# Power supply setting(s)



## bswartzwelder (Mar 8, 2012)

I just purchased a 0 to 30 VDC and 0 to 10 amp power supply. It has the capability to drive a load by setting either the voltage or the current as the limiting factor. If I use it to power an electrolytic cell, would it be best to set the output to a particular current and let the voltage fall where it may, or should I set it to a particular voltage and let the current vary in response to the load? What would you consider the best values to be for running an electrolytic cell? Also, once in operation, will the areas of plating which are touching each other transfer the gold, or does it need an occasional stir with a glass rod to better expose the gold plated areas?


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## butcher (Mar 9, 2012)

I would use constant voltage and varible current (the cells resistance will vary the current).


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## bswartzwelder (Mar 9, 2012)

Thank you, Butcher. Of course the opposite is true as well. Set a constant current and the voltage will vary with the changes in resistance. 

What current value would be a good starting point? Too high and it causes the solution to heat up (possibly overheat) and too low and it takes forever to deplate the gold.


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## Geo (Mar 9, 2012)

it depends on what material your working with. kadriver uses the 3.5v on his silver cell and i think rusty was using a much higher voltage in his copper parting cell (like 24v) if you are stripping silver in a deplating cell, i think the voltage is around 28v. at higher volts the amps should be lower.i use 12v on my gold cell at or below 8 amps.


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## Harold_V (Mar 10, 2012)

I witnessed a stripping cell that was operated by the US government, in New Mexico. I was informed that current falls off rapidly when gold has been stripped. The cell in question held a large amount of material (several pounds), and was operated @ 28 volts. When current fell off to near zero, it was assumed that the vast majority of the gold had been stripped. 

Note that the cell in question was water cooled, so base metals were not stripped. Running a sulfuric cell too hot will result in pulling base metals. 

Harold


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## butcher (Mar 10, 2012)

If I had a solution, either caustic base, or an acid, and put two dissimilar metals into this solution, I will have created a battery (electrochemical cell), this battery would have a certain voltage, determined by the two metals that we used, the current would be determined by the surface area of the metals we used.

Lets take a look at a redox reaction, reduction (a gain of electrons), oxidation a loss of electrons), now lets take a silver nitrate solution (AgNO3), and add a piece of solid metal copper (Cu), what happens in our example

Cu(s) + 2AgNO3(aq) ---> 2Ag(s) + Cu(NO3)2 (aq)
We see here the copper metal dissolves into solution as a nitrate, or copper ions, and replaces the silver that was dissolve in solution which now becomes a metal silver powder we call cement (or the cementing silver using copper process, this is a redox reaction. 

The solid copper Cu(s) is oxidized (loses electrons), and so it is the reducing agent in this reaction, or to say it another way it is causing the silver to be reduced, (or copper causes the dissolved silver in solution to gain electrons)
Oxidation of copper (loss of electrons):
Cu(s) --> Cu(NO3)2 (aq) + 2e-

Reduction of silver ions (which can be said the silver ions are the oxidizing agent in this reaction (they are causing the copper to be oxidized).

e- + AgNO3 --> 2 Ag(s)
The silver will try and plate onto the copper bar and fluff off into the bottom of the jar as silver metal, as more copper atoms lose electrons and dissolve into solution as copper ions.

Also note copper is higher in the reactivity series of metals, than silver is, so another way to discuss this reaction is a replacement reaction copper is replacing silver in solution.

Now lets make a battery or an electrochemical cell.

First lets take some nitric acid in a jar say 50% HNO3, and stick a large bar of silver into this solution, the silver will dissolve and saturate the solution with silver Ions creating silver nitrate AgNO3, once the nitric has dissolved all of the silver it can it will not dissolve any more of our bar of silver (unless we raised the temperature and more silver would go into solution), (and if we chilled the solution some silver nitrate crystals would form), so now we have a bar of silver sitting in this acid (at a constant room temperature so no more silver is being oxidized (not losing any more electrons), nothing further happens to our silver bar, so now we add a bar of copper, now since copper is less reactive than silver(or in our discussion here), copper is more electro negative than silver, the silver in solution will plate onto our copper bar as silver metal leaving the now free nitrate to dissolve more silver from our bar of silver, now if we added a wires and an LED we would be able to light the LED light(as long as we had enough surface area of our metals in solution to produce the current of 20 milliamps), with the voltage and external current flow(a little less than a half of a volt would be generated, by this battery (from the metals we chose), .From this circuit as electrons travel through our wire, and the current lighting our led lamp, the ions flowing through the solution (atoms of these metals and solution gaining and losing electrons redox reaction of our electrochemical cell.

The above cell was discussing a spontaneous reaction, but what happen if we had an impure silver bar, and a pure silver bar in this solution, and wanted to make pure silver, well the 95% silver in our impure silver anode is very close to the same metal as our pure silver cathode sitting in this saturated silver nitrate solution, so not much is going to happen as both metals are about the same materials and the same potential (one metal not more electronegative than the other, on not more reactive than the other, so with must force the reaction to occur, by using an external voltage source, an external battery or and external DC (direct current) power supply, to drive the reaction, the voltage (see battery charging) we would need would depend on the metals used, the current needed would depend on several factors but mainly the cells electrode surface area, the current through the cell would depend on the cells resistance (how much metal ions dissolved in the acidic solution, concentration of the acid, temperature of the cell and so on.

This is getting to be a long post, and I need to shorten it as the more I write the more mistakes I make in my post.

Lets look quickly a an car lead acid battery each cell in the battery is two volts (12 volt battery has six cells in series), this voltage is created by the two dissimilar metals in solution of sulfuric acid (well this is a bad example as we only have lead metal, a lead oxide electrode and a pure lead electrode), these dissimilar metals create the voltage in the cell as the redox reaction occurs, to charge the battery (force the reaction backwards) our charging voltage must overcome the cells natural tendency to want to react electrochemically in one direction (discharge of battery), so we must force the reaction backwards with a higher voltage, in this case our 12 volt battery is charged with 14.5 volts.

Note we can charge a six-volt battery with a 9-volt battery (as long as it had needed current capacity), by hooking the plus of both batteries together and the minus of both batteries together.


The voltage of the cell I would chose by the metals in solution (type of the cell, and its purpose), the current the cell would determine, and as stated in a post above (the current may need to also be limited to a certain amount allowed, so as not to overheat the cell, or cause excessive gas to escape solution) if it has a low resistance condition (example a short circuit within the cell).

Hope I did not make too many mistakes here.


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