Question #cfd9b

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Question #cfd9b

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Answer 1 · 2017-08-23T16:42:33

You wants the so-called $comproportionation$ $"comproportionation"$ of $iodide$ $"iodide"$ and $iodate$ $"iodate"$ ions to $elemental iodine$ $"elemental iodine"$ ?

Explanation:

$Iodide anion I(-I)$ $"Iodide anion I(-I)"$ is OXIDIZED to elemental iodine.....

$I^{-} \to \frac{1}{2} I_{2} + e^{-}$ $I^(-) rarr 1/2I_2 + e^(-)$ $(i)$ $(i)$

And $Iodate anion I(V+)$ $"Iodate anion I(V+)"$ is REDUCED to elemental iodine.....

$I O_{3}^{-} + 6 H^{+} + 6 e^{-} \to I^{-} + 3 H_{2} O$ $IO_3^(-) +6H^(+) + 6e^(-) rarr I^(-) + 3H_2O$ $(i i)$ $(ii)$

We adds $(i)$ $(i)$ and $(i i)$ $(ii)$ so as to eliminate the electrons.......i.e. $6 \times (i) + (i i) :$ $6xx(i) +(ii):$

$IO_3^(-) +cancel(6)5I^(-) + 6H^(+) + cancel(6e^(-)) rarr cancel(I^(-)) + 3I_2 + 3H_2O+cancel(6e^-)$

To give finally........

$IO_3^(-) +5I^(-) + 6H^(+) rarr 3I_2 + 3H_2O$

Mass and charge are balanced as is absolutely required. Please note that all I have done is to balance mass and charge according to the standard rules of oxidation; and you yourself can become very adept at this with practice. If there is an issue you want clarified, raise it, and someone will address your question.

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Question #cfd9b

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