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High oxide ion conductive ceria based solid electrolyte

机译:高氧化物离子传导性二氧化铈基固体电解质

摘要

PROBLEM TO BE SOLVED: To introduce an oxygen deficiency and to simultaneously improve oxide ion conductivity and reduction resistance by specifying the composition of a fluorite type ceria compound consisting of La, Sr, Ba, Ce and O and providing a specified effective index. SOLUTION: This high oxide ion conductive ceria-base solid electrolyte comprises a fluorite type ceria compound of the formula (La1-xMx)yCe1-yO2-z (wherein M is Sr or Ba, 1=x=0.5, 0.1y0.2 and 0.05z0.15). When the average ionic radius of all cations constituting the ceria compound, the average ionic radius of the cations except Ce, the ionic radius of Ce and the ionic radius of oxygen are represented by Rc, Rd, Rce and Ro, respectively, the solid electrolyte has an effective index of 0.925 to 1.20 represented by the equation (effective index)=(Rd/Rce)*(Rc/Ro), wherein the symbol * shows integration, Rd=RM*x+RLa*(1-x), Rc=Rd*y+Rce*(1-y), Ro=1.4*(1-z), Rce=0.97, RLa is 1.18 which is the ionic radius of La, RM is the average ionic radius of Sr and Ba and 1.187=Rd=1.30 .
机译:解决的问题:通过指定由La,Sr,Ba,Ce和O组成的萤石型二氧化铈化合物的组成并提供规定的有效指数,引入氧缺乏并同时提高氧化物离子电导率和抗还原性。解决方案:这种高氧化物离子传导性氧化铈基固体电解质包含式(La1-xMx)yCe1-yO2-z的萤石型氧化铈化合物(其中M为Sr或Ba,1 <= x <= 0.5,0.1 0.925至<1.20的有效折射率,其中符号*表示积分,Rd = RM * x + RLa *(1-x ),Rc = Rd * y + Rce *(1-y),Ro = 1.4 *(1-z),Rce = 0.97,RLa为1.18,这是La的离子半径,RM是Sr的平均离子半径, Ba和1.187 <= Rd <= 1.30。

著录项

  • 公开/公告号JP3598344B2

    专利类型

  • 公开/公告日2004-12-08

    原文格式PDF

  • 申请/专利权人 独立行政法人物質・材料研究機構;

    申请/专利号JP19980294620

  • 发明设计人 森 利之;池上 隆康;

    申请日1998-09-30

  • 分类号H01B1/06;C01F17/00;C04B35/50;H01M8/02;

  • 国家 JP

  • 入库时间 2022-08-21 22:26:21

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