Calculating Time for Electrolysis of Water

[PeetPb]

greetings

I'd like you to check my calculations. The task is to calculate the time needed to produce 10 moles of hydrogen from water by electrolysis with a current of 1Amp.
I first wrote the equation of the reaction that occurs on the cathode
2H2O+2e- = H2 +2OH-
Then I derived the equation for time from Faraday's law
t=(F*n*z)/I
Since there are two electrons exchanging , z=2, I evaluated the equation and get
t= 1929.7 seconds which is .536 hours.

Is this correct ?

thanx

 

[Borek]

Approach is OK, but somehow you are off by orders of magnitude.

Please note this should land in the homework section.

 

[PeetPb]

oh, right thanks , I somehow used the wrong constant ... so again when I plug the numbers along with the units I get
t=(96485.3*2*10 C mol)/(1 mol A) the moles cancel and I'm left with C/A which should be seconds. The actual number is about 536 hours ... is this right ? it seems to be a little lot to me ... I got the Faraday's constant from http://www.wolframalpha.com/input/?i=F&a=*C.F-_*Unit.dflt-&a=UnitClash_*F.*FaradayConstantValue--"

shall I rewrite the thread or would you move it there ?

thanx

 

[Borek]

Hundreds of hours sounds much better. And the thread is already in HW subforum.

 

[PeetPb]

I was just wondering ... shouldn't the time depend on the amount of water that we are electrolysing or on the surface area of the electrodes ?

 

[Borek]

I was just wondering ... shouldn't the time depend on the amount of water that we are electrolysing

Doesn't matter how much water is present - what does matter is how much water decomposed.

or on the surface area of the electrodes ?

As long as the current is constant, electrode size doesn't matter. If you were to compare two experiments, where we supply the same potential, but electrodes are different, then yes - larger electrode will produce more gas. But it will produce more gas because the current will be higher.

 

[PeetPb]

thank you very much ... I got it finally :)