I have difficulties finding the mechanism of the state solid synthesis of $\ce{ LiMn2O4}$ in literacy.

Starting from lithium acetate and manganese acetate, I've found the general equation easily:

$$\ce{Li(OAc) + 2 Mn(OAc)2 + 43/4 O2 -> LiMn2O4 + 10 CO2 + 15/2 H2O}$$

What I can't find is a mechanism of this reaction. I know that $\ce{Mn}$ is oxidized and $\ce{Li}$ keeps its oxidation state. Is $\ce{Mn3O4}$ a precursor in this reaction? Is $\ce{Mn2O3}$ a by-product?


1 Answer 1


$\ce{LiMn2O4}$ has been made by rotary kilning of $\ce{Li2CO3}$ (or $\ce{LiOH}$ or $\ce{LiOH.H2O}$) + $\ce{MnO2}$. I made hundreds of grams on a small scale and 1000 pounds on a larger scale about 25 years ago.

The fine powders are added to a rotary kiln (similar to making Portland cement) and heated to about 800 degrees C. The kiln rotates, stirring the mix; the lithium compound melts and reacts with the $\ce{MnO2}$ as the powdery mass gets mushed around and around. The product was used as the cathode material in a lithium battery.

The oxidation state of the manganese drops from +4 to +3.5. The significant process is that oxygen gets lost, so to balance the equation, take one lithium, two manganese and four oxygens on the product side to make $\ce{LiMn2O4}$; everything else from the raw material side goes away.

Perhaps $\ce{LiMn2O4}$ can be made in water from dissolved salts, but gaseous $\ce{O2}$ would not be a convenient reagent.


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