1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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: Mar 5, 2006
  8. Mar 5, 2006 #7
    It's all solved at www.ChemicalForums.com

    We figured it out over there, even thought the syntax for typing equations takes some practice to learn, it has a lot of people posting about chemistry!

    Here's a link to the Solution for this problem.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Chemistry Reduction Reaction Help Needed
Loading...