[CHEM] Accuracy of Electrolysis understanding check.
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Of course, I am gonna be a bit biased to the inputs of Chemistry veterans. Sorry :D
During the electrolysis of dilute sulphuric acid, which is technically, electrolysis of acidified water (or just water), I came across a question of why the volume of hydrogen gas produced is twice the volume of oxygen gas.
First thing that came to my mind (I am certain of the others too, or maybe I was too naive), is because of the dibasic nature of sulphuric acid.
To be exact, just check my answers.
My answer: The number of electrons required to reduce H+ ions into hydrogen gas is two times lesser than the number of electrons required to oxidise the OH- ions into oxygen gas. Thus, the same number of electrons would reduce twice the number of H+ ions into hydrogen gas than oxidise the number of OH- ions into oxygen gas.
Correct?
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During the electrolysis of dilute sulphuric acid, which is technically, electrolysis of acidified water (or just water)
Pure H20 cannot conduct electricity. There has to be ions dissolved in it such as H+ and SO42- that would cause the H+ and OH- of water to dissociate.
Doesnt the answer have to do with your ionic equation? 4OH- -> 2H20 + O2 + 4E-
2H+ + 2E- -> H2
I humbly think it has nothing to do with dibasic since that only affects the max amount of H+ that can be produced. Over here we are talking about every unit time, there is 2 times more hydrogen produced compared to oxygen.
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As stated, yeah, dibasic nature means naive thinking.
So I guess I'm correct heh.
Pure H2O is? I think you mean distilled water (a.k.a. DEIONISED water). Because water from the tap CAN conduct electricity, and I'm telling you this now so that you won't try it :D
Thanks!
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Garrick, David is right. You are carrying out electrolysis of water only. The sulfuric acid is only to allow a current to flow (required for electrolysis to get started).
To explain "why the volume of hydrogen gas produced is twice the volume of oxygen gas", you need to write the reduction half equation at the cathode and the oxidation half equation at the anode, and balance the electrons. Do it and you'll see the answer.
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So in all, it's about the charge per unit time right? Did anyone of you read my answer? I understand about the half equations, but what if I have to describe it...
4H+ + 4e- --> 2H2
4OH- --> 2H2O + O2 + 4e-
Been writing this for like 98496422146 times liao... oh yes, and of course, my "worded" answer...
My answer: The number of electrons required to reduce H+ ions into hydrogen gas is two times lesser than the number of electrons required to oxidise the OH- ions into oxygen gas. Thus, the same number of electrons would reduce twice the number of H+ ions into hydrogen gas than oxidise the number of OH- ions into oxygen gas.
Thanks! :D
EDIT: I just realised I have to quote Avogadro's Law in my answer, do I?
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Originally posted by Garrick_3658:
So in all, it's about the charge per unit time right? Did anyone of you read my answer? I understand about the half equations, but what if I have to describe it...
4H+ + 4e- --> 2H2
4OH- --> 2H2O + O2 + 4e-
Been writing this for like 98496422146 times liao... oh yes, and of course, my "worded" answer...
Thanks! :D
EDIT: I just realised I have to quote Avogadro's Law in my answer, do I?
Yes, correct, Garrick. Well done.
As to Avogadro, you don't actually have to specify the number 6.02 x 10^
23.
One way to phrase the answer could be :
Cathode :
4H+ + 4e- --> 2H2
Anode :
4OH- --> 2H2O + O2 + 4e-
Overall :
2H2O ---> 2H2 + O2
From the equations above, For every 4 moles of electrons passed from anode to cathode, 2 moles of hydrogen gas and 1 mole of oxygen gas will be evolved.
Any further qn?