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Let's say we have two copper electrodes as shown in the image, and there is a solution with copper nitrate. In this case, I know that copper will deposit onto the cathode and copper from the anode will be oxidized and enter solution when a voltage is applied across the two electrodes. However, what if instead of copper nitrate, there was just NaCl (or any other electrolyte that wouldn't cause any problems like precipitation) in solution instead? Would current still be able to flow, and if so, would the electrodes still lose and gain mass as they did before? I am fairly certain that current does flow, but I am not really sure if copper will still deposit onto the cathode and vice versa.

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It is a complex problem. Yes, NaCl solution is a conductor, so electricity will pass through the solution. Now in your situation both electrode are made of copper. Since electricity is passing through the solution, electrolysis must occur at both electrodes. You cannot have electrolysis only at one electrode.

Assuming there is sufficient salt in the solution, what will happen at the anode?

Chlorine gas will evolve. I will let you find relevant equations. Chlorine gas attacks copper very quickly and the anode starts to "degrade" quite fast, but will it pass into solution? Nope.

Let us see what is happening at the cathode?

Sodium ions cannot be reduced in water esp. on copper electrodes (on mercury they do), instead water will be reduced. What happens when water is reduced? It turns into hydrogen gas, which escapes, and leaves hydroxide ions behind. This process makes the solution basic.

Now coming back to the anode, since you have hydroxyl ions around, they precipitate copper ions. All you get is yellow orange junk in such a system. Cu(II) is blue so why an orange yellow mess? It is most likely due to Cu(I) compounds. Anode dissolves pretty fast but nothing happens to the cathode.

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  • $\begingroup$ Thanks for the answer! I see that you stated that at the anode, chlorine will be oxidized. Do you have an explanation for why chloride will be oxidized but the copper on the electrode will not be? $\endgroup$ – p-adict Mar 24 at 4:59
  • $\begingroup$ Good point, look at the electrode potential tables for Cu(2+)/Cu and Cl (-)/Cl2 half cells. Which species will have a higher tendency to oxidize at the $anode$? $\endgroup$ – M. Farooq Mar 24 at 5:45
  • $\begingroup$ @M. Farooq Hm, I would suppose the copper oxidation would occur sooner than chloride oxidation, comparing the standard potentials being almost 1 V apart. Like Cu -> Cu2+ + 2 e-, or Cu + 4 Cl- -> [CuCl4]2- + 2e- Unless current density is too high to be saturated by copper oxidation and with high enough voltage, chloride oxidation would start as well. I suppose oxidation to Cu(I) can happen as well. $\endgroup$ – Poutnik Mar 24 at 7:27
  • $\begingroup$ @Poutnik, the oxidation potential of Cu to Cu(II) would be -0.34 V and that of chlorine would be -1.36 V with respect to SHE. Since there is no Cu(II) to begin with, it is hard to say. Under standard conditions, copper oxidation should occur. $\endgroup$ – M. Farooq Mar 24 at 16:10

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