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    "Acidified FeO42- (aq) ions oxidise aqueous iodide ions ions, I-, to form aqueous Iodine, I2.

    Construct the half-equation for the oxidation of iodide ions to form iodine"
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    (Original post by boromir9111)
    "Acidified FeO42- (aq) ions oxidise aqueous iodide ions ions, I-, to form aqueous Iodine, I2.

    Construct the half-equation for the oxidation of iodide ions to form iodine"
    2I- -> I2 + 2e- ?
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    So we want I2 as the product basically right? so we have I- ----> I2 but we need two I's so it would be 2I- ----> I2 + 2e- can we put the 2e- there?
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    2I- > I2 + 2e-
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    (Original post by JacobM)
    2I- -> I2 + 2e- ?
    That's what i'm worried about can we put the 2e- there? because I2 is neutral and not charged. So that means 2- on the left and 2- on the right side cancel each other out?
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    (Original post by boromir9111)
    "Acidified FeO42- (aq) ions oxidise aqueous iodide ions ions, I-, to form aqueous Iodine, I2.

    Construct the half-equation for the oxidation of iodide ions to form iodine"
    I- -->I2
    So it went from -1 to 0
    Hence it lost an electron.

    I- --> I2 + e-

    Balance the equation to get:

    2I- -->I2 +2e-
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    Yeah, i need to go over half-equations..... thankfully there is chem guide and TSR. Thank you for your help!!!!
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    how come you have to balance the e? isn't this just 2I --> I2 e- because 2I is oxidation state of -1 and the I is 0, so isn't that only missing 1 electron to get to -1?
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    (Original post by lynnlynn123)
    how come you have to balance the e? isn't this just 2I --> I2 e- because 2I is oxidation state of -1 and the I is 0, so isn't that only missing 1 electron to get to -1?
    You do need to balance everything, including the electrons.
    There is one e- lost for each iodide ion, but 2 of them. The equation must have the same total charge on left hand side and right hand side.
 
 
 
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