xLi2MnO3-(1-x)LiNi13Co13Mn13O2 (x=0.5) layered complex cathode showing voltage hysteresis
二次根式练习题30道加答案过程
二次根式练习题30道加答案过程1.当a______时,a?2有意义;当x______时,2.当x______时,1有意义. x?315.计算:??11有意义;当x______时,的值为1. 2?22x?xab?11 xx3.直接写出下列各式的结果: 49=______;2=______;2=______;2=______; 2=______;[2]2=______.4.下列各式中正确的是. ??42??2?4?? 27?35.下列各式中,一定是二次根式的是. ?32 2?x6.已知2x?3是二次根式,则x应满足的条件是.x>0 x≤0 x≥-x>-3.当x为何值时,下列式子有意义? ?x; ?x2;x2?1; 7?x.8.计算下列各式:29.若?2?成立,则x,y必须满足条件______.10. ?112______;=______;4324?________.49?36=______;0.81?0.25=______;24a?a3=______.11.下列计算正确的是. 2?3? 2??6?42??312.化简5?2,结果是.?2-10 10 13.如果??,那么.x≥0 x≥ 0≤x≤ x为任意实数 14.当x=-3时,x2的值是.± - 93a6a2b?13a2?492?572x2y716.已知三角形一边长为,这条边上的高为cm,求该三角形的面积.17.把下列各式化成最简二次根式:=______;=______; 45=______; 48x=______;23=______;412=______;a5b3=______; 112?3=______.18.在横线上填出一个最简单的因式,使得它与所给二次根式相乘的结果为有理式:如:32与2. 2与______; 32与______; a 与______; 8a与______;6a2与______.19.?x?xx?x成立的条件是. x<1且x≠0 x>0且x≠1 0<x≤1 0<x<10.下列计算不正确...的是. 3116?72y3x?13x6xy 2??209x?2x21.下列根式中,不是..最简二次根式的是 A.B.C.12D.22.1625= 279=243= 27=5=23=34.当a=______时,最简二次根式与?可以合并.35.若a=+2,b=-2,则a+b=______,ab=______. 36.合并二次根式:?5x1111? ?0.125222?=______;23.把下列二次根式,27,,445,2,,,化简后,与2的被开方数相同的有_________;与的被开a?4ax=______. xx?y23xy37.下列各式中是最简二次根式的是. ab2?3方数相同的有______;与的被开方数相同的有______.4. ?313=______;7?548=______.25.化简后,与的被开方数相同的二次根式是.141626.下列说法正确的是.被开方数相同的二次根式可以合并与可以合并只有根指数为2的根式才能合并2与不能合并27.可以与合并的二次根式是.2aa127a3a28、9?7?5.29.??.30.?3??31.?.32.27?13?.33.12?3438.下列计算正确的是.2??5ab?5a??6?5x?4x?x39.等于.6?6??221 ??2240.?112? 1..42..3..44.? 5.2.46.4?6?3?2.47...78.49.2ba?3a3bab?.参考答案1.a?2,x?3..2.x>0,x=1.3.7;7;7;7;0.7;49.4.D.5.B.6.D..x≤1;x=0;x 是任意实数;x≥-7..18;6;15;6.9.x≥0且y≥0.10.;24;16. 42;0.45;11.B.12.A.13.B. 14.Ba2.b; 15.2;6;24;2x;2ab; 49;12;6xy32y. 16..217.2;;;4;632302?;; abab;18.;;;;19.C.20.C.21.C.453; ; ; 22; ; 53222;2;4.23.,2,,,422.24.3;?6.25.B.26.A. 7.C.28.2?329.30.1123??434.6.35.2,3.36.2;?.31.?32.?33.37.B.38.D.39.B. 042. 6?41.36?7.19?6143.7?44.2.45.84?6.446.?8.47.2?5..?1..?2.? 二次根式1.表示二次根式的条件是______.2.使x有意义的x的取值范围是______..若?有意义,则m =______.4.已知??y?4,则xy的平方根为______..当x=5时,在实数范围内没有意义的是. 1?x| 7?x2?3x4x?206.若|x?5|?2?0,则x-y的值是.--7.计算下列各式: ?2?1)2328.已知△ABC的三边长a、b、c均为整数,且a和b 满足a?2?b2?6b?9?0.试求△ABC的c边的长.9.已知数a,b,c在数轴上的位置如图所示:化简:a2?|a?c|?2?|?b|的结果是:______. 10.已知矩形的长为2,宽为,则面积为______cm2.11.比较大小:3______2;5______4;?22______?6. 12.如果nm是二次根式,那么m,n应该满足条件. mn>0m>0,n≥0 m≥0,n>0 mn≥0且m≠013.把4234根号外的因式移进根号内,结果等于. ? ?44414.计算:5?=______;8a3b.122ab2=______; ?2213?2;=______;3?=______.15.先化简,再求值:?a,其中a?5?12. 16.把下列各式中根号外的因式移到根号里面: a?1 a;?1y?1?17.已知a,b为实数,且??0,求a2008-b2008的值. 18.化简二次根式:17=______;18=______;?413=______. 19.计算下列各式,使得结果的分母中不含有二次根式: 1=______; 132______;2x2=______;y=______.0.已知≈1.732,则13≈______;27≈______.1.计算b1a?ab?ab等于.1ab2ab 11a2bab bab bab22.下列各式中,最简二次根式是.1x?yab x2? 5a2b23.?? ?a?ba?b24.已知:△ABC中,AB=AC,∠A=120°,BC?8,求△ABC的面积.25.观察规律:12?1?2?1,1?2?3?,12??2?3求值.122?7=______;1?=______;1n?1?n=______.26.238ab3与6ba2b无法合并,这种说法是______的.27.一个等腰三角形的两边长分别是2和3,则这个等腰三角形的周长为.2?4362?262?42?4或62?28.?.29.0??12?|5?|?230.a?a133a?12aa.31.2aba1a?bb?aa3b?2bab3.32.化简求值:3x1?4y?x?y,其中x=4,y=1x9.33.已知四边形ABCD四条边的长分别为,,.5和3,求它的周长.4.探究下面问题判断下列各式是否成立.你认为成立的,在括号内画“√”,否则画“×”.①2?23?22;②3?38?338;③4?4?4;④5?524?5524.1515你判断完以上各题后,发现了什么规律?请用含有n的式子将规律表示出来,并写出n的取值范围.请你用所学的数学知识说明你在中所写式子的正确性.35.设a??b??,则a2007b2008的值是______.36.的运算结果是. 0abab2abab37.下列计算正确的是. 2?a?ba??aba2?b2?a?ba?1a?a8.1?2.1?2?.100101.40.2?2.41.已知x??,y??,求值:x2-xy+y2.42.已知x+y=5,xy=3,求x?y的值.yx43.若b<0,化简?ab3的结果是______.44.若菱形的两条对角线长分别为和则此菱形的面积为______.45.若x??2,则代数式x2-4x+3的值是______.6.当a<2时,式子a?2,2?a,a?2,2中,有意义的有. 1个 2个 3个7.若a,b两数满足b<0<a且|b|>|a|,则下列各式有意义的是.a?bb?a a?b ab48abab5??ab?9.?8x4.50.已知:如图,直角梯形ABCD中,AD∥BC,∠A =90°,△BCD为等边三角形,且AD=2,求梯形ABCD的周长.二次根式基础练习一、选择题1.若3?m为二次根式,则m的取值为A.m≤3B.m<3C.m≥D.m>32.下列式子中二次根式的个数有⑴1;⑵3?3;⑶?x2?1;⑷8;⑸12;⑹3?x;⑺x2?2x?3.A.2个 B.3个 C.4个 D.5个3.当a?2a?2有意义时,a的取值范围是A.a≥B.a>C.a≠ D.a≠-24.下列计算正确的是①??4??9?6;②?4?9?6;③52?42?5?4??4?1;④52?42?52?42?1;A.1个 B.2个 C.3个 D.4个5.化简二次根式2?3得A.?B.5C.?D.306.对于二次根式x2?9,以下说法不正确的是A.它是一个正数 B.是一个无理数C.是最简二次根式D.它的最小值是37.把3aab分母有理化后得A.4bB.C.1 bD.b28.ax?by的有理化因式是A.x?yB.x?yC.ax?by D.ax?by9.下列二次根式中,最简二次根式是A.3a B.13C.D.10.计算:a1b?ab?ab等于A.1ab2abB.1ababC.1bab D.bab二、填空题11.当x___________时,?3x是二次根式.12.当x___________时,3?4x在实数范围内有意义. 13.比较大小:?32______?23.14.2ba?a18b?____________;252?242?__________.15.计算:3a?2b?___________.16b216.计算:ca2=_________________.17.当a=3时,则15?a2?___________.18.若x?2x?23?x?3?x成立,则x满足_____________________.三、解答题19.把下列各式写成平方差的形式,再分解因式:)计算:⑴?3?;⑵2?13?6;⑶131?23?;⑷x?10?1y?z.221.计算:⑴?220;⑵0.01?81; 0.25?144⑶12123ab1?2?1;⑷?.352bab22.把下列各式化成最简二次根式: abc27132?122 ⑴;⑵?252723.已知:x?24.参考答案:一、选择题 c3.a4b120?4,求x2?2的值.x1.A;2.C;3.B;4.A;5.B;6.B;7.D;8.C;9.D;10.A.二、填空题11.≤1314b;12.≤;13.<;14.,7;15.302ab;16.;17.32;a34318.2≤x<3.三、解答题19.⑴;⑵;⑶;⑷;20.⑴?243;⑵2;⑶?43;⑷10xyz; 33c2321.⑴?;⑵;⑶1;⑷;22.⑴33;⑵ ?2bc;23.18.4a420二次根式检测题一、选择题有意义,那么x的取值范围是 A.x?B.x?3C.x? D.x≥3 2.下列二次根式中,是最简二次根式的是新- 课-标- 第-一 -网 1.A.2xyB.ab23.1?2a,那么A.a<≥11 B.错误!24.下列二次根式,5.a的值为6.m?n的值是C.1D..D.8. )A.x?1B.x??1C.x≥1D.x≤?19.n的最小值是A. B.C. D.210.k、m、n为三整数,若错误!未找到引用源。
氧化还原反应习题及答案
mol·L-1 HI中的氢。
ቤተ መጻሕፍቲ ባይዱ
(1) 金属铁能置换Cu2+,而Fe3+溶液又能溶解铜。
【解释】根据对角线关系,将各标准电极电势φө按
由小到大排列后,电势表左下方的物质能和右上方的 物质发生反应。
Fe2+ + 2e- = Fe φөFe2+/Fe= - 0.440V Cu2+ + 2e- = Cu φөCu2+/Cu= 0.337V Fe3+ + e- = Fe2+ φөFe3+ /Fe2+ = 0.771V
②将氧化还原反应分为两个半反应,一个发生氧 化反应,另一个发生还原反应:
H2S → S Cr2O72- + H+ → Cr3+ + H2O
③分别配平两个半反应:
H2S - 2e-→ S + 2H+ Cr2O72- + 14H+ + 6e- → 2Cr3+ + 7H2O ④确定两个半反应得、失电子数的最小公倍数,将 两个半反应分别乘以相应系数,使其得、失电子数 相等,再将两个半反应合并为一个配平的氧化还原 反应的离子方程式,最后改写成化学方程式。
则原电池的电动势:Eө = φөCl2/Cl-- φөBr2/Br-
= 1.36- 1.07 = 0.29V △rGө = - z·F·E ө = -2×96500C·mol-1×0.29V
= -55970J ·mol-1 = -55.97 kJ ·mol-1 < 0
东北三省三校(哈师大附中、东北师大附中、辽宁省实验中学)2023-2024学年高三下学期第一次联合模
东北三省三校(哈师大附中、东北师大附中、辽宁省实验中学)2023-2024学年高三下学期第一次联合模拟考化学试卷学校:___________姓名:___________班级:___________考号:___________一、单选题1.“神舟”飞天,逐梦科技强国。
下列说法中正确的是A.神舟飞船返回舱系统复合材料中的酚醛树脂属于有机高分子材料B.空间站的太阳能电池板的主要材料是二氧化硅C.飞船返回舱表面的耐高温陶瓷材料属于传统无机非金属材料D.神舟飞船的推进系统中使用的碳纤维属于有机高分子材料2.下列化学用语或表述正确的是A.基态氧原子的轨道表示式:B.甲醛分子的空间填充模型:C.用电子式表示HCl的形成过程:D.钢铁发生吸氧腐蚀时的负极反应式:O2+2H2O+4e-=4OH-3.奥司他韦是一种口服活性流感病毒神经氨酸酶抑制剂,分子结构如图所示。
下列说法正确的是A.该分子含有4种官能团B.与互为同系物C.分子式为C16H28N2O4D.该分子可发生取代、加成、消去、氧化反应4.设N A为阿伏加德罗常数的值。
下列说法正确的是A.标准状况下,11.2L环庚烷中氢原子数目为7N AB.13g苯、乙炔的混合物中所含氢原子数目为N AC.2.4gMg在空气中燃烧生成MgO和Mg3N2,转移电子数目为0.1N AD.1mol NH4+中含有完全相同的N—H共价键的数目为3N A5.邻二氮菲能与Fe2+发生显色反应,生成橙红色螯合物,用于Fe2+检验,化学反应如下。
下列说法正确的是A.邻二氮菲的核磁共振氢谱有6组吸收峰B.元素的电负性顺序:N>H>C>FeC.每个螯合物离子中含有2个配位键D.用邻二氮菲检验Fe2+时,需要调节合适的酸碱性环境6.下列离子方程式正确的是A.将少量SO2通入Ca(ClO)2溶液中:SO2+Ca2++ClO—+H2O=CaSO4↓+Cl—+2H+ B.向乙二醇溶液中加入足量酸性高锰酸钾溶液:5+84MnO-+24+H=5+82+Mn+22H2OC.向饱和Na2CO3溶液中通入过量CO2:CO23-+CO2+H2O=2HCO3-D.向Fe(NO3)3溶液中加入过量HI溶液:Fe3++12H++3NO3-+10I—=Fe2++5I2+6H2O+3NO↑7.2022年度化学领域十大新兴技术之一的钠离子电池(Sodium-ion battery)是一种二次电6C+NaTMO2,下列说法错误的是池,电池总反应为:Na x C6+Na1-x TMO2 放电充电A.放电时正极反应式:Na1-x TMO2+xNa++xe—=NaTMO2B.钠离子电池的比能量比锂离子电池高C.充电时a电极电势高于b电极D.放电时每转移1mol电子,负极质量减少23g8.下列实验对应的现象及结论均正确且两者具有因果关系的是选项实验现象结论A向淀粉碘化钾溶液中通入足量Cl2溶液先变蓝后褪色不能证明Cl2氧化性强于I2B 向5mL0.1mol/L AgNO3溶液中先滴入5滴0.1mol/L NaCl溶液,再滴入5滴0.1mol/L KI溶液先产生白色沉淀后产生黄色沉淀Ksp(AgCl)>Ksp(AgI)C 蔗糖与浓硫酸混合搅拌,用湿润的品红试纸检验其气体产物蔗糖变黑,品红试纸褪色浓硫酸具有脱水性和氧化性D向K2Cr2O7溶液中滴加NaOH溶液溶液颜色由黄色变为橙色减小H+浓度,Cr2O27-转为CrO24-A.A B.B C.C D.D9.某种钾盐具有鲜艳的颜色,其阴离子结构如图所示。
2020—2021学年人教版(2019)化学选择性必修2第二章《分子结构与性质》测试题(含答案)
C.最高价氧化物对应水化物的酸性:W>Y
D.气态简单氢化物的热稳定性:Y>Z
9.厌氧氨化法(Anammox)是一种新型的氨氮去除技术。下列说法中错误的是
A.联氨(N2H4)中含有极性键和非极性键
B.过程II中N2H4发生了氧化反应
C.过程IV中 发生了氧化反应
A.f在元素周期表的位置是第3周期ⅢA族。
B.d、e常见离子的半径大小:r(O2-)>r(Na+)
C.g、h的最高价氧化物对应的水化物的酸性强弱是:HClO4>H2SO4
D.x2d2的结构式:
14.下列各组物质中,前者与后者不相同的是
A.反应热效应:稀硫酸和NaOH与Zn和稀盐酸
B.化学键类型:Mg3N2与Na2O2
(3)叠氮酸(HN3)是一种弱酸,可部分电离出H+和 。请写出两种与 互为等电子体的分子的化学式:__________________。
(4)NaN3与KN3相比,NaN3的晶格能________(填“>”、“=”或“<”)KN3的晶格能。
(5)某元素X形成的离子X+中K、L、M三个电子层均充满了电子。它与 形成晶体的结构如图所示。X+的符号是________,晶体中距离每个 最近的X+有________个。
2.下列关于化学键的说法中不正确的是
A.化学键是一种作用力
B.化学键可以使离子相结合,也可以使原子相结合
C.化学反应过程中,反应物分子内的化学键断裂,产物分子中的化学键形成
D.氢键也是一种化学键
3.CFCl3破坏臭氧层的过程如图所示,下列说法错误的是( )
A.过程Ⅰ:断裂C-Cl键
B.过程Ⅱ:O3+Cl=ClO+O2
第13讲陌生情境下氧化还原反应方程式的书写(原卷版)-备战2024年高考化学复习精讲精练
备战2024年高考化学【一轮·夯基提能】复习精讲精练第13讲陌生情境下氧化还原反应方程式的书写本讲复习目标1、掌握陌生氧化还原反应方程式的书写方法及技巧。
夯基·知识精讲书写复杂氧化还原反应方程式的步骤(1)“读”取题目中的有效信息,找出发生氧化还原反应的物质或离子。
(2)依据掌握的氧化还原反应规律合理地预测产物(这里要重视题目中的信息提示,或给出的生成物),写出主要反应物和生成物的化学式。
(3)根据氧化还原反应的守恒规律确定氧化剂、还原剂、还原产物、氧化产物的相应计量数。
(4)根据原子守恒和溶液的酸碱性,通过在反应方程式的两端添加H+、OH-或H2O的形式使方程式的两端的电荷守恒。
(5)再依据质量守恒,写出规范的方程式。
1.明确一些特殊物质中元素的化合价CuFeS2:Cu为+2,Fe为+2,S为-2;K2FeO4:Fe为+6;FePO4:Fe为+3;LiFePO4:Fe为+2;Li2NH、LiNH2、AlN:N为-3;N2H4:N为-2;Na2S2O3:S为+2;C2O2-4:C为+3;HCN:C为+2;N为-3;CuH:Cu为+1,H为-1;BH-4:B为+3、H为-1;FeO n-4:Fe为+(8-n);Si3N4:Si为+4,N为-3;MnO(OH):Mn为+3;2.认识并掌握常见的氧化剂、还原剂。
(1)强氧化性物质或离子:Na2O2、Fe3+、NO2、HNO3(浓)、HNO3(稀)、O2、O3、H2O2、H2SO4(浓)、Cl2、Br2、HClO、NaClO、Ca(ClO)2、KMnO4(H+)。
(2)强还原性物质或离子:Fe2+、Fe(OH)2、SO2(H2SO3、Na2SO3、NaHSO3、SO2-3)、H2S(Na2S、NaHS、S2-)、HI(NaI、I-)等。
3.书写关键——识记常见氧化剂、还原剂及产物预测(1)常见的氧化剂及还原产物预测(2)常见的还原剂及氧化产物预测4.熟记各氧化反应和还原反应得失电子数目【例题1】MnO2是重要的化工原料,由软锰矿制备MnO2的一种工艺流程如图:溶出时,Fe的氧化过程及得到Mn2+的主要途径如图所示:Fe Fe2+Fe3+MnO2Mn2+步骤II是从软锰矿中溶出Mn2+的主要反应,反应的离子方程式是__________________________。
化学高一必修二知识点总结(推荐13篇)
一、热量变化常见放热反应:1,酸碱中和2,所有燃烧反应3,金属和酸反应4,大多数的化合反应5,浓硫酸等溶解常见吸热反应:1,CO2+C====2CO2,H2O+C====CO+H2(水煤气)3,Ba(OH)2晶体与NH4Cl反应4,大多数分解反应5,硝酸铵的溶解热化学方程式;注意事项5二、燃料燃烧释放热量化学高一必修二知识点总结第2篇铝及其化合物的性质1.铝与盐酸的反应:2Al+6HCl=2AlCl3+3H2↑2.铝与强碱的反应:2Al+2NaOH+6H2O=2Na[Al(OH)4]+3H2↑3.铝在空气中氧化:4Al+3O2==2Al2O34.氧化铝与酸反应:Al2O3+6HCl=2AlCl3+3H2O5.氧化铝与强碱反应:Al2O3+2NaOH+3H2O=2Na[Al(OH)4]6.氢氧化铝与强酸反应:Al(OH)3+3HCl=AlCl3+3H2O7.氢氧化铝与强碱反应:Al(OH)3+NaOH=Na[Al(OH)4]8.实验室制取氢氧化铝沉淀:Al3++3NH3?H2O=Al(OH)3↓+3NH4+1、空气的成分:氮气占78%,氧气占21%,稀有气体占,二氧化碳占,其它气体与杂质占2、主要的空气污染物:NO2、CO、SO2、H2S、NO等物质3、其它常见气体的化学式:NH3(氨气)、CO(一氧化碳)、CO2(二氧化碳)、CH4(甲烷)、SO2(二氧化硫)、SO3(三氧化硫)、NO(一氧化氮)、NO2(二氧化氮)、H2S(硫化氢)、HCl(_)4、常见的酸根或离子:SO42-(硫酸根)、NO3-(硝酸根)、CO32-(碳酸根)、ClO3-(氯酸)、MnO4-(高锰酸根)、MnO42-(锰酸根)、PO43-(磷酸根)、Cl-(氯离子)、HCO3-(碳酸氢根)、HSO4-(硫酸氢根)、HPO42-(磷酸氢根)、H2PO4-(磷酸二氢根)、OH-(氢氧根)、HS-(硫氢根)、S2-(硫离子)、NH4+(铵根或铵离子)、K+(钾离子)、Ca2+(钙离子)、Na+(钠离子)、Mg2+(镁离子)、Al3+(铝离子)、Zn2+(锌离子)、Fe2+(亚铁离子)、Fe3+(铁离子)、Cu2+(铜离子)、Ag+(银离子)、Ba2+(钡离子)各元素或原子团的化合价与上面离子的电荷数相对应:课本P80一价钾钠氢和银,二价钙镁钡和锌;一二铜汞二三铁,三价铝来四价硅。
第二十一章 一元二次方程 单元测试(含答案) 2024-2025学年人教版九年级数学上册
第二十一章一元二次方程一、选择题(每题3分,共24分)1.在一元二次方程x2−2x−3=0中,一次项系数是( )A.1B.0C.−2D.−3 2.若x=−1是关于x的方程x2+ax=0的一个根,则a的值为( )A.1B.2C.3D.43.用配方法解方程x2-6x-1=0时,配方结果正确的是( )A.(x-3)2=10B.(x-3)2=8C.(x-6)2=10D.(x-3)2=1 4.一元二次方程x2−2x=0的解是( )A.x1=3,x2=1B.x1=2,x2=0C.x1=3,x2=−2D.x1=−2,x2=−15.一元二次方程x(x−1)=2(x−1)的解完全正确的是( )A.x=2B.x1=2,x2=1C.x1=−2,x2=1D.x1=3,x2=−1 6.若关于x的一元二次方程(k−1)x2−4x−1=0有实数根,则k的取值范围( )A.k>−3B.k≥−3且k≠1C.k>−3且k≠0D.k≤−37.若一元二次方程2x2+3x﹣6=0的两个根分别为x1,x2,则x1•x2的值等于( )A.﹣6B.6C.﹣3D.38.甲流病毒是一种传染性极强的急性呼吸道传染病,感染者的临床以发热、乏力、干咳为主要表现.在“甲流”初期,若有一人感染了“甲流”,若得不到有效控制,则每轮传染平均一个人传染x人,经过两轮传染后共有256人感染了“甲流”.则关于x的方程为( )A.x+x(x+1)=256B.x2+x=256C.1+x+x(x+1)=256D.(x+1)+(x+1)2=256二、填空题(每题4分,共20分)9.若方程(m−1)x2+6x−1=0是关于x的一元二次方程,则m的取值范围是 .10.用配方法解一元二次方程x2+6x+3=0时,将它化为(x+m)2=n的形式,则m−n的值为 .11.已知关于x的一元二次方程2m x2−4x+1−5n=0有两个相等的实数根,则2m+5n的值为 .12.已知三角形两边的长分别是2和5,第三边的长是方程x2-7x+10=0的根,则这个三角形的周长是 .13.已知m,n是方程x2+4x−3=0的两个实数根,则m2+5m+n+2024的值是 .三、计算题(共10分)14.解方程:(1)x2−4x−12=0;(2)x(x−9)=8(9−x).四、解答题(共46分)15.关于x的一元二次方程2x2−4x+(2m−1)=0有两个不相等的实数根.(1)求m的取值范围;(2)若方程有一个根为x=3+1,求m的值和另一根.16.已知关于x的一元二次方程x2−(m+2)x+m−1=0.(1)求证:无论m取何值,方程都有两个不相等的实数根;(2)如果方程的两个实数根为x1,x2,且x21+x22−x1x2=9,求m的值.17.为了提升居民生活质量,完善社区公共区域配套设施,今年夏天长春市在多个城区实施了旧城改造工程.已知某工程队在开始施工的7月份为某小区翻新道路12000m2,为了在入冬前完成道路翻新工程,之后加快了工程进度,结果9月份为该小区翻新道路14520 m2.(1)求这两个月该工程队工作效率的月平均增长率.(2)若10月份该工程队的工作效率按此增长率增长,估计到10月末该工程队能否完成该小区共55000m2的道路翻新任务?18.某超市销售一种衬衫,平均每天可售出20件,每件盈利40元.为了扩大销售、增加盈利,该超市准备适当降价,经过一段时间测算,发现每件衬衫每降价1元,平均每天可多售出2件.(1)若每件衬衫降价4元,平均每天可售出多少件衬衫? 此时每天销售获利多少元?(2)在每件盈利不少于 25元的前提下,要使该衬衫每天销售获利为1 200元,问每件衬衫应降价多少元?(3)该衬衫每天的销售获利能达到 1 300 元吗?如果能,请写出降价方案;如果不能,请说明理由.1.C2.A3.A4.B5.B6.B7.C8.C9.m≠110.−311.112.1213.202314.(1)解:x2−4x−12=0 x2−4x=12x2−4x+4=12+4(x−2)2=16x−2=±4即:x−2=4或x−2=−4∴x1=6,x2=−2(2)解:x(x−9)=8(9−x)解:x(x−9)−8(9−x)=0x(x−9)+8(x−9)=0(x−9)(x+8)=0即:x−9=0或x+8=0∴x1=9,x2=−815.(1)解:∵方程2x2−4x+(2m−1)=0有两个不相等的实数根,∴Δ=16−8(2m−1)=24−16m>0解得m<32;∵方程有一个根x=3+1,∴2×(3+1)2−4×(3+1)+(2m−1)=0解得m=−32,则2x2−4x−4=0,x2−2x−2=0∵x1+x2=2,∴x2=2−(1+3)=1−3,则x1=1+3,x2=1−3,即m的值是−32,另一根是1−3.16.(1)证明:Δ=[−(m+2)]2−4×1×(m−1)=m2+8,∵无论m取何值,m2+8>0,恒成立,∴无论m取何值,方程都有两个不相等的实数根;(2)解:∵x1,x2是方程x2−(m+2)x+m−1=0的两个实数根,∴x1+x2=m+2,x1⋅x2=m−1,∵x21+x22−x1x2=(x1+x2)2−3x1x2=9,∴(m+2)2−3(m−1)=9解得:m1=1或m2=−2.17.(1)解:设该工程队工作效率的月平均增长率为x,根据题意,得12000(1+x)2=14520.解这个方程,得x1=0.1,x2=−2.1(不合题意舍去).答:该工程队工作效率的月平均增长率为10%.(2)解:8月的工程量为:13200m2;10月的工程量为:15972m2;12000+13200+14520+15972=55692>55000.所以该工程队能完成该小区的道路翻新任务.18.(1)解:由题意可得,每件衬衫降价4元,平均每天可售出衬衫的数量为:20+4×2=28(件);此时每天获取的利润为(40-4)×28=1008(元);(2)解:设每件衬衫降价x元(0≤x≤15),由题意可得(20+2x)×(40-x)=1200,整理得x2-30x+200=0,解得x1=10,x2=20(舍),答:在每件盈利不少于25元的前提下,要使该衬衫每天销售获利为1200元,每件衬衫应降价10元;(3)解:该衬衫每天的销售获利不能达到1300元,理由如下:设每件衬衫降价y元,由题意可得(20+2y)×(40-y)=1300,整理得y2-30y+250=0,∵b2-4ac=302-4×1×250=-100<0,∴此方程没有实数根,即该衬衫每天的销售获利不能达到1300元.。
2024-2025学年上海曹杨二中高三上学期数学周测及答案(2024.09)
曹杨二中2024学年第一学期高三年级数学月考2024.09一、填空题(本大题共有12题,满分54分,第1-6题每题4分,第7-12题每题5分) 1.已知集合()()3,2A ,B ,=−∞=+∞,则A B ⋂= . 2.已知复数z 满足15i z =−(i 为虚数单位),则z = . 3.已知向量()()102,210a ,,b ,,==,则a ,b <>= .4.523x ⎫⎪⎭的二项展开式中的常数项为 .(结果用数值表示)5.设()y f x =是以1为周期的周期函数.若当01x <≤时,()2f x log x =,则32f ⎛⎫= ⎪⎝⎭.6.设m 为正实数.若直线0x y m −+=被圆()()22113x y −+−=所截得的弦长为m ,则m = .7.从一副去掉大小王的52张扑克牌中无放回地任意抽取两次。
在第一次抽到A 的条件下,第二次也抽到A 的概率为 .(结果用最简分数表示)8.设数列{}n a 前n 项和为n S 。
若()21n n S a n ,n N +=≥∈,则5S = . 9.已知,x y 为正实数,且1x y +=,则当21x y+取最小值时,x = . 10.设(),1a R f x lnx ax ∈=−+.若函数()y f x =的图像都在x 轴下方(不含x 轴),则a 的取值范围是 .11.已知{}n a 是严格增数列,且点()()1n n P n,a n ,n N ≥∈均在双曲线2231x y −=上。
设M R ∈,若对任意正整数n ,都有1n n P P M +>,则M 的最大值为 .12.设(){}2,235a R f x min x ,x ax a ∈=−−+−,其中{}min u,v 表示,u v 中的较小值.若函数()y f x =至少有3个零点,则a 的取值范围是 .二、选择题(本大题共4题,满分18分,第13-14题每题4分,第15-16题每题5分)13.已知a R ∈,则"1a >"是"11a<"的( ). A.充分非必要条件 B.必要非充分条件 C.充要条件 D.既非充分又非必要条件14.为研究某药品的疗效,选取若干名志愿者进行临床试验,所有志愿者的舒张压(单位:kPa )的分组区间为[)[)[)[)1213,1314,1415,1516,,,,,[]1617,.将其按从左到右的顺序分别编号为第一组,第二组,,第五组,下图是根据试验数据制成的频率分布直方图。
锂离子电池开题报告
武汉理工大学本科毕业论文(设计)开题报告题目锂离子电池正极材料Li2MnO3的掺杂改性院、系材料科学与工程研究院专业无机非金属材料科学与工程 10级学生姓名马娟学号 *************指导教师郝华1、研究背景锂离子电池是20世纪70年代以后发展起来的一种新型储能电池。
由于其具有高能量、寿命长、低能耗、无公害、无记忆效应以及自放电小、内阻小、性价比高、污染少等优点,锂离子电池在逐步应用中显示出巨大的优势,广泛应用于移动电话、笔记本电脑、摄像机、数码相机、电动汽车、储能、航天等领域。
特别是新能源汽车的开发与应用,要求具有高比能量的锂离子电池,而传统的正极材料难以满足能量密度的需要,因此迫切需要开发新型高比容量的锂离子电池正极材料。
高比容量,绿色环保,以及价格便宜都将是锂离子电池必不可少的因素。
正极材料作为整个电池的重要组成部分,直接影响电池的使用性能和制造成本。
近年来锂离子电池电极材料的研究和开发一直受到社会的广泛关注,其中正极材料的研究是对锂离子蓄电池研究和开发有着重要的价值。
目前使用的正极材料主要有 Li2CoO2,LiNi0.9Co0.lO2。
由于钴价格较锰将近贵到40倍,若将资源丰富、价格便宜、对环境污染小的锰用于阳极材料取代现在的钴,将会带来很大的经济效益。
层状结构Li2Mn03基正极材料以其理论容量高,环境友好以及原料价格便宜等优势得到广泛关注。
但该材料体系电导率低,制约了它的进一步应用。
制备正极材料的方法很多,而溶胶凝胶法由于其特有的优点备受关注。
溶胶凝胶法在配位化合物、纳米材料、金属簇合物的合成中已经得到了广泛的应用。
一般的合成方法中均采用两种或者两种以上的配合剂,将采用配合物低分子基团柠檬酸,且该物质对人体无害,目的在于减少有机物用量和环境污染,同时具有溶胶凝胶法合成材料的优点。
2、文献综述国内外对锂离子电池的研究进行了很长时间,锂离子电池也得到了广泛的应用,主要正极材料是Li2CoO2。
(word完整版)高考陌生情景中化学方程式的书写技巧与真题训练
高考陌生情景中化学方程式的书写技巧与真题训练陌生情景中的化学方程式书写,首先要根据材料中的信息写出反应物和生成物的化学式,然后再配平即可。
配平要遵循质量守恒定律(原子守恒、元素守恒、物料守恒),对于氧化还原反应还要遵循得失电子数相等即得失电子守恒规律,对于离子反应还要遵循电荷守恒规律。
【解题技巧】关键结合题目信息、流程图信息并熟记常见的氧化产物和还原产物判断生成物。
1. 熟记常见的氧化剂及对应的还原产物、还原剂及对应的氧化产物2.掌握书写信息型氧化还原反应的步骤(3步法)第1步:根据氧化还原顺序规律确定氧化性最强的为氧化剂,还原性最强的为还原剂;根据化合价规律及题给信息和已知元素化合物性质确定相应的还原产物、氧化产物;根据氧化还原反应的守恒规律确定氧化剂、还原剂、还原产物、氧化产物的相应化学计量数。
第2步:根据溶液的酸碱性,通过在反应方程式的两端添加H+或OH-的形式使方程式的两端的电荷守恒。
第3步:根据原子守恒,通过在反应方程式两端添加H2O(或其他小分子)使方程式两端的原子守恒。
3.氧化还原反应方程式的配平:步骤:①标出化合价变化了的元素的化合价。
②列变化:分别标出化合价升高数和化合价降低数③根据化合价升降总数相等确定发生氧化还原反应的物质的化学计量数。
④利用元素守恒,观察配平其他物质4.“补缺”的技巧(1)可能出现的情况:H+→H2O(酸性)、OH-→H2O(碱性)、H2O→H+(中性或酸性)、H2O→OH-(中性或碱性)(2)绝不可能出现的情况:H+→OH- 或者OH-→H+题型1 氧化还原反应型的化学方程式和离子方程式的书写1、氧化锌为白色粉末,可用于湿疹、癣等皮肤病的治疗。
纯化工业级氧化锌[含有Fe(Ⅱ)、Mn(Ⅱ)、Ni(Ⅱ)等杂质]的流程如下:提示:在本实验条件下,Ni(Ⅱ)不能被氧化;高锰酸钾的还原产物是MnO2。
反应②中除掉的杂质离子是__________,发生反应的离子方程式为_________________。
【初+中数学】+因式分解法课件+人教版九年级数学上册++
12.解方程:(2x + 1)2 − (2x + 1) − 12 = 0. [答案] x1 = 32,x2 = −2 13.已知三角形两边长分别是3和4,第三边长是方程x2 − 12x + 35 = 0
的根,试判断这个三角形的形状. [答案] 解方程x2 − 12x + 35 = 0,得x1 = 7,x2 = 5.当x = 7时, 3 + 4 = 7,此时不能构成三角形;当x = 5时,32 + 42 = 52,此时构成
么a的值为( C ) .
A.−7
B.1
C.7或−1
D.−2或3
5.若代数式x2 + 5x + 6与−x + 1的值相等,则x的值为( A ) .
A.−1或−5
B.−6或1
C.−2或−3
6.下列方程中,适合用因式分解法求解的是( D ) .
A.x2 + x + 2 = 0
B.x2 − 6x + 12 = 0
Байду номын сангаас
轻松达标
1.方程2x(5x − 4) = 0的解是( C ) .
A.x1
=
2,x2
=
4 5
B.x1
=
0,x2
=
5 4
C.x1
=
0,x2
=
4 5
D.x1
=
12,x2
=
4 5
2.一元二次方程x2 − x − 2 = 0的解是( D ) .
A.x1 = 1,x2 = 2
B.x1 = 1,x2 = −2
14.(2023·广州)解方程:x2 − 6x + 5 = 0. [答案] x1 = 1,x2 = 5
高中化学 2021届高三第三次模拟考试卷 化学(二) 教师版
(新高考)2021届好高三第三次模拟考试卷化 学(二)注意事项:1.答题前,先将自己的姓名、准考证号填写在试题卷和答题卡上,并将准考证号条形码粘贴在答题卡上的指定位置。
2.选择题的作答:每小题选出答案后,用2B 铅笔把答题卡上对应题目的答案标号涂黑,写在试题卷、草稿纸和答题卡上的非答题区域均无效。
3.非选择题的作答:用签字笔直接答在答题卡上对应的答题区域内。
写在试题卷、草稿纸和答题卡上的非答题区域均无效。
4.考试结束后,请将本试题卷和答题卡一并上交。
相对原子质量:H 1 C 12 N 14 O 16 Cl 35.5 Fe 56 Mn 55 Ba 137一、选择题(每小题2分,共20分,每小题只有一个选项符合题意。
)1.2020年11月10日,中国“奋斗者”号载人潜水器在马里亚纳海沟深度10909米处成功坐底并进行了一系列的深海探测科考活动。
下列说法正确的是 A .“奋斗者”号使用的锂离子电池工作时Li +向负极移动B .制造潜水器载人球舱的钛合金比纯金属钛具有更高的强度、韧性和熔点C .“奋斗者”号返回水面的浮力材料纳米级玻璃微珠可产生了达尔效应D .未来对海底“可燃冰”(主要成分为甲烷)的开采将有助于缓解能源危机 【答案】D【解析】A .锂离子电池工作时Li +向正极移动,故A 错误;B .因为合金比单独的组分金属具有更高的强度、更低的熔点,故B 错误;C .纳米级玻璃微珠并没有确定具体粒子大小,1nm ~100nm 会有丁达尔效应,小于lnm 不会有丁达尔效应,故C 错误;D .未来对海底“可燃冰(主要成分为甲烷)的开采将有助于缓解能源危机,故D 正确;故选D 。
2.下图是以秸秆为原料制备某种高分子化合物W 的合成路线:下列有关说法正确的是A .W 的化学式为(C 10H 16O 4)nB .Y 可以发生酯化反应和加成反应C .Y 和1,4-丁二醇通过加聚反应生成WD .X 具有两种不同的含氧官能团【答案】A【解析】A .根据W 的结构简式可知其化学式为(C 10H 16O 4)n ,A 正确;B .Y 中含有羧基,可以发生酯化反应,但不能发生加成反应,B 错误;C .Y 和1,4-丁二醇通过脱水缩合生成W ,C 错误;D .X 中的含氧官能团只有羧基一种,D 错误;答案为A 。
级数知识点总结和例题
n
n
lim
un +1 u 1 或 lim n +1 不易计算或不存在时,不能用此法 (见例 6 评注、例 7(5)、例 9) 。 n u n u n n
5.用根植审敛法 (1)若 1 ,则
设 lim n un
n
un 收敛;(2)若 1(或 ) ,则 un 发散;
n
思路二:求 s2 n ,而 s2 n +1 =s2 n +u2 n +1 ,则 lim sn s lim s2 n lim s2 n 1 s .(见例 2 解法
n n n
1) 2.用收敛级数的性质判定级数的敛散性 (1)要判定某一级数的敛散性,可根据级数的性质将该级数转化成敛散性已知的级数来讨 论,(见例 2 解法 2,例 4,例 5).需要掌握下面三个最常用级数的敛散性: 等比级数
x
们将一些简单函数间接展开成幂级数. 11.了解傅里叶级数的概念和狄利克雷收敛定理,会将定义在 [l , l ] 上的函数展开为傅 里叶级数,会将定义在 [0, l ] 上的函数展开为正弦级数与余弦级数,会写出傅里叶级数的和 函数的表达式.
11.2 基本题型及解题思路分析 题型 1 用级数敛散性的定义与性质判定级数的敛散性
例 5 (1991-研)已知级数
(1)n1 an 2 , a2n1 5 ,则级数 an =__________.
n 1 n 1 n 1
【分析】此题关键是弄清三个级数的一般项之间的关系。 解:因为 an 2a2 n1 (1)
n 1
an ,又 (1) n 1 an , a2 n 1 均收敛,故由收敛级数的
Ni2O3催化剂对NaClO分解产生活性氧的影响
Ni2O3催化剂对NaClO分解产生活性氧的影响景洁昆明理工大学化工学院摘要:采用混合法制备Ni2O3催化剂选择性催化次氯酸钠分解生成原子氧,该原子氧具有极强的活性,能增强次氯酸钠的氧化性。
考察了不同负载量催化剂、次氯酸钠浓度、pH值和温度对镍基催化剂分解次氯酸钠产生活性氧的反应速率的影响。
实验结果表明,次氯酸钠溶液浓度与反应速率成正比;而对于pH值来说,中性或弱酸性环境更有利于提高反应速率和产氧量;温度越高分解反应速率越快。
关键词:Ni2O3催化剂;催化氧化;次氯酸钠;活性氧一、前言本实验采用混合法制Ni2O3催化剂,Ni2O3催化剂在使用上具有普遍性;可用于制造高能电池;产品收率高、质量好;制备此催化剂反应条件温和;且对环境污染小,符合国家绿色的要求;制备价格低,经济型好;且此催化剂可以很好的是NaClO分解,使用量少。
次氯酸钠又称漂白液,次亚氯酸钠。
因其在酸性、弱碱性溶液中均具有较强的氧化性,常用作消毒剂、漂白剂。
浓溶液用于处理废水,可达到氧化分解污染物、漂白、杀菌消毒、脱臭的目的。
在处理过程中,常常由于次氯酸钠是不稳定的化合物,在温度较高或日光照射下容易发生光分解反应,易生成NaClO3、NaCl、HCl等,大大降低了次氯酸钠强氧化性的利用率,同时在分解过程中产生的HCl气体对环境造成了很大的危害。
因此,人们试图研究催化剂以改善次氯酸钠的氧化性。
贵金属催化剂虽然活性较高,但需有氧气存在的条件,且价格昂贵,成本高;而过渡金属氧化物催化剂,如镍氧化物催化剂,因其具有可变的d电子结构,很容易改变其价态,促进氧化还原循环,故而表现出较好的氧化性能。
谢少雄等、劳嘉葆研究发现:镍氧化物能够提高次氯酸钠的氧化速率、增强氧化程度。
林险峰等发现Fe2+能增强氧化剂的氧化性能。
但未发现其大规模使用。
本研究针对次氯酸钠在使用过程中存在的问题,采用混合法制备新型的镍基催化剂,并对其进行了一系列的表征,同时研究了助剂对催化活性的影响。
专题5.3 根据化学方程式的简单计算(十大题型)-九年级化学上册同步举一反三系列(人教版)
专题5.3根据化学方程式的简单计算题型梳理【题型1】根据化学方程式计算质量比或相对分子质量之比【题型2】化学计量数或部分化学式未知的化学方程式的计算【题型3】根据分步反应流程的化学方程式计算【题型4】根据反应微观示意图的化学方程式计算【题型5】化学方程式计算的解题过程分析【题型6】文字表述型的计算【题型7】折线图的计算【题型8】表格式的计算【题型9】根据反应前后实验记录的数据计算【题型10】判断是否达到某个标准的计算举一反三【知识点】1.根据化学方程式计算的依据:利用化学方程式能反映物质间的质量比,且质量比呈正比例关系。
2.根据化学方程式计算的步骤:(1)设未知数;(2)根据题意写出方程式;(3)根据化学方程式找出已知量与未知量的质量比;(4)列出比例式,并求出未知数;(5)简明地写出答案3.要领:步骤完整,格式规范,计算准确4.关键:化学式要正确;化学方程式要配平;准确计算相对分子质量;代入计算的质量必须是参加反应的纯净物的质量;单位必须统一。
【题型1】根据化学方程式计算质量比或相对分子质量之比1.已知化学反应:2A+B=2C+D,当10g A和8g B刚好完全反应后,若A、D的相对分子质量之比为5:3,则A .Cu 2O 中Cu 元素化合价为-2价C .理论上每16gCH 4参与反应,最终消耗64gO 10.一种利用太阳能分解水的原理如图所示。
下列说法不正确的是A .反应I 的化学方程式为I 2+SO 2+2H 2O=2HI+H 2SO 4B .反应Ⅲ属于分解反应C .反应I 、Ⅱ和Ⅲ在反应前后均有元素化合价发生改变D .理论上每生成2gH 2,同时生成32gO 211.工业上可利用Fe 3O 4和太阳能分解水制氢,流程如图。
下列说法不正确的是已知:①反应I :3422Fe O 6FeO+O ↑太阳能②反应Ⅱ:23423FeO+H O Fe O +H ↑加热A .上述流程中仅H 、O 元素的化合价发生了变化B .理论上,每产生32gO 2,消耗水的质量为36gC .相较电解水制氢而言,该方法具有节约能源,产品易分离的优点D .5.8g 可能由Fe 、FeO 、Fe 2O 3、Fe 3O 4中的两种或多种组成的混合物中,12.如图所示是一种新型合成氨的方法。
锂离子电池背景介绍及研究进展.ppt
3. Cathode performance
Fig. 2. Plateau voltage and capacity (see Fig. 1) for LiFePO4 [123,153–162] and LiCoO2 [163–167] with a charging voltage of 4.2V and discharge current of 1C.
• (3) substituting Li or Ti by other metal cations, such as Cr3+,
V5+, Mn4+, Fe3+, Al3+, Co3+, Ta5+, Cu2+;
Nb
No investigation was reported on the electrochemical characteristics of Nb-doped Li4Ti5O12 as an anode material.
The decrease in capacity with increasing discharge current is
generally smaller for LiCoO2 than for Li(Ni,Mn,Co)O2.
Fig. 8. Discharge capacity of LiFePO4 as a function of discharge rate.
0.1C.
The capacity of Li(Ni1/3Mn1/3Co1/3)O2 increases more than that of LiCoO2, suggesting that the kinetics of charge transfer and/or mass transport are slower in Li(Ni1/3Mn1/3Co1/3)O2 than in LiCoO2.
江苏省镇江市2023-2024学年高三上(期初考)-化学试题+答案
2023-2024学年高三(上)期初质量检测试卷·化学注意事项:1.本试卷分为选择题和非选择题两部分,共100分,考试时间75分钟。
2.请把选择题和非选择题的答案均填写在答题卷的指定栏目内。
可能用到的相对原子质量:H 1 Li 7 C 12 N 14 O 16 S 32 Cl 35.5 K 39 Cr 52 Mn 55 Fe 56一、单项选择题:共13题,每题3分,共39分。
每题只有一个选项最符合题意。
1.我国提出2030年碳达峰、2060年碳中和的目标。
下列关于2CO 的说法不正确是( )A.2CO 固体易升华B.2CO 是酸性氧化物C.2CO 为极性分子D.2CO 属于非电解质2.肼(24N H )是发射航天飞船常用的高能燃料,可通过反应3242NaClO 2NH N H NaCl H O +++═制备。
下列说法正确的是()A.NaClO 既含离子键又含共价键B.3NH 的电子式为HH :N :HC.24N H 的结构式为:||H H N HNH—═— D.Na +与Cl -具有相同的电子层结构3.实验室制取2Cl 的实验原理及装置均正确的是()A.制取2ClB.除去2Cl 中的HClC.收集2Cl D.吸收尾气中的2Cl 4.7N 、15P 、33As 、51Sb 是周期表中ⅤA 族元素。
下列说法正确的是( )A.原子半径:()()()N P As r r r >>B.酸性:34343H AsO H PO HNO >>C.第一电离能:()()()111N P As I I I >>D.ⅤA 族元素单质的晶体类型相同阅读下列材料,完成5~7题:第三周期元素的单质及其化合物具有重要用途。
如在熔融状态下,可用金属钠制备金属钾;2MgCl可制备多种镁产品;铝—空气电池具有较高的比能量,在碱性电解液中总反应为224Al 3O 4OH 6H O -+++═[]44Al(OH)-。
富锂锰基正极材料xLi2MnO3×(1-x)LiMO2的可控制备研究
目录1.前言2.电化学性能与结构3.合成方法3.1 水热法3.2 共沉淀法3.3 固相法3.4 溶胶-凝胶法4.电性能改善4.1 表面修饰4.2 形貌和结构特殊化改性4.3 共混改性4.4 预处理改性5.总结与展望1.前言随着科学技术的不断进步,微电子行业不断发展,电子产品、医疗设备、汽车电池、航天航空等领域对储能设备的要求也进一步提高,锂离子电池因其能量密度高、体积小、循环利用率高的特点而备受关注使用,但传统的正极材料LiCoO2成本高、容量低; 而LiNiO2也有苛刻的合成条件、较差的可逆性等一些缺点; LiFePO4虽然价格相对低廉一些,但其离子电导率较差,所以实际放电比容量只有160 mAh/g[1]。
这些锂离子电池正极材料很难满足高容量、高能量密度电子产品的需求,所以近几年锂离子电池的研究热点是锂离子电池的能量、功率等方面。
而富锂正极材料xLi2MnO3·(1−x)LiMO2 (M=Co, Fe, Ni1/2Mn1/2…)因其具有高比容量(200~ 300 mAh/g)、优秀的循环能力以及新的电化学充放电机制等优点而受到广泛关注,是目前正极商业化主流产品LiCoO2很好的替代品[2]。
富锂正极材料主要是由Li2MnO3与层状材料LiMO2 (M=Co, Fe, Ni1/2Mn1/2…)形成的固溶体。
1997 年Numata 等[3]率先报道了层状的Li2MnO3·LiCoO2 固溶体材料, 并早提出利用L i2AO3·LiBO2固溶体设计新电极材料。
研究发现当充电到 4.8 V时, 材料显示将近280 mAh/g 的初始放电比容量, 新的充放电机制带来了如此高的放电比容量, Mn在后面的电化学过程中同样参与氧化还原反应。
[4-6] Li2MnO3组分的存在使得该类材料在首次充电大于4.5V时出现平台,该平台对应于Li2MnO3组分中Li+脱出伴随O2p键氧化(净脱出形式为Li2O)。
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Electrochimica Acta 146(2014)79–88Contents lists available at ScienceDirectElectrochimicaActaj o u r n a l h o m e p a g e :w w w.e l s e v i e r.c o m /l o c a t e /e l e c t a c taElectrochemical study on x Li 2MnO 3-(1-x )LiNi 1/3Co 1/3Mn 1/3O 2(x =0.5)layered complex cathode showing voltage hysteresisMasahiro Kasai a ,∗,Shin Nishimura a ,Akira Gunji b ,Hiroaki Konishi b ,Xiaoliang Feng b ,Sho Furutsuki b ,Shin Takahashi ba International Research Center for Hydrogen Energy,Kyushu University,744Motooka,Nishi-ku,Fukuoka 819-0395,Japan bHitachi Research Laboratory,Hitachi,Ltd.,7-1-1Omika,Hitachi,Ibaraki 319-1292,Japana r t i c l e i n f o Article history:Received 1May 2014Received in revised form 13August 2014Accepted 13August 2014Available online 15September 2014Keywords:Lithium ion battery Cathode materials Layered cathodeExcess lithium compound OCP hysteresisa b s t r a c tThe layered complex cathode has attracted much interest recently due to its high capacity,which exceeds 250mAhg −1.On the other hand,it also shows large voltage hysteresis which causes energy loss when applied as a practical battery material.It has been reported that the hysteresis is an intrinsic bulk property,and the phenomenological model due to the migration of transition metal has been also proposed.Then,it is considered that to study the hysteresis for various compositions of transition metal is valuable to clar-ify the phenomenon.In this work,the layered complex cathode of 0.5Li 2MnO 3-0.5LiNi 1/3Co 1/3Mn 1/3O 2was investigated.The results are discussed in comparison with the previous reports for 0.5Li 2MnO 3-0.5LiMn 0.5Ni 0.5O 2.Investigating d Q /d V profilesrevealed that redox peaks at around 3.75V showed small dependency on composition of transition metal,while it seemed that a reduction peak at 3.26V was inde-pendent on the composition.Precise measurement of open circuit potential vs.capacity characteristics demonstrated that the energy dissipation during a cycle was abut 26kJmol −1.©2014Elsevier Ltd.All rights reserved.1.IntroductionRecently,there has been intensive investigation on the x Li 2MnO 3-(1−x )LiMO 2(M =Ni,Co,Mn)layered complex cath-ode [1,2].This material is very attractive as a high-energy-density cathode material for lithium ion batteries because it has a high capacity that exceeds 250mAhg −1at a charge voltage above 4.5V [3].However,the layered complex cathode shows a large hysteresis in the potential versus capacity curves.Cell potentials for the cor-responding capacity show different values between the charging and discharging process.Such differences in the potential decrease the energy efficiency of the battery because the discharge poten-tial is lower than the charge potential.It is considered that there are two possible causes for the hysteresis.One is the intrinsic electrochemical properties of the bulk material and the other is a large overpotential as a surface property.In general,every cath-ode material has a certain magnitude of overpotential,as shown in Fig.1(a).Therefore,the small deviation from the equilibrium potential is observed in the charge and discharge curves.When the overpotential is rather large,as shown in Fig.1(b),or the∗Corresponding author.Tel.:+81928023270.E-mail address:kasai.masahiro.987@m.kyushu-u.ac.jp (M.Kasai).capacity dependency of the equilibrium potential itself has a hysteretic characteristic as shown in Fig.1(c),hysteresis will appear in the charge/discharge curves.If the hysteresis is caused by the intrinsic bulk properties,then the strategy for improving the material will be optimization of the chem-ical composition or cation substitution;if it is caused by the overpotential,the optimization will include surface modifica-tion or changes in the particle structure.Therefore it is very important to clarify the origin of the hysteresis.The earliest work referring to the hysteresis in charge/discharge curves was reported by Ohzuku et al.[4]for x Li 2MnO 3-(1−x )LiMn 0.5Ni 0.5O 2(that is Li[Li 1/5Ni 1/5Mn 3/5]O 2in Dahn’s notation)cathode.They considered that the hysteresis could not be explained by kinetic effect because the hysteresis loop observed at 55◦C is larger than that at room temperature.However,the origin of the hysteresis was still unclear then.Croy et al.also investigated the hysteresis for the same cathode as Ohzuku and some other layered complex cathodes [5]They examined OCP vs.capacity characteristics vary-ing the lower and upper limit voltages,and investigated structural change during a charge/discharge procedure using an in-situ and ex-situ X-ray diffraction method.Croy et al concluded that the hysteresis was an intrinsic bulk property of the layered complex cathode,which accompanied by crystallographic change,and was caused by a small amount of lithium which experienced a hystere-/10.1016/j.electacta.2014.08.0730013-4686/©2014Elsevier Ltd.All rights reserved.80M.Kasai et al./Electrochimica Acta 146(2014)79–88SmallOverpotential ηEquilibrium PotentialP o t e n t i a lCapacity LargeOverpotential ηEquilibrium PotentialP o t e n t i a lCapacity HystereticEquilibrium PotentialP o t e n t i a lCapacity(a)(b)(c)Fig.1.Schematic illustrations of the relationships for the OCP and overpotential.(a)A case showing a small overpotential,(b)a case showing a large overpotential and (c)hysteretic OCP.sis about 1V in the site energy.They considered that the difference in the site energy was result from reversible migration of tran-sition metal ions,which could affect that of neighboring Li ions,between a metastable tetrahedral site in the lithium layer and the octahedral site in the transition metal layer.Gallagher et al.extended the idea to propose the phenomenological model cor-relating the reversible hysteresis and the irreversible voltage fade [6][7].According to the model,the migrated transition metal ions could return to the original octahedral sites when sufficient driv-ing force is applied.The reduction potential which acted as the driving force was observed as a 1eV hysteresis in a d Q /d V curve.It is considered that the elementary processes of the migration,when the external potential is applied to the electrode,are com-posed of simultaneous occurrence of lithium insertion and diffusion in the cathode material,valence change of the transition metal ions and subsequent structural change.All these process will show dependency of the transition metal composition to some extent [8].Therefore,it is expected that to study the hysteretic behavior for cathode materials having various composition of transition metals will give us important information about the phenomenon.The hysteresis of redox peaks in d Q /d V curves indicates the dif-ference of the site energy between the desertion and reinsertion state.On the other hand,the hysteresis width in the OCP versus capacity characteristics gives total dissipation of the Gibbs free energy during a cycle,which is consumed due to heat loss accompa-nied by an electrochemical reaction or irreversible structural decay etc.Generally speaking,the equilibrium potential of a lithium insertion material is determined by the difference of the Gibbs free energy between the reactant and the product.For exam-ple,in the case of the cathode material LiMO 2(M:transition metal),the equilibrium potential (in this paper,OCP is used to denote the equilibrium potential as they are equivalent)is given by E =−[G (LiMO 2)−G (x Li ++xe −+Li 1−x MO 2)]/nF .Here,G ()indicates the Gibbs free energy of the reactant or product of the corresponding lithium composition,given by x ,n is the number of electrons,and F is the Faraday constant.The OCP versus capacity characteristics are schematically shown in Fig.2(a)for the case in which the OCP is determined by composition x ,as described above.The OCP shows no hysteresis in the charge/discharge process,and no rapid change is observed in the OCP.A rapid decay of OCP occurs by elastic deformation,as shown in Fig.2(b).Hirai reported such behavior for Li 2/5Sn alloys [9].The potential drop was estimated by the change in the Gibbs free energy due to elastic strain as G /nF .It is considered that the potential change is reversible because the origin is elastic deformation.An OCP versus capacity characteristic accompanied by an energy loss during a cycle is also schematically shown in Fig.2(c).In such cases,hysteresis in the OCP versus capac-ity characteristics appears and the change in Gibbs free energy is given by G ABCD =−nF (E AB −E CD )during a cycle.The hysteresis observed for the layered complex cathode corresponds to the case shown in Fig.2(c).Zhu and Wang,and Dreyer et al.reported the hysteretic behavior of the OCP for a LiFePO 4cathode [10,11].They estimated the accommodation energy accompanied by phase sep-aration to be on the order of 10mV by applying the elastic/plastic deformation model for Ni–MH alloys [12].Croy et al.already reported properties for x Li 2MnO 3-(1-x )LiMn 0.5Ni 0.5O 2,and revealed that hysteresis width increased as the contents x of Li 2MnO 3component increased [13].However,dependency of the hysteresis width on the transition metal com-position is not still clear,as well as dependency of the redox peaks.There are various cathode materials in a ternary layered oxide system LiMO 2(M:Ni,Co,Mn),some of which are commercially available,for example LiNi 1/3Co 1/3Mn 1/3O 2,LiNi 0.5Co 0.2Mn 0.3O 2or LiNi 0.4Co 0.2Mn 0.4O 2.The LiNi 1/3Co 1/3Mn 1/3O 2cathode is the most popular and well-studied material among the family.In this work,we will start a series of systematic studies for dependency on the transition metal composition from the 0.5Li 2MnO 3-0.5LiNi 1/3Co 1/3Mn 1/3O 2cathode.Obtained electro-chemical characteristics are discussed in comparison with theCapacityCapacity CapacityO C PO C PO C PABCDΔG/nF(a)(b)(c)Fig.2.Schematic illustrations of three kinds of OCP vs.characteristics.(a)A conventional lithium insertion material,(b)OCP change by elastic lattice strain and (c)Hysteretic OCP.M.Kasai et al./Electrochimica Acta 146(2014)79–8881results of the previously reported 0.5Li 2MnO 3-0.5LiMn 0.5Ni 0.5O 2cathode.The characteristics of the LiNi 1/3Co 1/3Mn 1/3O 2cathode are also briefly described as a typical conventional insertion cathode.2.ExperimentalsThe x Li 2MnO 3-(1−x )LiNi 1/3Co 1/3Mn 1/3O 2(x =0.5)cathode powder was synthesized with spray-dried acetate precursors of the transition metals.The precursor powder was lithiated with LiOH at 850◦C for 12h in air.Electrodes were fabricated by mixing the powder with a solution of polyvinylidene difluoride (PVDF)in N-methyl-2-pyrrolidone,using carbon powder as a conductive agent in a cathode:carbon:PVDF ratio of 85:10:5wt%.The obtained slurry was coated on aluminum foil and pressed after vacuum drying.The typical electrode density was 2.2gcm −3and the thickness was approximately 50 m .Three-terminal electro-chemical cells were fabricated and measured in an Ar-circulated glovebox.Lithium metal foil was used as the counter and reference electrodes.Ethylene carbonate (EC)/ethyl methyl carbonate (EMC)(1/2,v/v)with 1M LiPF 6was used as the electrolyte.Four cells were fabricated for this study and all cells were initialized by charging at a current density of 0.11mA cm −2using the constant current–constant voltage (CCCV)method,followed by a CC dis-charge procedure at the same current density.The CV was set at 4.3,4.4,4.5,and 4.6V for each cell and the cut-off current density was 0.011mA cm −2.The cut-off voltage for the discharge process for all cells was 2.5V.The current density of 0.11mA cm −2corresponds to the 0.05C rate for the discharge capacity of the 4.6-V-charged cell.The GITT was used to investigate the OCP versus capacity characteristics after the initialization process.A battery tester (Hokuto SM8)with a voltage resolution of 0.3mV was used for the GITT measurements.A GITT pulse was applied to a cell for 20min followed by a 10h rest time.The OCP was determined as the potential 10h after the GITT pulse was terminated.Redox peaks were examined by using d Q /d V curves,which were obtained by numerically differentiating the constant-current region in the charge/discharge curves of the 2nd -mercially purchased LiNi 1/3Co 1/3Mn 1/3O 2cathode powder was examined in the same way as a reference material.3.Results and DiscussionA charge/discharge initialization procedure was conducted to stabilize the cathode material and confirm the initial capacity for the four upper-limit voltages before starting the GITT measure-ments.A LiNi 1/3Co 1/3Mn 1/3O 2cathode was tested at the same time for demonstrating a typical behavior of a conventional insertion cath-ode.Potential versus capacity curves for the first and second cycles are shown in Fig.3and 4respectively for various upper-limit volta-ges.As shown in Fig.3,there is no rapid change in the curve profile according to the upper-limit voltages for the LiNi 1/3Co 1/3Mn 1/3O 2cathode.Although potential hysteresis is observed for each 1st cycle due to the irreversible capacity,there is little hysteresis shown in the each 2nd cycle.The behavior is exactly corresponding to the Fig.1(a)described above.On the other hand,in the case of the layered complex cathode of 0.5Li 2MnO 3-0.5LiNi 1/3Co 1/3Mn 1/3O 2drastic change in the curve profile was observed.For upper-limit voltages of 4.3and 4.4V,the capacity is not very high,approximately 100mAhg −1.However,a rapid increase of the capacity is observed as the upper-limit voltage exceeds 4.5V,with the simultaneous appearance of a plateau region in the charge curve.The deviation between the first and second cycle curves quickly increases,which indicates an increase in the irreversible capacity for the first cycle.Here,it is considered that the observed irreversible capacity includes the energy loss due to a certain kind of conver-sion reaction of the cathode for the upper-limit voltages of 4.5and 4.6V.Therefore,the “coulombic efficiency”described in this work is a “nominal”value,which is obtained by simply dividing the dis-charge capacity by the charge capacity in the 1st cycle.However,we will use the notation for comparing the characteristics of the layered complex cathode with the conventional insertion cathode.It will be helpful to understand the difference between two kinds of cathodes.Hysteresis in the potential versus capacity curves in the second cycle is clearly observed especially for the upper-limit voltages of 4.5and 4.6V.Results of he GITT measurement will be described later to confirm the hysteresis by excluding the contri-bution of the overpotential.2.53.03.54.04.55.02.53.03.54.04.55.02.53.03.54.04.55.02.53.03.54.04.55.0P o t e n t i a l (V v s . L i /L i +)Specific capacity (mAhg -1)Fig.3.Charge and discharge potential curves of LiNi 1/3Co 1/3Mn 1/3O 2as a conventional insertion cathode,for various upper limit voltages (CV voltage)of 4.3,4.4,4.5,and 4.6V.Cells were measured at current density of 0.2mA cmˆ−2.82M.Kasai et al./Electrochimica Acta 146(2014)79–88P o t e n t i a l (V v s . L i /L i +)Specific capacity (mAhg -1)Fig.4.Charge and discharge potential curves of 0.5Li 2MnO 3-0.5LiNi 1/3Co 1/3Mn 1/3O 2layered complex cathode,for various upper limit voltages (CV voltage)of 4.3,4.4,4.5,and 4.6V.Cells were measured at a current density of 0.1mAcm −2.The limit voltage dependency of the capacity and coulombic effi-ciency for the 1st and 2nd cycles is shown in Fig.5by black squares.For the 1st cycle,the charge capacity increased from 107.4mAhg −1(at 4.4V)to 249.2mAhg −1(at 4.5V),finally reaching 308.1mAhg −1(at 4.6V)as the upper-limit voltage increased [Fig.5].The increase of the charge capacity was accompanied by a rapid decrease of the coulombic efficiency from 90.6(at 4.3V)and 88.0%(at 4.4V)to 75.9(at 4.5V)and 71.1%(at 4.6V),as shown in Fig.5(c).For the 2nd cycle,the charge and discharge capacity were very close to each other,as shown in Fig.5(d)and (e),respectively.Therefore,the coulombic efficiency exceeded 90%for all the upper-limit voltages shown in Fig.5(f).The results for LiNi 1/3Co 1/3Mn 1/3O 2cathode are also shown in the Fig.5by white triangles.A remarkable change in the coulombic efficiency is not observed accordingFig.5.Charge-discharge capacities and coulombic efficiency of the 1st and 2nd cycles for various upper-limit voltages.Results of the 0.5Li 2MnO 3-0.5LiNi 1/3Co 1/3Mn 1/3cathode are shown by black squares,and results of the LiNi 1/3Co 1/3Mn 1/3cathode are shown by white triangles.The observed irreversible capacity includes the energy loss due to an activation reaction of the cathode for the upper-limit voltages of 4.5and 4.6V.The “coulombic efficiency”described in this work was obtained by simply dividing the 1st discharge capacity by the 1st charge capacity.The notation was used for comparing the characteristics of the layered complex cathode with the conventional insertion cathode.M.Kasai et al./Electrochimica Acta146(2014)79–8883Fig.6.Obtained d Q/d V profiles of LiNi1/3Co1/3Mn1/3O2examined at different upper-limit voltages for various upper-limit voltages of4.3,4.4,4.5,and4.6V.The curves was obtained by differentiating the2nd charge/discharge curves.to the upper-limit voltages,both for the1st and the2nd cycle respectively.The efficiencies are about88%for the1st cycle and 99%for the2nd cycle.The result shows that the rapid decrease of the coulombic efficiency in the layered complex cathode is not due to a side reaction like decomposition of the electrolyte but due to a certain kind of material conversion reaction as called“activation”, which occurs when the charge voltage crosses4.5V.Figure6shows d Q/d V plots of the2nd charge/discharge cycle for the LiNi1/3Co1/3Mn1/3O2cathode.The upper-limit voltages were set at4.3,4.4,4.5and4.6V,while the lower voltage wasfixed at 2.5V,as same as the layered complex cathode.An oxidation peak and a reduction peak were observed at3.77and3.73V respectively for the upper-limit voltage of4.3V,as shown in Fig.6(a).If the midpoint of them indicates the equilibrium potential,the value is 3.75V and the overpotentialÁfor the redox reaction is20mV.There are no significant change observed when the voltage is increased up to4.6V,although small increase of the overpotential is shown for the upper limit voltages of4.5and4.6V[Fig.6(c),(d)].The obtained results indicate the typical behavior of the d Q/d V charac-teristics in a conventional insertion cathode.Measurement results of d Q/d V curves for the layered complex cathode of0.5Li2MnO3-0.5LiNi1/3Co1/3Mn1/3O2are shown in Fig.7(a)and(b)for upper limit voltages of4.3and4.4V.An oxidation peak at4.02V and a reduction peak at3.95V are observed as indicated by arrows.The equilibrium potential and overpotential are estimated as3.985V and35mV respectively.The redox peak profiles have a character of the conventional insertion cathode,while the equilibrium potential is about250mV higher than the results of the LiNi1/3Co1/3Mn1/3O2 cathode.As shown in Fig.7(c),once the cell is charged up to4.5V the oxidation peak shifts to3.87V and the reduction peak also shifts to 3.77V,while a broad reduction peak is formed at around3.36V.For the upper-limit voltage of4.6V shown in Fig.7(d),the whole shape of the peak profile is unchanged from the result for the voltage of 4.5V.It seemed that a pair of redox peaks at3.84and3.71V still show a character of the conventional insertion cathode.The oxi-dation peak is70mV higher than that of LiNi1/3Co1/3Mn1/3O2,and the reduction peak appears at the nearly same potential of3.71V. On the other hand,there is also a broad reduction peak observed at3.26V with simultaneous generation of a small oxidation peak at4.42V,having a character of a layered complex cathode.Ohzuku also reported d Q/d V characteristics,showing the peaks with two kinds of characters,of0.5Li2MnO3-0.5LiMn0.5Ni0.5O2cathode for the upper-limit4.7V[4](cf.Fig.5in Ohzuku et al.).The oxida-tion peak with a character of a conventional insertion cathode appeared at3.93V,which was160mV higher than the oxidation peak of LiMn0.5Ni0.5O2shown in the samefigure.The difference in the oxidation potential is two times larger than our work.The cor-responding reduction peak appeared at the nearly same potential of3.74V as the above-mentioned conventional insertion cathode. This behavior is similar with our result,although the correspond-ing peak positions are different.It is difficult to ascribe each peak to an appropriate electrochemical reaction right now.However, it is possible to ascribe the redox peaks at3.84and3.71V to the redox reaction Ni2+/Ni3+/Ni4+due to insertion/desertion of lithium ions,because the reaction occurs at3.75V for LiNi1/3Co1/3Mn1/3O2 cathode.The oxidation peak at4.42V was not clearly observed in this work as reported by Croy[5],rather in similar broad shape with Ohzuku.Recent study on charge compensation mechanisms for the layered complex cathode reported that all the transition metal ions are oxidized except Mn ions to a tetravalent state(i.e. Ni4+and Co4+)after the charging process to4.5V[16].Therefore, it is considered that the oxidation peak is possible to be ascribed to the oxidation process of Ni and Co ions.However,it seemed that the peak is sensitive not only to the chemical composition but also synthesis condition,because Ohzuku and Croy examined the cathode material with the same chemical composition.On the contrary,the reduction peak at3.26V is common to a cathode mate-rial in this work and the cathodes in the previous works.For the 0.5Li2MnO3-0.5LiMn0.5Ni0.5O2cathode,the peak was observed at 3.28V in Ohzuku’s work and at3.25V in the Croy’s work(cf.Fig.5in Croy et al.).Furthermore,a cathode material with a different com-position of0.5Li2MnO3-0.5LiNi0.375Co0.15Mn0.375O2cathode also84M.Kasai et al./Electrochimica Acta 146(2014)79–88d Q /d Vd Q /d VCell Potential (V vs. Li/Li +)Cell Potential (V vs. Li/Li +)Fig.7.Obtained d Q /d V profiles of 0.5Li 2MnO 3-0.5LiNi 1/3Co 1/3Mn 1/3O 2examined at different upper-limit voltages for various upper-limit voltages of 4.3,4.4,4.5,and 4.6V.The curves was obtained by differentiating the 2nd charge/discharge curves.showed the reduction peak at 3.25V (we can read it off with the Fig.7(b)in Croy et al.).The reduction peak was assigned to Mn 4+/Mn 3+by Croy et al [18].The corresponding oxidation occurs at about 4.3V showing a potential difference about 1V,which is caused by a structural change accompanying the transition metal migration [5].We are considering such potential shift will show a certain extent of dependency on the transition metal composition,although it was not observed for the cathode materials discussed in this work.Fur-ther study is necessary for wider range of chemical composition of the layered complex cathode.In this section,we presented the electrochemical char-acteristics for the layered complex cathode of 0.5Li 2MnO 3-0.5LiNi 1/3Co 1/3Mn 1/3O 2.The obtained redox curves showed two kinds of peaks.One pair is the oxidation peak at 3.84V and the cor-responding reduction peak at 3.71V,which has a character of the conventional insertion cathode,and another pair is the oxidation peak at 4.42V and the reduction peak at 3.26V showing a char-acter of the layered complex cathode.It seemed that the former three peaks showed small dependency on the composition of tran-sition meal,however the latter reduction peak at 3.26V appeared at almost exactly same potential.We are considering that to investi-gate the behavior of the redox reaction for more wide range of metal composition is necessary to reach the clarification of the hysteresis.In general,hysteresis is accompanied by energy dissipation.Therefore,precise estimation of the dissipation is also important to clarify the hysteresis.GITT measurement was examined to obtain OCP versus characteristic,which gives the dissipation.Figure 8shows complete GITT curves for upper-limit voltages of 4.3,4.4,4.5,and 4.6V.The 20min galvanostatic pulse is contin-ued until the charge voltage reaches the set upper-limit voltage.When the cell becomes unable to maintain a 20min galvanostatic current,the charge process is terminated and the discharge pro-cess is started in the same way until the cell potential reaches the lower-limit voltage of 2.5V.In the present test system,it was not possible to stop the GITT test automatically when the galvanostatic pulse was shorter than 20min.Therefore,several pulses shorter than 20min were incorporated in the original raw data for exper-imental convenience.The galvanostatic pulses that did not satisfy the defined measurement condition of 20min duration were elim-inated from Fig.8and the other results reported here.The relaxation procedure starts as soon as the galvanostatic pulse charge or discharge ends.Relaxation curves,which were extracted from the GITT test results for the 4.6V upper-limit,are shown according to several capacities in Fig.9(a)for the charge pro-cess and in Fig.9(b)for the discharge process.The point where the 20min galvanostatic current was cut off is indicated by an arrow.After the pulse duration,the relaxation procedure continued for 10h to the OCP.There is a large difference in OCP between charge and discharge capacities that are close to each other.For example,the top curves in Figs.9(a)and (b)show capacities of 215.9mAhg −1for the charge process and 215.4mAhg −1for the discharge process,with corresponding OCPs of 4.441and 4.097V,respectively,which is a difference of approximately 340mV.This result indicates that the dependency of the OCP on the capacity has a large width of hys-teresis.The relaxation rate was evaluated according to the potential change between 9and 10h.All the obtained values were below 2mVh −1,which is stable enough to determine the OCP character-istics.The dependency of the OCP on capacity is shown in Fig.10for the various upper-limit voltages.For the upper-limit voltages of 4.3and 4.4V [Fig.10(a)],the difference in the OCP for the same capacity was not significantly large.For example,at a capacity of ca.50mAhg −1,the difference in the OCP was ca.60mV for 4.3V and 100mV for 4.4V.However,for the upper-limit voltages of 4.5andM.Kasai et al./Electrochimica Acta 146(2014)79–88852.02.53.03.54.04.52.02.53.03.54.04.52.02.53.03.54.04.502004006008001000120014002.02.53.03.54.04.5Upper Limit 4.6 VUpper Limit 4.5 VP o t e n t i a l v s . (L i /L i +)/ VUpper Limit 4.4 Vt /hUpper Limit 4.3 VFig.8.GITT curves measured for upper limit voltages of 4.3,4.4.,4.5and 4.6V.The GITT measurements were performed after the 1st charge and discharge cycle for initialization up to the corresponding upper limit voltages.4.6V [Fig.10(b)],the difference in the OCP at around 135mAhg −1was 430mV for 4.5V and 450mV for 4.6V.The dependencies of the overpotential Áwere also obtained by the GITT curves,as shown in Fig.11.For example,point A,shown in the upper panel of Fig.11(a),indicates the OCP (E 0)using the notation in the inset,and points B and C indicate E t and E 1,respectively.In the discharge pulse of the GITT test,the measurement started from point A and then pro-ceeded to point B after a discharge pulse for 20min.The relaxation process of 10h is indicated as line BC,i.e.,the potential difference of B and C corresponds to the overpotential Áat the capacity.At around the capacity of 135mAhg −1,the overpotentials were 20mV for the charge process and 100mV for the discharge process in the case of the upper-limit voltage of 4.6V.It is obvious that these results also support the claim that the OCP hysteresis exceeding 400mV can-not be explained by the large overpotential of the layered complex cathode.We also investigated OCP versus capacity characteristics for LiNi 1/3Co 1/3Mn 1/3O 2as a conventional insertion cathode.The GITT pulse length was 20min,the same as the layered complex cathode test,and the upper-limit voltages were 4.3,4.4,4.5and 4.6V.However,the GITT current was set at 0.16mA cm −2and the relaxation interval was 5h because the relaxation rate of LiNi 1/3Co 1/3Mn 1/3O 2was faster than that of the layered complexcathode.The measurement was performed after an initializa-tion procedure consisting of a charge/discharge cycle up to the set upper-limit voltages,respectively.The results are shown in Fig.12(a)for the upper-limit voltages of 4.3and 4.4V,and in Fig.12(b)for the upper-limit voltages of 4.5and 4.6V.It is clearly seen that there is no hysteretic behavior observed for 4.3V.The result corresponds to the case described in Fig.2(a),and the OCP is decided only by the lithium composition,which is equivalent to the capacity.On the other hand,small hysteresis,showing the hysteresis width of abut 50mV,was observed for the upper-limit voltages of 4.4,4.5and 4.6V.Such small hysteresis was also observed for the layered complex cathode in this work for the voltage of 4.3V.Structural studies of a cathode material for various charged states have already been reported.For example,Nam et al.reported results for LiNi 1/3Co 1/3Mn 1/3O 2by in situ X-ray measurements [14].They showed that there is a lattice volume change from 3to 5%for capacities exceeding about 160mAhg −1.It is possible that such lattice strain causes the change in the OCP.For example,Hirai’s results showed an OCP drop of 0.528V caused by an elastic volume strain of 22%accompanied by the discharge process from ˇ-Sn to LiSn 2/5[9].The volume change of the layered complex cathode during the charge/discharge process,which wasFig.9.Relaxation curves obtained from the GITT measurements for (a)charge and (b)discharge processes.Each galvanostatic pulse was maintained for 20min and the rest time was 10h.The results are shown for the upper limit voltage of 4.6V in accordance with the capacity.。