Abstract
Oxidation of vanadyl sulfate by H2O2 involves multiple reactions at neutral pH conditions. The primary reaction was found to be oxidation of V(IV) to V(V) using 0.5 equivalent of H2O2, based on the loss of blue color and the visible spectrum. The loss of V(IV) and formation V(V) compounds were confirmed by ESR and 51V-NMR spectra, respectively. In the presence of excess H2O2 (more than two equivalents), the V(V) was converted into diperoxovanadate, the major end-product of these reactions, identified by changes in absorbance in ultraviolet region and by the specific chemical shift in NMR spectrum. The stoichiometric studies on the H2O2 consumed in this reaction support the occurrence of reactions of two-electron oxidation followed by complexing two molecules of H2O2. Addition of a variety of compounds--Tris, ethanol, mannitol, benzoate, formate (hydroxyl radical quenching), histidine, imidazole (singlet oxygen-consumption that also used V(IV) as the reducing source. This reaction requires concomitant oxidation of vanadyl by H2O2, favoured at low H2O2:V(IV) ratio. Another secondary reaction of oxygen release was found to occur during vanadyl oxidation by H2O2 in acidic medium in which the end-product was not diperoxovanadate but appears to be a mixture of VO3+ (-546 ppm), VO3+ (-531 ppm) and VO2+ (-512 ppm), as shown by the 51V-NMR spectrum. This reaction also occurred in phosphate-buffered medium but only on second addition of vanadyl. The compounds that stimulated the oxygen-consumption reaction were found to inhibit the oxygen-release reaction. A combination of these reactions occur depending on the proportion of the reactants (vanadyl and H2O2), the pH of the medium and the presence of some compounds that affect the secondary reactions.
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