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[求助]
鋰離子電池正負極材料首次充放電損失的容量的去向 已有1人參與
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扣電中若鋰離子電池正極材料首次充放電效率是90%,那么其余充的10%的電量去哪了呢,正極材料結(jié)構(gòu)的變化導致?lián)p失的電量占多少呢,副反應(如電解液的副反應消耗的電量)會占多少呢? 另外扣電中碳負極材料首次效率是90%,那么嵌入碳層間不再脫出的鋰離子占多少呢,電解液與負極的副反應所消耗的電量占多少呢? 請各位大神幫忙解答,萬分感謝。 |
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The situation is simple on anode: 10% is responsible for the irreversible reaction that consumes Li+, 90% is responsible for Li+ intercalation. Among the 10% irreversible, tiny amount is caused by Li+ stuck in graphite (such as defects or surface functionality reaction with Li etc), most were caused by reduction of solvents to form Li-salts, which is main ingredient of SEI. On cathode it's more comlicated and depends on cathode material. In layer TM oxides the first cycle will experience some structural transformation (such as NCA NMC or LMR), and this process will produce excess amount of Li+ that cannot go back to cathode. Sometimes this irreversible cancels out the irreversible produced at anode side. The oxidation of eelctrolytes also happen, which does not directly consume Li+, but some of the products are solible and diffuse to anode to reduce. The net result of whether Li+ is consumed or produced deoends on many factors. |
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