Equation for the reaction of iodide to iodine using hydrogen peroxide under acidic conditions? I'm building a home lab around All Lab, No Lecture Illustrated Guide to Home Chemistry Experiments and I'd like to understand the chemistry that I'm doing a bit better. For example, here's an iodine experiment.


  
  Weigh out 2.0 g of potassium iodide and transfer it to a test tube.
  Add about 1.5 mL of distilled water to the test tube and swirl to dissolve the potassium iodide.
  Add 1.5 mL of concentrated hydrochloric acid (or about 1.8 mL of hardware store muriatic acid) to the test tube and swirl to mix the
  solutions.
  Add about 10 mL of drugstore 3% hydrogen peroxide. The solution immediately turns dark brown as the iodide ions are oxidized to
  elemental iodine, which precipitates out.
  Swirl the test tube to suspend the iodine and pour the liquid through a funnel with a folded piece of filter paper to capture the
  iodine crystals.
  Rinse the iodine crystals on the filter paper several times with a few mL of distilled water. The rinse solution appears brown from
  dissolved iodine, but iodine is not very soluble in water, so you're
  not losing much of your yield.
  Spread out the filter paper on a watch glass or saucer and allow the crystals to dry thoroughly. Iodine gradually sublimates (passes
  directly from solid to gaseous form) at room temperature, so don't
  leave the crystals exposed to air any longer than necessary to dry
  them.
  Once the crystals are dry, transfer them to a sealed storage bottle or vial.
  


It's easy enough to follow and I know I have $$\ce{KI, H2O, HCl, H2O2}$$ with a result of $$\ce{I2\, +\, ???}$$ but then what?

I've had HS Chemistry, skipped Chem in college :(, so I know how to balance equations, convert from g or ml to moles, etc, but where can I learn to translate an experiment into equations without having to ask others (Textbooks or online resources are ideal, going and taking chem at a local uni. is a distant second, since I work fulltime)? 

Currently I'm sticking to inorganic chem, so that should simplify things a bit.

I'm currently studying EdX's Solid State Chemistry, so I hope that'll help some.
We know that this is a redox situation, where iodide ion is oxidized to iodine, and where hydrogen peroxide is reduced to water.

The oxidation equation is as such: $\ce{2I- -&gt; I2 + 2e-}$

The reduction equation is as such: $\ce{2H+ + H2O2 + 2e- -&gt; 2H2O}$

Therefore, the overal equation is as such: $\ce{2H+ + H2O2 + 2I- -&gt; 2H2O + I2}$

You can read more about redox reactions here.