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Balancing - Half-Cell method

  1. May 26, 2005 #1
    It aks me to balance some equations, using this method. It says to show both half-cell reactions, and identify them as oxidation or reduction.

    a) SO3^2 + MnO4- + H+ <----- Mn2+ + SO4^2- + H2O (l)

    b) Cl2(g) + OH- <----- Cl- + ClO3- + H2O(l)

    c) SO4^2- + I- + H+ <----- S2- + I2(s) + H2O(l)

  2. jcsd
  3. May 26, 2005 #2
    SO what have you done so far?

    Do you know how to check if a reaction is an oxidation of a particular species or a reduction (or none)?
  4. May 26, 2005 #3
    You need to show some work instead of just posting the question.

    Start by explaining the difference between oxidation and reduction.
  5. May 26, 2005 #4


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    This might help give you a clue as to the meaning of oxidation and reduction. (follow the link) :bugeye:

    Do you know anything about half-cell reactions?
  6. May 27, 2005 #5
    Well, an oxidized substance is one that has lost electrons in the process of getting oxidized and a reduction substance is one that has gained in the process of reduction.

    I dont understand the half-cell method though. I have looked at examples of balancing an equation using that method but I dont get it.

    So, after its balanced, using this method, I would know which is the oxidized and reduced substance but I dont know how to do the half-cell method..so if someone can show me at least a) I can do b and c.

  7. May 27, 2005 #6
    To balance equations by the half cell method you first need to break your equation up into the redox half-cell reactions.

    Assign your oxidation numbers to each compound. (Oxygen is -2, Hydrogen is +1...). The overall number of the compound is assigned already in the reaction equation. Take [tex]MnO_4^-^1[/tex] for example. You have 4 oxygen's which each have an oxidation number of -2 (total is -8). But the overall number is -1. So the oxidation number for Mn has to be +7 then to get the net -1.

    Do the same for the rest of the equation now.

    Once you have done this take note of the commonalities with each side of the reaction. Sulphur and Manganese are on both sides of the equation. Taking what you know of oxidation and reduction you should be able to determine which reaction is oxidation and which is reduction.

    Now that you have your two half reactions balance them by observation. Don't forget to balance the charge on both sides by adding electrons. The net doesn't have to be zero, but it does have to be the same on both sides.
    Finally take the two balanced half reactions and balance the electrons (they should be on opposite sides of the reaction due to redox) and put the equations back together to get your balanced equation.
    Last edited: May 27, 2005
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