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Homework Help: Chemistry Reduction Reaction Help Needed

  1. Mar 4, 2006 #1
    Chemistry Reduction Reaction Help Needed: MnO4- + S2- --> MnO2 + S8

    I was wondering if I had done this reduction reaction equation correctly:
    [tex]MnO_4^-+S^{2-}\xrightarrow{}MnO_2+S_8[/tex]

    After going through the reduction reaction problem solving process (I think) I get this equation as the final equation in the reaction:

    [tex]2MnO_4^-+3S^{2-}+4H_2O\xrightarrow{}2MnO_2+3S_8+8OH^-[/tex]

    Does this look correct? Can anyone check? Thanks.
     
    Last edited: Mar 4, 2006
  2. jcsd
  3. Mar 4, 2006 #2
    Here is a more detailed explanation of the answer I have

    Ok, starting with the original equation:

    [tex]MnO_4^-+S^{2-}\xrightarrow{}MnO_2+S_8[/tex]

    I then seperate into two "half-reactions":

    [tex]MnO_4^-\xrightarrow{}MnO_2[/tex]

    [tex]S^{2-}\xrightarrow{}S_8[/tex]

    Figureing the oxidation charge for the left side of the first "half-reaction":

    [tex]X-8+1=0\xrightarrow{}X=7[/tex]

    Therefore, the oxidation number of Mn on the left side is 7, right?

    Then I do the right side of the first "half-reaction":

    [tex]X-4=0\xrightarrow{}X=4[/tex]

    This means that the oxidation number of Mn on the right side is 4. Meaning that I have to add 3 electrons to the left side of the first "half-reaction" equation, right?:

    [tex]MnO_4^-+3e^-\xrightarrow{}MnO_2[/tex]

    Now, I added H+ ions to the left side to bring it's overall charge down to zero, because the right side has an overall charge of zero:

    [tex]MnO_4^-+3e^-+4H^+\xrightarrow{}MnO_2[/tex]

    Finally, I add H2O to even out the oxygens in the first "half-reaction" equation:

    [tex]MnO_4^-+3e^-+4H^+\xrightarrow{}MnO_2+2H_2O[/tex]

    Going on to the second of the two "half-reaction" equations:

    [tex]S^{2-}\xrightarrow{}S_8[/tex]

    The oxidation number for the left hand side of the second "half-reaction" equation is -2 and for the right side it is 0, right? So then I added 2 electrons to the right side of the second "half-reaction" equation:

    [tex]S^{2-}\xrightarrow{}S_8+2e^-[/tex]

    I then multiply the two results from the "half-reaction" equations in order to eliminate the electrons:

    [tex](MnO_4^-+3e^-+4H^+\xrightarrow{}MnO_2+2H_2O)*2=2MnO_4^-+6e^-+8H^+\xrightarrow{}2MnO_2+4H_2O[/tex]

    [tex](S^{2-}\xrightarrow{}S_8+2e^-)*3=3S^{2-}\xrightarrow{}3S_8+6e^-[/tex]

    Adding them together, I get:

    [tex]2MnO_4^-+3S^{2-}+8H^+\xrightarrow{}2MnO_2+3S_8+4H_2O[/tex]

    Finally, I add OH- to balance and eliminate the H2O's on the right side:

    [tex]2MnO_4^-+3S^{2-}+8H^++8OH^-\xrightarrow{}2MnO_2+3S_8+4H_2O+8OH^-[/tex]

    [tex]2MnO_4^-+3S^{2-}+8H_2O\xrightarrow{}2MnO_2+3S_8+4H_2O[/tex]

    [tex]2MnO_4^-+3S^{2-}+4H_2O^+\xrightarrow{}2MnO_2+3S_8+8OH^-[/tex]

    Does this seem reasonable?
     
  4. Mar 4, 2006 #3
    your final line: [itex] ... 4H_2O^{+} [/itex] ?

    and the sulfur is not balanced.
     
  5. Mar 4, 2006 #4
    How do I balance the equation then? Did I do the "half-reaction" equations correctly?

    Do you see where I made a mistake?

    Thanks again.
     
  6. Mar 4, 2006 #5
    Redoing the problem

    I did it again by starting with an actual balanced equation:eek:

    Ok, starting with the original equation (now balanced):

    [tex]MnO_4^-+8S^{2-}\xrightarrow{}MnO_2+S_8[/tex]

    I then seperate into two "half-reactions":

    [tex]MnO_4^-\xrightarrow{}MnO_2[/tex]

    [tex]8S^{2-}\xrightarrow{}S_8[/tex]

    Figureing the oxidation charge for the left side of the first "half-reaction":

    [tex]X-8+1=0\xrightarrow{}X=7[/tex]

    Therefore, the oxidation number of Mn on the left side is 7, right?

    Then I do the right side of the first "half-reaction":

    [tex]X-4=0\xrightarrow{}X=4[/tex]

    This means that the oxidation number of Mn on the right side is 4. Meaning that I have to add 3 electrons to the left side of the first "half-reaction" equation, right?:

    [tex]MnO_4^-+3e^-\xrightarrow{}MnO_2[/tex]

    Now, I added H+ ions to the left side to bring it's overall charge down to zero, because the right side has an overall charge of zero:

    [tex]MnO_4^-+3e^-+4H^+\xrightarrow{}MnO_2[/tex]

    Finally, I add H2O to even out the oxygens in the first "half-reaction" equation:

    [tex]MnO_4^-+3e^-+4H^+\xrightarrow{}MnO_2+2H_2O[/tex]

    Going on to the second of the two "half-reaction" equations:

    [tex]8S^{2-}\xrightarrow{}S_8[/tex]

    The oxidation number for the left hand side of the second "half-reaction" equation is -2 and for the right side it is 0, right? So then I added 2 electrons to the right side of the second "half-reaction" equation:

    [tex]8S^{2-}\xrightarrow{}S_8+2e^-[/tex]

    I then multiply the two results from the "half-reaction" equations in order to eliminate the electrons:

    [tex](MnO_4^-+3e^-+4H^+\xrightarrow{}MnO_2+2H_2O)*2=2MnO_4^-+6e^-+8H^+\xrightarrow{}2MnO_2+4H_2O[/tex]

    [tex](8S^{2-}\xrightarrow{}S_8+16e^-)*3=24S^{2-}\xrightarrow{}3S_8+48e^-[/tex]

    Adding them together, I get:

    [tex]2MnO_4^-+24S^{2-}+8H^+\xrightarrow{}2MnO_2+3S_8+4H_2O[/tex]

    Finally, I add OH- to balance and eliminate the H2O's on the right side:

    [tex]2MnO_4^-+24S^{2-}+8H^++8OH^-\xrightarrow{}2MnO_2+3S_8+4H_2O+8OH^-[/tex]

    [tex]2MnO_4^-+24S^{2-}+8H_2O\xrightarrow{}2MnO_2+3S_8+4H_2O+8OH^-[/tex]

    [tex]2MnO_4^-+24S^{2-}+4H_2O^+\xrightarrow{}2MnO_2+3S_8+8OH^-[/tex]

    Does this seem reasonable?
     
    Last edited: Mar 5, 2006
  7. Mar 5, 2006 #6
    Last edited by a moderator: Apr 22, 2017
  8. Mar 5, 2006 #7
    Last edited by a moderator: Apr 22, 2017
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