GSP - I think I understand the above for making the 'starter' solution. Let's say I make a 1 liter solution by disolving say 70 grams of Sterling in the nitric/h2o. This solution would be silver nitrate with copper in it? Or, would that be a mixture of silver nitrate and copper nitrate? In the next phase, the electrolysis phase, how much silver can I expect to percipitate from Sterling bars before having to make a new starter solution?
From reading below I'm kind of getting the idea that the starter solution can be 'recycled' enumerable times. Is this correct? The title of this thread first led me to believe that nitric acid was not needed - that all you had to do was start the electrolsys in a solution of copper nitrate. So, maybe my question would be - how much silver can you expect with each cycle? I have about 25 Toz's of Sterling that I want to process as cheaply as possible. If the process is to expensive then I'l just melt the sterling down and keep it as sterling bars.
Thanks,
Mike
Mike,
I could probably write at least 20 or 30 pages on the silver cell, but that's not going to happen. Instead, I'll just give you some non-technical information about what happens in the cell.
(1) Besides pure water, a basic silver cell contains 3 chemicals: Silver nitrate = AgNO
3, Copper Nitrate = Cu(NO
3)
2, and a little Nitric Acid = HNO
3. When in solution, each splits into positive and negative ions (charged species). For these three, you would have these 3 positive ions in solution: H
+, Ag
+, and Cu
+2. The only negative ion in solution would be the nitrate ion, NO
3-. All solutions must have an equal number of positive and negative charges. For each H
+ or Ag
+ ion present, there must be one NO
3- ion and, for each Cu
+2 ion, you would need
2 NO
3- ions. Note that I could have written Cu
++, instead of Cu
+2 - same thing.
(2) Note that there is a finite number of nitrate ions, NO
3-. If the solution is made up properly, with plenty of silver and not too much copper, only silver will deposit at the cathode - no copper. As the silver plates out, it is replaced by silver and copper ions from the sterling anode. However, since each copper ion requires 2 nitrate ions and, since there is only so much nitrate ion available, the silver in solution will decrease. There's more to it than that but, if you do the math, you'll find that, for each gram of copper dissolving at the anode, the amount of silver in solution is reduced by 3.4 grams. That's really all you need to know.
(3) To prevent copper from co-depositing with the silver, the silver concentration in the solution should be kept above about 25g/l and the copper below about 75g/l. There is some variation in this, but that is a good rule to follow. If the copper is low, you might get away with a little less silver
(4) Say you started a one liter solution by dissolving 70g of 90/10 sterling in a minimum amount of nitric acid. The silver concentration would be 63g/l and the copper 7g/l. This cell should work until the silver is reduced to 25g/l or a reduction of 63-25 = 38 grams of silver. To eliminate that much silver from the solution, 38/3.4 = 11.2 grams of copper would have been dissolved from the anode. In other words, this would happen after 11.2/.1 = 112 grams of sterling had been dissolved. Time-wise, if you were running only 1 amp through the cell, this would happen in about 25-30 hours of operation. At that point, you would have to remove part of the solution and replace it with enough silver nitrate solution (silver dissolved in nitric) to build up the silver in the solution to an acceptable level. If you add back fairly pure silver solution, you will also reduce the copper in the cell somewhat. The silver from the waste cell solution is cemented on copper and then dissolved in nitric to make fresh solution. To know how much cell solution to remove and how much silver solution to replace it with requires some slightly heady math.
(5) The moral of this story is that it's a lot of extra work to run sterling anodes in the cell because you have to constantly mess with it. If you first dissolved that sterling you were going to use as anodes in nitric acid, cemented the silver out with copper and rinsed it well, you would have about 99% silver. If you used the 99% silver as anodes, you would be able to run about 10 times more silver though the cell before you had to mess with the solution. And, whether you ran 90% or 99% silver through the cell, it would end up taking exactly the same amount of nitric acid, when all is said and done.
Chris
