《电化学基础》测试题

第四章《电化学基础》测试题一、单选题1．有一种新型的乙醇电池，它用磺酸类质子溶剂，在200 ℃左右时供电，乙醇电池比甲醇电池效率高出32倍且更安全。

电池总反应为C2H5OH＋3O2===2CO2＋3H2O，电池示意如图，下列说法不正确的是( )A．a极为电池的负极B．电池工作时，电流由b极沿导线经灯泡再到a极C．电池正极的电极反应为4H＋＋O2＋4e－===2H2OD．电池工作时，1 mol乙醇被氧化时就有6 mol电子转移2．能正确表示下列反应的离子方程式是（）A．Na2S水解：S2- +2H2O ⇌ H2S+2OH﹣B．向FeCl3溶液中加入Mg（OH）2：3Mg（OH）2+2Fe3+ = 2Fe（OH）3+3Mg2+C．用醋酸溶液除水垢：2H++ CaCO3=Ca2++ H2O+CO2↑D．用铜为电极电解饱和食盐水：2Cl-+ 2H2O Cl2↑+H2↑+2OH-3．某电动汽车配载一种可充放电的锂离子电池，放电时电池总反应为Li1－CoO2＋Li x C6===LiCoO2＋C6(x＋1)。

下列关于该电池的说法正确的是()xA．放电时，Li＋在电解质中由正极向负极迁移B．放电时，负极的电极反应式为Li x C6+ x e－===x Li＋＋C6C．充电时，若转移1 mol e－，石墨(C6)电极将增重7x gD．充电时，阳极的电极反应式为LiCoO2＋x e－===Li1－x CoO2＋x Li＋4．下列关于原电池的叙述中，正确的是（）A．电流从正极流出B．正极不断产生电子经导线流向负极C．负极发生还原反应D．电极只能由两种不同的金属构成5．下列说法中正确的是（）A．钢铁在潮湿空气中生锈属于电化学腐蚀B．电解水生成H2和O2的实验中，可加入少量盐酸或硫酸增强导电性C．同一可逆反应使用不同的催化剂时，高效催化剂可增大平衡转化率D．升高温度能使吸热反应速率加快，使放热反应速率减慢6．500 mL 1 mol/L的稀HCl与锌粒反应，下列措施不会使反应速率加快的是A．升高温度B．加入少量的铜粉C．将500 mL 1 mol/L的HCl改为1000 mL 1 mol/L的HClD．用锌粉代替锌粒7．下列关于如图所示原电池装置的叙述中，正确的是A．铜片是负极B．电流从锌片经导线流向铜片C．硫酸根离子在溶液中向正极移动D．锌电极上发生氧化反应8．对于下列实验事实的解释，不合理．．．的是A．A B．B C．C D．D9．我国科学家已研制出一种可替代锂电池的“可充室温Na-CO2电池”，该电池结构如图所示。

电化学基础测试题（1）第Ⅰ卷（选择题共60分）一、选择题（共15小题，每题只有一个正确的答案，每题2分，共30分）1．甲、乙两个电解池均以Pt为电极，且互相串联，甲池盛有AgNO3溶液，乙池中盛有一定量的某盐溶液，通电一段时间后，测得甲池中某电极质量增加2.16 g，乙池中某电极上析出0.24 g金属，则乙池中溶质可能是（）A．KNO3B．MgSO4 C．CuSO4D．Na2SO42．下列叙述正确的是（）A．原电池中阳离子向负极移动B．用铝质铆钉接铁板,铁易被腐蚀C．马口铁(镀锡)的表面一旦破损,铁腐蚀加快D．白铁(镀锌)的表面一旦破损,铁腐蚀加快3．在电解水制取H2和O2时，为了增强导电性，常常要加入一些电解质，最好选用（）A．NaOH B．HC1C．NaCl D．CuSO44．用惰性电极电解2L 1 mol·L－1的硫酸铜溶液，在电路中通过0.5mol 电子后，将电源反接，电路中又通过1mol 电子，此时溶液中c(H+)是(设溶液体积不变)A、1.5 mol·L－1B、0.75 mol·L－1C、0.5 mol·L－1D、0.25 mol·L－15．当两极都用银片作电极电解AgNO3溶液时，在阳极上的产物是（）A．只有银B．银和氧气C．只有氢气D．上述都不对6．按下图的装置进行电解实验：A极是铜锌合金，B极为纯铜，电解质中含有足量的铜离子。

通电一段时间后，若A极恰好全部溶解，此时B极质量增加7.68克，溶液质量增加0.03克，则A合金中Cu、Zn原子个数比为A 4︰1B 3︰1C 2︰1D 任意比7．某原电池总反应的离子方程式为2Fe3＋＋Fe＝3Fe2＋，不能实现该反应的原电池组成是（）A．正极为铜，负极为铁，电解质溶液为FeCl3溶液B．正极为碳，负极为铁，电解质溶液为Fe(NO3)3溶液C．正极为铂，负极为铁，电解质溶液为Fe2(SO4)3溶液D．正极为银，负极为铁，电解质溶液为CuSO4溶液8．以石墨棒作电极，电解氯化铜溶液，若电解时转移的电子数是3.01×1023，则此时在阴极上析出铜的质量是（）A．8g B．16gＣ．32g D．64g9．用石墨作电极，电解1mol·L－1下列物质的溶液，溶液的pH变大的是（）A．HCl B．H2SO4C．NaCl D．Na2SO410．关于电解NaCl水溶液，下列叙述正确的是A．若在阴极附近的溶液中滴入酚酞试液，溶液呈无色B．若在阳极附近的溶液中滴入KI溶液，溶液呈棕色C．电解时在阳极得到氯气，在阴极得到金属钠D．电解一段时间后，将全部电解液转移到烧杯中，充分搅拌后溶液呈中性11．用惰性电极电解＋n价金属的硝酸盐溶液，当阴极上析出m g金属时，阳极上产生560mL（标准状况）气体，此金属的相对原子质量应为（）A．10mn B．10m C．10n D．40mn12．镁铝合金在碱性溶液中开始反应缓慢,后反应加速,经分析是氧化膜及微电池作用的结果.下列叙述正确的是（）A．微电池负极为Mg B．微电池负极为AlC．铝的电极反应式为2H++2e－＝H2↑ D．镁的电极反应式为Mg－2e－＝Mg 2+13．用惰性电极电解下列溶液，一段时间后，再加入一定质量的另一物质中（括号内），溶液能与原来溶液完全一样的是（）A ．AgNO 3 [Ag 2O]B ．NaOH [NaOH]C ．NaCl [盐酸]D ．CuSO 4 [Cu(OH)2]14．用铂电极电解CuSO 4和KNO 3的混合液500mL,经过一段时间后,两极均得到标况下11.2L 气体,则原混合液中CuSO 4的物质的量浓度为（）A ．0.5mol·L -1B ．0.8mol·L -1C ．1.0mol·L -1D ．1.5mol·L -115．一种新型燃料电池，一极通入空气，另一极通入丁烷气体；电解质是掺杂氧化钇(Y 2O 3)的氧化锆(ZrO 2)晶体，在熔融状态下能传导O 2-。

第四章综合能力测试01(时间90分钟满分100分)可能用到的相对原子质量：H—1C—12N—14O—16F—19Na—23Mg—24 S—32Cl—35.5K—39Ca—40Zn—65Ba—137第Ⅰ卷(选择题，共48分)一、选择题(本题包括18小题，每小题只有一个选项符合题意，每小题3分，共48分) 1．下列过程需要通电才能进行的是()①电离②电解③电镀④电泳⑤电化学腐蚀A．①②③B．②④⑤C．②③④D．全部2．在相同温度下用惰性电极电解下列物质的水溶液，一段时间后溶液酸性增强或碱性减弱的是A．HCl B．NaOH ( )C．Na2SO4D．CuSO43．将等质量的A、B两份锌粉装入试管中，分别加入过量的稀硫酸，同时向装A的试管中加入少量CuSO4溶液。

如图表示产生氢气的体积V与时间t的关系，其中正确的是()4．如图所示，铜片、锌片和石墨棒用导线连接后插入番茄里，电流表中有电流通过，则下列说法正确的是()A．锌片是负极B．两个铜片上都发生氧化反应C．石墨是阴极D．两个番茄都形成原电池5．在理论上不能用于设计成原电池的化学反应是()A. 4Fe(OH)2(s)＋2H2O(l)＋O2(g)===4Fe(OH)3(s)ΔH<0B．CH3CH2OH(l)＋3O2(g)===2CO2 (g)＋3H2O(l)ΔH<0C．Al(OH)3(s)＋NaOH(aq)===NaAlO2(aq)＋2H2O(l)ΔH<0D．H2(g)＋Cl2(g)===2HCl(g)ΔH<06．(2011年泰安高二期末)用惰性电极电解下列溶液，一段时间后，停止电解，向溶液中加入一定质量的另一种物质(括号内)，能使溶液完全复原的是()A. CuCl2(CuO) B．NaOH(NaOH)C ．CuSO 4 (CuCO 3)D ．NaCl (NaOH) 7．如图所示，X 、Y 分别是直流电源的两极，通电后发现a极板质量增加，b 极板处有无色无味的气体放出，符合这一情况的是( ) a 极板 b 极板 X 电极 Z A 锌 石墨 负极 CuSO 4 B 石墨 石墨 负极 NaOH C 银 铁 正极 AgNO 3 D 铜 石墨 负极 CuCl 2 8．下图有关电化学的示意图正确的是( )9．惰性电极分别电解下列物质的水溶液一段时间后，氢离子浓度不会改变的是 ( )A ．NaClB ．CuSO 4C ．AgNO 3D ．Na 2SO 4 10．pH ＝a 的某电解质溶液中，插入两支惰性电极，通直流电一段时间后，溶液的pH>a ，则该电解质可能是( )A. NaOH B ．H 2SO 4 C ．AgNO 3 D ．Na 2SO 411．在铁制品上镀上一定厚度的铜层，以下电镀方案中正确的是 ( )A ．铜作阳极，铁制品作阴极，溶液中含Fe 2＋B ．铜作阴极，铁制品作阳极，溶液中含Cu 2＋C ．铜作阴极，铁制品作阳极，溶液中含Fe 3＋D ．铜作阳极，铁制品作阴极，溶液中含Cu 2＋12．镁燃料电池以镁合金作为电池的一极，另一极充入过氧化氢，电解质溶液是酸化的氯化钠溶液，放电时总反应方程式：Mg ＋2H ＋＋H 2O 2===Mg 2＋＋2H 2O 。

第4章《电化学基础》测试题一、单选题（每小题只有一个正确答案）1．图1是电解饱和氯化钠溶液示意图。

图2中，x轴表示实验时流入阴极的电子的物质的量，y轴表示（）A．n(Na＋) B．n(Cl－) C．c(OH－) D．c(H＋) 2．某科研小组利用甲醇燃料电池进行如下电解实验，其中甲池的总反应式为2CH3OH+3O2+4KOH=2K2CO3+6H2O，下列说法不正确的是（）A．甲池中通入CH3OH的电极反应：CH3OH-6e-+8OH-=CO32-+6H2OB．甲池中消耗560mLO2（标准状况下），理上乙池Ag电极增重3.2gC．反应一段时间后，向乙池中加入一定量Cu(OH)2固体，能使CuSO4溶液恢复到原浓度D．丙池右侧Pt电极的电极反应式：Mg2++2H2O+2e-=Mg(OH)2↓+H2↑3．将含有0.4mol Cu(N03)2和0.3 mol KCl 的水溶液 1 L，用惰性电极电解一段时间后，在一个电极上析出 0.1 mol Cu ，此时要将溶液恢复到电解前溶液一样，可加入一定量的（）A．CuCl2 B．CuO C．Cu（OH）2 D．CuCO34．对钢铁析氢腐蚀和吸氧腐蚀的比较，合理的是（）A．负极反应不同B．正极反应相同C．析氢腐蚀更普遍D．都是电化学腐蚀5．铜锌原电池（如图，盐桥中含KCl）工作时，下列叙述错误的是（）A．正极反应为：Cu2++2e–=Cu B．电池反应为：Zn+Cu2+=Zn2+ +CuC．在外电路中，电子从负极流向正极 D．盐桥中的K+移向ZnSO4溶液6．下列有关电化学的说法正确的是（）A．锌锰干电池工作一段时间后碳棒变细B．在海轮外壳上镶入锌块可减缓船体的腐蚀，是采用了牺牲阳极的阴极保护法C．铜锌原电池工作时，电子沿外电路从铜电极流向锌电极D．电解MgCl2饱和溶液，可制得金属镁7．某同学用如图所示的电化学装置电解硫酸铜溶液，有一个电极为Al，其它三个电极均为Cu，则下列说法正确的是（）A．电子方向：电极Ⅳ→→电极ⅠB．电极Ⅰ发生还原反应C．电极Ⅱ逐渐溶解D．电极Ⅲ的电极反应：Cu-2e-═Cu2+ 8．下列事实不能用电化学原理解释的是( )A．铝片不用特殊方法保护B．轮船水线下的船体上装一定数量的锌块C．纯锌与稀硫酸反应时,滴入少量CuSO4溶液后速率增大D．镀锌铁比较耐用9．在盛有稀H2SO4的烧杯中放入用导线连接锌片和铜片，下列叙述正确的是（）A．正极附近的SO42―离子浓度逐渐增大 B．电子通过导线由铜片流向锌片C．正极有O2逸出 D．铜片上有H2逸出10．某原电池装置如图所示，电池总反应为2Ag＋Cl2=2AgCl。

第四章《电化学基础》单元测试题一、单选题（每小题只有一个正确答案）1．利用反应 Zn+Cu2+= Zn2++ Cu 设计成原电池。

则该电池的负极、正极和电解质溶液选择合理的是（）A．Zn、 Cu、 ZnSO4 B． Cu、 Zn、 ZnCl 2C．Cu、 Zn、 CuCl2 D． Zn、 Cu、 CuSO42．控制适合的条件，将反应 2Fe3＋＋2I －2Fe2＋＋I 2设计成如下图所示的原电池。

下列判断不正确的是（）A．为证明反应的发生，可取甲中溶液加入铁氰化钾溶液B．反应开始时，甲中石墨电极上Fe3＋被还原，乙中石墨电极上发生氧化反应C．电流计读数为零时，在甲中溶入FeCl 2固体时，甲中的石墨电极为正极D．此反应正反应方向若为放热，电流计读数为零时，降低温度后乙中石墨电极为负极 3．某同学组装了如图所示的电化学装置，电极Ⅰ为Zn，其他电极均为Cu，电解质溶液都是 CuSO4 溶液，则下列说法正确的是（）A．电子移动：电极Ⅰ→电极Ⅳ→电极Ⅲ→电极ⅡB．电极Ⅰ发生还原反应C．电极Ⅳ逐渐溶解D．电极Ⅲ的电极反应： Cu－ 2e－=Cu2＋4．已知海水中含有的离子主要有Na+、 Mg2+、 Cl －、Br－，利用下图装置进行实验探究，列说法正确的是（）A． X为锌棒， K置于 M，可减缓铁的腐蚀，铁电极上发生的反应为2H＋＋ 2e－＝H2↑B．X 为锌棒， K置于 M，外电路电流方向是从 X到铁C．X 为碳棒， K置于 N，可减缓铁的腐蚀，利用的是外加电流的阴极保护法D．X 为碳棒， K置于 N，在 X电极周围首先失电子的是 Cl－5．下列解释事实的反应方程式不正确的是（）A.热的纯碱溶液可去除油污： CO32－ +2H2O H2CO3+2OH－B.钢铁发生吸氧腐蚀时，铁作负极被氧化： Fe－2e－＝ Fe2+C.以 Na2S为沉淀剂，除去溶液中的 Cu2+：Cu2++S2－＝CuS↓D.向电解饱和食盐水的两极溶液中滴加酚酞，阴极变红：2H2O+2e－==H2↑+2OH-6．化学与科学、技术、社会、环境密切相关。

《电化学基础》测试卷The electrochemical base test volume(1) basic practiceA,multiple choiceThe following processes need to be electrified to do so: ionization;(2) the electrolysis; (3) plating; (4) electrophoresis; The electrochemical corrosion;A,(1) (2) (3); B, (2) (3) (4); C, (2) (4) (5); D, all;At room temperature, the PH of 0. 1 molar of acid is:A, 1; B, > 1; C, < 1; D, can，t be sure;3, 1 liter contains NaOH under normal temperature of saturated salt water, the PH = 10, with platinum electrodes during electrolysis, when the cathode has a 11. 2 liter gases (standard conditions), the PH of the solution is close to (set after electrolytic solution volume is still 1 liter):A., 0; B, 12; C, 13; D, 14;The basic structure of the newly developed bromine-zinc battery in foreign countries is to use carbon rods as the poles, and the electrolyte is the zinc bromide solution. There are four electrode reactions:N - 2e = Zn = Zn + 2e 二ZnThe anode reaction and discharge when charged are:A, (4) (1); B, (2) (3); C, (3) (1); D, (2) (4);The colbay reaction is 2RC00K + 2H2OR -r + H2 + 2K0H, and the following statement is true (C)The products ofB. the products of hydrogen are produced in the cathode.C. the products of carbon are produced in the anode.D. The products of carbon are produced in the cathode.hydrogen are produced in the anode.Use 0.01 mo/litre H2S04 titrate 0. 01 mo/liter NaOH solution, neutralize and add water to 100m 1. If there is an error in the end, the ratio of 1 drop of H2S04 to 1 drop of H2S04 (1 drop to 0. 05 ml), and the ratio of C (H +) is:A: ten b. fifty c. fiveIn a beaker of saturated sodium carbonate, insert inert electrodes, keep the temperature constant, and after a certain amount of time:The PH of the solution will increaseB.The ratio of sodium ions to carbonate ions will be smallerC.The concentration of the solution gradually increases, and a certain amount of crystal is precipitated outThe concentration of the solution stays the same8, room temperature, the pH = 1 hydrochloride average into 2 portions, 1 add right amount water, the other one to join with the mo1ar concentration of hydrochloric acid after the same amount of NaOH solution, pH is increased by 1, belong to the water and the volume of NaOH solution is as follows:A.10 c. 11 d. 129, there are five bottles of solution are respectively (1) 10 ml of 0. 60 mo/NaOH aqueous solution (2) 20 ml mo / 0. 50 litres of sulfuric acid aqueous solution (3) 30 ml of 0. 40 the 40 ml/1 HC1 solution (4) the 0. 30 / d) aqueous solution (5) 50 ml mo / 0. 20 litres of sucrose solution. The order of the total number of ions and molecules in each of these bottles is:A, B, >, >, >, >, >C, >, >, >, >, >, >To connect Al and Cu slices to a wire, a group of HN03 is inserted into a dilute NaOH solution, which forms the original cel 1. In the two original batteries, the two are:A, Al, Al pills; B, Cu, Al pills; C, Al, Cu pills; D, Cu and Cu pills;11, the mass fraction of 0. 052 (5. 2%) of the NaOlI solution 1 liter (density of 1.06 g/ml) with a platinum electrode electrolysis, when the mass fraction of the NaOH solution changed 0. 010 stops electrolysis(1. 0%), should comply with the relationships of the solution is: The mass fraction of NaOHThe mass of the anode (grams)・ The mass of the cathode deposition.19152B0. 062 (6. 2%)152 a.0. 062 (6. 2%)0. 042 (4. 2%)1.29.4D0. 042 (4. 2%)9.41.212, a team of extracurricular activities, will cut a piece of galvanized iron in the conical flask, and drops into a small amount of salt water to soak, addend phenolphthalein try drops again, according to the figure device experiment, observation, a few minutes later the following phenomenon is impossible:A,B, the air bubble in the ducts;B,B, a water column formed in the trachea;C,metal shears red; A BZinc is corroded;13, a new type of fuel cell, it is porous nickel plate as the electrode in KOH solution, then access to polar ethane and oxygen respectively, the overall reaction is: 2 c2h6 o2 + 7 + 8 KOH k2co3 + 10 二4 h2o; The correct inference about the batteryis:A, the negative reaction is: 14H20 + 702 + 28e -.B: after a period of discharge, the PH around the negative pole rises. C, 1 molc2h6, and 14 moles of electrons transferred on the circuit.D,the concentration of KOH in the process of discharge is essentially the same;14,contains two kinds of solute in a solution of NaCl and - H2S04, the amount of substance ratio of 3:1, the mixed solution with graphite electrode electrolysis, according to the electrode product, can be clearly divided into three stages・ The following statements are incorrect:A, the cathode starts from the beginning and only comes out of H2; B, the anode comes out of the C12, and then it comes out of 02.The final stage of C and electrolysis is electrolytic water; D, the PH of the solution goes up, up to 7.The anode is made of iron, and copper is an electrolyte of the sufficient amount of NaOH solution, and after a period of time, two moles of Fe (OH) 3 solid are obtained, and the water is consumed in this room:A, , 3 mol. B, 4 mol. C, 5 mol; D, 6 mol;16, 100 ml, 2 mo/liter of NaOH solution, 100 ml, 2 mo/1 - H2S04 solution and mix a certain amount of ammonia, the solution obtained make phenolphthalein solution show pale red, the relationship between ion concentration in the solution is correct:A, C (S042 —)二C (Na +) > C (NH4 +) >・B, C (Na +) > C (S042 -)> C (NH4 +)・C, C (NH4 +), >, C (S042 —)二 C (Na +), >, C (OH C(H +)・C (H +) + C (NH4 +) + C (Na +)二C (OH -)+ 2C (S042 -);The PH value of the acid and acetic acid are diluted to the original n times m, and the PH of the two solutions is still the same, and the relationship between n and m isA, m = n B, m > n C, m < n D, m? nThe volume of hydrochloric acid with the same volume, the same pH, NaOH, and NH3 H20 is VI, V2, and V3, and the three relationships are:A, VI, >, V2 bl, V3, B, V3, V2, > VI, C, V3, >, V2, V2, VI, V2, VI, V2, VI, V2, VI, V2, all the way to V2Second, fills up the topicHydrogen peroxide (H202) and water are extremely weak electrolytes, but H202 is more acidic than H20.(1)if you view H202 as a di-acid, write the ionization equation in the water:(2)because of the weak acidity of H202, it can form positive salt with strong base, and also form acid salt under certain conditions・ Write the chemical equation of the salt in the form of H202 and Ba (OH) 2:(3)hydroelectricity is dissociated from the generation of H30 + and OH 一called water・Like water, hydrogen peroxide has a very weak since I ionization, it accidentally ionization equation is as follows:;The industrial wastewater containing cr2o72-acid industrial waste water is used in the following way: the appropriate amount of NaCl is added to the industrial waste water, and the mixing is uniform・Using Fe to electrolyte the electrode, over a period of time there are Cr (OH) 3 and Fe (OH) 3 deposition; The waste water reaches the discharge standard. Try to answer:(1)electrode reaction in electrolysis: anode cathode(2)write an ion reaction equation that changes Cr2072 to Cr3 + •■(3)how does the precipitation of the Cr (OH) 3, Fe (OH) 3 deposit occur in the electrolysis process?It is known that the ionizing degree of ammonia is equivalent to the degree of ionization of acetic acid, which is equal to the same temperature, and the solution of ammonium chloride is acidic・ Now, in a small amount of Mg (OH) 2 suspension, add the right amount of saturated ammonium chloride solution, and the solid is completely dissolved. A classmate，s explanation is:It's Mg (OH) 2.(2) NH4 + + H20NH3? H20 + H +; H + OH 二H20;Because NH4 + hydrolysis is acidic, H + and OH - react to form water, causing the reaction to balance right and dissolve・The interpretation of classmate b is:It's Mg (OH) 2. NH4 plus OH minus NH3? H20.Because NH4 cl is ionized by an NH4 plus the OH, which is ionized by Mg (OH) 2, which produces a weak electrolyte NH3? H20, which causes the equilibrium between the reaction, right, and Mg (OH) 2 is dissolved ・(1) c student can，t sure which students reasonable explanation, then choose one of the following reagents, to prove that a, b two classmates explain only a right, he chooses reagent is: (fill in the number)A,NH4N03; B, CH3C00NH4; C, Na2C03; D, NH3? H20.(2)please explain why the classmate made the choice・(3) in the suspension of Mg (OH) 2, the selected reagent is put into the suspension of Mg (OH) 2. As a result of this, the interpretation of the classmate is more reasonable (the or 〃b〃)；Completing the NH4C1 saturation solution makes the ion equation for Mg (OH) 2 dissolved:.22. From H +, Na +, Cu2 +, Ba2 +, Cl 一，S042 一ion, choose appropriate ion of an electrolyte, electrolyte solution according to the following requirements to electrolysis (are inert electrode):(1)the electrolyte is reduced in electrolyte and the amount of water is constant・(2)the quality of the electrolyte is kept constant while electrolyte, the amount of water is reduced, and the electrolyte used is:(3)the quality of electrolytes and water changes when electrolysis, and the electrolyte used is;At room temperature, 0. 01 molCH3C00Na and 0. 004 mollICl dissolve in water and make 0. 5 L mixed solution. Judge:There is a particle in the solution;The amount of matter in the solution that has two particles must be equal to 0.01 moles, and they're sum;In this case, it's n (CH3COO 一) + n (OH -) -n (H +)二mol.Three, calculation problem24.Under a certain temperature to a certain volume density of 1. 15 g/ml of sodium chloride solution with inert electrodes electrolytic, set exactly the electrolysis of sodium chloride and no other reaction occurs, the oxygen mass fraction of 80% in the solution, try to calculate:(1)the ratio of the amount of solute and solvent in an electrolyte solution.(2)the concentration of the substance of the original sodium chloride solution・25.To 8 grams of divalent metal oxide solid dilute sulphuric acid is added to make it completely dissolved, known to the consumption of sulfuric acid volume is 100 ml, insert the platinum electrode in the solution of electrolysis, electricity after a certain period of time, collected on an electrode 224 ml (standard conditions) of oxygen, in another 1. 28 grams of the metal electrode・(1)the name of the metal oxide is determined by calculation.(2)the concentration of the substance in the solution of sulfuric acid (the volume of solution) is calculated by 100 ml.(2) ability testingA,multiple choiceAs people's quality of life improves, the problem that the battery must be focused on is raised to the agenda, which is the primary reasone the metal material of the battery caseB.prevent the contamination of soil and water by heavy metal ions such as mercury, cadmium and leadC.not to corrode other substances from the electrolyte that leaks inthe batteryRecycle the graphite electrodesThe following statement is trueZinc reacts with dilute sulphuric acid to make hydrogen,Adding a small amount of copper sulfate can accelerate the reaction. When the coating is broken, the white iron (galvanized iron) is more corrosive than the tin plate・ When plating, it is necessary to place the plated in the cathode of the electrolyte・When smelting aluminum, the aluminum oxide is added to the liquid crystal and it becomes the molten body and then the surface of the steel surface is easily corroded to produce Fe203 nh20A. in b. by c. with d. inThe right image is an electrolytic CuC12 solution, where c and d are graphite electrodes・ The following are the right onesA. a is negative, b is positive B・a is anode and b is cathodeIn the process of electrolysis, d electrode mass increases d. In the process of electrolysis, the concentration of chloride ions remains constantElectrolysis with inert electrodes・The following statement is correctA. electrolytic dilute sulphuric acid solution, which is essentiallyelectrolytic water, so the solution p H remains unchangedThe solution of sodium hydroxide is to consume OH -，so the pH of the solution decreasesC. electrolysis of sodium sulphate, which is a ratio of 1 to 2 in the cathode and anodeD.Electrolytic chlorinated copper solution, the ratio of the amount of matter in the cathode and anode is 1:1The correct description of the device in the right picture is correct A: this is an original batteryThis is an electrolytic NaOH solutionC. P is positive, and the electrode reaction is 2H + + 2e -二H2 arrowThe d. f. e is negative, and its electrodes react to it: 40H 一二2H20 + 02What happens when you get the reaction Cu + 2H20 二Cu (OH) 2 and the H2 arrowA.the copper is the anode of the original battery, the carbon is the anode of the original battery, and the sodium chloride is the electrolyte solutionB.copper zinc alloy is chemically corroded in damp aire copper to make Yin, electropositive, and electrolyte sodium sulphate solutione copper to make Yin, electropositive, electrolysis of copper sulfate solution 7. Hydrogen fuel cell is a kind of high performance battery, total reaction of 112 + 02 二2 1120, electrolyte solution of KOH solution, in the following concerning the describe of the battery is not correctA.H 2 is negative, 02 is positiveB. The PH of electrolyte solution is increasing at workThe negative reaction: 2h2-4e - + 40H - 二4H20When iron rods are connected with graphite rods, they are immersed in 0. 01 mol/L of salt solution, which may occurAn oh~b in the vicinity of an iron bar is corrodedRelease the C12 D. On the graphite rodUse the inert electrode electrolyte to electrolyte the Na2C03 solution, and if the temperature remains constant, it will be over a period of timeThe ratio of the pH of the solution to the c (Na +) and c (C032 -)is greaterC. The concentration of the solution is large, and the crystal is precipitated out・ The concentration of the solution is constant10. With a platinum electrode electrochemical a certain concentration of aqueous solution of the following materia1. After the electrolysis, add right amount of water to the rest of the electrolyte, can make the solution and electrolyticbefore is the sameA. gnoc3b・ cA, B, and C are three kinds of metals, in which case A and B are immersed in A dilute H2S04. (2) the amount of electrolytic material concentration in the same A and C of the mixed solution of salt, on the cathode precipitation C first, (using inert electrode), determineB, B, B, B, B, B, B, B, B, B, B, B, B, B, B, B, B, B, B, B, B, B, B, B, B, C, B, B, B, B, B, B, B, C, B, B, B, B, C, B, B, C, their reducing strength for sequenceB, B, B 12. With graphite electrode, electrolytic silver nitrate solution in the anode collected 0. 4 g of oxygen, and electrolytic acid generated 250 ml of a sodium hydroxide solution, is the concentration of a solution of sodium hydroxideA. 0. 20 moles of IT. 0. 15 moles of 1-1 c. 10 moles of 1-1. 0. 05 moles of 1~113.The solid-oxide fuel cell was developed by Westinghouse・ It is an electrolyte from zirconium oxide, a solid electrolyte that allows oxygen ions (02 -) to pass through during high temperatures・ The battery，s working principle is shown in the diagram, where the porous electrode a and b are not involved in the electrode reaction・ The following is trueA.there's A very strong battery in the reaction of 02B.there is an H2 participating in the positive electrode of the BThe corresponding electrode for c a is 02 + 2H20The total reaction equation of the battery is 2H2 + 02 二2H2014. With a platinum electrode electrolysis of NaCl and the mixture of CuS04 solution, when circuit through the 4 mol, electronic power, Yin and Yang both produce 1. 4 mol, gas, after electrolytic solution volume is 4 1, the final pH value of electrolyte nearestA. four b. two c. 13 d. 14There is a button microcell with an electrode of Ag20 and Zn. Electrolytic cell is KOH solution, so commonly known as silver-zinc battery, the battery electrode reaction to zinc + OH - 2 + 2 二e zinc (OH) 2, Ag20 + H20 + 2 二2 e ag + OH - the following claims, the right is for (1) zinc anode, Ag20 (2) for the positive discharge, pH near the anode rise (3) discharging, near the cathode solution pH value lower (4) anion in the solution to the positive direction, cation to negative directionA. in b. by c. in d. inWhen using graphite as an electrode for electrolysis of 3 mol/LNaCl dnd 0.4 molar 2 (S04) 3, the following curve is correct ()Second, fills up the topic17. Molten salt fuel cells with high power generation efficiency, therefore, available Li2CO3, Na2C03 and molten salt mixture of electrolyte, CO as the anode gas, a mixture of air and C02 gas for cathode to gas and work under 650 °C of the f uel cell system, complete the relevant battery reactive: reactive: battery anode reaction: 2 CO + 2 co32 — > 4 C02 + 4 e -・The cathode reaction:, total reaction: battery.Add the Na2S04 solution to the u-tube, add a few drops of litmus test liquid, insert the inert electrode, and turn on the power suppl y.(1)the observable phenomenon is observed after a period of time.(2)to cut off the power supply, remove the electrodes will be in the u-tube solution in a small beaker, stirring, observable phenomenon is that the reason for this is that・If (3) to switch to the device electrolysis with a few drops of phenolphthalein try drops Na2C03 saturated solution of anode was observed after a period of time, the cathode・ 19. A research study to explore the influencing factors of steel rust, design the following solutions: A, B, C, D four little flask respectively into dry chicken wire, soaked in salt water of fine wire, fine wire soaked in water, salt water and fine wire, and make the wire completely immersed in salt water, and then to assemble as shown in the four sets of devices, every once in A while measuring the height of the catheter in the water rise, result in the following table data (listed in the table for the catheter in the height of water rise/cm)・The rise of water at different timesTime/hour0. 51.01.52.02.53.0A bottle (dry wire)B (iron wire with salt water)0.41.23.45.67.69.8C (the wire with water in it)(the wire that is fully immersed in the salt water)If you are the group leader, please answer the following questions: (1)why the water in the catheter rises・(2)in these experiments,The speed of wire rust is in the order of large to smal1.(3)the factors that affect the rust of steel are mainly.There are two different points of view about the "pH change in the electrolytic CuC12 solution":The idea is that the "theorist" thinks that the pH of the solution after electrolytic CuC12 is elevated・Point 2 it is, "experimental" after repeated several times, precise experimental determination, prove that the electrolytic CuC12 solution pH change as shown in the curve relationship with・ Please answer the following questions:(1)of CuC12 before electrolytic solution pH value at A point location is (with ion equation)・(2)the theoretical basis of the theory is that it isIt is.(3)^experimental^ experimental conclusion is that they are "precise experiment" is described through the accurate determination of the pH of the solution.(4)what do you think? Your point of view of the reason is that (briefly) through chemistry.A battery is a device that can charge and discharge repeatedly, with a battery that occurs when charging and dischargingThe reaction is: Ni02 + Fe + 2H20 Fe (OH) 2 + Ni (OH) 2Use this battery for electrolytic M (N03) x solution:(1)the anode (inert) of the electrolytic pool shall be connected (the serial number)・A.NIf the battery works for a period of time, it consumes 0. 36 grams of water・(2)the calculation of the relative atomic mass of M (N03) x when the M (N03) x solution increases to mg(using m and x).(3)this battery electrolysis contains 0. 01 moles of CuS04 and 0.01 moles of NaCl, and the anode produces gasL (standard condition), the solution of electrolysis after electrolysis is released to IL, the pH of the solution is.Three, calculation problem22.To 8 g of divalent metal oxide solid adding rare - 1I2S04, make its just completely dissolved, known by sulfuric acid volume is 100ml, insert the platinum electrode in electrolysis in the solution obtained, electricity after a certain period of time, collected on an electrode 224 ml (standard conditions) of oxygen, in another electrode on the metal precipitation 1. 28 g.(1)the name of the metal oxidation is determined by calculation.(2)the concentration of the substance in the solution of sulfuric acid (the volume of solution) is calculated by lOOmL・The basic chemistry foundation of chapter four(1)basic practiceone23456co DDB9 001415161718CBa.CDCCDBC(1)H02 - H++ 022 -(2)H202 + Ba (OH) 2 = Ba 02 + 2 H20.2H202 + Ba (OH) 2 = Ba (H02) 2+2 H20;(3) 2 H202 H02 - + H302 +(1)the anode・ Cathode: 2H + + 2e -二H2 arrow(2)Cr2072 - + 6Fe2 + + 14H + 二2Cr3 + + + 7H20(3)as the reaction progresses, the H + concentration decreases, the oh-concentration increases, and eventually the Cr (OH) 3, Fe (OH) 3 precipitates・21,(1) B; (2) the CH3C00NH4 solution appears neutral, whichcan prove which classmate is correct・⑶ b;22,(1) CuC12, HC1; (2) H2S04, Na2S04 (3) NaCl, BaC12, CuS0423,7 (1) ; (2) CH3COOH + CH3COO 二0. 01 moles; 0. 006 mol, (3);24,(1) 1:10 (2) 4.48 molar(1)the concentration of copper oxide (2) sulphuric acid is 0. 2 molar ・[please note: 224 ml (standard condition) oxygen is less than 8 grams with 1.28 grams・This shows that MS04 is not fully electrolytic!(2)ability testing oneC 広 DCO O02 + 2C02 + 4e - 二2C032-2C0 + 02 二2C02(1)the area is red near the anode and blue near the cathode・ The solution is also purple, and by electron conservation, the amount of H + and OH minus of the Yin and Yang is exactly the same, while the Na2S04 solution is not hydrolyzed and is neutra1.(2)the red is getting lighter; The red gradually deepens・(1)when iron rusts, it reacts with oxygen in the air, depleting oxygen and reducing the gas pressure in a small beaker.(2) B > C > A = D・(3)contact with oxygen; The presence of water; There is an electrolyte (or a salt) that is present, and at the same time the iron rusts the fastest (or from the condition of the battery that makes up the original battery)・(1)Cu2 + hydrolysis, the solution is acidic, Cu2 + + 2H20 Cu (OH) 2 + 2H +・(2)when electrolysis, the Cu2 + is separated by the cathode, and because the concentration of Cu2 + decreases, the hydrolysis equilibrium moves to the left, so the H + concentration decreases and the pH increases・(3)the pH of the solution is lower, pH.(4)the "experimentalisls" are correct because, while electrolysis, the anode is dialyse and C12 is dissolved in water to form hydrochloric acid・(1) A (2) 50mx (3) 0. 16& 222.(1)the metal is M, n (02)二0.01 mol, and the amount of matter measured by M is 0. 02 moles, according to the mo 1 ar mass of M, and the metal oxide is Cu0・(2)in accordance with the Cu ~ - H2S04,N of H2S04 is equal to 0. 02 moles・ one。

第四章《电化学基础》测试题一、选择题（每小题有1～2个选项符合题意）1．下列有关金属腐蚀与防护的说法正确的是A．纯银器表面在空气中因化学腐蚀渐渐变暗B．当镀锡铁制品的镀层破损时，镶层仍能对铁制品起保护作用C．在海轮外壳连接锌块保护外壳不受腐蚀是采用了牺牲阳极的阴极保护法D．可将地下输油钢管与外加直流电源的正极相连以保护它不受腐蚀2．钢铁在潮湿的空气中会被腐蚀，发生的电池反应为：2Fe＋2H2O＋O2 = 2Fe2＋＋4OH－。

以下说法正确的是A．负极发生的反应为：Fe－2e－= Fe2＋B．正极发生的反应为：2H2O＋O2＋2e－= 4OH－C．原电池是将电能转变为化学能的装置D．钢柱在水下部分比在空气与水交界处更容易腐蚀3．下列装置或操作能达到实验目的的是4．我国首创的海洋电池以铝板为负极，铂网为正极，海水为电解质溶液，空气中的氧气与铝反应产生电流。

电池总反应为：4Al+3O2+6H2O = 4Al（OH）3，下列说法不正确的是A．正极反应式为：O2+2H2O+4e -=40H -B．电池工作时，电流由铝电极沿导线流向铂电极C．以网状的铂为正极，可增大与氧气的接触面积D．该电池通常只需更换铝板就可继续使用5．将纯锌片和纯铜片按图示方式插入同浓度的稀硫酸中一段时间，以下叙述正确的是A ．两烧杯中铜片表面均无气泡产生B ．甲中铜片是正极，乙中铜片是负极C ．两烧杯中溶液的pH 均增大D ．产生气泡的速度甲比乙慢6． LiFePO 4电池具有稳定性高、安全、对环境友好等优点，可用于电动汽车。

电池反应为：FePO 4+Li LiFePO 4，电池的正极材料是LiFePO 4，负极材料是石墨，含Li ＋导电固体为电解质。

下列有关LiFePO 4电池说法正确的是A ．可加入硫酸以提高电解质的导电性B ．放电时电池内部Li ＋向负极移动C ．充电过程中，电池正极材料的质量减少D ．放电时电池正极反应为：FePO 4+Li ++e -=LiFePO 47．镍镉（Ni —Cd ）可充电电池在现代生活中有广泛应用。