Chemical Forums

Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: bopll on September 29, 2010, 06:45:37 PM

Title: Redox half-reaction in an acidic solution.
Post by: bopll on September 29, 2010, 06:45:37 PM

The problem: complete the half reaction.

C₂O₄^2-  :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 C₂O₄^2- instead of just floating around like in the following reaction?

6 e^- + 14 H⁺ + Cr₂O₇^2-  :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

Title: Re: Redox half-reaction in an acidic solution.
Post by: Borek on September 29, 2010, 06:57:59 PM

Oxalic acid is a weak one.

Title: Re: Redox half-reaction in an acidic solution.
Post by: bopll on September 29, 2010, 08:03:15 PM

Isn't chromic acid a weak acid too?

Title: Re: Redox half-reaction in an acidic solution.
Post by: Borek 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).