I having trouble understanding the equation of the cellular respiration. The thing that bothers me is the number of $\ce{H_2O}$ molecules. Generally, cellular respiration is written thus :

$\ce{C_6H_{12}O_6 + 6O_2 + 6H_2O -> 6CO_2 + 12H_2O}$

Yet studying the reaction on a molecular level, I realized that the water molecules just don't add up to be six! (on the left side) What I obtained from my textbook and Wikipedia is that (per 1 glucose molecule)

  • (Glycolysis) (in the reaction that 2-Phosphoglycerate(2PG) forms Phosphoenolpyruvate(PEP)) 2 molecules of $\ce{H_2O}$ are out of the reaction
  • (TCA cycle) (in the reaction that combines Oxaloacetate and Acetyl-CoA to form Citrate) 2 molecules of $\ce{H_2O}$ are into the reaction
  • (TCA cycle) (in the reaction that Citrate form Cis-Aconitate) 2 molecules of $\ce{H_2O}$ are out of the reaction
  • (TCA cycle) (in the reaction that Cis-Aconitate form D-Isocitrate) 2 molecules of $\ce{H_2O}$ are into the reaction
  • (TCA cycle) (in the reaction from Fumarate to Malate) 2 molecules of $\ce{H_2O}$ are into the reaction

These sums up to a net of (only) 2 molecules of $\ce{H_2O}$ into the process of glycolysis and TCA cycle... additional 6 molecules of $\ce{H_2O}$ comes from oxidative phosphorylation. Then, where can the remaining 4 molecules of $\ce{H_2O}$ possibly be?

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    $\begingroup$ Can you point to a source that uses the equation in that form? Everywhere that I have looked has no water on the LHS and 6 water on the RHS, and that is the equation that I'm familiar with. Incidentally in your accounting you missed the loss of 1 water at the aldolase step I believe. $\endgroup$
    – Alan Boyd
    Commented Mar 29, 2014 at 9:11
  • $\begingroup$ My (possibly) outdated school textbook states the equation like what I wrote before... but I don't think it's trustworthy.(It's weird in many ways) I think the equation with only 6 water molecules is correct. So,(may I ask) is there a water molecule used in the aldolase step? two of them? $\endgroup$ Commented Apr 20, 2014 at 14:18
  • $\begingroup$ @AlanBoyd, I don't think H2O is involved in the aldolase reaction. Aldolase is a lyase, not a hydrolase; see chem.qmul.ac.uk/iubmb/enzyme/EC4/1/2/13.html $\endgroup$
    – Roland
    Commented Jul 25, 2016 at 20:42

1 Answer 1


First thing to make clear is that net $6 \, \ce{H_2O}$ go out of the reaction.($12 \,\ce{H_2O}-6\,\ce{H_2O}$)

Let me tell you my calculation, you should then be able to figure out what went wrong.

For the Left hand Side, $6\,\ce{H_2O}$ are accounted here :

  • $2\,\ce{H_2O}$ go in conversion of 2-Phosphoglycertae to phosphoenolpyruvate.
  • $2\,\ce{H_2O}$ in TCA from conversion of Oxoloacetate to Citrat by combining with Acetyl CoA.(One for each Acetyl CoA molecule)
  • The aconitase -cis-aconitase-isocitrate water molecules cancel out. Equal no go in and out.
  • $2\,\ce{H_2O}$ in TCA from conversion of fumarate to Malate.

For the right hnd side, if we calculate the no. of NADH formed = 10 and assume that each participates in respiratory chain, we have $10\,\ce{H_2O}$ from there. The other $2$ will be from the $2$ FADH produced.

So we have 6 on the LHS and 12 on the RHS. Hope this makes sense !

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    $\begingroup$ I'm afraid that per a NADH or a FADH2, these yield only 2 electrons, reducing only 1/2 water molecule. So, only 6H2O produced at the respiratory chain $\endgroup$ Commented Apr 20, 2014 at 14:22
  • $\begingroup$ @ChanheeJeong, No, 2 electrons reduce 1/2 O2 molecule, so forms 1 H2O. So this answer is correct I think. $\endgroup$
    – Roland
    Commented Jul 25, 2016 at 21:15

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