Does anybody know how to balance this redox equation?

Question

(I3)^ -1 -----> (I)^ -1 + (IO3)^ -1 , under acidic conditions. .....plzz show ur work !

Answer 1

First off, your crafty perfesser wants to give you a hard time. He knows (but you, poor innocent, do not) that iodine, I2, plus iodide ion, I-, form a species that is triiodide ion, I3-

There is confusion among capital I, lower case l, and the number 1.
From here out, let the symbol for iodine be *eye*.

Let us balance it by half-reactions, without the obstruction of triiodide.

5*eye*2 + 10e- ===> 10 *eye*^-1

That is the reduction of iodine to iodide

6H2O + *eye*2 ===> 2*eye*O3- + 10e- + 12 H+

That is the oxidation of iodine to iodate. The combined equation is:

6H2O + 6*eye*2 ===> 2*eyeO3- + 10*eye*- +12H+

So what about the *eye3-? There should be 6 more *eye- on the right to convert all that 6*eye*2 to 6*eye3^- and the 10*eye*- on the right to 16*eye*-

What your crafty perfessor didn't explain is that there are three kinds of iodine: The kind that gets oxidized, the kind that gets reduced, and the kind that does neither.

Answer 2

Yes, i do.

Do your own homework. As a hint, separate the oxidation and reduction components to balance on their own. At the end, make sure that atoms and charges balance. It's in (aqueous?) acid, so this means you can add certain molecules and ions to help you.