按如图装置进行实验，并回答下列问题：（1）锌极为
\u9ad8\u4e2d\u5316\u5b66\u9898\uff0c\u950c\u6781\u4e3a\u8d1f\u6781\uff0c\u7535\u6781\u53cd\u5e94\u5f0f\u4e3a\uff0c \u94dc\u6781\u4e3a\u6b63\u6781\uff0c\u7535\u6781\u53cd\u5e94\u5f0f\u4e3a\uff0c\u77f3\u58a8\u68d2C1\u7535\u6781\u53cd\u5e94\u5f0f\u4e3a\uff0c\u77f3\u58a8\u68d2C2A\u4e3a\u539f\u7535\u6c60 Zn\u4e3a\u8d1f\u6781\uff1aZn-2e=Zn2+ Cu\u4e3a\u6b63\u6781\uff1aCu2+ +2e=Cu
B\u4e3a\u7535\u89e3\u6c60C1\u4e3a\u9634\u6781\uff1a2H+ +2e=H2 C2\u4e3a\u9633\u6781\uff1a2Cl- +2e=Cl2\u915a\u915e\u6eb6\u6db2\u53d8\u7ea2
\u603b\u7535\u89e3\u53cd\u5e942KCl+2H2O=2KOH+H2+Cl2
\u7531\u7535\u5b50\u5b88\u6052\u53ef\u77e5\uff1a
\u5f53C2\u6781\u6790\u51fa224mL\u6c14\u4f53\uff08Cl2\uff09\u65f6\uff0c\u950c\u7684\u8d28\u91cf\u51cf\u5c110.65g\uff0cCuSO4\u6eb6\u6db2\u7684\u8d28\u91cf\u53d8\u5316\u4e86\u51cf\u5c11\u4e860.64g
\uff081\uff09\u7532\u80fd\u81ea\u53d1\u7684\u8fdb\u884c\u6c27\u5316\u8fd8\u539f\u53cd\u5e94\uff0c\u5219\u7532\u662f\u539f\u7535\u6c60\uff0c\u4e59\u662f\u7535\u89e3\u6c60\uff0c\u950c\u6613\u5931\u7535\u5b50\u4f5c\u8d1f\u6781\u3001Cu\u4f5c\u6b63\u6781\uff0cA\u662f\u9633\u6781\u3001B\u662f\u9634\u6781\uff0cB\u7535\u6781\u4e0a\u94dc\u79bb\u5b50\u653e\u7535\u751f\u6210\u94dc\uff0c\u7535\u6781\u53cd\u5e94\u5f0f\u4e3aCu2++2e-=Cu\uff0c\u6240\u4ee5B\u7535\u6781\u4e0a\u6790\u51fa\u7684\u7269\u8d28\u662fCu\uff0c\u6545\u7b54\u6848\u4e3a\uff1a\u8d1f\uff1bCu\uff1b\uff082\uff09\u4e59\u88c5\u7f6e\u4e2dA\u7535\u6781\u4e0a\u6c2f\u79bb\u5b50\u653e\u7535\u3001B\u7535\u6781\u4e0a\u94dc\u79bb\u5b50\u653e\u7535\uff0c\u6240\u4ee5\u7535\u6c60\u53cd\u5e94\u5f0f\u4e3aCuCl2 \u7535\u89e3 . Cu+Cl2\u2191\uff0c\u6545\u7b54\u6848\u4e3a\uff1aCuCl2 \u7535\u89e3 . Cu+Cl2\u2191\uff0e
| （1）A能自发的进行氧化还原反应，能将化学能转化为电能，活泼性较强的Zn为负极，电极反应式为Zn-2e - =Zn 2+ ；活泼性较弱的Cu为正极，石墨棒C 1 与正极相连为阳极，阳极上氯离子失电子生成氯气，其电极反应式为：2Cl - -2e - =Cl 2 ；石墨棒C 2 为阴极，阴极上电解水生成氢气和氢氧根离子，所以阴极附近有气泡，溶液变红， 故答案为：负；Zn-2e - =Zn 2+ ；阳；2Cl - -2e - =Cl 2 ；有气泡，溶液变红； （2）石墨棒C 2 为阴极，阴极上电解水生成氢气和氢氧根离子，其电极方程式为：4H 2 O+4e - =2H 2 ↑+4OH - ，当C 2 极析出224mL气体（标准状态）时，生成氢气的物质的量为n（H 2 ）=
故答案为：0.65． |
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