Need help with redox reactions

I'm trying to work out a chemistry lab report, and I don't get it at all. For example, one of the reactions is I-(aq) + NO3-(aq) --> I2(aq) + NO2(g). Can someone please explain how to balance this using the half-cell method?

half reactions:

I-(aq) --> I2(aq)
NO3-(aq) --> NO2(g)

balance in acid:

2I-(aq) --> I2(aq) + 2e-
e- + 2H+ + NO3-(aq) --> NO2(g) + H2O

equalize electrons and add:

4H+ + 2I-(aq) + 2NO3-(aq) --> I2(aq) + 2NO2(g) + 2H2O

More redox:

http://www.chemteam.info/Redox/Redox.htm…

i think it should be NO and not NO2 in the products side.
well it doesn't matters,we will balance the given red-ox reaction step wise.

Also the reaction medium is not specified (acidic/basic or neutral)

We will assume acidic medium for our convenience.

Step 1: separate the oxidation and reduction half

I-(aq) ---> I2(aq)

NO3-(aq) ---> NO2(g)

Step 2: Balance that atom whose oxidation no. is changing. (Iodine and Nitrogen in this case)

2I-(aq) ---> I2(aq)

NO3-(aq) ---> NO2(g)

Step 3: Now Balance the oxygen atoms by adding moles of H2O on the reqd side. And then balance the hydrogen atoms by adding H+ ions on reqd side.

2I-(aq) ---> I2(aq)

NO3-(aq) ---> NO2(g) + H2O (due to one less O atom in products side)

2H+ + NO3-(aq) ---> NO2(g) + H2O (due to two less H atoms on reactant side, after adding H2O)

Step 4: Now add the reqd moles of electrons so as to balance the net charge on the reactant and products side.

2I-(aq) ---> I2(aq) + 2e (since there were 2 I- on products side which contributed 2 -ve charges)

this half reaction is overall neutral.

e + 2H+ + NO3-(aq) ---> NO2(g) + H2O
this half reaction is also overall neutral.

Step 5: now add both the half reactions eliminating the electrons.