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

A. If you plate at 100 percent efficiency (and you may be able to approach it with some gold baths), Faraday's Law says 96,485 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.

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.

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 96,485 Ampere-seconds (96,485 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,

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 may be 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,

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, each electron seeks instant 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 of 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,