1. First  read this one (it is about Li as cathode material):

https://saurorja.org/2011/08/14/specific-capacity-of-cathode-materials/

2. Ask an other student what Mw/n means (it is called equivalent) and how it is calculated for redox reactions (or Google).

1. n in Faraday law is not the number of Li ions, but the number of electron changed by a quantity of 1 mol  ; n=1 for Li+  and LiFePO4, but n=3 for Li3Fe2(PO4)3

Concerning LiFePO4 and Li3Fe2(PO4)3, look at Fe oxidation state (it can vary only from +2 to +3). Concerning Sn, you should look at thee phase diagram of Li-Sn system (which compounds exist in this system).

Cornel Radu : thank you so much for your help.. It's very useful.

Please read this paper : http://www.sciencedirect.com/science/article/pii/S037877531100694X

"They said 2 mol of Li+ can be intercalated"  then I calculated the capacity n=2 for Li3Fe2(PO4)3 not 3 and I got the same theoretical capacity of 128 mAh/g.

Oleg Brylev : I do agree with you.. we can look at the oxidation state.. But why for LiFePO4, n=1.. while the oxidation state of Fe = +2 or +3?I've calculated for LiFePO4, if n=1, I got the same value of capacity with the literature. Concerning Li3Fe2(PO4)3, n = 2.. Please help me, if I want to calculate the theoretical capacity of LiCuPO4.. n=1 or will be more than 1 (oxidation state of Cu = +1 or +2)..

Thanks all..

In Li3Fe2(PO4)3 there are 2 Fe atoms, for LiCuPO4, n=1. We should look at the difference between two extreme oxidation states of transition metal.

I understood.. Thanks..

My question again, this patent (http://www.google.com/patents/US20020039681) showed that the capacity of LiCuPO4 is around 400 mAh/g, as you can see in the following figure.. while If I calculated the theoretical capacity of LiCuPO4 (n=1, Mw=165.4584 g/mol), I got the capacity around 162 mAh/g, so if the capacity more than 162 mAh/g, there's the formation of lithium rich or more than 1 mol of Li+ can be intercalated? I am curious..

If you got the value of ca. 400 mAh/g, about 2.5 moles of Li can be intercalated. As the authors of the patent claim, you can go to metallic Cu and Li3PO4.

A. You have to know  an initial and a final state for the electrode material.

Li3Fe2(PO4)3 + 2Li = Li5Fe2(PO4)3

where two moles of Fe(3+) are reduced to Fe(2+). and two moles of Li are oxidized . Thus 2 mol of electrons are changed per formula unit (n=2 to be used in Faraday's law). This exercise is theoretical (if the stoichiometry is obeyed).

B.        LiCuPO4  + 2 Li = Li3PO4 + Cu

1 mol Cu(2+) is reduced to Cu(0) per formula unit   (n=2)

C.    Think also to a limit case (Li doped, not only substituted compound): 

Li1-2x Cu1+x PO4 + (2+2x) Li = Li3PO4 + (1+x)Cu

n= 2+2x

Oleg Brylev and Cornel Radu : Thank you very much for your valuable answer...

Regards,

Vinsensia

Depends if it is half cell or full cell. I half cell there is always more Li than in a full cell.