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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Brad on October 23, 2005, 12:48:07 AM

Title: Is the following logic/math for electrolysis of water to gas correct
Post by: Brad on October 23, 2005, 12:48:07 AM
I need to know if these figures are correct.
Below are some of my notes concerning electrolysis of water.
Please forgive the simplistic layout of my math and logic, this is done to keep the facts clear in my mind. Absolute accuracy is not nessary, "close enough for goverment work" is fine for my current experiments. All corrections are welcomed..  8)

1 mole = 1 Faraday = 96500 coulombs

1 mole of water = 18cc/grams~

With electrolysis, it takes 2 electrons to decompose 1 water molecule

Example: 96500 coulombs of water x 2 electrons = 193000 coulombs of electrons needed at 100% efficiency.

“one mole of any gas at STP (standard temperature and pressure) will occupy a volume of approximately 22.4 Liters.  From this, we now see that 2 mol X 22.4 = 44.8 L of hydrogen gas will react with 1 mole 22.4 L of oxygen gas (67200cc total gas) to produce 36.02988 grams/cc of liquid water”.  (2 moles of water)  

Note: 67200 cc gas divided by 36 cc water =1866cc~ total gas per 1cc water.
1cc of water divided by 1866cc (total gas) = .0005359cc of water per cc of gas

From the above statements, how many cc of gas  (STP) can be produced by 1 mole of water?
67200 cc/2 = 33600 cc total gas per mole (18 cc) of water

How many Coulombs of electrons to produce 1 cc of gas from water at 100% efficiency?
193000 Coulombs  (96500 x 2) divided by 33600 cc gas=5.75~ coulombs

note: 193000 Coulombs/18 cc water=10722 coulombs needed per  cc of water  if 100% efficient

Another way to calculate:
How many coulombs of electrons to produce 1 cc gas from water at 100% efficiency?
10722 coulombs~ (electrons ) divided by 1866 cc  (gas)=5.75~ coulombs to produce 1cc of gas

Calculate gas production from a capacitor:

TO CALCULATE JOULES PROVIDED BY CAPACITOR ----
Voltage x capacitance = coulombs
 COULOMBS x VOLTS = JOULES DIVIDED BY 2 = TOTAL JOULES PROVIDED BY CAPACITOR

.
Example: 3f capacitor x 15v= 45 coulombs x 15v =675 joules/ 2=337.5 joules.
 
If the cap used is rated 3f @ 15v  (45 coulombs) and is discharge into a “100%” efficient electrolysis device, how much gas would be produced?
45 coulombs divided by 5.75 coulombs (electrons needed to produce 1 cc gas)= 7.83 cc total gas

From the above example, how many joules are needed to produce 1 cc of gas if 100% efficient? 337.5 joules divided by 7.83 cc gas = 43.1 joules per cc of gas at 100% efficiency.

Again, thanks for any help Brad 8) 8) 8)