March 28, 2024, 03:33:42 PM
Forum Rules: Read This Before Posting


Topic: most stable Vanadium species  (Read 11497 times)

0 Members and 1 Guest are viewing this topic.

Offline 1stplace

  • Regular Member
  • ***
  • Posts: 12
  • Mole Snacks: +0/-0
most stable Vanadium species
« on: April 06, 2008, 09:36:35 PM »


Determine the ultimate fate of a small piece of vanadium metal placed in contact with a 1 M [H+] solution in the presence of 1 atm of Hydrogen gas at 25C.

I know that this can be solved with the frost diagram method but I have 2 concerns.....In the half reactions, there are 2 ways that the vanadium metal can go: to V(2+) and to V2O5 directly. So, I don't know how to graph that.

My second concern is that is there a quicker way of solving this?

THanks a lot

Offline Structuralist

  • Regular Member
  • ***
  • Posts: 17
  • Mole Snacks: +6/-0
Re: most stable Vanadium species
« Reply #1 on: April 08, 2008, 08:14:43 AM »
Take in care the electromotive force of the 2 possible cells, the first one with V(+2) and the second with V2O5 formed.
Which one is bigger?

Offline 1stplace

  • Regular Member
  • ***
  • Posts: 12
  • Mole Snacks: +0/-0
Re: most stable Vanadium species
« Reply #2 on: April 08, 2008, 01:18:33 PM »
Right Right :P, Thank you very much

I have another question....The second part of the question asks for the most stable vanadium species in a pH=4 solution...I know that I can use the Nernst equation to find the potentials for all these reactions and calculate which is the most stable species but do I have to consider the effects of the pH on the 2H+ + 2e- => H2 equation? Since all these vanadium reactions proceed with 2H+ + 2e- => H2, I have to consider that too, right?

Ex. E3 turns into -0.136 V in a pH=4 solution and since the 2H+ + 2e- => H2's potential turns to -0.237 V, the overall reaction of E3 turns to be -0.136 + 0.237 = 0.101 V. Is this correct?

Offline Structuralist

  • Regular Member
  • ***
  • Posts: 17
  • Mole Snacks: +6/-0
Re: most stable Vanadium species
« Reply #3 on: April 08, 2008, 03:34:19 PM »
You have to use the Nernst equation because the concentration of the H+ will be 10^-4.
Do you think that the H+ concentration will influence the 2H+ + 2e- => H2 equation?
Use the Nernst equation that's enough in this case :)

Offline 1stplace

  • Regular Member
  • ***
  • Posts: 12
  • Mole Snacks: +0/-0
Re: most stable Vanadium species
« Reply #4 on: April 10, 2008, 08:01:06 PM »
Okay :P, thanks a lot

Can you help me with the last part of this question? (there are many parts)

Determine the exact pH range, under which a 1 molar solution of VO2(+) would be stable if all other conditions are standard. Hint: the range is from acidic to slightly basic. Show your work.

For this, I am not sure how to approach this problem. Can you help me?

Thanks a lot

Offline Structuralist

  • Regular Member
  • ***
  • Posts: 17
  • Mole Snacks: +6/-0
Re: most stable Vanadium species
« Reply #5 on: April 11, 2008, 04:29:16 PM »
in this case you should use the Nernst's equation were you have to find two limit concentrations of H+ that make up the conditions needed
After that you will calculate the pH for the two H+ values, therefore you will find the pH range.

Offline 1stplace

  • Regular Member
  • ***
  • Posts: 12
  • Mole Snacks: +0/-0
Re: most stable Vanadium species
« Reply #6 on: April 12, 2008, 09:23:35 AM »
Um....I don't really know how to set that up, can you give me an example?

Thanks a lot

Offline Structuralist

  • Regular Member
  • ***
  • Posts: 17
  • Mole Snacks: +6/-0
Re: most stable Vanadium species
« Reply #7 on: April 12, 2008, 11:19:40 AM »
As you can see there are several outcomes when you put V in a cell.
The cell will have specific emf for each reaction.
Calculate the range of the emf when VO2+ will be most stable. After that consider that the concentration of H+ is not constant using the Nernst's equation calculate the pH range in which you will have the conditions needed. 

Sponsored Links