Redox and redox equilibria

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I am currently investigating the iodine clock reaction involving hydrogen peroxide. I don't understand how to work out if the overall equation is redox, especially in view to oxidation numbers? I also understand that the overall reaction can be split into 'half-equations' but am not sure how to go about this.
210208

Corrie writes ....
 
Iodide (Ox. No. = -1) is oxidised to iodine (Ox. No. = 0), losing electrons, and thus acting as a reducing agent:
 
I^-(aq) -> I₂(aq) + e-
 
You need to balance this half-equation, including balancing charges.
 
Hydrogen peroxide (Ox. No. for O, unusually, is -1 here) is reduced to water (Ox. No. for O = -2), gaining electrons and acting as an oxidising agent:
 
H₂O₂ + H⁺ + e^- -> H₂O
 
Again you need to balance this half-equation, including balancing charges.
 
Now combine the two half-equations together so that the electrons lost = electrons gained, and therefore cancel out. To do this, you may have to multiply one or both of the half-equations through, so that the same number of electrons is involved in each.
 
For instance, of one half-equation involves 2e^- and the other 3e^-, then you'll have to multiply the first one by 3 and the second one by 2 before combining them.

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updated: 21 February 2008