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Number of electrons to balance redox reaction

  1. Nov 29, 2015 #1
    1. The problem statement, all variables and given/known data

    What number of electrons to balance the equation
    ##I_2(s)+OH^-(aq)\rightarrow IO_3^-(aq)+H_2O(l)##

    3. The attempt at a solution
    I see that the oxidation state of I increases from 0 to +5
    But, I don't see any reduction at the reaction..
    So, I don't understand how to balance the reaction
    And I don't know how to get the number of required electrons to balance the reaction
     
  2. jcsd
  3. Nov 29, 2015 #2

    Borek

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    Staff: Mentor

    There is no reduction because this is only a half reaction, where electrons are listed as a reagent. Just like Fe(II) to Fe(III) oxidation can be written as

    Fe2+ → Fe3+ + e-
     
  4. Nov 29, 2015 #3
    So, the number of electron required to balance the equation is 5 electron, right? Since the oxidation state increases by 5
     
  5. Nov 29, 2015 #4

    Borek

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    Staff: Mentor

  6. Nov 29, 2015 #5
    Balance all atoms
    ##I_2(s)+12OH^-(aq)\rightarrow 2IO_3^-(aq)+6H_2O(l)##


    charges in left : -12
    charges in right: -2

    So, there should be 10 electrons added to the right side of reaction, right??
     
  7. Nov 29, 2015 #6

    Borek

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    Staff: Mentor

    Yes, that's correct.

    I2 + 12OH- → 2IO3- + 6H2O + 10e-
     
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