What is the absolute change in oxidation number of vanadium in the given reaction?.
• Finding the absolute change in oxidation number of vanadium in the given reaction. • Computing the oxidation states for V2O5. • Finding rules for computing the oxidation numbers for the atoms in V2O5. • No rules found. Since no rules were found for O and one other atom, the oxidation state of O is -2. The oxidation state O(x) of V in V2O5 can be computed from the formula O(x) = C – (Z(x) / S(x)), where C is the compound charge, S(x) is the subscript of atom x and Z(x) is the sum of the products of the other atoms in the compound and their subscripts; the oxidation state of V in V2O5 is thus 0 – (-10 / 2) = 5. • Therefore, the oxidation states for V2O5 are O:-2 and V:5. • Computing the oxidation states for the ionic parts of V2O5. • Therefore, the oxidation states for V2O5 are O:-2 and V:5. • Computing the oxidation states for Cl2. • The oxidation number of an element is 0. • Therefore, the oxidation number of Cl2 is 0. • Therefore, the oxidation states for Cl2 are Cl:0. • Computing t
