Question

Question #d0f6f

Answer 1

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

Explanation:

`"Oxidation reaction:"` `Fe(II)` is OXIDIZED to `Fe(III)`

`FeI_2 rarrFe^(3+) +2I^(-) +e^(-)` `(i)`

`"Reduction reaction:"` `"Iodate ion"` is REDUCED to `I^(+I)`

`stackrel(V+)IO_3^(-) + 6H^+ + 4e^(-) rarrI^(+) +3H_2O` `(ii)`

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. `4xx(i) + (ii)`:

`4FeI_2 +IO_3^(-) + 6H^+ rarr4Fe^(3+) +8I^(-) +I^(+) +3H_2O`

And this is in fact balanced with respect to mass and charge. But we could add `1xxCl^-` to each side of the chemical equation....

`4FeI_2 +IO_3^(-) + 6H^+ +Cl^(-)rarr4Fe^(3+) +8I^(-) +ICl +3H_2O`

And to make it even simpler, I could remove `8xxI^-` FROM EACH SIDE of the EQUATION:

`4Fe^(2+) +IO_3^(-) + 6H^+ +Cl^(-)rarr4Fe^(3+) +ICl +3H_2O`,

i.e. a four electron reduction with respect to `IO_3^-`, and a four electron oxidation with respect to 4 equiv of ferrous ion.