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Topic: Reduction of Ammonium Metavanadate.  (Read 10419 times)

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Offline zilalti

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Reduction of Ammonium Metavanadate.
« on: November 19, 2007, 04:05:31 PM »
I recently carried out an experiment where Ammonium Metavanadate in H2SO4 was reduced using Sodium Sulphate and re-oxidized by titrating with a Kmno4 solution. I'm trying to work out the oxidation state that the Vanadium was reduced to but im having a little trouble with the calculations.

To work this out im taking the number of moles of Kmno4 titrated, multiplying by 5 to give the number of moles of electrons provided. Then diving this number by the number of moles of the original ammonium metavanadate used to give the number of times it was reduced. However when I do this I seem to be getting ridicolously large numbers.

Does this sound about right for working out the oxidation state.

Thanks

Offline hmx9123

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Re: Reduction of Ammonium Metavanadate.
« Reply #1 on: November 19, 2007, 10:31:56 PM »
Start by writing out the half reactions that are occuring.

Offline zilalti

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Re: Reduction of Ammonium Metavanadate.
« Reply #2 on: November 20, 2007, 03:55:15 AM »
Never mind I already failed  :(

Offline puppy8800

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Re: Reduction of Ammonium Metavanadate.
« Reply #3 on: November 26, 2007, 04:35:09 AM »
notice the color, it may help you
VO2+ blue
V3+ green
V2+ violet

Offline shelanachium

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Re: Reduction of Ammonium Metavanadate.
« Reply #4 on: December 04, 2007, 03:17:30 AM »
Did you mean sodium sulphite? Didn't think the sulphate was a reducing agent. Reduces as [SO3]2- + H2O -> [SO4]2- + 2H+ + 2e. Should help with your equations.

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