reaction between Iron(ii) and Vanadium(v)

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.
> 
> 
> 




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?


 3