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Offline funboy

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electrolysis question
« on: December 19, 2006, 07:44:21 PM »
By the electrolysis of water, 11.2 L of oxygen at STP was prepared

What charge was required

Using Faradays Law

1 mol of a gas at STP = 22.4 L
Therefore 11.2 L of a gas at STP = 1/2 mole

H2O --> 2H+ + 1/2O2 + 2e-

So 2H+ and 1/2 O2 accompanied by 2 X (9.64 X 10^4) electrons yeilds 1 mol of H2O

(1/2 mole of O would require 2 moles of electrons or 2 X (9.64 X 10^4)

Given that I = Q/T

All I know is Q

Question says nothing about T and I need to calculate I

What am I doing wrong

Offline Borek

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Re: electrolysis question
« Reply #1 on: December 19, 2006, 07:48:55 PM »
What charge was required

(...)

Question says nothing about T and I need to calculate I

What do you need I for if the question - as you have entered it - asks for charge?
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Offline funboy

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Re: electrolysis question
« Reply #2 on: December 19, 2006, 11:18:04 PM »
By the electrolysis of water, 11.2 L of oxygen at STP was prepared

What charge was required

Using Faradays Law

1 mol of a gas at STP = 22.4 L
Therefore 11.2 L of a gas at STP = 1/2 mole

H2O --> 2H+ + 1/2O2 + 2e-

So 2H+ and 1/2 O2 accompanied by 2 X (9.64 X 10^4) electrons yeilds 1 mol of H2O

(1/2 mole of O would require 2 moles of electrons or 2 X (9.64 X 10^4)

Charge required is 2 moles which equals 1.93 X 10^5

Question B
If a current of 0.5A was used, how long did it take

I = Q/T

.5 = 1.93 X 10^5 / X
.5x = 1.93 X 10^5
x = 9.65 X 10 ^4

x / 3.6 X 10^3 = 26.8

Are these answers correct?

Am I close??

Thanks

Chris

Offline Borek

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Re: electrolysis question
« Reply #3 on: December 20, 2006, 04:17:44 AM »
Are these answers correct?

Not without units. 26.8 of what? Otherwise OK.
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Offline Jabus

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Re: electrolysis question
« Reply #4 on: January 03, 2007, 03:06:58 PM »
with I = Q/t

I don't see why the x is used...couldn't it just be:

t = 1.93 x 10^5 / 0.5A

which is about 386000 seconds and divide by 3600 to get hours so it took roughly 107 hours?

Offline pepe02ar

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Re: electrolysis question
« Reply #5 on: January 04, 2007, 06:14:22 PM »
Hello. Those results are wrong. Remeber that the Water cannot conduce elecricity without the precense of a compound wich could transmit electricity. You need (for an electrolisys) a ionic salt (for example) disolved in the Water (an electrolite solution). The Faraday's Laws apllies to the quantity of electrolite descomposed in the electrolisis. Bye.

Offline Borek

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Re: electrolysis question
« Reply #6 on: January 04, 2007, 06:36:48 PM »
Hello. Those results are wrong. Remeber that the Water cannot conduce elecricity without the precense of a compound wich could transmit electricity. You need (for an electrolisys) a ionic salt (for example) disolved in the Water (an electrolite solution). The Faraday's Laws apllies to the quantity of electrolite descomposed in the electrolisis. Bye.

You are right that there is an ionic substance needed. But the rest of your post is wrong - put some NaOH into water and do the calculations to see that you will get exactly the same result, and that NaOH will not get consumed - so in fact you will be electrolysing water.
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Offline jennielynn_1980

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Re: electrolysis question
« Reply #7 on: January 11, 2007, 03:26:46 PM »
with I = Q/t

I don't see why the x is used...couldn't it just be:

t = 1.93 x 10^5 / 0.5A

which is about 386000 seconds and divide by 3600 to get hours so it took roughly 107 hours?

I did the question the same way as Jabus.  Is this a correct way of doing it?

Offline Jabus

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Re: electrolysis question
« Reply #8 on: January 11, 2007, 06:13:09 PM »
After looking into it, I'm pretty sure our result is the correct one Jennie as funboys result would be in C/s. Since the question asks for 'how long did it take' I'm fairly sure our answer is correct. But I suppose someone double checking wouldn't hurt :D

Offline Borek

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Re: electrolysis question
« Reply #9 on: January 11, 2007, 06:50:36 PM »
Fact the funboy solved the question using x instead of t is not important - replace x with t and it will work OK.

However, he did a mistake in his math which I have not catched earlier. Note how his result is exactly 4 times too low (4*26.8=107.2). That's not a coincidence.
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Offline jennielynn_1980

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Re: electrolysis question
« Reply #10 on: January 11, 2007, 08:04:35 PM »
Thanks for clearing that up Borek :)  I knew he was using x instead of t but I wasn't sure how exactly he got his answer of 26.8 that's all.

Offline pepe02ar

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Re: electrolysis question
« Reply #11 on: February 02, 2007, 04:24:11 PM »
Hello. Sorry by the late. Ok. I was wrong, I doesn't take care that details :'(. Ok. We got this redox:
 2 H2O ---- 2 H+ + O2 + 4e-
The Faraday's definition consist that a Faraday (unit of charge) es equal to the charge of ONE MOLE of electrons. In this reaction we can see that four electrons by 2 molecules of Water there are in the change, or there are four electrons by one molecule of Oxygen. So, for one mole of Oxygen, we got four moles of electrons, or, wich is the same, 4 F (four Faraday's). The problem says that 1/2 mole of Oxygen was obtained (in STP), and ask for the charge requeride. Make your own conclusions.

P.D.: Isn't beautifull, the chemistry ? ;D
P.P.D.: Sorry by my bad english.

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