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    Hi everyone,

    basically i was going through the f325 jan2013 thread, and the 7 mark question about vanadium was being discussed.

    a member posted a series of questions (copy and paste) that i think may have been from the ocr textbook, because the post was in response to a previous poster about how these kind of questions were not even in the text book. I was hoping if someone knew where these questions came from, as in, was it from the ocr chemistry text book, or elsewhere?

    33.Vanadium can exist in a number of different oxidation states. One compound of vanadium is ammonium vanadate(V) and this contains the ion VO3–. This can be reduced to V2+ in several steps, using zinc metal and aqueous sulphuric acid.
    (a) 25.0 cm3 of 0.100 mol dm–3 ammonium vanadate(V) is completely reduced to V2+(aq) using zinc and aqueous sulphuric acid. The resulting solution is titrated with 0.0500 mol dm–3 MnO4–(aq) and 30.0 cm3 is required to oxidise the V2+(aq) back to VO3–(aq).
    The half equation for acidified MnO4– acting as an oxidising agent is shown below.
    MnO4– + 8H+ + 5e– ----> Mn2+ + 4H2O
    Show that the vanadium has changed oxidation state from +2 to +5 in this titration.






    [4]

    (b) Suggest an equation for the oxidation of V2+(aq) to VO3–(aq) by MnO4–(aq) under acid conditions.





    ................................ ................................ ................................ ........................
    [2]
    [Total 6 marks]

    33. (a) Moles V2+ = 25.0 × 0.100 / 1000 = 0.0025 mols 1
    Moles MnO4– = 30.0 × 0.0500 / 1000 = 0.00150 mols 1
    1 mole of MnO4- changes its Oxidation State by 5 to change
    the Oxidation State of 1.67 moles of V2+ 1
    Oxidation State of V2+ changes by 5 / 1.67 = 3 1


    (b) 3MnO4– + 5V2+ + 3H2O → 3Mn2+ + 5VO3– + 6H+
    (1 mark for correct species, 1 mark for balanced) 2
    [6]




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