Specific Capacity of Cathode Materials

Recently I read some news about Lithium Ion batteries and the author mentioned that Lithium has the highest Specific Capacity of 3861mAh/g. I was very curious to know where this magic number came from. After trying to understand more about that, finally I found the answer. That was 12th class electrochemistry. The calculation is given below.

Specific Capacity = (N x F) / (Atomic Weight)
N = Valency of the Material
F = Faraday constant = 96485 Coulombs/Mole

If this has to be expressed in terms of current, divide that by 3600
F = 26.801Ah/Mole

This is how the equation works for Lithium. Faraday constant is nothing but the total charge of Avogadro number of electrons. Since Lithium has a Valency of 1 and atomic weight of 6.94g/Mole, every 6.94g of Lithium would give out Valency times Avogadro number of electrons when ionized. So, to find out the Specific Capacity

Specific Capacity = (1 Valency x 26.801Ah/g x 1000mA/A) / 6.94g/Mole
= 3861mAh/g

That means, when 1 gram of Lithium metal ionizes to Li+ ions, it gives out 3.861Ah of electricity. This is just the current, but not the energy. If we are interested to calculate the energy density of reaction, it is required to know the Standard Electrode Potential of each material. Since it is 3.03V for Lithium, the total energy density of ionization of Lithium is
3.03V x 3.861Ah/g * 1000g/kg = 11701Wh/kg.
This comes very close the values of hydrocarbons.

But, remember this is just a half reaction taking place in the cell. We did not consider the anode reaction at all. So, total Specific Capacity of a complete cell would be much lower and it also depends upon the anode, electrolyte, other chemicals used in the cell etc.

Table1: Specific Capacity of Cathode Materials

Reaction Atomic No: Atomic Weight Valency Specific Capacity (mAh/g)
H/H(+) 1 1.008 1 26588
Li/Li(+) 3 6.94 1 3861
Na/Na(+) 11 22.990 1 1166
Mg/Mg(2+) 12 24.310 2 2205
Al/Al(3+) 13 26.980 3 2980
K/K(+) 19 39.10 1 685
Ca/Ca(2+) 20 40.08 2 1337
Zn/Zn(2+) 30 65.39 2 820
Pb/Pb(2+) 82 207.2 2 259

The following bar graph summarizes the specific capacity of all Metals given above.
Specific Capacity of Metals


Published by Anand Sivaram (आनन्दः )

twitter.com/anand_sivaram https://www.linkedin.com/in/anandsivaram/

8 thoughts on “Specific Capacity of Cathode Materials

  1. I’m not sure where you’re getting your info, but good topic. I needs to spend some time learning much more or understanding more. Thanks for magnificent information I was looking for this info for my mission.

  2. hi~There is one error in your text.
    The Specific Capacity is correct calculated as below:

    = (1 Valency x 26.801Ah/g x 1000mA/A) / 6.94g/Mole
    = 3861mAh/g=3.861 Ah/g

    But in your following word the specific capacity is error with 3.681Ah and the calculation of gravimetric energy density is error. The correct energy density should be 3.03V x 3.861Ah/g * 1000g/kg = 11701 Wh/kg.

  3. Thanks for pointing out those mistakes. I incorrectly given 3.681 instead
    of 3.861 there.
    I corrected the post with the values you have given.

  4. is there any formulea to convert discharge capacity in mAh/g to capacitance in F/g

  5. its informative …can please anyone explain how to calculate specific capacity of anode material such as LiC6

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