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Description of experiment
Below follows a plain text transcript of the selected
experiment.
Needed compounds: ----------------- sodium formiate : HCOONa sodium sulfite hepta hydrate : Na2SO3 . 7H2O hydrogen peroxide : H2O2 sodium hydroxide : NaOH formic acid : HCOOH sulphuric acid : H2SO4 vanadyl sulfate : VOSO4.5H2O vanadium pentoxide : V2O5
Class: ------ elem=V redox
Summary: -------- Without the help of a strong acid, vanadium pentoxide is not capable of oxidizing formic acid, not even when heated. When some sulfite is added, then incomplete reduction of the V2O5 can be observed. With the help of a strong acid and when in solution, vanadium (V) is capable of oxidizing formic acid, but only very slowly. With formic acid and vanadium (IV), apparently a coordination complex is formed. Another explanation is given below at the end of the description.
Description: ------------ Sequence 1: ----------- Add some solid V2O5 to HCOOH (85% by weight): No reaction.
Heat liquid and gently boil for a few minutes: No visible reaction. The solid V2O5 remains at the bottom, with a colorless liquid on top of it.
Add some solid Na2SO3.7H2O to the still warm liquid: Liquid starts foaming a little and it becomes yellow/green. Part of the vanadium pentoxide is reduced now. The yellow/green color is caused by a mixture of blue [VO]2+ and yellow [VO2]+ or one of its condensation products.
Sequence 2: ----------- Dissolve some V2O5 in a solution of NaOH (some heating is required to do so): A colorless and clear liquid is obtained.
Add some H2O2 (10% by weight): Liquid becomes yellow. It looks like a strongly acidic pervanadyl solution, but this is a strongly alkaline peroxo-vanadium (V) solution.
Add some dilute H2SO4 (2 mol/l): Liquid becomes dark brown, some gas is evolved.
Add a large excess amount of HCOOH: No visible change. After one hour, the liquid still is dark brown and clear.
Wait one day: Liquid has changed color, it now has a hard to describe color, something best described as a mixture between mossgreen/grey/brown. The liquid is clear.
Two days later: Liquid is moss green, with a bluish hue.
Heat part of the liquid and boil gently: Liquid becomes green/blue, more green than blue. The color is not really brilliant, it is a somewhat dull color, certainly not to be compared with a bright cyan color.
Keep both the cold part and the boiled part of the liquid for two weeks: Both liquids shift more towards blue, but the bright blue color of vanadyl ([VO]2+) is not obtained, a dull green compound remains in the color of the liquid and the intensity of the color is somewhat higher than the intensity of a plain vanadyl solution.
Sequence 3: ----------- Prepare an acidic solution of vanadyl sulfate and sulphuric acid and add some sodium formiate to this: The solid partly dissolves, liquid remains bright blue, also when heated for a while.
Add some H2O2 (10%) to the still hot liquid: Liquid at once becomes yellow and foams fairly strongly for a short while. The solid now completely dissolves and the liquid becomes clear.
Further heating of the liquid: Liquid remains yellow and more gas is evolved.
Add some sulphuric acid (appr. 2 mol/l): Liquid becomes brown/red and remains clear. Some gas is evolved at a low rate. This evolution of gas remains for all 20 minutes, during which an eye was kept on the liquid. Every time, when the liquid is shaken, the evolution of gas becomes faster, but it slows down again, when the liquid is put to rest again. The bubbling can be heard clearly.
Wait for two days: The liquid has become much lighter brown.
Wait another two days: The liquid has become green.
Wait another 10 days: The liquid has become blue, but not the original bright blue of plain vanadyl.
Possible explanation: --------------------- When formic acid is involved in reduction of vanadium (V), then there apparently is a need for H2O2. Without H2O2 no reduction can be observed. The reduction product, however, appears to be not pure vanadyl, put a blue compound or mixture. A possible explanation may be the following:
- vanadium (V) is not capable of oxidizing HCOOH. - vanadium (V) forms peroxo compounds with H2O2 - peroxo compounds are capable of oxidizing HCOOH, forming vanadyl very slowly - part of the peroxo compound decomposes again to plain vanadium (V), which together with the blue of vanadyl gives a blue/green liquid. - the remaining part of the strange color is due to traces of peroxo compound.
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