Suppose you are testing a coin cell comprised of LiCoO_2 cathode material. The l

Question

Suppose you are testing a coin cell comprised of LiCoO_2 cathode material. The loading amount of the cathode material is about 1.7 mg. What will be the current drawn from the cell with the C-rates of 2C? (The theoretical specific capacity of LiCoO_2 is ~274 mAh/g). Calculate the maximum theoretical specific energy (energy/weight) of the cell with graphite anode and LiCoO_2 as a cathode material according to the reaction: C_6 + 2LiCoO_2 rightarrow LiC_6 + 2Li_0.5CoO_2; delta G = 357 kJ/mol, E = 3.7 V

Explanation / Answer

Given loading amount of cathode material = 1.7mg = 0.0017g

Given theoretical specific capacity = 274mAh/g = .274 Ah/g

Therefore, theoretical capacity of loading material = 0.0017 * 0.274 = 0.0004658 Ah

We know,

current drawn from cell = theoretical capacity of material * C rating.

Hence, current drawn = 2 * 0.0004658 = 0.0009316 Ampere

Answer = 0.9316 mA

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