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What's the relationship between amp-minutes and metal consumption for gold?


A discussion started in 2004 and continuing through 2017 . . .

(2003)

Q. How can I calculate the estimated consumption of metal from a gold bath when plating with insoluble anodes using the readings on the amp-minute meter?
I'd like to be able to predict when to replace metal without having to titrate or do AA.

John Nelson
- Leesburg, Virginia, US


(2003)

A. If you plate at 100 percent efficiency (and you may be able to approach it with some gold baths), Faraday's Law says 96,487 ampere-seconds will deposit one gram equivalent weight. The Metal Finishing Guidebook has a table in the back which does the conversions for you and factors in the atomic weight, usual oxidation state, and density of commonly plated metals and gives you the weight and thickness you can expect to deposit per ampere-hour.

This will probably not free you of all need to titrate or do AA analysis, but it can reduce the required frequency because you will be replacing gold in rough proportion as you use it, so it will be longer before the bath drifts out of the control range.

Ted Mooney, finishing.com
Teds signature
Ted Mooney, P.E. RET
finishing.com
Pine Beach, New Jersey


Electronic scale

(2003)

A. Invest in a nice little scale. Weight some parts before plating and then after. Keep track of what is plated. You will have better control of your tank and better costs of gold on goods.

Jon Quirt
- Minneapolis, Minnesota



Gold Bath Calculation for Alumina Substrate Plating

December 13, 2016

Q. We are having issue calculating time and current based on various substrates we are manufacturing. In an attempt to explain, we plate multiple 2x2 alumina substrates in a gold bath. What we are trying to understand (not being chemists or people familiar with plating) is how to calculate current and time to coat different volumes of boards with 50 microinches of gold.

Example: if I have 4 2x2 substrates that would be a surface area of 32 square inches (4 sq in per side times 8 sides), if my tank is 20 gallons with a density of 1 troy oz/gal of 24k gold what current and time should I plate to achieve the 50 microinches.

What we need to be able to do is make this calculation for different surface areas. We have researched the finishing handbook as suggested in prior posts but we keep getting confused as it appears that the surface area does not enter into the equation.

Any help would be greatly appreciated !! Thanks in advance.

Brian Alexander
design engineer - Manchester, New Hampshire, USA


December 2016

Hi Brian. The relationship is very simple in concept: your rectifier/power supply moves electrons from the anode to the cathode, which causes a corresponding number of gold ions to be reduced by those electrons and deposited onto the cathode as gold atoms. If your gold plating solution is K+Au+(CN)-2, one electron will reduce the gold from this ion into an atom of gold deposited on the cathode.

The number of electrons your rectifier applies is expressed in Amperes, and Faraday's Law says 96,487 Ampere-seconds (96,487 Coulombs) will deposit one gram equivalent of gold. There is an "Electrochemical Equivalents" chart in every issue of the Metal Finishing Guidebook which does most of the math conversions for you, incorporating the atomic weight, oxidation state, and density of metals, so that you can just directly read that 6.2 amp-hours will deposit 0.001" of gold on one square foot of surface.

For gold plating we generally think in amp-minutes rather than amp-hours, so 372 amp-minutes will deposit 0.001" of gold on one square foot of surface. That means that to deposit 50 microinches will require 18.6 amp-minutes per square foot.

Up to this point we're working from first principles, but beyond this point we start getting into empirical stuff: you can't plate at infinite speed because the ions can't migrate through the solution and boundary layer that fast, so you have to reduce the applied current to a point where experience shows that it will work. Depending on the gold plating process you are using, this might be between about 5 and 20 Amps/square foot of surface area plated.

Sometimes some of the electricity which you apply ends up pulling hydrogen out of the plating solution water rather than depositing gold, i.e., the efficiency may not be 100%; it can be anywhere from 30% for electronics plating to over 95% for decorative plating. The vendor of your gold plating solution can tell you what process you are using and its typical acceptable current range and efficiency.

But it is conventional and best to adjust the current rather than the time when the load size varies. For example, if two minutes of plating time at 20 Amps satisfactorily plates a surface area of one square foot, then it would be best to handle a load of two square foot by plating at 40 Amps for two minutes. Good luck.

Regards,

pic of Ted Mooney
Teds signature
Ted Mooney, P.E. RET
finishing.com
Pine Beach, New Jersey


simultaneous December 14, 2016

thumbs up signThank you for taking the time to answer me -- greatly appreciated -- I may have more questions further down the road once I absorb this all a bit more.

Brian

Brian Alexander [returning]
- Manchester, New Hampshire USA


December 14, 2016

A. Hi Brian
There is not nearly enough information in your question to attempt a helpful answer. What is the purpose of the deposit - not all gold plating is the same and pure gold is rare for technical applications. What chemistry are you using? Simply dissolving GPC in water is not good enough.
What are you plating onto, certainly not directly onto alumina? And what pre-treatments are you using - and many more questions.
A couple of observations:
20 gallons seems to be a vast tank for such a small job. The gold in solution is worth a lot, particularly as you are inexperienced and liable to contaminate it. Start small!
Gold plating chemicals are purchased from specialised suppliers who are always willing to give specific technical advice; just ask. Brewing your own plating solution is a recipe for sleepless nights and expensive failures.
If you are determined to tackle a very steep learning curve solo, I would suggest that you set up a simple copper plating tank and practice. The mistakes will be much cheaper.

geoff smith
Geoff Smith
Hampshire,
       England



December 14, 2016

Q. This is a system that has at one time been successful however the process experts have left and it has fallen to electronic design engineers to "get a handle" on it.

The bath is a cyanide based bath with a pH of 6.44. copper is 101.8 ppm, Nickel is 22.6 ppm. Specific gravity is 1.2324 gm/ml and Baume is 27.34 deg, gold is .959 troy oz/gal. We are not attempting to plate gold to bare alumina - that would be silly. The substrates are metalized with sputtered nichrome (various thickness based on ohms/square), nickel (approx. 1000 Angstroms) and a base layer of gold (750-1000 angstroms).

We have learned a lot in a couple of days and are close but any more insight would be appreciated from those more knowledgeable than us. Thanks in advance.

Brian Alexander [returning]
- Manchester, New Hampshire USA



March 27, 2017

Q. Earlier in this thread, it states: "

"For gold plating we generally think in amp-minutes rather than amp-hours, so 372 amp-minutes will deposit 0.001" of gold on one square foot of surface. That means that to deposit 50 microinches will require 18.6 amp-minutes per square foot."

Can this equation be interpreted to mean that it takes 372 Amp-minutes/144 to deposit .001" of gold on one square inch of surface? Thanks.

john_talcott
John Talcott
band instrument repair - Springville, Utah USA


March 2017

A. Hi John. If the plating bath operated at 100% current efficiency, yes. If you see almost no fizzing from the generation of hydrogen, your plating bath is probably quite efficient and this number is not far off -- 90 to 95% efficiency is possible for decorative plating. If you see abundant fizzing, though, your plating bath is low efficiency and will require proportionately more time. "Nobody" plates .001" thick though.

Regards,

pic of Ted Mooney
Teds signature
Ted Mooney, P.E. RET
finishing.com
Pine Beach, New Jersey


March 28, 2017

Q. Thanks for the quick response. We are trying to calculate the time to gold plate 5.5 square inches on flute headjoints to a thickness of 2.5 microns, and every bit of information helps.
As a layman at this I would have thought that to see all the bubbles meant that it was working efficiently. It seems I need to read up on the production of hydrogen in the process.

john_talcott
John Talcott [returning]
- Springville, Utah


March 2017

A. Hi again. When your power supply pumps electrons through the copper wiring, removing them from the anode and pushing them onto the cathode, they don't just endlessly pile up. Rather, those electrons seek neutralization by reducing the positively charged dissolved ions of gold adjacent to the cathode, converting them to neutrally charged gold atoms which deposit on the cathode.

But in fact, some percentage of those electrons will, instead, convert neutral charged water (2H2O) molecules into H2^ gas +20H - ions instead of reducing gold ions to gold atoms.

The percentage of the current which does the useful work of depositing gold in called the efficiency.

At 100 percent efficiency, as we've just agreed, 372/144 amp-minutes would deposit a thickness of .001 inch (25.4 microns) inch onto an area of 1 square inch, so it would take 5.5 times that or 12.9 amp-minutes to deposit 25.4 microns on 5.5 square inches. So 2.5 micron thickness would take 12.9 x 2.5/25.4 = 1.27 ampere minutes of current.

Next you have to estimate (hopefully with help your your gold plating process supplier) what is the maximum current density you can sustain without burning (too much electricity wasted on destructive evolution of hydrogen) or other plating problems. If we guess at an allowable current density of 10 amperes/ft2, you should plate 5.5 square inches at a rate of 10 Amps x 5.5 in2/144 in2 = 0.38 Amps, so the plating time would be 1.27 Amp-minutes divided by 0.38 Amps = 3.34 minutes. If the cathode current efficiency was 90%, the plating time would be 3.34/.9 = 3.7 minutes. It could take twice that long, or half that much time, but this puts you into the ballpark that it's a few minutes, not a few seconds or a few hours. Good luck.

Regards,

pic of Ted Mooney
Teds signature
Ted Mooney, P.E. RET
finishing.com
Pine Beach, New Jersey


simultaneous March 28, 2017

A. Hi John
Ted is correct, but efficiency is only part of your problem.
Gold density is 19.2 g/cc but electrodeposited gold is only about 17 depending on alloy and co-deposited organics resulting in a little thicker deposit than calculated.
If you are only looking to control the gold usage or consistent deposit this is not particularly relevant.
But if you are looking for an accurate 2.5 micron you must have a method of measuring thickness - and don't forget that the deposit will be greater at corners etc.
I suggest that your best bet is to take advice from your gold supplier who should have the necessary kit to measure thickness accurately and then base your plating conditions on that.
P.S. I hope you are not using a pure gold deposit on wearing parts like instrument keys. It is far too soft and will wear rapidly.

geoff smith
Geoff Smith
Hampshire,
      England



March 28, 2017

A. If it's a hard, bright, acid gold, I would think it must be alloyed with a small amount of nickel or cobalt and is therefore not 24K, although close - typically around 99.7% pure.

In my experience, these acid golds are not 100% efficient. The hard acid gold baths that I worked with were only about 30-40% efficient. I would ask the vendor what the typical efficiency is for the bath you are using. A way to determine efficiency yourself is to plate a small brass or copper panel of known surface area (say, 3 or 4 in2), using the same current density you are using on your parts. By weighing the panel before and after plating, the efficiency can be calculated. From all this, you can calculate the amp-min requirements for the thickness you desire.

Chris Owen
- Benton, Arkansas, USA

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