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Wierd reaction?

  1. Jan 23, 2007 #1
    In my textbook, it had

    2Cu2+(aq) + 4I-(aq) -> CuI(s) + I2(aq)

    How does that work? First of all it's not balanced. It should be 2CuI(s). But how does CuI(s) form. It should be CuI+.

    The book did say excess 4I- was usd to remove Ag+ since it was talking about back titrations. But the reaction as stated dosen't involve Ag. Has the book made mistakes?
  2. jcsd
  3. Jan 24, 2007 #2


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    No, it's correct (except for the missing 2).

    Cu(+2) is reduced to Cu(+1) and I(-1) is oxidized to I(0).
  4. Jan 24, 2007 #3
    I see. Thanks
  5. Jan 26, 2007 #4


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    Staff: Mentor

    Note: this reaction is interesting as its equilibrium is moved to the right thanks to the very low solubility of CuI.
  6. Jan 27, 2007 #5
    And the solubility is so low that CuI precipitates out as a solid.
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