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The chemistry of vanadium is noteworthy for the accessibility of the four adjacent oxidation states 2–5. In an aqueous solution, vanadium forms metal aquo complexes of which the colors are lilac [V(H 2 O) 6] 2+, green [V(H 2 O) 6] 3+, blue [VO(H 2 O) 5] 2+, yellow-orange oxides [VO(H 2 O) 5] 3+, the formula for which depends on pH. Vanadium ...
From left: [V(H 2 O) 6] 2+ (lilac), [V(H 2 O) 6] 3+ (green), [VO(H 2 O) 5] 2+ (blue) and [VO(H 2 O) 5] 3+ (yellow).. Vanadium compounds are compounds formed by the element vanadium (V). The chemistry of vanadium is noteworthy for the accessibility of the four adjacent oxidation states 2–5, whereas the chemistry of the other group 5 elements, niobium and tantalum, are somewhat more limited to ...
Formal charges in ozone and the nitrate anion. In chemistry, a formal charge (F.C. or q*), in the covalent view of chemical bonding, is the hypothetical charge assigned to an atom in a molecule, assuming that electrons in all chemical bonds are shared equally between atoms, regardless of relative electronegativity.
The conservation of charge results in the charge-current continuity equation. More generally, the rate of change in charge density ρ within a volume of integration V is equal to the area integral over the current density J through the closed surface S = ∂V, which is in turn equal to the net current I:
There are two rates which determine the current-voltage relationship for an electrode. First is the rate of the chemical reaction at the electrode, which consumes reactants and produces products. This is known as the charge transfer rate. The second is the rate at which reactants are provided, and products removed, from the electrode region by ...
For Faraday's first law, M, F, v are constants; thus, the larger the value of Q, the larger m will be. For Faraday's second law, Q, F, v are constants; thus, the larger the value of (equivalent weight), the larger m will be. In the simple case of constant-current electrolysis, Q = It, leading to
Vanadium(IV) oxide or vanadium dioxide is an inorganic compound with the formula VO 2.It is a dark blue solid. Vanadium(IV) dioxide is amphoteric, dissolving in non-oxidising acids to give the blue vanadyl ion, [VO] 2+ and in alkali to give the brown [V 4 O 9] 2− ion, or at high pH [VO 4] 4−. [4]
Vanadium(III) sulfate is the inorganic compound with the formula V 2 (SO 4) 3. It is a pale yellow solid that is stable to air, in contrast to most vanadium(III) compounds. It slowly dissolves in water to give the green aquo complex [V(H 2 O) 6] 3+. The compound is prepared by treating V 2 O 5 in sulfuric acid with elemental sulfur: [2]