✔ 最佳答案
Preparation of V(V) solution in a strongly acidic medium:
Place one spatula measure of solid ammonium vanadate(V), NH4VO3, in a beaker and added about 25 mL of 1 M sulphuric acid and about 5 mL of concentrated sulphuric acid. Stir until a yellow solution is obtained. This yellow solution is V(V) solution which contains dioxovanadium(V) ions.
VO3^-(aq) + 2H^+(aq) = VO2^+(aq) + H2O(l) .. (reversible)
Preparation of V(II) solution :
To a half of the V(V) solution obtained above, add zinc power, a little at a time with stirring. Heat if necessary. When the solution has become violet, filter the mixture. The violet solution obtained is the V(II) solution which contains vanadium(II) ions.
2VO2^+(aq) + 3Zn(s) + 8H^+(aq) → 2V^2+(aq) + 3Zn^2+(aq) + 4H2O(l)
Preparation of V(IV) solution :
To some of V(V) solution obtained above in a test tube, add equal volume of aqueous SO2 and shake. Then boil the solution carefully in a fume cupboard in order to remove excess sulphur dioxide. The blue solution obtained is the V(IV) solution which contains oxovanadium(IV) ions.
2VO2^+(aq) + SO2(aq) → 2VO^2+(aq) + SO4^2-(aq)
or 2VO2^+(aq) + SO3^2-(aq) + 2H^+ → 2VO^2+(aq) + SO4^2-(aq) + H2O(l)
Preparation of V(III) solution:
Mix equal volumes of V(IV) solution and V(II) solution both obtained above. The green solution obtained is the V(III) solution which contains vanadium(III) ions.
VO^2+(aq) + V^2+(aq) + 2H^+(aq) → 2V^3+(aq) + H2O(l)
2010-01-05 23:12:43 補充：
I've never used aqueous SO2 in laboratory, and it is not commercially available.
2010-01-05 23:14:06 補充：
In my experience, V(IV) solution is made by adding a spatula measure of Na2SO3 to V(V) solution, filtering if necessary, and then boiling the solution to remove excess SO2. This is a method of preparing V(IV) solution recommended by ILPAC (Independent Learning Project for Advanced Chemistry).
2010-01-05 23:27:29 補充：
V(IV) solution can also be made by reducing V(V) solution with KI(aq).
2VO^2+(aq) + 2I^-(aq) + 4H^+ → 2V^3+(aq) + I2(aq) + 2H2O(l)
The brown I2(aq) can be removed by added Na2S2O3(aq).
2S2O3^2-(aq) + I2(aq) → S4O6^2-(aq) + 2I^-(aq)
However, it is mixed with a reducing agent, I^- ions.
