1. ## Broken Up

I have 2 litres of pure water that I electrically separate into hydrogen and oxygen. I know that water is made up of H2O and so I expect to produce twice as much hydrogen as oxygen. However, upon examination, I find that the volume of hydrogen gas produced is NOT twice the volume of the oxygen gas produced.

Why?

2. ## Re: Broken Up

I am trying to see where the trick is in the question and , as yet, I have failed but, to get my brain working:

<span style="background-color: #FFFF00; color: #000000; font-weight: bold"><font color=yellow>Your measuring apparatus is inaccurate.
The measurements wewe not taken at the same temperature and pressure.
The collecting vessels were not big enough to contain the electrolytic products at the same pressure. The volume of water stipulated would produce 2666.7 cubic decimetres of hydrogen gas and 1333.35 cubic decimetres of oxygen gas at RTP (approx).</font color=yellow></span hi>

3. ## Re: Broken Up

At a guess <span style="background-color: #FFFF00; color: #000000; font-weight: bold"><font color=yellow>the hydrogen atoms are smaller than the oxygen atoms thus, although there may be twice as many, they do not result in twice the volume.</font color=yellow></span hi>

4. ## Re: Broken Up

The size of gaseous atoms/molecules is negligible compared to their separation.
Avogadro's Hypothesis - Equal volumes of all gases at constant temperature and pressure contain the same number of molecules. For ideal gases (and hydrogen and oxygen are ideal at RTP), 1 mole of gas occupies 24 cubic decimetres (approx).

5. ## Re: Broken Up

I can only "guess" at the discrepancy, but can think of several reasons:
<span style="background-color: #FFFF00; color: #000000; font-weight: bold"><font color=yellow>If they are both gasses and at similar temperatures I would expect to get a 1 mole of hydrogen H2 and 0.5 mole of oxygen (O2), whihc would have twice the moles of Hydrogen over oxygen and hence twice the volume (you wil not get twice the mass)

I can only "guess" that you have them at different temperatures or you have collected them both at cryo temperatures so that they are both liquids or 1 is a liquid and 1 a gas or they are stored at different pressures.

The true given is the the moles and molecules will be 2:1, volumes, weights, etc will vary depending on the conditions. The masses will also be constant, but it will not be 2:1 (you will 16 times more mass of O2 than H2)</font color=yellow></span hi>

Steve

6. ## Re: Broken Up

Question: <span style="background-color: #FFFF00; color: #000000; font-weight: bold"><font color=yellow>Is that fact that you are using 'pure' water relevant? Not that it would affect the composition of the water, just a bit difficult to achieve, I would guess...</font color=yellow></span hi>?

7. ## Re: Broken Up

To be honest, I cannot now remember why I mentioned 'pure' - I think it was so that people don't go down the track of saying that impurities are releasing additional H.

8. ## Re: Broken Up

<P ID="edit" class=small>(Edited by Timbo on 06-Aug-04 09:40. Clarification of some items)</P>More info:

This is all done at room temperatures using standard high school equipment that collects the H at one electrode (is that the right name) and the O at the other. There is no spillage, seepage, etc - I collect all of the gas that comes off the top of the water. You don't need to separate all the liquid to gas - the volume of liquid mentioned is a red herring.

However, you do not get a 2:1 relationship by volume or by mole in the gas collected. Why?

9. ## Re: Broken Up

<span style="background-color: #FFFF00; color: #000000; font-weight: bold"><font color=yellow>The solubilties of the two gases in water are different. Slightly more oxygen will dissolve in the water as it moves from its electrode to the surface than hydrogen, hence the 2:1 ratio will not be exact.</font color=yellow></span hi>

10. ## Re: Broken Up

Yup - that is it! Well done! <img src=/S/trophy.gif border=0 alt=trophy width=15 height=15>

11. ## Re: Broken Up

I don't know this answer: but I would think that the error in the volume ratio due to the solubilty of the oxygen would not be a lot larger than the error from assuming that they are both ideal gasses. How much of a increase in the ratio from 2:1 is expected due to solubility?

So, if you are being picky about the difference in solubilities, I would think that we shouldn't ignore the fact that neither of them are ideal gasses so the whole concept of even "expecting" them to have the same volumes is a moot point.

If the gasses are not ideal, and even if the same number of molecules were collected of each (granted you will have actually have less for the oxygen due to solubility), they would not have the expected ratio of volumes since the number of molecules of each actually have different volumes. In reality, gasses are not ideal and don't follow the ideal gas law.

Steve

12. ## Re: Broken Up

I don't get it.

If you convert <big>all</big> of the water into Hydrogen and Oxygen then how can anything remain <span style="background-color: #FFFF00; color: #000000; font-weight: bold"><font color=yellow>dissolved</font color=yellow></span hi>?

StuartR

13. ## Re: Broken Up

Perfectly true, but Tim wrote "You don't need to separate all the liquid to gas - the volume of liquid mentioned is a red herring." The explanation only works(?) with his proviso.

14. ## Re: Broken Up

Yes you said:
"You don't need to separate all the liquid to gas - the volume of liquid mentioned is a red herring", but your answer actually requires that you MAY NOT separate all the liquid into gas. It requires that you measure it before it is completely finished to have dissolved hydrogen and oxygen in the water and this was never hinted at nor explicitly stated:
"I have 2 litres of pure water that I electrically separate into hydrogen and oxygen" to most of us implies that it is all separated and nowhere hints at "only partial completion" or "during the separation"

Rob's answer implied that the ratio would be greater than 2:1 during the electrolysis (though at the end it would be 2:1!) and you imply that is correct.

Is this the answer that you found for this? Do you have a source? It seems to me, based on the solubility constants, that the ratio should be less than 2:1 during the electrolysis: the molar solubility of the oxygen is not quite twice the hydrogen, but the partial pressure of the hydrogen should be double the oxygen so the hydrogen should keep more molecules in the water than the oxygen (though the oxygen will have a lot more mass dissolved due to molecular weight differences).

Steve

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