Redox Combustion Reactions Consider a combustion reaction like the complete combustion of ethane. Here is what I have for the complete combustion as well as the oxidation and reduction:
$$\ce{\overset{-3}{C}_2\overset{+1}{H}_6 + \frac{7}{2} \overset{0}{O}_2 -&gt; 2 \overset{+4}{C}\overset{-2}{O}_2 + 3 \overset{+1}{H}_2\overset{-2}{O}}$$
$$
\begin{align}
\ce{2 H2O + C2H6 &amp;-&gt; 2 CO2 + 10 H+ + 10 e-}\tag{ox}\\
\ce{\frac{1}{2} O2 + 10 H+ + 10 e- &amp;-&gt; 5 H2O}\tag{red}
\end{align}
$$

Why is it that when writing the reduction half equation for the reduction of oxygen that we don't consider oxygen being reduced into carbon dioxide? i.e. why isn't the reduction equation something of the form $\ce{O2 -&gt; H2O + CO2}~?$

Also the oxidation reaction suggests that water is required initially for the reaction to proceed, is this true?


Edit
I realise my reduction half equation is not balanced and so I now have this equation:
$$\ce{\overset{-3}{C}_2\overset{+1}{H}_6 + \frac{7}{2} \overset{0}{O}_2 -&gt; 2 \overset{+4}{C}\overset{-2}{O}_2 + 3 \overset{+1}{H}_2\overset{-2}{O}}$$
$$
\begin{align}
\ce{2 H2O + C2H6 &amp;-&gt; 2 CO2 + 10 H+ + 10 e-}\tag{ox}\\
\ce{\frac{7}{2} O2 + 14 H+ + 14 e- &amp;-&gt; 7 H2O}\tag{red}
\end{align}
$$
which is clearly not correct as the reduction and oxidation half equations do not add together to give the combustion reaction and the reduction equation does not have the $\ce{CO2}$ product as well. I guess my question now stands as how might I write (or begin to write) the half equations for a combustion reaction?
Edit 2
I realise I wrote my equations wrong above
$$
\begin{align}
\ce{4 H2O + C2H6 &amp;-&gt; 2 CO2 + 14 H+ + 14 e-}\tag{ox}\\
\ce{\frac{7}{2} O2 + 14 H+ + 14 e- &amp;-&gt; 7 H2O}\tag{red}
\end{align}
$$
But my question still remains. Why do we not consider the reduction of $\ce{O2}$ into $\ce{CO2}$ as well?
Your redox equations look awkward (and wrong, too) because you are trying apply the half-reactions method as if it  were happening in aquatic media to the combustion reaction, which effectively occurs in the gas phase.
Of course, there should be no requirements for water to be present.
As it's been noted in the comments, ionic half-reactions method is not the most elegant way of find coefficients in this case.

If you really want to use it, I'd suggest to write the elements undergoing redox reactions separately from the corresponding compounds (and yes, we do consider oxygen being reduced into carbon dioxide too):

$$
\begin{align}
\ce{2 \overset{-3}{C} &amp;→ 2 \overset{+4}{C} + 14 e-} &amp;|\cdot 2 \tag{ox} \\
\ce{\overset{0}{O}_2 + 4 e- &amp;→ 2 \overset{-2}{O}} &amp;|\cdot 7 \tag{red} \\
\hline
\ce{4 \overset{-3}{C} + 7 \overset{0}{O}_2 &amp;→ 4 \overset{+4}{C} + 14 \overset{-2}{O}}
\end{align}
$$

resulting in a balanced equation:

$$\ce{2 C2H6 + 7 O2 → 4 CO2 + 6 H2O}$$

Note that in general there is an average oxidation number of carbon involved, and it might as well be fractional.