A vandium electrode is oxidized electrically. If the mass of the elctrode decreases by 114 mg...
A vandium electrode is oxidized electrically. If the mass of the elctrode decreases by 114 mg during the passage of 650 coulombs, what is the oxidation state of the vanadium product?
Homework Answers
we know that
accoriding to faradays first law of elecrtolysis
mass is given by
m= Q x M / F x z
here
z is the number of electrons transferred
given
charge Q = 650
F = faradays constant = 964585
Molar mass of vanadium (M) = 50.94 g/mol
given
mass = 114 mg
so
we get
m= Q x M / F x z
114 x 10-3 = 650 x 50.94 / 96485 x z
z = 3
so
3 electrons are transferred from vanadium to form a product
so vanadium becomes V+3
so
the oxidation state of the vanadium product is +3
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