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Li-ion batteries used in automobiles typically use a

Chapter , Problem 20.80

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QUESTION:

Li-ion batteries used in automobiles typically use a LiMn2O4 cathode in place of the LiCoO2 cathode found in most Li-ion batteries. (a) Calculate the mass percent lithium in each electrode material? (b) Which material has a higher percentage of lithium? Does this help to explain why batteries made with LiMn2O4 cathodes deliver less power on discharging? (c) In a battery that uses a LiCoO2 cathode approximately 50% of the lithium migrates from the cathode to the anode on charging. In a battery that uses a LiMn2O4 cathode what fraction of the lithium in LiMn2O4 would need to migrate out of the cathode to deliver the same amount of lithium to the graphite anode?

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QUESTION:

Li-ion batteries used in automobiles typically use a LiMn2O4 cathode in place of the LiCoO2 cathode found in most Li-ion batteries. (a) Calculate the mass percent lithium in each electrode material? (b) Which material has a higher percentage of lithium? Does this help to explain why batteries made with LiMn2O4 cathodes deliver less power on discharging? (c) In a battery that uses a LiCoO2 cathode approximately 50% of the lithium migrates from the cathode to the anode on charging. In a battery that uses a LiMn2O4 cathode what fraction of the lithium in LiMn2O4 would need to migrate out of the cathode to deliver the same amount of lithium to the graphite anode?

ANSWER:

Problem 20.80

Li-ion batteries used in automobiles typically use a LiMn2O4 cathode in place of the $$ \mathrm{LiCoO_{2}} $$ cathode found in most Li-ion batteries.

(a) Calculate the mass percent lithium in each electrode material?

(b) Which material has a higher percentage of lithium? Does this help to explain why batteries made with LiMn2O4 cathodes deliver less power on discharging?

(c) In a battery that uses a $$ \mathrm{LiCoO_{2}} $$ cathode approximately 50% of the lithium migrates from the cathode to the anode on charging. In a battery that uses a $$ \mathrm{LiMn_{2}O_{4}} $$ cathode what fraction of the lithium in $$ \mathrm{LiMn_{2}O_{4}} $$ would need to migrate out of the cathode to deliver the same amount of lithium to the graphite anode?

                                                          Step by step solution

Step 1 of 3

(a)

From the given,

Mass of Li in one mole of $$ \mathrm{LiMn_{2}O_{4}} $$ = 6.94 g

Molar mass of $$ \mathrm{LiMn_{2}O_{4}} $$ = 180.81 g

Let’s calculate the mass percentage of Li in $$ \mathrm{LiMn_{2}O_{4}} $$

                               $$ \mathrm{Mass\,percent\,of\,Li\,in\,LiMn_{2}O_{4}}=\frac{\mathrm{Mass\,of\,Li}}{\mathrm{Molar\,mass\,LiMn_{2}O_{4}}}

                                                                                          $$ =\frac{\mathrm{6.94 \,g}}{\mathrm{180.81\,g}} \times 100 =3.8 \%

Therefore, mass percentage of lithium in $$ \mathrm{LiMn_{2}O_{4}} $$ is 3.8 %.

From the given,

Mass of Li in $$ \mathrm{LiCoO_{2}} $$ = 6.94 g

Molar mass of $$ \mathrm{LiCoO_{2}} $$ = 97.87 g

Let’s calculate the mass percentage of “Li” in $$ \mathrm{LiCoO_{2}} $$.

                                 $$ \mathrm{Mass\,percentage\,of\,Li} = \frac{ \mathrm {Mass\,of\,Li}} { \mathrm{
Molar\,mass\,of}\, \mathrm {LiCoCO_{2}}}

                                                                           $$ =\frac{6.94 \mathrm{g}}{97.87 \,\mathrm{g}}\times 100 \% = 7.09 \%

Therefore, mass percent of Li in $$ \mathrm{LiCoO_{2}} $$ is 7.09

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