Help with chem!!! (balancing redox reactions in acidic solution)?

Question

When the following reaction is balanced in acidic solution, what is the coefficient of I2?

IO3- + I- -----> I2

the correct answer is 3..but i cant figure out how to get there

This is what I have so far:

10e- +12H+ +2IO3- ----> I2 + 6H2O

and

2I- ---> I2 + e-

Can someone please tell me what I'm doing wrong?? Thanks!

Answer 1

2 IO3)-  & 10 electrons taken --> 1 I2  

2 iodines in 2 IO3 goes from +5 each to I2  with I @ zero each, a change of 5 each)

2 I-  -->  I2  & 2 electrons lost

(2 iodide's I @ -1 each goes to 2 iodines I @ zero each, change of 1 each)

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to balance the electrons taken & lost , we will five times the iodide:

2 IO3)-  & 10 electrons taken --> 1 I2  

10 I-  --> 5 I2  & 10 electrons lost

then combine the two:

2 (IO3)-1 & 10 I-  --> 5 I2  & 1 I2

which  becomes:

2 (IO3)-1 & 10 I-  --> 6 I2  

you have done the H+'s & H2O just fine:

2 (IO3)-1 & 10 I- & 12 H+  --> 6 I2  & 6 H2O

whoops, almost forgot to cut everything in half:

(IO3)-1 & 5 I- & 6 H+  --> 3  I2  & 3 H2O

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so you were doing great...

but your "2I- ---> I2 + e-"  releases 1 e- per iodide, that's a total of 2 e- released