Li-ion batteries used in automobiles typically use a $\operatorname{LiMn}_{2} \mathrm{O}_{4}\( cathode in place of the \)\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 $\operatorname{LiMn}_{2} \mathrm{O}_{4}$ 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 \(\operatorname{LiMn}_{2} \mathrm{O}_{4}\) cathode, what fraction of the lithium in \(\mathrm{LiMn}_{2} \mathrm{O}_{4}\) would need to migrate out of the cathode to deliver the same amount of lithium to the graphite anode?

Step by step solution

01

Calculate the molar mass of LiCoO₂ and LiMn₂O₄

In order to find the mass percentage of lithium in each electrode material, we should first calculate the molar mass of each compound. For LiCoO₂, its molar mass is calculated as follows: Molar mass = 1 Li + 1 Co + 2 O = (1 × 6.939) + (1 × 58.933) + (2 × 16.00) = 97.872 g/mol For LiMn₂O₄, its molar mass is calculated as follows: Molar mass = 1 Li + 2 Mn + 4 O = (1 × 6.939) + (2 × 54.938) + (4 × 16.00) = 180.815 g/mol

02

Calculate the mass percent of lithium in each electrode material

Now, we can find the mass percent of lithium in each material using the formula: Mass percent = (Mass of Li in the formula / Molar mass of the compound) × 100 For LiCoO₂: Mass percent = (6.939 / 97.872) × 100 = 0.0709 × 100 = 7.09% For LiMn₂O₄: Mass percent = (6.939 / 180.815) × 100 = 0.0384 × 100 = 3.84%

03

Compare the materials and discuss power delivery

From Step 2, we can see that LiCoO₂ has a higher mass percentage of lithium (7.09%) compared to LiMn₂O₄ (3.84%). This indicates that batteries made with LiMn₂O₄ deliver less power on discharging compared to batteries made with LiCoO₂ since they have a lower lithium content.

04

Calculate the fraction of lithium migration for LiMn₂O₄

It is given that in a LiCoO₂ battery, approximately 50% of the lithium migrates from the cathode to the anode on charging. We need to find what fraction of lithium in LiMn₂O₄ needs to migrate out of the cathode to deliver the same amount of lithium to the graphite anode. Let x be the fraction of lithium that needs to migrate from LiMn₂O₄. Since both batteries have to deliver the same amount of lithium, we can use the following equation: Li migration (LiCoO₂) = Li migration (LiMn₂O₄) 0.5 (Mass percent of Li in LiCoO₂) = x (Mass percent of Li in LiMn₂O₄) x = (0.5 × 7.09%) / 3.84% = 0.924 So, around 92.4% of the lithium in the LiMn₂O₄ cathode needs to migrate out of the cathode to deliver the same amount of lithium to the graphite anode.

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