Question #53a26

1 Answer
Dr. Hayek
Sep 15, 2015

The balanced equation is as follows:

Explanation:

#IO_3^-# is the oxidizing agent, it gets reduced into #I_2#.
#I^-# is a reducing agent, it gets oxidized also into #I_2#.

You should write two half equations:
Reduction: 2#IO_3^-# + 12#H^+# + 10#e^-# ----------> #I_2# + 6#H_2#O
Oxidation: 2#I^-# ----------> #I_2# + 2#e^-#

You should multiply the oxidation half equation by 5 and sum both half equations to get:
Reduction: 2#IO_3^-# + 12#H^+# + 10#e^-# ----------> #I_2# + 6#H_2#O
Oxidation: x5 (2#I^-# ----------> #I_2# + 2#e^-#)

RedOx: 2#IO_3^-# + 10#I^-# + 12#H^+# ----------> 6#I_2# + 6#H_2#O

Now, since all coefficients are multiple of 2, we can divide all reaction by 2 to get the final answer as:
RedOx: #IO_3^-# + 5#I^-# + 6#H^+# ----------> 3#I_2# + 3#H_2#O

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