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Half equation for S2O32-/S4O62- couple

  1. May 14, 2008 #1
    I need to write the half equation for the S2O32-/S4O62- redox couple but i'm having trouble because they have the same ioic value.

    I've come up wth two solutions but am happy with nether...

    S2O32- + 3H2O + S2= S4O62- + 6H+ + 6e-

    or

    2S2O32- = S4O62- + 2e-

    I know that the second one is wrong because the charges don't balance, but i'm not sure that the first is right either because we generally only add water, or electrons to the equations, so i'm not sure that I can add the S2.

    am i just being obtuse or?!?
     
  2. jcsd
  3. May 14, 2008 #2

    symbolipoint

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    Try ignoring any use of water. Just concentrate on the stoichiometry and the use of electron(s).
     
  4. May 15, 2008 #3
    the charges DO balance in the second one.

    2S2O32- = -4

    S4O62- + 2e- = -4
     
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