Question

Question #cfd9b

Answer 1

You wants the so-called `"comproportionation"` of `"iodide"` and `"iodate"` ions to `"elemental iodine"`?

Explanation:

`"Iodide anion I(-I)"` is OXIDIZED to elemental iodine.....

`I^(-) rarr 1/2I_2 + e^(-)` `(i)`

And `"Iodate anion I(V+)"` is REDUCED to elemental iodine.....

`IO_3^(-) +6H^(+) + 6e^(-) rarr I^(-) + 3H_2O` `(ii)`

We adds `(i)` and `(ii)` so as to eliminate the electrons.......i.e. `6xx(i) +(ii):`

`IO_3^(-) +cancel(6)5I^(-) + 6H^(+) + cancel(6e^(-)) rarr cancel(I^(-)) + 3I_2 + 3H_2O+cancel(6e^-)`

To give finally........

`IO_3^(-) +5I^(-) + 6H^(+) rarr 3I_2 + 3H_2O`

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.