Abstract
The kinetics of reduction of vanadium(V) by iodide ions in aqueous solution have been investigated in the presence of oxygen at 25 °C and ionic strength 0·4M(NaClO4). The results show that the reaction occurs through two parallel paths (i), the first being oxygen-independent, the second involving a molecule of O2. The term b(s–1)v=a+b[O2](i) is given by equation (ii). The mechanism of the oxygen-dependent path involves formation of a peroxovanadium(V)b=(3·3 ± 0·5)× 103[H4VO4+][I–]2[H+]2(ii) intermediate, [VO(O2)]+, which, rapidly reacting with iodide ions, regenerates vanadium(V), VO2+, ions.
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