Potassium chlorate from potassium chloride and hydrogen peroxide
Do not attempt! Peroxide is not a strong enough oxidant to oxidize chloride ion to chlorate. Neither is oxygen. However, peroxide can oxidize chloride to poisonous chlorine gas.

Redox reactions are spontaneous if the two half reactions would produce a positive potential in a galvanic cell. Most redox half reactions have measured standard electrode potentials against a known standard so that they can be compared. 

The reaction you want to do is:

$$ \ce{3H2O2 + Cl- -&gt; 3H2O + ClO3-}$$

The two half-reactions are:


Oxidation:   $\ce{Cl- +3H2O -&gt; ClO3- +6H+ +6e-}$
Reduction:   $\ce{H2O2 +2H+ +2e- -&gt; 2H2O} \ \ \ \ E^ \circ =  +1.78 \text{ V}$


The reduction potential of hydrogen peroxide is on the table, but the reduction potential for the other reaction is not. However, the other reaction is the sum of two reactions on the table, so we can use Hess's Law. 

$$\ce{ 2ClO3- + 12H+ + 10e- -&gt;Cl2 + 6H2O} \ \ \ \ E^\circ = +1.49 \text{ V}$$
$$\ce{Cl2 +2e- -&gt; 2Cl-} \ \ \ \ E^\circ = +1.36 \text{ V}$$
$$\ce{2ClO3- 12H+ + 12e- -&gt; 2Cl- + 6H2O}  \ \ \ \ E^\circ = +2.85 \text{ V}$$

Since standard potentials do not scale by stoichiometric factor (it's in the Nernst equation instead), the value we just found also works for our oxidation reaction. However, since this reaction is reverse of the "reduction half-reaction", we need a negative sign:


Oxidation:   $\ce{Cl- +3H2O -&gt; ClO3- +6H+ +6e-}\ \ \ \ E^ \circ =  -2.85 \text{ V}$
Reduction:   $\ce{H2O2 +2H+ +2e- -&gt; 2H2O} \ \ \ \ E^ \circ =  +1.78 \text{ V}$
Overall: $\ce{3H2O2 + Cl- -&gt; 3H2O + ClO3-}\ \ \ \ E^ \circ =  -1.07 \text{ V}$


This reaction is not spontaneous. Peroxide cannot oxidize chloride, at least not to the chlorate ion. Note that peroxide can and does oxidize chloride to poisonous chlorine gas in the presence of acid. This redox is spontaneous and will happen.  Do not attempt!


Reduction:   $\ce{H2O2 +2H+ +2e- -&gt; 2H2O} \ \ \ \ E^ \circ =  +1.78
   \text{ V}$ 
Oxidation:   $\ce{2Cl- + 2e- -&gt; Cl2} \ \ \ \ E^ \circ = 
   -1.36 \text{ V}$ 
Overall: $\ce{ 2Cl- + H2O2 + 2H+ -&gt; Cl2 + 2H2O} \ \ \ \ E^ \circ =  +0.42 \text{ V}$


Peroxide will oxidize chlorine gas to chlorate, but who wants to handle chlorine?