# Question #cfd9b

Aug 23, 2017

You wants the so-called $\text{comproportionation}$ of $\text{iodide}$ and $\text{iodate}$ ions to $\text{elemental iodine}$?

#### Explanation:

$\text{Iodide anion I(-I)}$ is OXIDIZED to elemental iodine.....

${I}^{-} \rightarrow \frac{1}{2} {I}_{2} + {e}^{-}$ $\left(i\right)$

And $\text{Iodate anion I(V+)}$ is REDUCED to elemental iodine.....

$I {O}_{3}^{-} + 6 {H}^{+} + 6 {e}^{-} \rightarrow {I}^{-} + 3 {H}_{2} O$ $\left(i i\right)$

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

$I {O}_{3}^{-} + \cancel{6} 5 {I}^{-} + 6 {H}^{+} + \cancel{6 {e}^{-}} \rightarrow \cancel{{I}^{-}} + 3 {I}_{2} + 3 {H}_{2} O + \cancel{6 {e}^{-}}$

To give finally........

$I {O}_{3}^{-} + 5 {I}^{-} + 6 {H}^{+} \rightarrow 3 {I}_{2} + 3 {H}_{2} O$

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.