Question #d0f6f

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anor277
Aug 30, 2017

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.

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