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When we reduce Vanadium (V) using iron(II), why are we adding 85% Phosphoric acid? After the above reduction, if we are going to titrate the mixture with Potassium permanganate, why are we adding dipotassiumpersulfate to the mixture before the titratioin?

Here's the procedure:
To 5 ml of vanadium (V) solution, add 2 ml of 6 M sulfuric acid, 1 ml of phosphoric acid and 0.20 g of hydrated ferrous sulfate to reduce vanadium (V). Then swirl the solution and allow it to stand for 3 minutes. Then add 0.23 g of dipotassiumpersulfate and let to stand for 5 minutes. Stir the contents vigorously and titrate with 0.02 standard potassium permanganate.

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    $\begingroup$ Could you provide either 1) a link to your procedure or 2) more detail about it? These questions are hard to answer out of context. $\endgroup$ – Ben Norris Jan 16 '14 at 17:20
  • $\begingroup$ Procedure : to 5 ml of Vanadium (v) solution, add 2 ml of 6M sulfuric acid, 1ml of Phosphoric acid and 0.20g of hydrated Ferrous sulfate to reduce Vanadium (v). then, swirl the solution and allow it to stand for 3 minutes. then add 0.23 g of dipotassiumpersulfate and let to stand for 5 minutes. stir the contents vigorously and titrate with 0.02 standard potassium permanganate. $\endgroup$ – user4191 Jan 17 '14 at 14:42
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I think I realize (a part of) the answer, concerning the conditions of a redox titration of vanadium(V), for which a metavanadate, such as $\ce{NH4VO3}$ is typically used as starting material. Under acidic conditions, i.e. in the presence of 85% $\ce{H3PO4}$, the colourless $\ce{VO2+}$ is formed.

$\ce{VO3- + 2 H+ -> VO2+ + H2O} $

Again under acidic conditions, the latter species is reduced by Fe(II) to the blue $\ce{VO^{2+}}$.

$\ce{VO2+ + Fe^{2+} + 2 H+ -> VO^{2+} + Fe^{3+} + H2O} $

Is is conceivable that the peroxodisulfate is used to oxidize the excess of Fe(II), which otherwise would be oxidized by the permanganate?

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