Electrochemistry Power Supplies & Cell Size

[kurt]

How do you determine the size (liter capacity) of an electrolytic cell you can run off of a particular power supply &/or the anode to cathode size (surface area) for optimum performance in the cell based on the power supply

In the “Copper Nitrate Cell” thread (started by Rusty) on Pg 3 Butcher makes this statement – “it is the current that is important in determining how much copper is transferred from anode to the cathode in the cell, and surface are will play a big part in that figure”

So if I am understanding this right the volts you run a cell at (whether silver or copper) are a predetermined set value based on the electro potential of the metal --- but the amp out put determines metal ion transfer potential from anode to cathode – which in turn determines surface area (size) of anode/cathode for optimum transfer of metal ions – which in turn determines cell size (liter capacity)

Another words (if I understand this right) if your anode & cathode surface area are to small you don’t get optimum transfer because you don’t utilize the full amp out put potential of the power supply - & on the other hand – if the anode & cathode surface area are to large you don’t get optimum transfer because you are trying to spread the amps out over to large a surface area.

Am I on the right track ? --- If so – how do I calculate this based on my power supply amp out put

And then how does that equate to weight of metal transferred per hour

(this is assuming the recomended 4" - 4 3/4" anode/cathode spacing & around 60 grams dissolved metal per liter elecrolyte)

Kurt

 

[philddreamer]

Hi Kurt!
I followed GSP's advised on the matter and had great success. I was expecting 28g per hour deposition and ended with 1.2 T oz per hour.
I decided on the surface area of the anodes first, then the size of the cathodes required to match the anodes; followed by the corrent and size of cell...
You can read about it at:
http://goldrefiningforum.com/phpBB3/viewtopic.php?f=50&t=11281&p=110538&hilit=half+gallon+silver+cell#p110212

Kadriver and others have more information.
Take care!
Phil

 

[mikeinkaty]

For a power supply I would highly recommend one of these (current ebay item # 221186451356). A Variac capable of producing up to 20 amps. Put a 20 amp bridge rectifer on it available from Radio Shack (cheap). Also get a 0-20 amp meter and a 0-10 volt meter and monitor both the current and the voltage continuously. Put a 20 amp in-line fuse in the circuit (also at Radio Shack).

For SS cathode size use the figure of 30 Ma / sq cm as the highest current density you should use. So, if the cathode is 500 sq cm, then don't go over 15 amps. Keep the voltage between the anode and cathode between 1.5 and 2.5 volts. The more anode area you have will enable you to get more current and stay within the 1.5-2.5V limit. The more current you can use the faster the electrolysis. Keep the anode-cathode distance to no more than about 4".

The cathode should be contiinuous Stainless Steel like Kadriver's bowl. If any metal edges are in the solution most of the deposition will be on these edges causing long feather like crystals (not good). With a large SS bowl say 12" diameter and 6+" deep you should be able to produce 20+ ounces of large nodules before having to do a cleanout. SS is not a good conductor of electricity so you will need to connect the negative lead to several places on the cathode. Put several turns of #12 copper wire around the outside, if you use a bowl or a bucket. A bowl will be easier to clean out than a bucket but can be more easily spilled.

Here is a pic of my Variac's. The cell is the first one I made and I did get >200 ozt crystals from it. But, I had to keep the current down to prevent the feather crystals from growing.
http://www.mikeinkaty.com/DSCN0085a.JPG

Mike

 

[jmdlcar]

Hi,

Will these meter work or not? This is just an idea.

http://www.ebay.com/itm/3-1-2-Red-LED-DC-20V-Digital-Voltag-Volt-Panel-Meter-/350627171831?pt=LH_DefaultDomain_0&hash=item51a3020df7

http://www.ebay.com/itm/DC-0-20A-3-1-2-Digital-Red-LED-Amp-current-meter-panel-/350627224481?pt=LH_DefaultDomain_0&hash=item51a302dba1

Added this

http://www.ebay.com/itm/20A-75mV-DC-Current-Shunt-Good4-Digital-Analog-meter-/230411314796?pt=LH_DefaultDomain_0&hash=item35a595a66c

Jack

 

[element47.5]

Jim, those should work, but on the ammeter, you will need to supply a shunt...a "bypass" resistor that takes most of the current away from going through the meter. No way can you put 20 amps through "chip stuff". The ad mentions a "75 mv, 20 amp" shunt. In the old days, these were literal bars of metal, almost short-circuits, to again, conduct the overwhelming bulk of the current to be measured around the meter. You can probably get some idea of what a meter shunt looks like by googling "meter shunt" in google images.

Using ohm's law, E = IR, (volts = amps * ohms) you would need a .075 volts = 20 amps * 1.5 ohm resistor.
Again using ohms law, P = I*I*R (power = current [squared] * ohms) = 20*20*1.5 = a 600 watt resistor!!! That would be freaking huge.
The problem is that anything you yourself came up with would be cumbersome and have questionable accuracy. You would want to enclose it, it could get red hot if you made it out of nichrome wire or something similar. Upon making an enclosure for such, you would not have a compact little doodad any more.

Looking at other offerings after searching for "meter shunt" on ebay, others have this shunt. All I can tell you is Do NOT buy one of those ammeters without getting the proper shunt!!! Don't even think about it!

 

[jmdlcar]

Jim, those should work, but on the ammeter, you will need to supply a shunt...a "bypass" resistor that takes most of the current away from going through the meter. No way can you put 20 amps through "chip stuff". The ad mentions a "75 mv, 20 amp" shunt. In the old days, these were literal bars of metal, almost short-circuits, to again, conduct the overwhelming bulk of the current to be measured around the meter. You can probably get some idea of what a meter shunt looks like by googling "meter shunt" in google images.

Using ohm's law, E = IR, (volts = amps * ohms) you would need a .075 volts = 20 amps * 1.5 ohm resistor.
Again using ohms law, P = I*I*R (power = current [squared] * ohms) = 20*20*1.5 = a 600 watt resistor!!! That would be freaking huge.
The problem is that anything you yourself came up with would be cumbersome and have questionable accuracy. You would want to enclose it, it could get red hot if you made it out of nichrome wire or something similar. Upon making an enclosure for such, you would not have a compact little doodad any more.

Looking at other offerings after searching for "meter shunt" on ebay, others have this shunt. All I can tell you is Do NOT buy one of those ammeters without getting the proper shunt!!! Don't even think about it!

The 3nd item is shunt for that meter I think.

http://www.ebay.com/itm/20A-75mV-DC-Current-Shunt-Good4-Digital-Analog-meter-/230411314796?pt=LH_DefaultDomain_0&hash=item35a595a66c

Jack

 

[resabed01]

I think your math is off...

For a shunt to have potential difference of 0.075v @ 20a it would need a resistance of 0.00375 ohms. It would have to dissapate 1.5w at 20A

A chunk of 14 gauge wire of the correct length will work.

 

[jmdlcar]

I found some more. Are these better to use?

http://www.ebay.com/itm/1-DC-10V-Analog-Panel-Volt-Voltage-Meter-Voltmeter-Gauge-85C1-White-0-10V-DC-/121002503910?pt=LH_DefaultDomain_0&hash=item1c2c4faee6

http://www.ebay.com/itm/1-DC-20A-Analog-Panel-AMP-Current-Meter-Ammeter-Gauge-85C1-White-0-20A-DC-/110951975829?pt=LH_DefaultDomain_0&hash=item19d540cb95

Jack

 

[mikeinkaty]

I found some more. Are these better to use?

http://www.ebay.com/itm/1-DC-10V-Analog-Panel-Volt-Voltage-Meter-Voltmeter-Gauge-85C1-White-0-10V-DC-/121002503910?pt=LH_DefaultDomain_0&hash=item1c2c4faee6

http://www.ebay.com/itm/1-DC-20A-Analog-Panel-AMP-Current-Meter-Ammeter-Gauge-85C1-White-0-20A-DC-/110951975829?pt=LH_DefaultDomain_0&hash=item19d540cb95

Jack

Yes, those will do just fine. I prefer the analog kind anyway but I'm older than dirt! I get more of a 'feel' for what's happening by watching the needle swing.

Mike

 

[element47.5]

resa, you are correct, I scrambled terms. The required R would be 1/20th of .075 as you stated.

A length of heavyish wire would work but it would yield a device difficult to calibrate. And if too high a resistance, would blow up the meter under high current conditions.

Best is to use the correct shunt. I'll stick with that admonition.

 

[mikeinkaty]

The problem with a Variac that has 120-130 volt output is that you only have to turn the knob a smidgen to get the 1.5-2.5V needed for the cell. A way to change that is to find a 120V Variac with only about 10-20 volt output or get a 120V input transformer with a 12 volt output rated at the amps you want. So, the two 120 volt output leads from the variac go to the input side of the transfromer and the output of the transformer go to the full vave rectifer. Then you will have much better sensitivity with the knob on the variac. I use my two as is and have no problems. Just gotta turn the knob very slow and watch the voltmeter and the ampmeter.

There is a good looking red variac now on Ebay for $90 +$20 shipping.

Mike

 

[resabed01]

A length of heavyish wire would work but it would yield a device difficult to calibrate. And if too high a resistance, would blow up the meter under high current conditions.

Best is to use the correct shunt. I'll stick with that admonition.

No need to calibrate if the math was done and the correct measurements were made, remember it's intended purpose here. Even if you were to feed 40A through, the resulting voltage on the meter would be 0.15v, hardly enough to blow it up.

Just for future reference for those that might want to save a few bucks. To make a shunt of 0.00375 ohms you'd need a piece of 14 gauge wire 17.8"long. If you wanted to have 1% precision, you could add a inch to that then add a variable resistor in parallel (not series) to the shunt to trim the meter.

 

[element47.5]

You'd have a difficult time measuring to that precision with ordinary meter leads unless your meter can zero out the resistance of the test leads....which good ones can do.

 

[mikeinkaty]

Something I forgot. If you put a 10 amp ampmeter in the circuit then you should us a 10 amp in-line fuse to protect the meter from a dead short. If the crystals grow into the anode this can happen. If you run 15 amps through those cheap 10 amp chinese meters they might melt or worse start a fire.

Mike

 

[kurt]

Phil/Mike – thanks for the replies – they helped get me head in the right direction but didn’t quite answer my question --- I may not have worded my question right (my fault)

What I am looking for is the math to figure out what “size” cell you can run based on the current out put (amp potential) of any given power supply & the yield expected from a cell built to the size of power supply potential

I have been doing research on this for around a month now & the problem has been finding bits & pieces of info here & there – as well as numbers all over the map --- Examples of numbers all over the map

(1) voltage = 1.5 – 3.5

(2) current density = 20 – 40 ma/sq Cm (30 ma/sq Cm = .19 amp/sq 1”) to .35 amp/sq 1”

Here is the best I have been able to put it together using numbers said to be “standards” for running a silver cell

(1) 60 gr silver/liter electrolyte (call it 2 ozt/liter)
(2) space anode – cathode 4” – 4 & ¾” apart (call it 4 & 3/8”)
(3) run cell @ 3 volts
(4) 3 volts provides .35 amp current density/sq 1” of anode
(5) .35 amp/sq 1” = about .06 ozt silver deposited/hr (or about 1.866 gr/hr)
(6) Figure only about 80% efficiency of power supply

So if you had say a 10 amp power supply @ 80% = 8amps available @ .35 amp/sq 1” = 2.85 sq 1”/amp = about 22.857 sq 1” of anode surface area

Lets round that up to 25 sq 1” of anode assuming you can get a little better the 80% from your power supply – that’s an anode 5” X 5” (surface area)

You need just a little over 61 cubic inches for a liter of electrolyte – so – based on a 4 & 3/8” space between anode & cathode lets figure about 5” deep of electrolyte & the cell would need to be “at least” 1” larger on each side of anode - so for a 5” X 5” anode the cell would need to be at least 7” X 7” X 5” deep electrolyte = 245 cubic 1” divided by 61 cu1”/liter = 4.01 liter cell size & a 4 liter cell would need about 8 ozt silver for electrolyt

So – assuming the “standard” numbers I have used are correct – a 10amp power supply should part about 1.5 ozt/hr with a 25 sq 1” anode in a 4 liter cell figured @ about 80% power supply efficiency

From there you should be able to figure power supply amp needed + anode size + cell size to = any given per hour silver production desired

Open to comment

Kurt

 

[mikeinkaty]

Moebius Cell (vertical cathode and vertical anode)
Most of what I've read say 1.5-2.5V. I have a 78 sq" SS cathode and I prefer 4 to 5 amps. This give me good crystals to work with in a reasonable time. I am retired and a hobbyiest to boot so time is not a critical thing for me. At these settings my cell will 'consume' 5ozt of 990 silver in about 7 hours. My prototype cell had a cathode with exposed edges in the solution. The 'edge effects' of SS caused my cell to start producing long spidery web type crystals growing off the edges and they are harder to work with than nice big grains or nodules. If you have no exposed edges then all the crystals stay on the cathode and get more bulky, not longer. Also, voltages out of the 1.5 - 2.5V range might cause other metals to be deposited on the anode if any are present in the anode silver. My spacing between cathode and anode is more like 3". I have covered the exposed edges on the cathode in my 2nd generation cell with plastic report binder strips.

Thum Cell (horizontal cathode and horizontal anode)

I don't have one but most of what I've read say these need voltages in the 3-5V range because of the larger cathode -anode distance. I think the rules for amperage might be the same as above. (30ma/cm² But then again to get the crystals you like you may need something smaller.
-----

Each cell is different so you need to start with parameters in the ball park then adjust them as you gain more experience with that cell. Record everything that goes in, comes out, and the time required. Even record what you observed during the disolve and during the copper or salt drop and the filtering. Hang a clipboard close to your workstation. Eventually the need for such elaborate record keeping will dimish.

I started with 8ozt of 990 silver per three liters in my electrolyte. That allowed me to process about 120 ozt of 990 anodes recovering slightly less than that in 99.99 silver crystals. If you run 980 silver then it would only have been able to process about 60 ozt before the silver concentration got to low. I put 10 ozt of 990 silver in my last 3 liter electrolyte because one really does not generally know how pure their anodes will be. Look for my Silver Concentration spreadsheet under DATA.

Mike

 

[kurt]

Mike – thanks for the reply – I understand all that as it applies to refining as a hobby were time & production is not important – I have been hobby refining for going on 4 years now & have used computer power supplies to run my cells & my last cell was set up with a variable
(0 – 10 volt 0 -10 amp) power supply

However – about 7 month ago I started refining as a legitimate business so time & production are now important factors

Therefore being able to calculate anode size to power supply/cell size to meet production needs is important (I have all the SS to make as many & as big a cells as I want/need so the cell(s)them selves are the cathode)

As I said in my last post – there is plenty of info on cells – but its in bits & pieces here & there & numbers very depending on the bit or piece you find & read --- your own calculation of 30 ma/sq cm is an example of different numbers – 30 ma/sq cm = about .19 amp/sq inch (current density) but other info runs current density as high as .35 amps/sq inch

I “think” I have pieced the info together for making these calculations & (whether I have it right or wrong) I am hoping this thread will bring the info for making this calculation all together in one place for the benefit of others as well as me

Kurt

 

[mikeinkaty]

Kurt - If I ran my cell at 15 amps the electrolyte would get VERY hot. I could probably run the current that high with continuous SS and still get good crystals but I couldn't deal with such high temperature with the glass container. I did try my present cell at 10 amps and the electrolyte started getting hot. Plus, I started getting the edge effect crystals again. And, I don't really know what high temps would do in regard to quality crystal growth. Remember all those Gypsum crystals found in that cave? They said those were formed at 200+ degrees or thereabouts.

My next cell will again be of the Mobieus type but I'm going to try a SS stock pot with vertical sides. 9" tall and 7.5" wide. I'm going to try cutting off the bottom of a plastic jug or bottle and slide it down to the bottom so that only the sides of the vessel will act as the cathode. What I'd like to do is paint the bottom of the pot then slide in the plastic that has several small holes drilled. My thoughts are to be able to harvest the crystals by just sliding the plastic out of the pot. I don't yet know what type paint would be best and/or practical. It might work or it might not. For me, half the fun is in the experimentation.

Mike

 

[mikeinkaty]

According to this: http://www.savitapall.com/Redox/Notes/Quantitative Aspects of Electrolytic Reactions.pdf

About 4.6 ozt per 6 hrs per 6 amps. or 18.4 ozt in 12 hours at 12 amps.

Mike

 

[kurt]

I recently received some PMs from another forum member asking some questions about this thread – The problem is that I don’t really have the answers as I my self am trying to piece it together from bits & pieces of info found here & there

So I asked if it was OK to post his PMs here on the forum in hopes of sparking more in put

The following few post are from our private discussion – which I have edited (as best I could) to what’s relative to this thread

Kurt