The vanadic ion V 3+ forms green salts and is a good reducing agent. In a neutra

Question

The vanadic ion V3+ forms green salts and is a good reducing agent. In a neutral aqueous solution, it is transformed into the colorless vanadic hydroxide ion, V(OH)4+. The salt, vanadic sulfate V2(SO4)3 , can be oxidized in aqueous solution according to the following incomplete and unbalanced equation fragment:

V2(SO4)3(aq) + ... V(OH)4+(aq) +SO42-(aq) + ...

(a) What is the oxidation number of vanadium on the left hand side of the equation fragment? What is the oxidation number of vanadium on the right hand side?

(b) Using the equation fragment as a guide, complete and balance the reaction for the oxidation of vanadium sulfate using common species that commonly participate in aqueous redox reactions of electrolytes (e.g. hydronium ions, hydroxide ions, water molecules, electrons). Add only species that are absolutely necessary.

(c) Suppose you are given 0.540 g of an unknown substance X and have a 0.200 M aqueous solution of vanadium sulfate. Suppose each molecule of X can accept one electron. You would like to know the molecular weight of X, so you add the vanadium sulfate solution to X in a beaker a little bit at a time and stir thoroughly while watching. What kind of color change do you expect to see as you increase the vanadium sulfate solution? If the solution changes color after 15.0 mL have been added to substance X, what is the molecular weight of X?

Explanation / Answer

a) in V2(SO4)3   V= +III oxidation state.

b) V2(SO4)3 + 8H2O ------> 2V(OH)4 + 3H2SO4 +2H+ +2e-

c) after adding solution volume is 15 ml contain 0.540 g X with 0.2M

0.2M= (0.540/ MW )x (1000 / 10)

MW= 270

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