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Topic: Electrolysis  (Read 6774 times)

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Offline Rutherford

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Electrolysis
« on: March 28, 2012, 01:43:06 PM »
For how long do you need to pass direct current (I=0.5A) through 50cm3 of a HCl solution (c=0.02mol/dm3) so that the pH is going to be 2?
I know that the pH will reduce because H+ ions take part in the electrolysis. I calculated that 5*10-4 H+ ions need to take part in it. Now I wrote the electrolysis process:
K-:2H++2e--->H2
A+:2Cl--->Cl2+2e-
I used Faradey's law: nFz=It, t=nFz/I, because two electrons take part in the electrolysis I used for z=2 and I get that t=193s which is two times bigger than the right answer. I don't know where I've mistaken. Is it z?

Offline Borek

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Re: Electrolysis
« Reply #1 on: March 28, 2012, 03:18:19 PM »
z is 2 if you are looking for amount of H2 produced, but 1 when calculating amount of H+ removed from the solution.
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Offline Rutherford

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Re: Electrolysis
« Reply #2 on: March 28, 2012, 03:36:58 PM »
I don't understand. Is it related to the coefficients in the equation?

Offline Borek

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Re: Electrolysis
« Reply #3 on: March 28, 2012, 03:41:30 PM »
Just look at the reaction equation - there is one electron consumed for each H+ reduced (2 electrons per two H+) and two electrons consumed per each H2 produced.
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Offline Rutherford

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Re: Electrolysis
« Reply #4 on: March 29, 2012, 08:04:11 AM »
Thanks Borek.

Offline Rutherford

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Re: Electrolysis
« Reply #5 on: March 30, 2012, 09:36:12 AM »
I've done many problems correct thanks to your advice, but I came to this one:
100cm3 of KOH solution (c=0.01mol/dm3) is electrolysed for 20 hours (I=1A) using Pt electrodes. Calculate the concentration of OH- ions after the electrolysis.

I figured out that the n of the OH- ions remains the same-->1*10-3mol. The cathode process is:
K-:2H2O+2e--->2OH-+H2
Need to calculate the mass of water that is lost during the process. nFz=It n=It/(Fz). z is again the problem. For 2 moles of water, 2 electrons are spent, so I thought that for 1 mole of water 1 electron is spent and I used that z=1. I got two times more moles than I should, to get the right answer.

Offline Olympiad_Tutor

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Re: Electrolysis
« Reply #6 on: March 30, 2012, 11:31:56 AM »
what happens at Anode?
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Offline Rutherford

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Re: Electrolysis
« Reply #7 on: March 30, 2012, 12:43:05 PM »
4OH--->2H2O+O2+4e-
The first one should be multiplied by 2. Still can't figure out what to do now.

Offline Olympiad_Tutor

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Re: Electrolysis
« Reply #8 on: March 30, 2012, 04:23:50 PM »
Try combining your half-reactions.
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Offline Rutherford

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Re: Electrolysis
« Reply #9 on: March 31, 2012, 05:51:56 AM »
K-:4H2O+4e--->4OH-+2H2
A+:4OH--->2H2O+O2+4e-
4 moles of water are spent, 2 moles are created. How can I combine these two reactions? Calculate the mass of the spent water and then the mass of the created water and substract them?

Offline Borek

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Re: Electrolysis
« Reply #10 on: March 31, 2012, 06:22:29 AM »
4 moles of water are spent, 2 moles are created.

So what is the overall change?
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Offline Rutherford

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Re: Electrolysis
« Reply #11 on: March 31, 2012, 06:55:58 AM »
2 moles are spent, but how much is z? For those 2 moles z=4, for 1 mole z=2?

Offline juanrga

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Re: Electrolysis
« Reply #12 on: March 31, 2012, 08:00:12 AM »
I don't understand. Is it related to the coefficients in the equation?

z is charge of ion under study (z=1 for H+, for instance) not a stoichiometric coefficient
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Offline Rutherford

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Re: Electrolysis
« Reply #13 on: March 31, 2012, 10:18:17 AM »
What would it be for the 2 H2O moles?

Offline Olympiad_Tutor

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Re: Electrolysis
« Reply #14 on: March 31, 2012, 10:55:26 AM »
4e- are needed to decompose 2 H2O.

Combine the half-reactions aiming to cancel the electrons.  Even though the electrons are cancelled out, they are still responsible for the reaction.  Therefore you need to plug in 4 for moles of electrons into Faraday's law.
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