[mooliak]

Hi all,
I have some history in practical electrochemistry, and have, I think, a reasonable uderstanding of the basics of Redox reactions and ½ cells. Recently, I have been thinking about a project, and it has me in a complete quandry. There’s probably something obvious that I’m missing, but here goes.

To set the scene, take the classic galvanic cell situation in which we have 2 separate compartments. One contains Zn with a molar electrolyte, and the other Cu with same. The Zn will be oxidised, and the Cu reduced with the appropriate potential. We use a salt bridge to connect the 2 compartments in order to combat the build up of Zn ions in one cell, and the depletion of Cu in the other, which would slow down and eventually stop the reaction.

Now imagine an electrolytic cell pair which both contain Cu, and a molar electrolyte. Now, we have an external power supply which supplies the necessary p.d. to cause the reduction of Cu in one cell, and the corresponding Cu oxidation in the other. If we link the 2 with a salt bridge, this should keep going for quite a while- it is a valid and stable situation.

The power supply ‘expects’ to see electrons leaving the negative pole, and this does indeed happen. They are consumed in reducing the Cu ions. Also, electrons are arriving at the positive pole due to being donated by the oxidising Cu at the anode. This doesn’t need to be a theoretical type cell in which we are looking for full reversibility, and therefore an almost zero current flow, as we can afford to push a bit to get some volume transacting. The requirements are being met. The current flow is being fully maintained by the redox reactions, and is provided by nF electrons flowing, directly connected to the Cu redox rate.
Instead of a salt bridge, we can use a small pump to circulate the electrolyte to even out the concentrations, because the 2 solutions don’t need to be separated for chemical reasons. The reason for the separation is an operational one.

Comparing this to a commercial copper electrolyser, how is it that I can’t get out of my head that current density is all about electrode spacing, and that the above won’t happen. I can’t find a problem with running this reaction in the above way. Your arguments would be greatly appreciated.

Edit. Following comment by Borek. Thanks for your interest.

My problem is that I think I have argued soundly that the separate copper electroylsis reaction will work continuously. However, instinctively, and reading articles on copper refining, there is no evidence of such a cell operating, and all the information seems to indicate that unless the electrodes are in close proximity, there will be little if any current flow. Current flow seems to be taken as a phenomenon which is controlled by a potential gradient between 2 opposite electrode plates, and governed by solution resistance. It is taken as a flow 'through' the electrolyte, like a resistor in a normal circuit. Thus, if 2 separate tanks were linked by a pipe, the resistance would be so high that it would effectively not work. I think I have argued that the current flow is not through the electrolyte, but by the 2 redox reactions at the electrode surfaces.  I can't find a mechanistic argument which will resolve this issue.

I hope this clarifies the issue, as this is driving me nuts ! I think I,m just going to need to set up the experiment and see what happens. Although that is a perfectly valid scientific thing to do, it seems a little crude that it will be because I can't put a conclusive looking hypothesis together.

Thanks in advance.

[Borek]

I can’t get out of my head that current density is all about electrode spacing, and that the above won’t happen.

Please elaborate, no idea what your problem is.

[Enthalpy]

If the metal you harvest differs from the material you feed, for instance if you electrolyse alumina into aluminium, then the cell needs a minimum voltage (plus losses) and consequently a minimum energy per deposited metal mount.

But if you feed in "the same" metal as you harvest, for instance in copper refining, there is no theoretical minimum voltage drop nor energy investment resulting from the couples.

Only technological losses.

"Only"... 1kg copper needs 3MC or 850A for an hour, and a plant shall produce many tons. Hence such a plant has many cells in parallel, with very broad and tall electrodes, closely spaced - and it still consumes much power.

You relate the metal deposition speed with the speed of the ions, which results from the electric field and the "ion mobility" tabulated for instance in the Handbook of Chemistry and Physics, and compute an electric power that is very uncomfortable, even after geometrical optimization.

I suspect (but know too little for it) that more loss mechanisms exist, not only concentration gradients and competing electrolysis reaction. Maybe Cu+ mixes in between, whereas we write only Cu and Cu++. H+ must also be reduced temporarily, aided by temperature if needed, and reverting to Cu and H+ later means a loss. And certainly many more.

[Wastrel]

A salt bridge is supposed to reduce mixing between compartments, not encourage it.