Oxidation of hydroxyurea with oxovanadium(V) ions in acidic aqueous solution

Hydroxyurea (HU) effectively reduces vanadium(V) into vanadium(IV) species (hereafter VV and VIV species, respectively) in acidic aqueous solution via the formation of a transient complex followed by an electron transfer process that includes the formation and subsequent fading out of a free radical, U (U ≡ H2N–C(=O)N(H)O). The electron paramagnetic resonance (EPR) spectra of U in H2O/D2O solutions suggest that the unpaired electron is located predominantly on the hydroxamate hydroxyl-oxygen atom. Visible and VIV–EPR spectroscopic data reveal HU as a two-electron donor, whereas formation of U, which reduces a second VV, indicates that electron transfer occurs in two successive one-electron steps. At the molarity ratio [VV]/[HU] = 2, the studied reaction can be formulated as: 2 VV + HU → 2 VIV + 0.98 CO2 + 0.44 N2O + 1.1 NH3 + 0.1 NH2OH. Lack of evidence for the formation of NO is suggested to be a consequence of the slow oxidation of HNO due to the too low reduction potential of the VV/VIV couple under the experimental conditions used.The nuclear magnetic resonance (51V-NMR) spectral data indicate protonation of (H2O)4VVO2+, and the protonation equilibrium constant was determined to be K = 0.7 M−1. Spectrophotometric titration data for the VV–HU system reveal formation of (H2O)2VVO(OH)U+ and (H2O)3VVOU2+. Their stability constants were calculated as K110 = 5 M−1 and K111 = 22 M−2, where the subscript digits refer to (H2O)4VVO2+, HU and H+, respectively.