ÌâÄ¿ÄÚÈÝ

ÏÂÃæÊdz£¼ûµÄµç»¯Ñ§×°ÖÃͼ£º
 
ͼ1                 Í¼2                 Í¼3
Çë»Ø´ðÒÔÏÂÎÊÌ⣺£¨1£©Í¼1ÖÐÍ­°åÉÏÌúí¶¤´¦ÈÝÒ×ÉúÐ⣬³ÆÎª              ¸¯Ê´£¬±»¸¯Ê´µÄ½ðÊôÊÇ         £¬Ô­µç³ØµÄ×Ü·´Ó¦·½³ÌʽÊÇ                          ¡£
£¨2£©Í¼2ÖÐa¡¢bÊǶà¿×ʯīµç¼«£¬¶Ï¿ªK2£¬±ÕºÏK1Ò»¶Îʱ¼ä£¬¹Û²ìµ½Á½Ö»²£Á§¹ÜÄÚ¶¼ÓÐÆøÅݽ«µç¼«°üΧ£¬b¼«Éϵĵ缫·´Ó¦Ê½Îª                   £¬OH-Ïò          £¨Ìîa»òb£©¼«Òƶ¯¡£È»ºó¶Ï¿ªK1£¬±ÕºÏK2£¬¹Û²ìµ½µçÁ÷¼ÆAµÄÖ¸ÕëÓÐÆ«×ª¡£b¼«Éϵĵ缫·´Ó¦Ê½Îª                         £¬OH-Ïò               £¨Ìîa»òb£©¼«Òƶ¯¡£
£¨3£©Í¼3ÖУ¬X¡¢Y¶¼ÎªÊ¯Ä«µç¼«£¬ÔÚUÐ͹ÜÁ½²à·Ö±ð¼ÓÈëÒ»µÎ×ÏɫʯÈïÊÔÒº£¬ÔòͨµçºóX¸½½üÏÔ______É«£¬Y¸½½üÏÔ_______É«¡£

£¨1£©ÎüÑõ     Fe      2Fe+O2+2H2O=2Fe(OH)2 
£¨2£©     2H++2e£­=H2      a     H2+2OH¡ª£«2e£­=2H2O    b  £¨3£©À¶  ºì

½âÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø