I wanted to calculate the theoretical capacity of NiMnO3, which will be use as anode material for LIB. It is very hard to get it through literature. So
I have calculated it like this:
NiMnO3 + 6Li + 6e = Ni + Mn + 3Li2O
n x f / M.W x 3600
= 6 x 96485 / 161.63 x 3600 =994.9 mAh/g
MW of NiMnO3 = 161.63 g/mol
Electron involved = 6 e-
Please suggest.
Thanks
Dear, you may find useful information in links below:
pubs.acs.org/doi/abs/10.1021/ef400797r
pubs.acs.org/doi/abs/10.1021/jp9017184
pubs.rsc.org/en/content/articlelanding/2015/ee/c5ee00930h
https://www.researchgate.net/.../How_did_they_calculate_Silicons_Theoretical_capacit...
I recommend you follow the link given to you by Mr. Mosab.
and study it carefully.
It is incorrect, certainly.
If we assume that this substance will work as a cathode material, then most likely nickel and manganese can not be reduced to the oxidation state "0", as you implicitly supposed. This would mean the decomposition of the initial compound and the appearance of the pure metals. Most likely final oxidation state of metalls will be +2 (this must be verified experimentally). So, you will get one third of your result (331.6 mAh/g, if your calculation was correct)... Those, the final formula will be Li2NiMnO3 (or Li2O + NiO + MnO - there are few else variants exist).
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