Chemistry - Potassium chlorate from potassium chloride and hydrogen peroxide

Solution 1:

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:

$$ \ce{3H2O2 + Cl- -> 3H2O + ClO3-}$$

The two half-reactions are:

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

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.

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

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

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

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

Solution 2:

Potassium chlorate ($\ce{KClO3}$) has covalent bonds between the chloride and the oxygen, while potassium chloride ($\ce{KCl}$) lacks those covalent bonds to oxygen.

Hydrogen peroxide decomposes into water ($\ce{H2O}$) and oxygen gas ($\ce{O2}$) while oxidizing (LEO: Loss of Electrons is Oxidation) its environment. Peroxide is useful primarily for sucking electrons from a solution, which is very different from forming covalent bonds.

If you wished to dissolve solid potassium in water to make KOH peroxide would likely perform nicely as the peroxide would suck electrons from the solid K and thereby help it form ions ($\ce{K+}$) which would combine with $\ce{OH-}$ from the water to from KOH.