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Oxidation related question

  1. Nov 2, 2004 #1
    Hi All,

    [itex]\mathrm{Fe^{3+}(aq) + SCN^{-} (aq) \rightleftharpoons FeSCN^{2+}(aq)}[/itex]

    I add some [itex]\mathrm{NaHSO_{3}(s)}[/itex] to the above reaction !

    This makes the above chemical solution sour, consequently in a sour solution [itex]\mathrm{HSO^{-}_{3}}[/itex] will reduce Fe^{3+} to Fe^{2+}, meanwhile [itex]HSO_{3}^{-}[/itex] is itself oxidated to [itex]SO^{2-}_{4}[/itex].

    a/ The color of the original resolution changes then adding [itex]\mathrm{NaHSO_{3}(s)}[/itex] to yellow.

    b/ By adding [itex]\mathrm{NaHSO_{3}(s)}[/itex] to the orginal solution the equilibrium direction changes, such that it runs from right to left.

    c/ I need help writing the reaction between [itex]HSO^{-}_{3}[/itex] and [itex]SO^{-}_{4}[/itex].

    d/ Which influence does the adding of [itex]HSO^{-}_{3}[/itex] have on the concentration of [itex]Fe^{3+}[/itex]

    I hope that there someone out there who is willing to assist me since this is my last post !

    Sincerely
    Fred
     
  2. jcsd
  3. Nov 2, 2004 #2

    chem_tr

    User Avatar
    Science Advisor
    Gold Member

    Hello,

    Let me first do the redox thing in atomic level, since it's easier than dealing with a bunch of atoms.

    We know that iron(III) is reduced to iron(II), and in turn, sulfite (S4+) is oxidized to sulfate (S6+):

    [tex]S^{4+} \longrightarrow S^{6+}+2e^- [/tex]
    [tex]2Fe^{3+}+2e^-\longrightarrow 2Fe^{2+} [/tex]
    --------------------------------
    [tex]S^{4+}+2Fe^{3+} \longrightarrow S^{6+}+2Fe^{2+}[/tex]

    In the real redox, you'll need to put some other correcting factors (water and hydroxide, I think) to balance additional oxygens.

    Now that we've got rid of the redox reaction, we may look other things. When you add some sodium bisulfate to the medium, a redox reaction occurs; and so the thiocyanate complex changes from [itex]\displaystyle Fe(SCN)^{2+}[/itex] to [itex]\displaystyle Fe(SCN)^+[/itex]; this gives a totally different color. Iron(II) sulfate is also formed from the reaction, and the color may have also come from this one, so I don't think that the equilibrium shifts to the left, but I may be wrong; I don't insist on that.
     
    Last edited: Nov 2, 2004
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