Question #d0f6f

Aug 30, 2017

This is a redox reaction (and we will see why!), and we separate the reaction into individual oxidation reduction couples.......

Explanation:

$\text{Oxidation reaction:}$ $F e \left(I I\right)$ is OXIDIZED to $F e \left(I I I\right)$

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

$\text{Reduction reaction:}$ $\text{Iodate ion}$ is REDUCED to ${I}^{+ I}$

$\stackrel{V +}{I} {O}_{3}^{-} + 6 {H}^{+} + 4 {e}^{-} \rightarrow {I}^{+} + 3 {H}_{2} O$ $\left(i i\right)$

And for each half equation, charge and mass are balanced ABSOLUTELY, as indeed they must be if we purport to represent chemical reality. I add them in such a way as to eliminate the electrons from the final equation, i.e. $4 \times \left(i\right) + \left(i i\right)$:

$4 F e {I}_{2} + I {O}_{3}^{-} + 6 {H}^{+} \rightarrow 4 F {e}^{3 +} + 8 {I}^{-} + {I}^{+} + 3 {H}_{2} O$

And this is in fact balanced with respect to mass and charge. But we could add $1 \times C {l}^{-}$ to each side of the chemical equation....

$4 F e {I}_{2} + I {O}_{3}^{-} + 6 {H}^{+} + C {l}^{-} \rightarrow 4 F {e}^{3 +} + 8 {I}^{-} + I C l + 3 {H}_{2} O$

And to make it even simpler, I could remove $8 \times {I}^{-}$ FROM EACH SIDE of the EQUATION:

$4 F {e}^{2 +} + I {O}_{3}^{-} + 6 {H}^{+} + C {l}^{-} \rightarrow 4 F {e}^{3 +} + I C l + 3 {H}_{2} O$,

i.e. a four electron reduction with respect to $I {O}_{3}^{-}$, and a four electron oxidation with respect to 4 equiv of ferrous ion.