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kxch123

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[求助] 锂离子电池正负极材料首次充放电损失的容量的去向 已有1人参与

扣电中若锂离子电池正极材料首次充放电效率是90%,那么其余充的10%的电量去哪了呢,正极材料结构的变化导致损失的电量占多少呢,副反应(如电解液的副反应消耗的电量)会占多少呢?
另外扣电中碳负极材料首次效率是90%,那么嵌入碳层间不再脱出的锂离子占多少呢,电解液与负极的副反应所消耗的电量占多少呢?
请各位大神帮忙解答,万分感谢。
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kangxu

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【答案】应助回帖

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感谢参与,应助指数 +1
kxch123: 金币+10, ★★★很有帮助 2015-07-09 17:48:43
kxch123: 金币+10, ★★★很有帮助 2015-07-09 17:48:44
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.
2楼2015-07-09 04:12:46
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kxch123

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引用回帖:
2楼: Originally posted by kangxu at 2015-07-09 04:12:46
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 ...

Thank you very much!

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3楼2015-07-09 17:48:20
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