Question #53a26

1 Answer
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: 2IO_3^- + 12H^+ + 10e^- ----------> I_2 + 6H_2O
Oxidation: 2I^- ----------> I_2 + 2e^-

You should multiply the oxidation half equation by 5 and sum both half equations to get:
Reduction: 2IO_3^- + 12H^+ + 10e^- ----------> I_2 + 6H_2O
Oxidation: x5 (2I^- ----------> I_2 + 2e^-)

RedOx: 2IO_3^- + 10I^- + 12H^+ ----------> 6I_2 + 6H_2O

Now, since all coefficients are multiple of 2, we can divide all reaction by 2 to get the final answer as:
RedOx: IO_3^- + 5I^- + 6H^+ ----------> 3I_2 + 3H_2O