Chemical Reaction for making Acetic Acid from Glycerol inside Fermenter
Short answer: under anaerobic conditions, it is impossible to produce acetate as the sole product from glycerol with wild-type E. coli. 

Explanation: Under anaerobic conditions (and assuming no substrates for anaerobic respiration have been added), a significant concern is redox balancing. Since electrons are not transferred to oxygen (as they are under aerobic conditions), the number of electrons in the substrates must equal the number of electrons in the products. 

Glycerol as a substrate presents problems in this regard, because it is slightly more reduced than typical hexose substrates (eg glucose). Furthermore, the desired product acetic acid is more oxidized than glycerol (or hexoses for that matter).

Looking at the core metabolic pathways of E. coli, we immediately see that 1 molecule of glycerol is converted to 1 molecule of acetate. E coli is unable to utilize all three carbon atoms of glycerol make 3 molecules of acetate from 2 molecules of glycerol. The third carbon atom of glycerol is lost as either formic acid or $\ce{CO2}$ (with concomitant production of hydrogen gas). So we can write an initial stoichiometry of 

glycerol ($\ce{C3H8O3}$) + $\ce{H2O -&gt;}$ acetic acid ($\ce{C2H4O2}$) + formic acid ($\ce{CH2O2}$) + 4 electrons + 4 $\ce{H+}$. 

[Note that the formic acid can be converted to $\ce{H2}$ and $\ce{CO2}$. The exact ratio of hydrogen to formate will depend on conditions.]

Unfortunately, there is no pathway for disposing of the excess electrons. You would need to add a respiration substrate such as fumarate (converted to succinate) or nitrate or nitrite (both can be reduced to ammonia).

One alternative approach would be to ferment the glycerol to a mixture of ethanol and formic acid or hydrogen gas and then chemically oxidize the ethanol to acetic acid. The balanced stoichiometry of that fermentation is

glycerol ($\ce{C3H8O3}$) $\ce{-&gt;}$ ethanol ($\ce{C2H6O}$) + formic acid ($\ce{CH2O2}$)

or

glycerol ($\ce{C3H8O3}$) $\ce{-&gt;}$ ethanol ($\ce{C2H6O}$) + $\ce{H2 + CO2}$