Illustrating the Oxidation States of Mn & V

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Illustrating the Oxidation States of Mn & V
© KCl


© KCl
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

Ion (hydrated)VO3- / VO2+VO2+V3+V2+
Oxidation State+5+4+3+2
TestObservationsSummary of reaction
NH4VO3 + acidwhite solid dissolved to a yellow solution

V(V) + Znsolution changes from yellow via green to blue, then to green and finally deep blue, gas evolves with a rotten-egg odour

V(II) + manganate(VII)solution changes from violet to green, then to blue and finally yellow

1.> V(V) + sulphite
2.> Add V(II)solution changes from yellow via green to blue

blue solution turns green

V(V) + iodide...
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