[samiam]

Hello,
my teacher gave us this challenging question and dares us to try it..he said it wont be on any examination but..Im still curious as to how this question is done.
 

Q: an aqueous sodium chloride solution is electrolyzed using a current of 2.00 A. How much time is required to collect 20.0 L of chlorine gas if the gas is collected over water at 20.0 degrees Celsius and the total pressure is 770.0 Torr


he said assume water is already saturated with chlorine so no more dissolves.  Vapor pressure of water is 17.54 torr .. use


he gave us 5 options and said if we got any one of them we probably did it right

1) 3.9710^4 s
2) 7.94*10^4 s
3)1.58*10^3 s
4) 7.94 s
5) 7.94*10^4 min

**My attempts I think we can find pressure of chlorine by subtracting by vapor pressure of water?:s than we can use pv=nrt?
I got stuck

can someone give me the solution thanks!

[bromidewind]

PV=nRT isn't the right route to take at solving this. I'm not sure what your physics background is or what you've covered concerning physics in chemistry, but this relies on Faraday's Laws of Electrolysis. I'm assuming that the current of 2.00 A is constant since it doesn't specify otherwise. The temperature and pressure are only given to show that it's at STP so nothing special needs to be accounted for.

Faraday's laws can be summarized (assuming constant current) as n = (I*t/F)*(1/z), where n = moles, I = current, t = time, F = Faraday's constant (96,485 C/mol), and z = electrons transferred per ion. Since chlorine is only transferring one electron, (1/z) = 1. Solving for t, we get t = (F*n)/I. n is simply the number of moles in 20.0 L of Cl₂ which can be found by multiplying the volume by the density of Cl₂ (3.20 g/L), giving us 64.0 grams and finding the number of moles (64.0 g * mol/35.453 g), giving us 1.81 moles. Plugging everything in, we get t = (96,485 * 1.81) / 2.00 = 87,319 s or 8.73*10⁴ s. Your professor may have used different numbers than what I used or did it completely differently, but it's pretty close to answer #2. Hope this helps!