Marshall wrote:
> OK, I am trying to figure out what current we should be running on the
> copper/silver alloy for my daughter's pool. What I am getting doesn't seem
> right, can anyone find my error?
>
> If we use 10 mA, that is .01 coulomb per second. It takes 96,500 coulombs to
> free a mol of silver, and a mol of silver is 107 grams. Thus it takes 902
> coulombs to free 1 gram of silver. 10 mA will free .0011 grams of silver per
> second or 1.1 mg of silver per second.
>
> That would put 1.1 ppm of silver into a liter of water per second (This seems
> awful high). The pool contains about 5,000 gallons I believe, which is about
> 20,000 liters. At this rate the pool would get about .055 ppb of silver a
> second, or about 198 ppb of silver an hour.
>
> Could this be right? It only takes about 10 mA of current one hour or two to
> produce sufficient silver for the pool? This does not seem right at all. Can
> anyone find an error in my math?
Yes, there is an error in your math. The following was prepared by Dr. Maass to
clarify the math.
The formula for electrolytic deposition is
Weight = M I t
------
n F
where M = molecular or atomic wt.
I = current (amps = coul/sec)
t = time in seconds
n = # of moles of electrons transferred ( = 1 for Ag)
F = Faraday's constant = 96485 coul/mole
10 mA = 0.01 coul/sec, therefore
weight = (108)(0.01)(3600sec/hr) = 0. 0402 g of Ag/hr
-------------------------
(1)(96485)
which is 0.011 mg/sec
I believe Marshall got 1.1 mg/sec, or two orders of magnitude higher.
frank key
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