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I know how to solve redox reactions, but this one is really confusing me since I can't really write half reactions for it.

$$\ce{ClO3- -> Cl2 + O2}$$

The answer is apparently:

$$\ce{2H+ + 2ClO3- -> 5/2 O2 + Cl2 + H2O}$$

This makes sense. I understand that 12 electrons would be lost from oxygen if it all went to $\ce{O2}$ but only 10 electrons are gained by the two chlorine(V)'s so you have to use a water to make the charge balance work out. I just don't see how you'd come to this conclusion on your own.

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This is tricky and one way to think about it that I can offer is to treat $\ce{ClO3-}$ as a combination of $\ce{Cl^5+}$ and $\ce{O^2-}$. Since the problem is that the starting material is undergoing disproportionation with different elements being reduced/oxidised, if you could split the starting material into the two different elements, then you could circumvent the issue. Obviously the bonding in the chlorate ion is not ionic but this procedure allows you to construct "half-equations" and then an overall equation.

I would not suggest that you write this in an exam.

The two half-reactions are then:

$$\begin{align} \ce{Cl^5+ + 5e-} &\longrightarrow \ce{1/2Cl2} \tag{1} \\ \ce{O^2-} &\longrightarrow \ce{1/2 O2 + 2e-} \tag{2} \end{align}$$

$[(1) \times 2] + [(2) \times 5]$ gives

$$\ce{2 Cl^5+ + 5O^2- -> Cl2 + 5/2 O2}$$

In order to "re-form" your chlorate, you need three oxide ions per Cl ion, so you need to add one more oxide to both sides:

$$\ce{2 Cl^5+ + 6O^2- -> Cl2 + 5/2 O2 + O^2-}$$

Now you can get back your chlorate:

$$\ce{2 ClO3- -> Cl2 + 5/2 O2 + O^2-}$$

Now it's highly unlikely that there are going to be free oxide ions hanging around, so we just protonate them by adding $\ce{2 H+}$ on both sides. The oxide ion simply becomes water:

$$\ce{2 ClO3- + 2H+ -> Cl2 + 5/2 O2 + H2O}$$

I understand that 12 electrons would be lost from oxygen if it all went to $\ce{O2}$ but only 10 electrons are gained by the two chlorine (V)'s so you have to use a water to make the charge balance work out.

Yeah, if all six oxygens became $\ce{O2}$ then they would lose 12 electrons. Therefore, one of the oxygens does not become $\ce{O2}$; it becomes $\ce{H2O}$ and remains as oxygen(-2). It is not about charge balance.

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  • $\begingroup$ What I meant by "to make charge balance work out" was that if you didn't use a water to balance out the electrons then your net equation wouldn't have the protons on the left so the charge balance wouldn't work out. $\endgroup$ – gannex Apr 30 '16 at 0:08
  • $\begingroup$ Thanks a bunch orthocresol! I was trying to do exactly what you're doing but I didn't break the "half reactions" down to their fundamental level like you did (I was writing the numbers of each "ion" that come from ClO3 instead of just one ion to the half diatomic molecule). This seems to be the key here, so I will just remember to do that tomorrow :) This has been very helpful thank you! $\endgroup$ – gannex Apr 30 '16 at 0:08
  • $\begingroup$ @gannex I see what you mean by the charge balance, it's just a different way of looking at it I suppose. Anyway no problem. Again though I must reiterate that this is just a trick that allows you to get the half reactions - it has no physical interpretation nor does it reflect what's actually going on in the system. So be careful with it. $\endgroup$ – orthocresol Apr 30 '16 at 1:08
  • $\begingroup$ I guess you could do this: (12 H+ + 2ClO3- + 10 e= Cl2 +6H2O) +[2 Clo3 - + 10e= Cl2 + 3O2 + 12 e] 5 = 12 clo3 +12H+ =6 cl2 + 15 o2+6 H2o $\endgroup$ – gannex Apr 30 '16 at 1:52
  • $\begingroup$ @gannex Yeah, I suppose you could do that too. $\endgroup$ – orthocresol Apr 30 '16 at 18:45
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There is no real reason not to do it the traditional way. As a reminder: You do not have to oxidise oxides from chlorate, you can intermediately assume you are oxidising oxygen from water molecules. That allows you to formulate an oxidation half-equation without having to deal with cumbersome additional elements. But first the reduction:

$$\begin{align}\ce{2 \overset{\mathrm{+V}}{Cl}O3- + 10 e- \phantom{\ + 12 H+} &-> \overset{\mathrm{\pm0}}{Cl}_2}\tag{Red1}\\ \ce{2ClO3- + 10 e- + 12 H+ &-> Cl2}\tag{Red2}\\ \ce{2ClO3- + 10 e- + 12 H+ &-> Cl2 + 6 H2O}\tag{Red3}\end{align}$$

(In all these half-equations the first step is balancing oxidation states with electrons, the second is balancing charge with protons and the third is balancing mass with water.) Here goes for the oxidation:

$$\begin{align}\ce{2 H2O &-> O2 + 4e-}\tag{Ox1}\\ \ce{2 H2O &-> O2 + 4 e- + 4 H+}\tag{Ox2}\end{align}$$

(Here, we are done after two steps.)

And now, simply add up the two equations making sure the number of electrons is balanced; we need $2 \times (\text{Red3})$ and $5 \times (\text{Ox2})$ followed by cancelling out:

$$\begin{align}\ce{4 ClO3- + 24H+ + 10 H2O &-> 2 Cl2 + 12 H2O + 5 O2 + 10 H+}\tag{Redox1}\\ \ce{4 ClO3- + 4 H+ &-> 2 Cl2 + 2 H2O + 5 O2}\tag{Redox2}\\ \ce{4 ClO3- + 4 H+ &-> 2 Cl2 + 5 O2 + 2 H2O}\tag{Redox3}\end{align}$$

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