Illustrating_the_Oxidation_States_of_Mn

Illustrating the Oxidation States of Mn & V
© KCl /fatherofchemistry
Procedure
© KCl /fatherofchemistry
Discussion
Part A: Making Mn(VI) from Mn(VII) and Mn(IV)
1.> Explain why only one of the three mixtures reacted to give green Mn(VI). [ANS] By Le Chatelier's Principle, only the alkaline medium will shift the equilibrium to right and yield green MnO42-.
2.> What happened when acid was added to Mn(VI)? Explain.
[ANS] The solution changes from green to purple again. The acid removes OH- on the left side and the decrease in [OH-] favours the
backward reaction to form MnO4- (purple).
Part B: Making Mn(III) from Mn(II) and Mn(VII)
1.> Explain what happened when the Mn(III) solution is diluted.
[ANS] When Mn3+ is diluted, i.e. more water is added on right side, the equilibrium will shift leftward, purple MnO4- will form.
Part C: Making Mn(III) from Mn(II) and Mn(IV)
1.> What is different about the conditions of this experiment (part C) compared with last (part B) which makes its success less likely?
[ANS] The condition of part C is different from part B in which it's in an alkaline medium in contrary to acidic medium in part B. The
OH- will combine with Mn3+ to form insoluble Mn(OH)3 which unfavours this reaction.
Part D: Vanadium
blue, then to green and
violet to green, then to
1.> How would you explain the first appearance of a green colour in the solution? [ANS] The yellow VO2+ is partially reduced to blue VO2+ by zinc. The presence of these 2 ions in about comparable concentration in
the solution results in a green colour.
2.> What are the subsequent changes in colour and why do these changes occur? [ANS] Vanadium(V) is subsequently reduced by zinc dust in acidic medium, which is a strong reducing agent.
The yellow vanadium(V) is reduced to blue vanadium(IV). Vanadium(IV) is continuously reduced to green vanadium(III) and finally violet vanadium(II).
3.> What did you observe when you added iodide ions to vanadium(V)? What caused this colour?
[ANS] Yellow solution turns brown since iodide is oxidize by vanadium(V)
to give brown iodine. Vanadium(V) is reduced to give
vanadium(IV).
4.> Why did you add sodium thiosulphate?
[ANS] To decolourize brown I2 formed - avoid disturbing the original colour of blue vanadium (IV) ion formed.
2S2O32-+ I2→S4O62-+ 2I-
5.> Why does reduction with iodide not give the same result as reduction with zinc? [ANS] The reduction will stop when vanadium(V) ion is reduced to vanadium(IV) by iodide, since the Eø value of the reaction between iodide and vanadium(IV) is negative which means this reaction is less likely to success.
On the other hand, Eø value between Zn and VO2+, VO2+, V3+ & V2+ are all positive which favours for further oxidation of VO2+
and others.
6.> What did you observe when you added sulphite ions to acidified vanadium(V) solution? Does this result correspond with a
prediction made using the Eø values?
[ANS] The solution changes from yellow to blue.
Vanadium is reduced to vanadium(IV), not vanadium(III) predicted from Eø values.
Since Eø value neglects the activation energy required for the conversion between 2 species. The activation energy required
for the reduction of vanadium(IV) by sulphite may be so high that such reduction will not proceed.
7.> How would you set about finding a suitable oxidizing agent for the oxidation of vanadium(II) to vanadium(III) and no further?
Does one appear in the table above?
[ANS] In the experiment, both zinc, sulphite and iodide are reducing agents. Manganese(VII) is a very strong oxidizing agent that can oxidize V2+ to VO2+ directly. The weakest oxidizing agent that can only oxidize V2+ to V3+ and no further is VO2+. VO2+ cannot
further oxidize V3+ to VO2+ since the Eø value is 0.。