyte used is in a molten state compared to in an aqueous state?

  • $\begingroup$ At the cathode, metallic sodium will be created. At the anode, Cu will get dissolved. $\endgroup$ – Maurice Sep 14 '20 at 16:47
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    $\begingroup$ Missing water in molten salt is quite a difference. $\endgroup$ – Poutnik Sep 14 '20 at 17:48
  • $\begingroup$ Be aware of distinguishing galvanic and electrolytic cells. $\endgroup$ – Poutnik Sep 17 '20 at 7:10

A galvanic cell with the same electrodes ( copper ones in this case ) is usable as a galvanic cell only if cathodic and anodic electrolyte differs in the concentration of electroactive components.

E.g. if we can use 2 different concentrations of $\ce{CuSO4}$ solution, separating them by a diaphragm or a salt bridge. The electrode with more concentrated solution around would become cathode with the more positive potential and vice versa.

Putting 2 copper electrodes into homogeneous solution of $\ce{NaCl}$ ( or molten one ) does not form a galvanic cell, unless some minor undefined differences create some residual voltage.

In case you did mean an electrolytic cell with moltan salt instead, anodic copper would be dissolving.

At the cathode, only metallic sodium would depose initially. Since some copper ions reach the cathode, copper would start priority deposition. Sodium deposition would depend on the applied electrode potential.


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