Answer:
92.4%
Explanation:
Assume 100g of the cathode material. The LiCoO2 cathode has 7.09 g of Li. If half of Li migrates during charging, 3.55g Li migrates. 100g of the LiMn2O4 cathode contain 3.84g of Li. The fraction of lithium in LiMn2O4 that would need to migrate out of the cathode to deliver the same amount of lithium to the graphite anode:
(3.55g/ 3.84g ) * 100% = 92.4%