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Topic: Redox half-reaction in an acidic solution.  (Read 2513 times)

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Offline bopll

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Redox half-reaction in an acidic solution.
« on: September 29, 2010, 06:45:37 PM »
The problem: complete the half reaction.

C2O42-  :rarrow: CO2 (acidic solution)

the answer manual gives the following:

H2C2O4  :rarrow: 2 CO2 + 2 H+ + 2 e-

that's dandy and all, but why on earth did the hydrogens bond to the C2O42- instead of just floating around like in the following reaction?

6 e- + 14 H+ + Cr2O72-  :rarrow: 2 Cr3+ + 7 H2O

the methodology in the second one makes sense to me, but not the first one.  what's the difference in the two? They're done in two completely different ways.  And each different way produces a different answer.  

thanks

Offline Borek

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Re: Redox half-reaction in an acidic solution.
« Reply #1 on: September 29, 2010, 06:57:59 PM »
Oxalic acid is a weak one.
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Offline bopll

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Re: Redox half-reaction in an acidic solution.
« Reply #2 on: September 29, 2010, 08:03:15 PM »
Isn't chromic acid a weak acid too?

Offline Borek

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Re: Redox half-reaction in an acidic solution.
« Reply #3 on: September 30, 2010, 02:51:14 AM »
Good question. As far as I remember it is much stronger one.

Perhaps it is just a matter of convention then. I have no problems with writing oxalate oxidation equation in both forms (as oxalate and as an acid). As it is usually oxidized in very low pH it is mostly protonated, so the situation is relatively simple. With chromic acid it is much more complicated - it gets protonated to some extent, but it also polymerizes (dichromate is only a first step).
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