Question #cfd9b

1 Answer
Aug 23, 2017

Answer:

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.