Please balance this redox equation for me

Question

I(sub)2 NaOH = NaIO(sub)3 NaI H(sub)2O

In this equation, Iodine was both reduced and oxidized. So how to balance? Please explain. Thank you.

Answer 1

I2   &  NaOH = NaIO3   &  NaI   &  H2O

oxidation: I2 --> 2NaIO3 & 10 electrons lost (two I's @ zero to I+5 each, is 10 electrons lost)

reduction: I2  & 2 electrons taken --> 2 NaI (two I's @ zero to I-1 each, is 2 electrons taken)

to balance the electrons 5 times the reduction:

oxidation: I2 --> 2NaIO3 & 10 electrons lost

reduction:5 I2  & 10 electrons taken --> 10 NaI

now combine them:

6 I2 & NaOH --> 2 NaIO3 & 10 NaI  & H20

balance the Na's:

6 I2 & 12 NaOH --> 2 NaIO3 & 10 NaI  & H20

balance the H's:

6 I2 & 12 NaOH --> 2 NaIO3 & 10 NaI  & 6 H20

that's your answer

(the check on your work ,,, do the oxygens balance... yes... then it is done)