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How to balance equations using the half cell method?

  1. Jun 1, 2005 #1
    SO32- +MnO4-1+H+1 <-> Mn2+ +SO42- +H20(l)
     
  2. jcsd
  3. Jun 2, 2005 #2
    Step 1 Divide the skeleton equation into half-reactions.

    Step 2 Balance atoms other than H and O.

    Step 3 Balance oxygen atoms by adding H2O to the side that needs O.

    Step 4 Balance hydrogen by adding H+ to the side that needs H.

    Step 5 Balance the charge by adding electrons.

    Step 6 Make the electrons gained equal to the electrons lost and then add the two half-reactions.

    Step 7 Cancel anything that is the same on both sides.
     
  4. Jun 2, 2005 #3

    Gokul43201

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  5. Jun 2, 2005 #4
    this problem may seem a bit more tricky than the redox problems that you are used to since it may not be obvious which is getting reduced/oxidized based upon the superscripts alone.

    The half-reactions are:

    [tex]SO_3^{2-} + H_2O \rightarrow SO_4^{2-} + 2H^+ + 2e^-[/tex]
    [tex]MnO_4^{-1} + 8H^+ + 5e^- \rightarrow Mn^{+2} + 4H_2O[/tex]

    Make your electrons equal by multiplying the half rxns to make the number of electrons 10 for both, then add the half rxns up and you should get the correct answer.
     
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