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Home Q & A Board Science-Related Homework Help Chemistry (Moderator: Laser_3)
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ilikepoop ilikepoop
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12 years ago
Here is the procedure:

Obtain 25 mL of ammonium metavanadate solution (in 2 M H2SO4) in a 50-mL Erlenmeyer flask. The solution is yellow, which indicates that the vanadium is in the form of VO2+. What is the oxidation state of vanadium in this solution?
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#1
12 years ago
Be careful, since there are two vanadium ions written as VO2+. One of them is (VO2)+, or vanadyl(V), and the other is VO(2+), which is vanadyl(IV). Vanadyl(V) is yellow and vanadyl(IV) is blue, so I guess you meant (VO2)+. THe oxidation state of vanadium plus the oxidation states of the two oxygen atoms (2 × -2 = -4) must equal the charge of the ion (+1), so the vanadium atom is in its +5 oxidation state.
wrote...
#2
12 years ago
+5
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