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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.