ÌâÄ¿ÄÚÈÝ
£¨10·Ö£©µç½âÔÀíÔÚ»¯Ñ§¹¤ÒµÖÐÓй㷺ӦÓá£ÓÒͼ±íʾһ¸öµç½â³Ø£¬×°Óеç½âÒºa£»X¡¢YÊÇÁ½¿éµç¼«°å£¬Í¨¹ýµ¼ÓëÖ±Á÷µçÔ´ÏàÁ¬¡£Çë»Ø´ðÒÔÏÂÎÊÌ⣺
![]()
(1) ÈôX¡¢Y¶¼ÊǶèÐԵ缫£¬aÊDZ¥ºÍNaClÈÜÒº£¬ÊµÑ鿪ʼʱ£¬Í¬Ê±ÔÚÁ½±ß¸÷µÎÈ뼸µÎ·Ó̪ÊÔÒº£¬Ôò µç½â³ØÖÐX¼«Éϵĵ缫·´Ó¦Ê½Îª ¡£
¢Ú¼ìÑé¸ÃYµç¼«·´Ó¦²úÎïµÄ·½·¨ÊÇ
¢Û¸Ã·´Ó¦µÄ×Ü·´Ó¦·½³ÌʽÊÇ£º
(2) ÈçÒªÓõç½â·½·¨¾«Á¶´ÖÍ£¬µç½âÒºaÑ¡ÓÃCuSO4ÈÜÒº£¬Ôò
¢Ù Xµç¼«µç¼«·´Ó¦Ê½ÊÇ £¬
¢Ú Yµç¼«µÄ²ÄÁÏÊÇ ¡£
£¨1£©¢Ù2H+ +2e¨C == H2¡ü£¨2·Ö£©
¢Ú°ÑʪÈóµÄKIµí·ÛÊÔÖ½·ÅÔÚYµç¼« ¸½½ü£¬ÊÔÖ½±äÀ¶É«¡££¨2·Ö£©
¢Û2NaCl
+ 2H2O
2NaOH + H2¡ü + Cl2¡ü £¨2·Ö£©
£¨2£©¢ÙCu2+ +2 e¨C== Cu£¨2·Ö£©¢Ú´Ö Í£¨2·Ö£©
¡¾½âÎö¡¿¿¼²éµç»¯Ñ§µÄÓ¦Óá£
£¨1£©¢ÙXµç¼«ºÍµçÔ´µÄ¸º¼«ÏàÁ¬£¬×÷Òõ¼«£¬ÈÜÒºÖеÄÇâÀë×ӷŵçÉú³ÉÇâÆø£¬µç¼«·´Ó¦Ê½ÊÇ2H+ +2e¨C == H2¡ü¡£
¢ÚYµç¼«ÊÇÑô¼«£¬ÈÜÒºÖеÄÂÈÀë×ӷŵ磬Éú³ÉÂÈÆø£¬¶øÂÈÆø¼ìÑéÇ¿Ñõ»¯ÐÔ£¬¾Ý´Ë¿ÉÒÔ¼ø±ð£¬¼´°ÑʪÈóµÄKIµí·ÛÊÔÖ½·ÅÔÚYµç¼« ¸½½ü£¬ÊÔÖ½±äÀ¶É«¡£
¢ÛÒõ¼«»¹²úÉúÇâÑõ»¯ÄÆ£¬ËùÒÔ×ܵķ´Ó¦Ê½ÊÇ2NaCl + 2H2O
2NaOH + H2¡ü + Cl2¡ü¡£
£¨2£©´Ö;«Á¶Ê±£¬´Öͺ͵çÔ´µÄÕý¼«ÏàÁ¬£¬×÷Ñô¼«£¬Ê§È¥µç×Ó£¬ËùÒÔYµç¼«ÊÇ´ÖÍ£»´¿ÍºÍµçÔ´µÄ¸º¼«ÏàÁ¬£¬×÷Òõ¼«£¬ÈÜÒºÖеÄÍÀë×ӷŵ磬ËùÒÔXµç¼«µÄµç¼«·´Ó¦Ê½ÊÇCu2+ +2 e¨C== Cu¡£