按图装置进行实验，并回答下列问题（1）判断装置的名称：A池为 按图装置进行实验，并回答下列问题（1）判断装置的名称：A池为...
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\uff081\uff09\u88c5\u7f6eA\u7b26\u5408\u539f\u7535\u6c60\u7684\u6784\u6210\u6761\u4ef6\u5c5e\u4e8e\u539f\u7535\u6c60\uff0cB\u7b26\u5408\u7535\u89e3\u6c60\u7684\u6784\u6210\u6761\u4ef6\uff0c\u5c5e\u4e8e\u7535\u89e3\u6c60\uff0c\u6545\u7b54\u6848\u4e3a\uff1a\u539f\u7535\u6c60\uff1b\u7535\u89e3\u6c60\uff1b\uff082\uff09\u539f\u7535\u6c60\u4e2d\u6d3b\u6cfc\u91d1\u5c5e\u950c\u505a\u8d1f\u6781\uff0c\u53d1\u751f\u5931\u7535\u5b50\u7684\u6c27\u5316\u53cd\u5e94\uff0c\u5373Zn-2e-=Zn 2+ \uff0c\u6545\u7b54\u6848\u4e3a\uff1a\u8d1f\uff1bZn-2e-=Zn 2+ \uff1b\uff083\uff09\u77f3\u58a8\u68d2C 1 \u548c\u539f\u7535\u6c60\u7684\u6b63\u6781\u76f8\u8fde\uff0c\u662f\u9633\u6781\uff0c\u6c2f\u79bb\u5b50\u5728\u8be5\u6781\u53d1\u751f\u5931\u7535\u5b50\u7684\u6c27\u5316\u53cd\u5e94\uff0c\u53732Cl - -2e-=Cl 2 \u2191\uff0c\u77f3\u58a8\u68d2C 2 \u548c\u539f\u7535\u6c60\u7684\u8d1f\u6781\u76f8\u8fde\uff0c\u662f\u9634\u6781\uff0c\u6c22\u79bb\u5b50\u653e\u7535\uff0c\u6240\u4ee5\u6c22\u79bb\u5b50\u6d53\u5ea6\u51cf\u5c0f\u800c\u6c22\u6c27\u6839\u6d53\u5ea6\u589e\u5927\uff0c\u78b1\u6027\u589e\u5f3a\uff0c\u4f7f\u915a\u915e\u53d8\u7ea2\uff0c\u6545\u7b54\u6848\u4e3a\uff1a\u9633\uff1b2Cl - -2e-=Cl 2 \u2191\uff1b\u7535\u6781\u4e0a\u6709\u6c14\u6ce1\u4ea7\u751f\uff0c\u65e0\u8272\u6eb6\u6db2\u53d8\u7ea2\u8272\uff1b\uff084\uff09C 2 \u6781\u53d1\u751f\u7684\u7535\u6781\u53cd\u5e94\u4e3a\uff1a2H + +2e - \u2192H 2 \u2191\uff0c\u6790\u51fa224mL\uff08\u53730.01mol\uff09\u6c14\u4f53\uff08\u6807\u51c6\u72b6\u6001\uff09\u65f6\uff0c\u8f6c\u79fb\u7535\u5b50\u6570\u662f0.02mol\uff0c\u6240\u4ee5\u950c\u6781\u8f6c\u79fb\u7535\u5b50\u4e5f\u4e3a0.02mol\uff0c\u6839\u636eZn-2e-=Zn 2+ \uff0c\u950c\u6781\u8f6c\u79fb\u7535\u5b50\u4e3a0.02mol\u65f6\uff0c\u4f1a\u67090.01mol\u91d1\u5c5e\u950c\u51cf\u5c11\uff0c\u5373\u950c\u7684\u8d28\u91cf\u51cf\u5c110.65g\uff0c\u6545\u7b54\u6848\u4e3a\uff1a0.65\uff0e
（1）装置A符合原电池的构成条件属于原电池，B符合电解池的构成条件，属于电解池，故答案为：原电池；电解池；（2）原电池中活泼金属锌做负极，发生失电子的氧化反应，即Zn-2e-=Zn2+，故答案为：负；Zn-2e-=Zn2+；
（3）石墨棒C1和原电池的正极相连，是阳极，氯离子在该极发生失电子的氧化反应，即2Cl--2e-=Cl2↑，石墨棒C2和原电池的负极相连，是阴极，氢离子放电，所以氢离子浓度减小而氢氧根浓度增大，碱性增强，使酚酞变红，故答案为：阳；2Cl--2e-=Cl2↑；电极上有气泡产生，无色溶液变红色；
（4）C2极发生的电极反应为：2H++2e-→H2↑，析出224mL（即0.01mol）气体（标准状态）时，转移电子数是0.02mol，所以锌极转移电子也为0.02mol，根据Zn-2e-=Zn2+，锌极转移电子为0.02mol时，会有0.01mol金属锌减少，即锌的质量减少0.65g，故答案为：0.65．
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