Reduction of acetaldehyde from rectus and sinister faces

I am asked to consider the reduction of acetaldehyde from the re and si faces. I am also asked whether the added hydrogen (labeled H\* in my picture) is pro-R or pro-S in both cases. In my attached picture I replaced the added hydrogen (H\*) with D to determine the absolute configuration. 


If I did this correctly, addition of hydrogen from the si face results in a pro-S product with respect to the added hydrogen (H\*). And addition of hydrogen from the re face results in a pro-R product with respect to the added hydrogen. Am I correct? 


![enter image description here](https://i.stack.imgur.com/oe04i.jpg)




Just to keep it slightly more realistic from a chemistry point of view, let's consider the reduction of deuterated acetaldehyde (1-Deuterioethanal):


![Reduction of deuterated acetaldehyde from both the re and si face.](https://i.stack.imgur.com/vs71f.jpg)


When we add from the *si* face, we get an (*R*) center, and if we add from the *re* face, it's the other way around (as it should be!) and we have (*S*) configuration.


Now, what if we reduce "normal" acetaldehyde? We then have to decide which of the proton substituents gets the higher priority when trying to assign pro-*R* or pro-*S*. If you decide that the added proton has higher priority (as you did in your question) then you get an equivalent of a pairwise exchange of the substituents (in comparison to the structures that I drew), and the stereocenter is exactly the opposite of what I drew in the figure above.


From a synthetic point of view it is much cheaper to use hydrogen gas for the reduction and have the isotope label already reside on the starting material.


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