I am puzzeled by the answer. Acetyl-CoA when enter citric acid cycle, it produces 3NADH, 1FADH2, total 4 reducing equivalents. If each one makes 1H2O, then 1acetyl-CoA should make 4H2O. Why some textbooks make it 2H2O. In addition, almost like all (or many) do not balance oxygen; they say entering 16O2 and making 16H2O, so how?
Thank you.
Acetyl-CoA has three carbon-hydrogen and one carbon-carbon bond, and it is electrons 'held' in these bonds that are available for biological oxidation. Thus we say that acetyl-CoA contributes four reducing equivalents. In the TCA cycle, these are transiently passed to NAD+ (to give NADH) and to FAD (to give FADH2): the cofactors may be considered intermediate electron carriers.
It takes 4 electrons (2 reducing equivalents) to make 2 water molecules from 1 O2 (or 4 electrons to reduce 1 oxygen 'atom' to 1 water molecule), but this takes place in the cytochrome oxidase reaction of the respiratory redox chain (with reduced cytochrome C as electron donor), not in the TCA cycle.
It is important to realize that water contributes no reducing equivalents to cellular respiration, and that hydration reactions are neither oxidation or reductions. For water to contribute reducing equivalents, it would have to be 'split', as it is in photosynthesis.
In addition, oxygen is not consumed or produced in the TCA cycle.
The assertion that water can pass electrons to NAD+ in the TCA cycle is an infamous 'textbook error', notably advocated by Efraim Racker in A New Look at Mechanisms of Bioenergetics, and my opinion on this (for what it is worth) is given in this SE answer.
For the role of water in the TCA cycle and the controversies surrounding it, I would recommend the letters by (i) D.E. Atkinson and (ii) Herreros & Garcia-Sancho on TCA cycle confusion published in TIBS and kindly made available by David. (These references are not indexed by Pubmed and are apparently not available from the TIBS website)
