Sciencemadness Discussion Board

what determines thickness when copper plating?

CHRIS25 - 13-4-2012 at 08:51

I copper plated a pair of tweasers this morning, this is for practical purposes for picking up things inside acidic solutions so that impurities don't get in, while it was a success I was wondering how or what determines the thickness of a plating.

In this instance I dissolved:
25grams copper sulphate and
4ml concentrated sulphuric acid in 125 ml de-ionised water.

And used 6volts with 0.3 amps for two and half hours.

solo - 13-4-2012 at 09:09

Instrument used to determine copper plating thickness is a Fischercope X-Ray, accomplishing a margin of error less than 1% using certified thickness foil standards.

As for your question, the included link will provide some answers,

http://www.thinktink.com/stack/volumes/volvi/copplate.htm

CHRIS25 - 13-4-2012 at 09:18

OK Solo I had a quick read of what you kindly gave me, but this is really not for my purposes. It is commercial and I just thought that there might be a "homebrew" method that would help determine thickness. Thanks anyway.

solo - 13-4-2012 at 09:43

Prior to modernization the methods used could be called home brewed,...see,

http://www.nmfrc.org/subs/history/nov1953c.cfm

Nicodem - 13-4-2012 at 11:23

Quote: Originally posted by CHRIS25  
And used 6volts with 0.3 amps for two and half hours.

You can always make a simple estimation by using these approximations (obviously not real, but good for average values):
- the copper layer is equally thick all over the surface
- the density of the copper in the layer equals that of copper metal (8.94 g)
- no water was reduced during the plating (no hydrogen formation at the cathode)

Now, if you can measure the surface area of the plated object, you can calculate the (average) thickness of the plating by using the Faraday constant and the electric charge you say you used (2700 As).

CHRIS25 - 13-4-2012 at 12:15

Kind of you to help me here guys, but you got to be kidding me. I need a degree in physics to understand the faraday constant after having just read it. I think I will give this part of plating a miss. I wasn't really looking for mathematical atomic accuracies.

regards, Chris.

m1tanker78 - 13-4-2012 at 14:19

Assuming no losses (as Nicodem outlined), approximately 880mg of copper would have plated onto the tweezers. You can roughly figure the surface area by finding how much water the tweezers displace. You'd then know how much surface area was plated with how much copper.

I believe your homework just got a little easier. ;)

Tank

watson.fawkes - 14-4-2012 at 06:47

Quote: Originally posted by CHRIS25  
Kind of you to help me here guys, but you got to be kidding me. I need a degree in physics to understand the faraday constant after having just read it. I think I will give this part of plating a miss. I wasn't really looking for mathematical atomic accuracies.
Look, if you ask a question on a board about science, don't get your test tubes in a twirl when someone responds with a scientific answer. Your attitude here is, frankly, disgusting to me, as it exemplifies the kind of know-nothing and k3wlish attitude that gives amateur science a bad name. If you find an answer not useful to you, at least be gracious about it, rather than pulling your own consumerist string and parroting "science is hard".

watson.fawkes - 14-4-2012 at 06:51

Quote: Originally posted by m1tanker78  
You can roughly figure the surface area by finding how much water the tweezers displace.
That gives you a volume, not a surface area. You'd need the geometric constant relating length to volume in order for displacement to be of any use. For example, for spheres that constant in 4/3 π. You could get that with a computer model, where it could be virtually measured, but that's hardly worth it in this case.