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A problem about oxdizing power

  1. Aug 16, 2015 #1
    • You have to show your attempts at answering the question, this is a forum policy. Also, all homework like questions should go to homework forum.
    the three complexes CrO42- Mno42- FeO42-
    which is the order for their oxidizing power? And why?
  2. jcsd
  3. Aug 16, 2015 #2
    The oxidizing power depends on the redox potential of the redox couples considered, the higher the redox potential, the more the metal will give oxygen atoms. So if you know the redox potential of the couples including those 3 complexes , you know which one is the more oxidizing. Of course one complex can be included in two or more couples, it usually depends on the solution's pH.
    I'm not sure of the order of the oxidizing power here, but MnO42- is the most currently used for oxidations. In organic chemistry, we also use chromium, but under its anhydride form CrO3 with an organic solvant, not under its complex form. And I never heard about using FeO42- for oxidations.
    Hope I helped you.
  4. Aug 16, 2015 #3
    Thanks for your reply.
    But I still need to rank their order without data and need to explain my answer.
    I already know their oxidation state are +6. And then I don't know how to proceed.
    Last edited: Aug 16, 2015
  5. Aug 16, 2015 #4
    Okay, so without data you may use the fact that the more an element is electronegative, the higher will be is oxidizing power. And knowing the atomic number of the three elements considered (24Cr, 25Mn, 26Fe), you can say that the more electronegative is Fe, followed by Mn and then Cr, so the order for the oxidizing power of those elements : Fe>Mn>Cr.
    If we suppose that because they are under "the same complex form" (XO42- where X is the element), the oxidizing power order remains the same, I would class them like that :
    FeO42- > MnO42- > CrO42-
  6. Aug 16, 2015 #5
    Oh I see thank you.
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