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    Siobhan634's Avatar
    Siobhan634 Posts: 3, Reputation: 1
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    #1

    Oct 21, 2013, 03:08 PM
    How many electrons are transferred in the following reaction?
    How many electrons are transferred in 3I2(s) + 2MnO2(s) + 8OH-(aq) --> 6I-(aq) + 2MnO4-(aq) + 4H2O(l)?
    ma0641's Avatar
    ma0641 Posts: 15,675, Reputation: 1012
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    #2

    Oct 21, 2013, 03:31 PM
    Equation doesn't make too much sense as written. However, we do not do homework for you. Show us your work.
    Siobhan634's Avatar
    Siobhan634 Posts: 3, Reputation: 1
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    #3

    Oct 21, 2013, 03:56 PM
    The oxidation state of iodine goes from 0 to -1. Since there are 6 iodine ions, the amount of electrons transferred is -1 x 6= -6.
    The oxidation state of 2MnO2 goes from 0 to -1. Because there are two of these ions in the products, 2 x -1 = -2. Therefore there are 2 electrons that have been transferred.
    Total electrons gained by reactants= -8
    The oxidation state of OH is -1. During reaction, oxidation state changes to 0. Since there are 8 OH ions, total oxidation state= 8 x 1 = +8

    Thus, there is a total amount of 8 electrons that have been transferred (from reactants to products) during this reaction.

    I have worked this out, however I don't know if it is correct or not.
    Your help would be greatly appreciated.
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    ma0641 Posts: 15,675, Reputation: 1012
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    #4

    Oct 21, 2013, 07:40 PM
    When I got my degree in 1963 we used."valence " as the electrons. It's been a while since I balanced equations but your math appears correct -at least in the newer terminology.
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