Hboomans
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8 electron reduction
I have a recipe here it is an 8 electron reduction. where one uses 8 faradays to reduce 160grams.
Now is my question if I want to reduce 20grams is there still 8 faradays needed or will this be less?
Does the amount you want to reduce affect the faradays needed for the reduction to completion?
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Texium
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You can't expect to get a real answer when you come in here talking about "recipes" and using very vague language.
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PHILOU Zrealone
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F= Na x e(-)
F = Faraday constant
Na = Avogadro's number
e(-)= electron charge
You use 8 electrons (e(-)) per molecules what implies 8 x F/mole...
So your compound is 160g/mole and 20g is 1/8th of a mole...
20g will thus require 8 times less than 8F what is 1F!
[Edited on 1-7-2014 by PHILOU Zrealone]
PH Z (PHILOU Zrealone)
"Physic is all what never works; Chemistry is all what stinks and explodes!"-"Life that deadly disease, sexually transmitted."(W.Allen)
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Nicodem
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Thread Moved
1-7-2014 at 10:52 |
Artemus Gordon
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According to Wikipedia, one faraday equals 96485.3399 coulombs. Let's call it 100,000 C to make the math easier.
A current flow of 1 Ampere equals 1C/second. So if you have a power supply that can deliver 10 amps at the voltage you need to drive your reaction, it
would take 10,000 seconds to deliver 1 faraday, which is a bit more than 2 and 3/4 hours.
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