Question #d0f6f

1 Answer
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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