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Doubts about a redox equation

  1. Aug 17, 2015 #1
    It's not clear to me how I can balance the following equation, most of all because of K which I don't know how to deal with:

    [tex]\mathrm{SO_2 + K_2Cr_2O_7 \to Cr_2(SO_4)_3 + SO_3}[/tex] [this is the text of the exercise]

    It's acidic solution for sulfuric acid, so the equation can be better written so:

    [tex]\mathrm{SO_2 + K_2Cr_2O_7 + H_2SO_4 \to Cr_2(SO_4)_3 + SO_3 + H_2O + K}[/tex] (I assume K has to be added, preserving the charge in both members)

    To use the half-reaction method I write the ionic form:

    [tex]\mathrm{SO_2+K^+ + Cr_2O_7^{2-}+H^+ \to Cr^{3+}+SO_3+H_2O + K}[/tex]

    Writing and balancing three half-reactions (can it be?) I get:

    [tex]\mathrm{4SO_2 + K_2Cr_2O_7+3H_2SO_4 \to 4SO_3 + Cr_2(SO_4)_3+2K+3H_2O}[/tex]


    Is it correct?

    Thank you.
     
  2. jcsd
  3. Aug 17, 2015 #2

    Borek

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    Staff: Mentor

    Hint: K+ is just a spectator, and SO3 in water means just SO42-.
     
  4. Aug 17, 2015 #3
    Ok, your hint was foundamental to me.
    I have already thought about K+ being spectator but I "forgot" that
    [tex] \mathrm{ SO_3 + H_2O\to H_2SO_4\to 2H^+ + SO_4^{2-} }[/tex] (right?) so I didn't know what to do without K in the second member.


    [tex] \mathrm{SO_2 + K_2Cr_2O_7 + H_2SO_4\to Cr_2(SO_4)_3 + SO_3 + H_2O + 2K^+ }[/tex]
    [tex] \mathrm{SO_2 + K_2Cr_2O_7 \to Cr_2(SO_4)_3 + 2K^+ }[/tex]
    ionic form: [tex] \mathrm{SO_2 + Cr_2O_7^{2-}\to Cr^{3+} + SO_4^{2-} }[/tex]

    Balancing the two half-reactions:

    [tex] \mathrm{3SO_2 + Cr_2O_7^{2-} + 2H^+\to 3SO_4^{2-} + 2Cr^{3+} + H_2O }[/tex]

    And finally:

    [tex] \mathrm{3SO_2 + K_2Cr_2O_7 + H_2SO_4\to K_2SO_4 +Cr_2(SO_4)_3 + H_2O }[/tex]


    I'm not so sure though. Is it correct?

    Thank you for your reply.
     
    Last edited: Aug 17, 2015
  5. Aug 17, 2015 #4

    Borek

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    Staff: Mentor

    Looks OK to me.
     
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