I'm dealing with a voltaic cell constructed like this:
voltaic cell
(From “Electrochemistry”, chapter 19 from the book Principles of General Chemistry (v. 1.0).)

I understand that as oxidation occurs at the $\ce{Zn}$ electrode, the electrode loses mass as $\ce{Zn}$ dissolves into $\ce{Zn^{2+}}$ in the solution and eventually will bond with $\ce{Cl-}$ coming from the salt bridge. But on the $\ce{Cu}$ cathode side, how could it gain mass if initially the $\ce{Cu}$ electrode lost mass after the $\ce{Cu^{2+}}$ ions dissolved in the solution? I get that the $\ce{Cu^{(2+)}}$ ions are re-attracted to the $\ce{Cu}$ electrode once electrons flow from anode to cathode, but wouldn't that mean that the $\ce{Cu}$ will retain the same mass once the cell is complete?

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    $\begingroup$ The idea is that the copper ions are already in the solution prior to the electrodes being introduced $\endgroup$ – Dan Burden Apr 1 '17 at 23:40
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    $\begingroup$ Please always attribute your image sources, and always check that you are allowed to use them in the first place. After that, please upload them to the stack exchange server at imgur (via the built in dialogue) so that they remain accessible and don't suffer link rot. $\endgroup$ – Martin - マーチン Jun 1 '17 at 10:53
  • $\begingroup$ There is an error in the diagram for this post. The oxidation of zinc to Zn2+ is depicted as the process forming Zn2-. The reaction is correct, but the figure shows the wrong charge for the ion produced. ![enter image description here](i.stack.imgur.com/2pIos.jpg) $\endgroup$ – Christopher Davis Aug 3 '18 at 17:20

As you have figured out, the zinc electrode is undergoing oxidation since zinc is a stronger reducing agent than copper.

$$\ce{Zn -> Zn^2+ + 2e-} $$

Thus, the zinc electrode becomes thinner as shown in your image. It is oxidizing into soluble $\ce{Zn^2+}$ ions and thus its mass decreases.

In the right solution, $\ce{Cu^2+}$ ions are being reduced to metallic copper which deposits on the copper cathode

$$\ce{Cu^2+ + 2e^- -> Cu} $$

The $\ce{Cu^2+}$ ions are already there to begin with. The copper electrode isn't dissolving to form $\ce{Cu^2+}$ ions to begin with (the copper electrode does not react with the $\ce{Cu(NO3)2}$ solution), so there isn't any loss in mass of the copper electrode initially. Rather, during the operation of the cell, the reduction of copper(II) ions forms metallic copper, which causes the copper electrode to increase in mass and gets thicker, as seen in the image.

  • $\begingroup$ Hey. Do you know why the need of NO3- in the glass? $\endgroup$ – Lifeforbetter Apr 7 '20 at 22:38

So with the cathode, there were no $\mathrm{Cu}^{2+}$ ions that initially dissolved in the solution. Rather, they were there from the beginning and once the copper metal cathode was added and the cell completed, the copper ions in the solution would then form a solid on the cathode, and thus it would gain mass.


To make a galvanic cell you need two containers, which "initially contains the $\ce{Zn^2+}$ in one container and the other contains $\ce{Cu^2+}$." After that you've dipped those electrodes so that electrons can flow through the cell through the conducting mediums. And therefore, the electrolyte can change to solid electrode and vice versa (as can be seen on $\ce{Cu^2+}$ and $\ce{Zn^2+}$, respectively).


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