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Oxidation number in acidic medium

  1. Jan 31, 2014 #1
    Hi everyone,
    In my school I was given a problem. It's that Cr2O7-2 + H2S in a acidic medium. And my teacher told that Chromium becomes Cr+3 .

    I want to know is there such a way to calculate the oxidation number in acidic medium. Or is it required the experience?
  2. jcsd
  3. Jan 31, 2014 #2


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    Staff: Mentor

    First, experience. When it fails, there are always Pourbaix diagrams.
  4. Jan 31, 2014 #3
    I've recently been trying to teach myself this stuff. Somebody let me know if this is way off base but this is a method that seems to work for me.

    A neutral molecule has an overall oxidization number of zero, and a charged radical or ion has an overall oxidization number equal to it's charge.

    So Cr2O7(2-) has an overall oxidization number of -2. The way I've been dealing with ones like this is to just assume the number of the simple atom (oxygen) and thereby deduce the number of the less obvious one (transition metal in this case).

    Let the oxidization number of O be -2 and that of Cr be "n". Then,

    [tex] 2n + 7\times(-2) = -2[/tex]

    Solving for "n" the oxidization state of the Cr atom in Cr2O7(2-) must be +6.

    If this becomes Cr(3+) on the right hand side of the balanced equation then the oxidization number of Cr is decreasing from 6 to 3, therefore Cr is being reduced in this process.

    Disclaimer. I'm still on my Chemistry "L" plates :). So anyone please feel free to critic this answer if I'm wrong.
  5. Jan 31, 2014 #4


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    Staff: Mentor

    You are for two reasons - one, it doesn't address the OP problem (which is how to predict ON of a product of the reaction), two, it is wrong as oxidation number is a property of an atom, not of the molecule. Cr2O72- has a _charge_ of -2, not oxidation number. Cr in Cr2O72- has oxidation number of +6 (which you calculated correctly).

    Please note oxidation numbers are only an accounting device, handy when balancing redox reactions, but there is no measurable property of atoms that can be connected with ON. Which is why I prefer half reactions approach to balancing redox.
  6. Jan 31, 2014 #5
    Ok. So is it still correct to say that the sum of the oxidization numbers (of the individual atoms) in a molecule must equal the charge, -2 in the above example?

    To be honest I don't even know what the reaction products would be for that reaction, apart from the [itex]Cr^{\,3+}[/itex] as that was given.

    Does the OP need to complete a balanced equation something like as follows?

    [tex]Cr_2O_7^{\,\,2-} + \, ? H_2S \, + \, ?? H_3O^{+} \rightarrow 2 Cr^{\,3+} + \, ???[/tex]
  7. Jan 31, 2014 #6
    I see. So it's up to my experience. I'll have to find more equations and solve them. Thanks for help guys.

    Yes the overall charge is equal to -2
    Cr is +6
    So, 12 + (-14)=-2
  8. Jan 31, 2014 #7
    @chemnoob balancing is not a problem for me as long as I know the oxidation number.
  9. Feb 1, 2014 #8


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    Staff: Mentor

    Yes, that part is perfectly OK.

    Trick is, sometimes assigning individual ONs is quite difficult. Try S2O32-.

    Yes. In such cases hydrogen sulfide gets oxidized to elemental sulfur (when the oxidizer is mild) or to SO42- (when the oxidizer is strong). The important part is

    Cr2O72- + H2S -> Cr3+ + SO42-

    as all other molecules required for balancing (H2O, H+ and OH-) are always present in water solutions and they can be used freely whenever needed to balance charge, hydrogen and oxygen.
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