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-