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| | #1 (permalink) |
| New Member HSC: 2009 Gender: Female
Join Date: Oct 2008
Posts: 23
Last Activity:
18 Nov 2009, 9:49 AM ![]() | You can hide this advertisement by registering. hii'm having a little trouble understanding the half equations for the electrolysis of water: from the table of standard potenstials the two equations involving water are: 02 + 4H+ +4e- --> 2H2O (E0 = 1.23V) 2H2O + 2e- --> H2 + 2OH- (E0 = -0.83V) for the half equations i was always told that the most positive (ie 1.23V) equation stayed the same and the other one reversed and changed signs....so this should then become: 02 + 4H+ +4e- --> 2H2O (E0 = 1.23V) H2 + 2OH- --> 2H2O + 2e- (E0 = 0.83) but every text book i own and all the websites i've searched have it the other way around....why? any help would be appreciated |
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| | #2 (permalink) | ||
| Senior Member HSC: 2005 Gender: Male Location: Newcastle
Join Date: Jun 2004
Posts: 246
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Yesterday, 11:11 PM ![]() | Re: electrolysis of water Quote:
O2 + 4 H+ + 4 e- --> 2 H2O (+1.23 V) 4 H2O + 4 e- --> 2 H2 + 4 OH- (-0.83 V) 2 H2O --> O2 + 4 H+ + 4 e- (-1.23 V) 4 H2O + 4 e- --> 2 H2 + 4 OH- (-0.83 V) --- 2 H2O --> O2 + 2 H2 (-0.40 V) Quote:
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| | #6 (permalink) | |
| New Member HSC: 2009 Gender: Female
Join Date: Oct 2008
Posts: 23
Last Activity:
18 Nov 2009, 9:49 AM ![]() | Re: electrolysis of water Quote:
sorry no, i just hope they are good questions and texts, if not i know that most people doing the exam will feel the same way about it since everyone wouldn't have much practice on belonging section 1 | |
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