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    2IO3–+10I–+12H+ →6I2+6H2O

    where does the10 I- and 6 I2 come from in this half equation ?

    and where are the electrons ?

    how would i know when to use this?
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    (Original post by esmeralda123)
    2IO3–+10I–+12H+ →6I2+6H2O

    where does the10 I- and 6 I2 come from in this half equation ?

    and where are the electrons ?

    how would i know when to use this?
    The half-equation is not correct in that all coefficients can be divided by 2

    IO3 + 5I + 6H+ → 3I2 + 3H2O

    The reduced species is the iodate, IO3

    and the oxidised species is the iodide ion, I

    Both become iodine, I2

    In order to produce a half equation for the iodate you have to add hydrogen ions on the LHS and water on the RHS

    2IO3 + 12H+ --> I2 + 6H2O

    you now have to add in electrons to balance electronically:

    2IO3 + 12H+ + 10e --> I2 + 6H2O

    Now do the same for the iodide ions (easier)

    2I- --> I2 + 2e

    MTB 5 in the latter reaction to make the electrons the same and then add together

    2IO3 + 12H+ + 10e --> I2 + 6H2O
    10I- --> 5I2 + 10e
    --------------------------------------------------------- add
    2IO3 + 12H+ + 10I- --> I2 + 6H2O+ 5I2

    gather terms and divide through by 2

    IO3 + 5I + 6H+ → 3I2 + 3H2O
 
 
 
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