Let's start with I2. Note IO3- is singly charged, that's to prevent confusion.

I2     -->  IO3-

Firstly, balance the number of I needed.

I2    --> 2IO3-

Secondly, balance the number of O needed with H2O

I2 + 6H2O --> 2IO3-

Thridly, balance the H needed by adding H+

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

Forth, balance the charge by adding electrons

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

If it's an alkaline condition, we would have to add OH- to balance H+, take away the extra H2O on both side followed by adding electrons.

Do the same for HOCL,
You would get

HOCL + H+  + 2e --> Cl- + H2O -- (2)

Last step, we summarise the equation by balancing the electrons. In this case, equation (2) needs 5 times the electrons. So

5HOCL + I2 + H2O + --> 7H+  +  5Cl- + 2IO3-

Oops, thought that's what he meant