Author Topic: Caculating Silver PPM by Weight  (Read 639 times)

jimwilson042745

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Caculating Silver PPM by Weight
« on: May 02, 2020, 08:03:02 PM »
For some time now I have struggled with the formula of Faradays Law of Electrolysis being inaccurate when calculating the ppm of silver in DW. I have weighed the silver anode wire before and after a brew time of 20 minutes @ 15mA to find that only 10 mg of silver left the anode not 20! Why is there a discrepancy? What am I missing?

jimwilson042745

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Re: Caculating Silver PPM by Weight
« Reply #1 on: May 02, 2020, 11:09:34 PM »
Thank you for replying. Why is the cathode weight important? So you are saying that some silver plates out on the cathode? If so then that is silver that's not in the water. It may have left the anode but it didn't go into the water. ? I'm not trying to be combative or argumentative. I am just trying to make sense/understand of what I am weighing against what Faradays is saying. If weighing a gold anode to get an accurate or close to accurate ppm is accurate, why is weighing silver any different. My scale weighs to 1 mg and i have ordered a lab grade scale to confirm. Please what am I missing?
« Last Edit: May 02, 2020, 11:16:01 PM by jimwilson042745 »

jimwilson042745

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Re: Caculating Silver PPM by Weight
« Reply #2 on: May 03, 2020, 02:51:24 AM »
Thank you. Just to be clear, all the silver that leaves the anode does not get deposited into the water. Do I understand that to be correct, because some gets deposited on the cathode? Is the oxide that attaches to the anode part of the silver that leaves the anode during the electrolysis process?

jimwilson042745

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Re: Caculating Silver PPM by Weight
« Reply #3 on: May 03, 2020, 04:01:14 PM »
THANK YOU.

Offline imcool

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Re: Caculating Silver PPM by Weight
« Reply #4 on: May 03, 2020, 05:29:35 PM »
please move this to silver section of the form, its in Gold section of forums Thanks

jimwilson042745

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Re: Caculating Silver PPM by Weight
« Reply #5 on: May 03, 2020, 05:59:35 PM »
Can you tell me how to do that. Thanks