ÌâÄ¿ÄÚÈÝ

£¨14·Ö£©Ä³¿ÎÍâ»î¶¯Ð¡×éͬѧÓÃͼһװÖýøÐÐʵÑ飬ÊԻشðÏÂÁÐÎÊÌâ¡£

£¨1£©¢Ù Èô¿ªÊ¼Ê±¿ª¹ØKÓëaÁ¬½Ó£¬ÔòÌú·¢ÉúµÄÊǵ绯ѧ¸¯Ê´ÖеĠ            ¸¯Ê´£»
¢Ú Èô¿ªÊ¼Ê±¿ª¹ØKÓëbÁ¬½Ó£¬Ôò×Ü·´Ó¦µÄÀë×Ó·½³Ìʽ                          ¡£
£¨2£©Ã¢Ïõ£¨»¯Ñ§Ê½ÎªNa2SO4¡¤10H2O£©£¬ÎÞÉ«¾§Ì壬Ò×ÈÜÓÚË®£¬ÊÇÒ»ÖÖ·Ö²¼ºÜ¹ã·ºµÄÁòËáÑοóÎï¡£¸ÃС×éͬѧÉèÏ룬Èç¹ûÄ£Ä⹤ҵÉÏÀë×Ó½»»»Ä¤·¨ÖÆÉÕ¼îµÄ·½·¨¡£ÓÃÈçͼ¶þËùʾװÖõç½âÁòËáÄÆÈÜÒºÀ´ÖÆÈ¡ÇâÆø¡¢ÑõÆø¡¢ÁòËáºÍÇâÑõ»¯ÄÆ£¬ÎÞÂÛ´Ó½ÚÊ¡ÄÜÔ´»¹ÊÇÌá¸ßÔ­ÁϵÄÀûÓÃÂÊÀ´¿´¶¼¸ü¼Ó·ûºÏÂÌÉ«»¯Ñ§ÀíÄî¡£
¢Ù ¸Ãµç½â²ÛµÄÑô¼«·´Ó¦Ê½Îª£º                                       £¬´Ëʱͨ¹ýÒõÀë×Ó½»»»Ä¤µÄÀë×ÓÊý               £¨Ìî¡°´óÓÚ¡±»ò¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±£©Í¨¹ýÑôÀë×Ó½»»»Ä¤µÄÀë×ÓÊý¡£
¢Ú ÖƵõÄÇâÑõ»¯ÄÆÈÜÒº´Ó³ö¿Ú£¨Ñ¡Ìî¡°A¡±¡¢¡°B¡±¡¢¡°C¡±¡¢¡°D¡±£©       µ¼³ö¡£
¢Û Èô½«ÖƵõÄÇâÆø¡¢ÑõÆøºÍÇâÑõ»¯ÄÆÈÜÒº×éºÏΪÇâÑõȼÁϵç³Ø£¬Ôòµç³Ø¸º¼«µÄµç¼«·´Ó¦Ê½Îª                                                 ¡£ÒÑÖªH2µÄȼÉÕÈÈΪ285.8 kJ/mol£¬Ôò¸ÃȼÁϵç³Ø¹¤×÷²úÉú36 g H2Oʱ£¬ÀíÂÛÉÏÓР             kJµÄÄÜÁ¿×ª»¯ÎªµçÄÜ¡£

£¨1£©¢Ù ÎüÑõ   ¢Ú 2Cl¡ª +2H2O2OH¡ª + H2¡ü+ Cl2¡ü
£¨2£©¢Ù 4OH¡ª4e¡ª=2H2O + O2¡ü  Ð¡ÓÚ ¢Ú D   ¢Û H22e¡ª+ 2OH¡ª=2H2O      571.6

½âÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨2008?¶«Ý¸Ä£Ä⣩ij¿ÎÍâ»î¶¯Ð¡×éͬѧÓÃÓÒͼװÖýøÐÐʵÑ飬ÊԻشðÏÂÁÐÎÊÌ⣮
£¨1£©Èô¿ªÊ¼Ê±¿ª¹ØKÓëaÁ¬½Ó£¬ÔòB¼«µÄµç¼«·´Ó¦Ê½Îª
Fe-2e=Fe2+
Fe-2e=Fe2+
£®
£¨2£©Èô¿ªÊ¼Ê±¿ª¹ØKÓëbÁ¬½Ó£¬ÔòB¼«µÄµç¼«·´Ó¦Ê½Îª
2H++2e-=H2¡ü
2H++2e-=H2¡ü
£¬×Ü·´Ó¦µÄÀë×Ó·½³ÌʽΪ
2Cl-+2H2O
 µç½â 
.
 
2OH-+H2¡ü+Cl2¡ü
2Cl-+2H2O
 µç½â 
.
 
2OH-+H2¡ü+Cl2¡ü
£¬ÓйظÃʵÑéµÄÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨ÌîÐòºÅ£©
¢Ú
¢Ú
£®
¢ÙÈÜÒºÖÐNa+ÏòA¼«Òƶ¯     ¢Ú´ÓA¼«´¦ÒݳöµÄÆøÌåÄÜʹʪÈóKIµí·ÛÊÔÖ½±äÀ¶   ¢Û·´Ó¦Ò»¶Îʱ¼äºó¼ÓÊÊÁ¿ÑÎËá¿É»Ö¸´µ½µç½âÇ°µç½âÖʵÄŨ¶È   ¢ÜÈô±ê×¼×´¿öÏÂB¼«²úÉú2.24LÆøÌ壬ÔòÈÜÒºÖÐתÒÆ0.2molµç×Ó
£¨3£©¸ÃС×éͬѧģÄ⹤ҵÉÏÓÃÀë×Ó½»»»Ä¤·¨ÖÆÉÕ¼îµÄ·½·¨£¬ÉèÏëÓÃÓÒͼװÖõç½âÁòËá¼ØÈÜÒºÀ´ÖÆÈ¡ÇâÆø¡¢ÑõÆø¡¢ÁòËáºÍÇâÑõ»¯¼Ø£®
¢Ù¸Ãµç½â²ÛµÄÑô¼«·´Ó¦Ê½Îª
4OH--4e-=2H2O+O2¡ü
4OH--4e-=2H2O+O2¡ü
£®´Ëʱͨ¹ýÒõÀë×Ó½»»»Ä¤µÄÀë×ÓÊý
СÓÚ
СÓÚ
£¨Ìî¡°´óÓÚ¡±»ò¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±£©Í¨¹ýÑôÀë×Ó½»»»Ä¤µÄÀë×ÓÊý£®
¢ÚÖƵõÄÇâÑõ»¯¼ØÈÜÒº´Ó£¨Ìîд¡°A¡±¡¢¡°B¡±¡¢¡°C¡±¡¢¡°D¡±£©
D
D
³ö¿Úµ¼³ö£®
¢Ûͨµç¿ªÊ¼ºó£¬Òõ¼«¸½½üÈÜÒºpH»áÔö´ó£¬Çë¼òÊöÔ­Òò
H+·Åµç£¬´Ù½øË®µÄµçÀ룬OH-Ũ¶ÈÔö´ó
H+·Åµç£¬´Ù½øË®µÄµçÀ룬OH-Ũ¶ÈÔö´ó
£®
¢ÜÈô½«ÖƵõÄÇâÆø¡¢ÑõÆøºÍÇâÑõ»¯¼ØÈÜÒº×éºÏΪÇâÑõȼÁϵç³Ø£¬Ôòµç³ØÕý¼«µÄµç¼«·´Ó¦Ê½Îª
O2+2H2O+4e-=4OH-
O2+2H2O+4e-=4OH-
£®
ij¿ÎÍâ»î¶¯Ð¡×éͬѧÓÃÈçͼװÖýøÐÐʵÑ飬ÊԻشðÏÂÁÐÎÊÌ⣺
£¨1£©Èô¿ªÊ¼Ê±¿ª¹ØKÓëaÁ¬½Ó£¬ÔòA¼«µÄµç¼«·´Ó¦Ê½Îª
2H++2e-¨TH2¡ü
2H++2e-¨TH2¡ü
£®
£¨2£©Èô¿ªÊ¼Ê±¿ª¹ØKÓëbÁ¬½Ó£¬ÔòB¼«µÄµç¼«·´Ó¦Ê½Îª
2H++2e-¨TH2¡ü
2H++2e-¨TH2¡ü
£¬×Ü·´Ó¦µÄÀë×Ó·½³ÌʽΪ
2Cl-+2H2O
 µç½â 
.
 
2OH-+H2¡ü+Cl2¡ü
2Cl-+2H2O
 µç½â 
.
 
2OH-+H2¡ü+Cl2¡ü
£®ÓйØÉÏÊöʵÑ飬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨ÌîÐòºÅ£©
¢Ú
¢Ú
£®
¢ÙÈÜÒºÖÐNa+ÏòA¼«Òƶ¯
¢Ú´ÓA¼«´¦ÒݳöµÄÆøÌåÄÜʹʪÈóKIµí·ÛÊÔÖ½±äÀ¶
¢Û·´Ó¦Ò»¶Îʱ¼äºó¼ÓÊÊÁ¿ÑÎËá¿É»Ö¸´µ½µç½âÇ°µç½âÖʵÄŨ¶È
¢ÜÈô±ê×¼×´¿öÏÂB¼«²úÉú2.24LÆøÌ壬ÔòÈÜÒºÖÐתÒÆ0.2molµç×Ó
£¨3£©¸ÃС×éͬѧģÄ⹤ҵÉÏÓÃÀë×Ó½»»»Ä¤·¨ÖÆÉÕ¼îµÄ·½·¨£¬ÄÇô¿ÉÒÔÉèÏëÓÃÈçͼװÖõç½âÁòËá¼ØÈÜÒºÀ´ÖÆÈ¡ÇâÆø¡¢ÑõÆø¡¢ÁòËáºÍÇâÑõ»¯¼Ø£®
¢Ù¸Ãµç½â²ÛµÄÑô¼«·´Ó¦Ê½Îª
4OH--4e-=2H2O+O2¡ü
4OH--4e-=2H2O+O2¡ü
£®´Ëʱͨ¹ýÒõÀë×Ó½»»»Ä¤µÄÀë×ÓÊý
СÓÚ
СÓÚ
£¨Ìî¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±£©Í¨¹ýÑôÀë×Ó½»»»Ä¤µÄÀë×ÓÊý£®
¢Úͨµç¿ªÊ¼ºó£¬Òõ¼«¸½½üÈÜÒºpH»áÔö´ó£¬Çë¼òÊöÔ­Òò
H+·Åµç£¬´Ù½øË®µÄµçÀ룬OH-Ũ¶ÈÔö´ó
H+·Åµç£¬´Ù½øË®µÄµçÀ룬OH-Ũ¶ÈÔö´ó
£®
¢ÛÈô½«ÖƵõÄÇâÆø¡¢ÑõÆøºÍÇâÑõ»¯¼ØÈÜÒº×éºÏΪÇâÑõȼÁϵç³Ø£¬Ôòµç³ØÕý¼«µÄµç¼«·´Ó¦Ê½Îª
O2+2H2O+4e-¨T4OH-
O2+2H2O+4e-¨T4OH-
£®

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø