（2019）新人教版高中化学选择性必修一第四章第2节电解池第1课时 电解原理 进阶检测（原卷+解析版）.rar

第二章第 2 节 电解池第 1 课时 电解原理【随堂检测】1. . .“”(1)电解.电.电.(.)(2).电解.电.电.(.)(3).电解.电.电.(.)(4)电解.电.(.)(5)电解.原.(.)(6).电.电解.时.(.)(7).电.电解. Na.e.=Na(.)(8).电解. 2H2O.4e.=O2.4H.(.)(9).电.电解. 4Ag.2H2O4Ag.4H.O2.(.)= = = = =电解 (10).电解. Cu.2HClCuCl2.H2.(.)= = = = =电解 2.电解.原理.(.)A.电解.电.B.电解.原.C.电解.时.原.D.原电.电解.电时.电.电.3.电.电解. 3.1 .(.)A.2Cu2.2H2O2Cu.4H.O2. B.Cu2.2Cl.= = = = =电解 Cu.Cl2.= = = = =电解 进阶检测step by stepC.2Cl.2H2O2OH.H2.Cl2. = = = = =电解 D.2H2O2H2.O2.= = = = =电解 4.电解.电.电解.0.1 molL.1CuCl2.Cl2Cu.0.1 molL.1NaCl .Cl2H2.0.2 molL.1CuSO4.O2Cu.0.2 molL.1CuSO4.Cu2.Cu. NaClCl2Na.(.)A.电.Cu2.H.Na.B.电.Cl.OH.SO24C.时.电D.电解.5.电解. c.d .电.(1)a 电._.b 电._.(2)c 电._.电._._.d 电._.电._._.(3)电解._(.“.”.“.” .)._.电解._.【基础再现】1.(.)A.原电.电解.B.C.电.2H2O.4e.=4H.O2.Zn.2e.=Zn2.D. H. H.2.电.电解.电.电. a . b .(.)A.电. a .电. b .B.电. a .电. b .C.电. a .电. b .D.电.电.3. CO2.电.原. CH4.(.)A.电.B.电.电. CO2.8HCO.8e.=CH4.8CO.2H2O 323C.时. n(KHCO3).D.时. pH .4.电解.理解. H.Cu2.Na.SO.24Cl. 5 .电解.电解.(1).电.电解.电解.电解._.电解._.(2).电.电解.电解.电解._.电解._.(3).电解.电解.电解._.电解._.5. X .电.Y . c.d .Z . e.f .电. d .(1).b .电._.(.“.” “.” “.”.“.” .).Z . e ._. Y.Z .电. d_e(.).(2). c .电._. Y ._. Z . e .电._.【能力提升】1. 2HCl.2Ag=2AgCl.H2.(.)2.电.电解.时. H2. O2.电解.时.(.电解.).(.). pH . pH . pH .A. B.C. D.3. Cu2O .电解. OH. Cu2O . 2Cu.H2OCu2O.H2.(.)= = = = =通电 A.电.B. pH .C.D. 2Cu.2OH.2e.=Cu2O.H2O4. CO2.电.原. CH4.(.)A.电.B.电.电. CO2.8HCO.8e.=CH4.8CO.2H2O 323C.时. n(KHCO3).D.时. pH .5.A.B.C .电解.电.Na.K.Cu2.SO.OH.24. 1 . A . B . C .电.电.时. c 电. pH .电解时. t . 2 .(1)M .电._(.“.”.“.”).电. b .电._.(2)._.(3).电解. B .时电解. _.(4).时.原._.第二章第 2 节 电解池第 1 课时 电解原理【随堂检测】1. ( (“”(1)电解(电(电()(2)(电解(电(电()(3)(电解(电(电()(4)电解(电()(5)电解(原()(6)(电(电解(时()(7)(电(电解( Na(e(=Na()(8)(电解( 2H2O(4e(=O2(4H()(9)(电(电解( 4Ag(2H2O4Ag(4H(O2()= = = = =电解 (10)(电解( Cu(2HClCuCl2(H2()= = = = =电解 答案(1)(2)(3)(4)(5) (8)(1(2(电解(原理()A(电解(电(B(电解(原(C(电解(时(原(D(原电(电解(电时(电(电(答案(D解析(电(电(电(3(电(电解( 3(1 (进阶检测step by step()A(2Cu2(2H2O2Cu(4H(O2(= = = = =电解 B(Cu2(2Cl(Cu(Cl2(= = = = =电解 C(2Cl(2H2O2OH(H2(Cl2(= = = = =电解 D(2H2O2H2(O2(= = = = =电解 答案C解析( 4 L( 1 molL(1(3 mol ( 1 mol(电(第(电(电(电解(Cu2(2Cl(Cu(Cl2(时( 2.5 mol(第(= = = = =电解 电(电(电解( 2Cu2(2H2O2Cu(4H(O2(第(= = = = =电解 (电(电(电解( 2H2O2H2(O2(= = = = =电解 4(电解(电(电解(0.1 molL(1CuCl2(Cl2Cu(0.1 molL(1NaCl (Cl2H2(0.2 molL(1CuSO4(O2Cu(0.2 molL(1CuSO4(Cu2(Cu( NaClCl2Na()A(电(Cu2(H(Na(B(电(Cl(OH(SO24C(时(电D(电解(答案(D解析(电()(电(电(Cu2(H(Na( A (电(Cl(OH(电(OH(SO( B (电解(电(24(电(电(时(电( C (电解(电(电解(电解( D (5.(电解( c(d (电(1)a 电(_(b 电(_(2)c 电(_(电(_(_(d 电(_(电(_(_(3)电解(_(“(”(“(” ()(_(电解(_(答案(1)(2)(2Cl(2e(=Cl2(2H(2e(=H2(原(3)(2HClH2(Cl2(= = = = =电解 【基础再现】1()A(原电(电解(B(C(电(2H2O(4e(=4H(O2(Zn(2e(=Zn2(D( H( H(答案(B解析(电解(电解( 2H2O(4e(=4H(O2(原电( 2H(2e(=H2(2.(电(电解(电(电( a ( b ()A(电( a (电( b (B(电( a (电( b (C(电( a (电( b (D(电(电(答案(B解析电( a (H2O 电( H(电( H2O (电(c(OH()(电( b (H2O 电( OH(电(c(H()(A (B (电(a ( H2(电( b ( O2(H2( O2( 2 ( C(D (3( CO2(电(原( CH4()A(电(B(电(电( CO2(8HCO(8e(=CH4(8CO(2H2O 323C(时( n(KHCO3)(D(时( pH (答案(C解析(电(电(A (电(原(电( CO2(8HCO(8e(=CH4(8CO 3(2H2O(B (时(23C (电(原(电(电(2H2O(4e(=O2(4H( H( pH (D (4(电解(理解( H(Cu2(Na(SO(24Cl( 5 (电解(电解(1)(电(电解(电解(电解(_(电解(_(2)(电(电解(电解(电解(_(电解(_(3)(电解(电解(电解(_(电解(_(答案(1)HCl(CuCl2(HCl(CuCl2(2)Na2SO4(H2SO4(H2O(3)CuSO4(NaCl(CuO ( CuCO3(HCl解析(1)(电(电解(电解( H(Cu2( Cl(电解( HCl(CuCl2(2)(电解(H(OH(电(电解( Na2SO4(H2SO4(3)(电解( CuSO4,2CuSO4(2H2O2H2SO4(2Cu(O2(原( CuO(= = = = =电解 ( CuCO3(电解( NaCl,2NaCl(2H2O2NaOH(H2(Cl2(原(= = = = =电解 HCl(5( X (电(Y ( c(d (Z ( e(f (电( d (1)(b (电(_(“(” “(” “(”(“(” ()(Z ( e (_( Y(Z (电( d_e()(2)( c (电(_( Y (_( Z ( e (电(_(答案(1)(2)(2Cl(2e(=Cl2(2NaCl(2H2O2NaOH(H2(Cl2(Cu(2e(=Cu2(= = = = =电解 解析d ( d (电( 2H2O(2e(=H2(2OH(c (电( 2Cl(2e(=Cl2(Y (电解 NaCl (2NaCl(2H2O2NaOH(Cl2(H2(电( a (b (Z ( f (e (= = = = =电解 电( Cu2(2e(=Cu(Cu(2e(=Cu2(电( e(d(【能力提升】1( 2HCl(2Ag=2AgCl(H2()(C解析(Ag ( HCl (原( 2HCl(2Ag=2AgCl(H2(电解(Ag (电( Ag (H(电(原(电解( H(电( C(2(电(电解(时( H2( O2(电解(时(电解()()( pH ( pH ( pH (A( B(C( D(C(解析：(电(电解时( H2( O2(电解(电解(电解H2SO4(时(电解(H2SO4(pH (电解( NaOH (时(电解(NaOH (pH (电解( Na2SO4(时(电解(Na2SO4(pH ( NaOH (Na2SO4(电解(3( Cu2O (电解( OH( Cu2O ( 2Cu(H2OCu2O(H2()= = = = =通电 A(电(B( pH (C(D( 2Cu(2OH(2e(=Cu2O(H2O(D解析：(电(电(原( A ( 2Cu(2e(2OH(=Cu2O(H2O(pH ( B (2Cu(2e(2OH(=Cu2O(H2O( C (电解(电(时(电(电(原(电(电(2Cu(2e(2OH(=Cu2O(H2O( D (4( CO2(电(原( CH4()A(电(B(电(电( CO2(8HCO(8e(=CH4(8CO(2H2O 323C(时( n(KHCO3)(D(时( pH (答案(C解析(电(电(A (电(原(电( CO2(8HCO(8e(=CH4(8CO 3(2H2O(B (时(23C (电(原(电(电(2H2O(4e(=O2(4H( H( pH (D (5(A(B(C (电解(电(Na(K(Cu2(SO(OH(24( 1 ( A ( B ( C (电(电(时( c 电( pH (电解时( t ( 2 (1)M (电(_(“(”(“(”)(电( b (电(_(2)(_(3)(电解( B (时电解( _(4)(时(原(_(答案(1)(4OH(4e(=2H2O(O2(2)2CuSO4(2H2O2Cu(O2(2H2SO4= = = = =电解 (3)(电解(4)(解析(1)( c 电( c (电( Cu2(2e(=Cu( c 电( b (a (M (N ( Cu2( SO( B (24( pH (电解时( t ( A ( KOH ( NaOH(C ( Na2SO4( K2SO4(KOH ( NaOH (电解 H2O( b ( OH(电( 4OH(4e(=2H2O(O2(2)(电解 CuSO4( 2CuSO4(2H2O2Cu(O2(2H2SO4(3)(= = = = =电解 电解( H2SO4(电解(