Faraday Effect? Full Terminology and Explanation?
Anyone got a good, full terminology of the basis of the Faraday Effect in plating? I know I read a pretty good explanation for it, but I don't remember where.
Thanks in AdvanceMatthew Stiltner
- Toledo, Ohio
I'm assuming you are referring to "Faraday's Law" which relates to plating, rather than the "Faraday Cage Effect" which is applicable to electrostatic spraying.
Picture one atom (ion) of Ni++ floating around in a nickel plating bath. To turn that one ion of Ni++ into one atom of metallic nickel, Ni0, will require 2 electrons because '--' will neutralize the '++' charge of the ion.
You could also look at it the other way, and say that if you have an atom of nickel in your anode basket and you take two electrons from it, it will become one ion of Ni++.
Nickel always ionizes in this 'plus 2' state, whereas chrome ionizes 'plus 3' (trivalent) or 'plus 6' (hexavalent). Copper ionizes at 'plus 1' in a cyanide bath and 'plus 2' in an acid bath. Tin ionizes at 'plus 2' in an acid bath and (hopefully, if you film the anodes) 'plus 4' in an alkaline bath.
So far, so good? Then Faraday's Law is nothing more than applying a conversion factor--
To deposit one gram equivalent weight will require 96,487 ampere-seconds of electricity. That is, if you multiply the amps that are flowing by how many seconds they flow for, and divide by 96500, you have the number of gram equivalent weights that are plated out. A gram equivalent weight is the atomic weight (from the atomic table) divided by the number of electrons needed to neutralize one ion of the metal in question.
In the case of nickel, the atomic weight is 58.69 and the ionization state is 'plus 2'. So 96500 amp-seconds will deposit 29.345 grams of nickel.
The above assumes '100 percent efficiency', that is, none of the electricity is wasted in ionizing the H in water to H+. Say that your nickel bath is 95 percent efficient. Then 96500 amp-seconds (let's call that number of amp-seconds 'a Faraday') will deposit 29.345 grams x .95, or 27.88 grams.
Ted Mooney, P.E. RET
Pine Beach, New Jersey
You knew what I meant :-) Too bad I had my terminology all messed up. You explained it exactly as I wanted to see it too. Much thanksMatthew Stiltner
- Toledo, Ohio
I'm working on the Faraday effect experiment and I want some help on this topic. Or any related topics such as polarization, hall effect,...etc.
THANK YOU FOR ALL.Ashraf Abdallah Al-quraineh
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