CHRIS25
National Hazard
Posts: 951
Registered: 6-4-2012
Location: Ireland
Member Is Offline
Mood: No Mood
|
|
what determines thickness when copper plating?
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
International Hazard
Posts: 3972
Registered: 9-12-2002
Location: Estados Unidos de La Republica Mexicana
Member Is Offline
Mood: ....getting old and drowning in a sea of knowledge
|
|
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
It's better to die on your feet, than live on your knees....Emiliano Zapata.
|
|
CHRIS25
National Hazard
Posts: 951
Registered: 6-4-2012
Location: Ireland
Member Is Offline
Mood: No Mood
|
|
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
International Hazard
Posts: 3972
Registered: 9-12-2002
Location: Estados Unidos de La Republica Mexicana
Member Is Offline
Mood: ....getting old and drowning in a sea of knowledge
|
|
Prior to modernization the methods used could be called home brewed,...see,
http://www.nmfrc.org/subs/history/nov1953c.cfm
It's better to die on your feet, than live on your knees....Emiliano Zapata.
|
|
Nicodem
Super Moderator
Posts: 4230
Registered: 28-12-2004
Member Is Offline
Mood: No Mood
|
|
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).
…there is a human touch of the cultist “believer” in every theorist that he must struggle against as being
unworthy of the scientist. Some of the greatest men of science have publicly repudiated a theory which earlier they hotly defended. In this lies their
scientific temper, not in the scientific defense of the theory. - Weston La Barre (Ghost Dance, 1972)
Read the The ScienceMadness Guidelines!
|
|
CHRIS25
National Hazard
Posts: 951
Registered: 6-4-2012
Location: Ireland
Member Is Offline
Mood: No Mood
|
|
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
National Hazard
Posts: 685
Registered: 5-1-2011
Member Is Offline
Mood: No Mood
|
|
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
Chemical CURIOSITY KILLED THE CATalyst.
|
|
watson.fawkes
International Hazard
Posts: 2793
Registered: 16-8-2008
Member Is Offline
Mood: No Mood
|
|
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
International Hazard
Posts: 2793
Registered: 16-8-2008
Member Is Offline
Mood: No Mood
|
|
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.
|
|
Nicodem
|
Thread Split 15-4-2012 at 06:06 |
Nicodem
|
Thread Split 15-4-2012 at 06:12 |