How does sodium in ethanol reduce carbonyl compounds?
Sodium is a powerful reducing agent. It wants to donate an electron to things that are electron deficient. 

The mechanism of the reduction of aldehydes and ketones by sodium would be similar to that of the Bouveault-Blanc reduction. Ethanol is the source of the protons (at least early in the reaction). The dots in the reactions below signify radicals (unpaired electrons), which are represented to the best mchem and MathJax's ability (which is not great).


  Step 1 - one electron transfer from sodium to carbonyl group:


$$\ce{Na. + RCH=O-&gt;Na+ + RCH.-O^-}$$


  Step 2 - abstraction of a proton from ethanol:


$$\ce{CH3CH2OH + RCH.-O^- &lt;=&gt; CH3CH2O- + RCH.-OH}$$
This step is an equilibrium step, and is driven toward products by the second electron transfer (which consumes the $\ce{RCH.-OH}$.


  Step 3 - one electron transfer from a second sodium atom to the organic radical:


$$\ce{Na. +RCH.-OH -&gt; Na+ + RCH:^{-}{}-OH}$$


  Step 4 - second proton transfer from ethanol:


$$\ce{CH3CH2OH + RCH:^{-}{}-OH -&gt; CH3CH2O- + RCH2OH}$$

This reaction, while technically an acid-base equilibrium, essentially goes to completion because of the vast disparity in acidity of $\ce{C-H}$ and $\ce{O-H}$.

The net reaction is:

$$\ce{2Na + RCH=O + 2CH3CH2OH -&gt; RCH2OH + 2CH3CH2ONa}$$
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