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In acidic solution, what does $\ce{ClO3-}$ reduce into and, most importantly, why? It's $$\ce{6e- +ClO_{3}^- +6H^+ -> Cl^- + 3 H_2O}.$$

Now, with the gift of foresight, why was the equation for its oxidation not instead $$\ce {2e^- +ClO_{3}^- +2H^+ \rightarrow Cl^- + O_2 + H_2O}?$$ Obviously it balances and charge is conserved. However, the latter equation would seem more likely: less charges have to be pulled apart (so less work has to be done to move these charges), and the bond energy of bonds in a water molecule is almost twice that in an oxygen molecule?

Where is there a flaw in my logic and does there exist a rough set of rules to determine how things will split up under oxidation or reduction?

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    $\begingroup$ The chlorine atom in $ClO_3^-$ is actually being reduced in this reaction since it gains electrons and goes from a +5 to a -1 oxidation state. $\endgroup$ – kaliaden Mar 22 '13 at 20:18
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    $\begingroup$ and the bond energy ... in a water molecule is almost twice that in an oxygen molecule This indicates that the first equation is preferred. $\endgroup$ – LDC3 Aug 29 '14 at 4:42
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The first equation is correct because $\ce{ClO3-}$ is reduced to $\ce{Cl-}$ and oxygen maintains its -II oxidation state in both educts and products. As chlorine changes its oxidation state from +V in chlorate to -I in chloride, $6 \ \mathrm{e^-}$ are required for this reduction.

In your second equation, chlorate is reduced to chloride as well; however, only $2 \ \mathrm{e^-}$ are explicitly supplied. The remaining electrons would have to come from the concurrent oxidation of oxygen from -II in $\ce{ClO3-}$ to 0 in $\ce{O2}$, and this is unlikely to happen, as oxygen is more electronegative than chlorine.

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