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    2NO + 12H+(positive) + 10I-(negative) ------> 2NH4+ + 2H2O + 5I2 (5iodine2)

    The question is: Identify the species formed by oxidation in this reaction.

    the answer is I2 (iodine2)

    but how did they achieve this?

    I understand that

    10I- ------> 5I2 + e

    (Iodine loses electron to become more positive and thus is reduced)

    But that is 5I2 not I2...
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    (Original post by xXxiKillxXx)
    2NO + 12H+(positive) + 10I-(negative) ------> 2NH4+ + 2H2O + 5I2 (5iodine2)

    The question is: Identify the species formed by oxidation in this reaction.

    the answer is I2 (iodine2)

    but how did they achieve this?

    I understand that

    10I- ------> 5I2 + e

    (Iodine loses electron to become more positive and thus is reduced)

    But that is 5I2 not I2...
    The '5' is just the coefficient of the balanced equation and plays no part in the actual transfer of electrons.

    Work out the oxidation state of I-

    Work out the oxidation state of I2

    Now you can see where the electrons have gone from and to...
 
 
 
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