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Calculating moles of copper lost during electroplating



(-----) January 30, 2008

Hi I'm Abner Bucol, I'm teaching Chemistry in a local university. I just want to verify formula on how to calculate moles lost and gain from two copper coins after electroplating. I'm only using a simple set-up with 2 coins immersed in copper sulfate-sulfuric acid electroplating solution. I'm also using a 9-V electric source and the plating process lasts 45 minutes.

Abner Bucol
hobbyist - Dumaguete, Oriental Negros, Philippines
^


January 29, 2008

Hello, Abner. If the process were 100 percent efficient, then one Faraday of electricity (96,485 ampere-seconds) would transfer one gram equivalent weight. In standard simple salt solutions the oxidation state of copper is two, so the weight of copper transferred by one Faraday would be the gram molecular weight of copper divided by two. Note that most plating is not 100 percent efficient. 75 percent might be as good a guess as any for acid copper plating. Good luck.

Ted Mooney, finishing.com
Ted Mooney, P.E.
Striving to live Aloha
finishing.com - Pine Beach, New Jersey
^


January 31, 2008

In addition to what Ted explained I noticed you do not mention current. If you are not reading current (amps) you are totally lost. Voltage doesn't mean anything. You don't need to know volts. You need current multiplied by time (amperes-seconds). Manually measuring is not very accurate, for precise calculations get a digital amp-hour meter of the right size.

Guillermo Marrufo
Monterrey, NL, Mexico
^

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