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Chemistry Forums for Students => Analytical Chemistry Forum => Topic started by: plu on February 11, 2006, 08:51:52 PM

Title: vanadium species in solution
Post by: plu on February 11, 2006, 08:51:52 PM
Hello.  Again, I find myself stuck with a simple analytical problem  :'(   I am given a table of reduction potentials for various vanadium half-reactions relating to the speciation of vanadium in aqueous solutions and I am asked to find the most stable vanadium-containing species at pH 4 with all other conditions being standard.  I calculated the potential of the hydrogen half-cell to be -0.236 at pH 4.  To identify the most stable vanadium species, would I simply look for the half-reaction that has a value closest to -0.236?

The next question I am given asks to determine the pH range over which a 1 molar solution of VO2+ would be stable if all other conditions are standard.  The reduction potential of VO2+ is 0.991.  I calculated the pH that would be required to make the hydrogen half-cell potential 0.991 and found this to be 0.218.  However, the question gives a hint that the required pH range is from acidic to slightly basic.  I've checked my calculations over and they are all correct.  I am thinking that my mistake here is a conceptual one.  What conceptual mistake have I made? :-[
Title: Re:vanadium species in solution
Post by: Borek on February 13, 2006, 05:07:22 PM
Quote from: plu on February 11, 2006, 08:51:52 PM
I am given a table of reduction potentials for various vanadium half-reactions relating to the speciation of vanadium in aqueous solutions

Can you show it?
Title: Re:vanadium species in solution
Post by: plu on February 14, 2006, 05:17:30 PM
Here is the chart:
2 H+ + 2 e–  ->  H2                                  Eo = 0.000 V
V2+ + 2 e–  ->  V                                      Eo = –1.175 V (1)
V3+ + e–  ->  V2+                                    Eo = –0.255 V (2)
VO2+ + 2 H+ + e–  ->  V3+ + H2O                 Eo = 0.337 V (3)
VO2+ + 2 H+ + e–  ->  VO2+ + H2O               Eo = 0.991 V (4)
V2O5 + 6 H+ + 2 e–  ->  2 VO2+ + 3 H2O       Eo = 0.957 V (5)
V2O5 + 10 H+ + 10 e–  ->  2 V + 5 H2O         Eo = –0.242 V (6)
Title: Re:vanadium species in solution
Post by: Borek on February 14, 2006, 06:29:48 PM
At first sight - stable will be all compounds with halfreaction potential higher than that calculated for H+/H2.

However, note that some of the reactions include H+ - thus their potentials are pH dependent, I suppose that can be the missing point in your calculations.
Title: Re:vanadium species in solution
Post by: plu on April 15, 2006, 10:54:05 AM
Sorry to resurrect an old post but I recently looked over this problem again and realized that I still can't find a solution!  Would there be anybody out there that can give me some insight on this problem?  The question is posed as such: "Given the above half-reaction potentials, find the pH range under which a 1 molar solution of VO2+ is stable.  All other conditions are standard."  Kind thanks
Title: Re: vanadium species in solution
Post by: wereworm73 on May 19, 2006, 11:33:05 PM
Looks hard, but I'll take a stab at it.


H2 --> 2 H+ + 2 e-
2 VO2+ + 4 H+ + 2 e- --> 2 VO2+ + 2 H2O


E = Eo - 0.05916/2 (log [VO2+]2/[VO2+]2[H+]4)

0 = 0.991 - 0.05916/2 (log 1/[H+]4)

Solving for [H+], I get 4.212 x 10-9 M  (pH= 8.376)

I'm not 100% sure if that's correct, but it is slightly basic like the hint says.