ÌâÄ¿ÄÚÈÝ
͵¥Öʼ°Æä»¯ºÏÎïÔÚ¹¤ÒµÉú²úºÍ¿ÆÑÐÖÐÓÐÖØÒª×÷Óã®
£¨1£©ÒÑÖª£º2Cu2O£¨s£©+O2£¨g£©¨T4CuO£¨s£©¡÷H=-292kJ?mol-1
2C£¨s£©+O2£¨g£©¨T2CO£¨g£©¡÷H=-221kJ?mol-1
Çëд³öÓÃ×ãÁ¿Ì¿·Û»¹ÔCuO£¨s£©ÖƱ¸Cu2O£¨s£©µÄÈÈ»¯Ñ§·½³Ìʽ£º______£»
£¨2£©ÏÖÓÃÂÈ»¯Í¾§Ì壨CuCl2?2H2O£¬º¬ÂÈ»¯ÑÇÌúÔÓÖÊ£©ÖÆÈ¡´¿¾»µÄCuCl2?2H2O£®ÏȽ«ÆäÖÆ³ÉË®ÈÜÒº£¬ºó°´Èçͼ²½Öè½øÐÐÌá´¿£®ÒÑÖªCu2+¡¢Fe3+ºÍFe2+µÄÇâÑõ»¯Î↑ʼ³ÁµíºÍ³ÁµíÍêȫʱµÄpH¼ûϱí
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
¢ÙÏÖÓÐÑõ»¯¼ÁNaClO¡¢H2O2¡¢KMnO4£¬X¼ÓÄÄÖֺã¬ÎªÊ²Ã´£¿______£»¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪ______£®
¢ÚÈÜÒºIIÖгýCu2+Í⣬»¹ÓÐ______½ðÊôÀë×Ó£¬ÈçºÎ¼ìÑéÆä´æÔÚ£®
¢ÛÎïÖÊY²»ÄÜΪÏÂÁеÄ______
a£®CuOb£®Cu£¨OH£©2c£®CuCO3d£®Cu2£¨OH£©2CO3e£®CaOf£®NaOH
¢ÜÈôÏòÈÜÒº¢òÖмÓÈë̼Ëá¸Æ£¬²úÉúµÄÏÖÏóÊÇ______£®

£¨1£©ÒÑÖª£º2Cu2O£¨s£©+O2£¨g£©¨T4CuO£¨s£©¡÷H=-292kJ?mol-1
2C£¨s£©+O2£¨g£©¨T2CO£¨g£©¡÷H=-221kJ?mol-1
Çëд³öÓÃ×ãÁ¿Ì¿·Û»¹ÔCuO£¨s£©ÖƱ¸Cu2O£¨s£©µÄÈÈ»¯Ñ§·½³Ìʽ£º______£»
£¨2£©ÏÖÓÃÂÈ»¯Í¾§Ì壨CuCl2?2H2O£¬º¬ÂÈ»¯ÑÇÌúÔÓÖÊ£©ÖÆÈ¡´¿¾»µÄCuCl2?2H2O£®ÏȽ«ÆäÖÆ³ÉË®ÈÜÒº£¬ºó°´Èçͼ²½Öè½øÐÐÌá´¿£®ÒÑÖªCu2+¡¢Fe3+ºÍFe2+µÄÇâÑõ»¯Î↑ʼ³ÁµíºÍ³ÁµíÍêȫʱµÄpH¼ûϱí
| ½ðÊôÀë×Ó | Fe3+ | Fe2+ | Cu2+ |
| ÇâÑõ»¯Î↑ʼ³ÁµíʱµÄpH | 1.9 | 7.0 | 4.7 |
| ÇâÑõ»¯ÎïÍêÈ«³ÁµíʱµÄpH | 3.2 | 9.0 | 6.7 |
¢ÙÏÖÓÐÑõ»¯¼ÁNaClO¡¢H2O2¡¢KMnO4£¬X¼ÓÄÄÖֺã¬ÎªÊ²Ã´£¿______£»¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪ______£®
¢ÚÈÜÒºIIÖгýCu2+Í⣬»¹ÓÐ______½ðÊôÀë×Ó£¬ÈçºÎ¼ìÑéÆä´æÔÚ£®
¢ÛÎïÖÊY²»ÄÜΪÏÂÁеÄ______
a£®CuOb£®Cu£¨OH£©2c£®CuCO3d£®Cu2£¨OH£©2CO3e£®CaOf£®NaOH
¢ÜÈôÏòÈÜÒº¢òÖмÓÈë̼Ëá¸Æ£¬²úÉúµÄÏÖÏóÊÇ______£®
£¨1£©ÒÑÖª£º¢Ù2Cu2O£¨s£©+O2£¨g£©=4CuO£¨s£©¡÷H=-292kJ?mol-1
¢Ú2C£¨s£©+O2£¨g£©=2CO£¨g£©¡÷H=-221kJ?mol-1
¸ù¾Ý¸Ç˹¶¨ÂÉ£¬¢Ú-¢ÙµÃ4CuO£¨s£©+2C£¨s£©=2CO£¨g£©+2Cu2O£¨s£©£¬¡÷H=+71kJ?mol-1
¼´2CuO£¨s£©+C£¨s£©=CO£¨g£©+Cu2O£¨s£©¡÷H=+35.5kJ?mol-1
¹Ê´ð°¸Îª£º2CuO£¨s£©+C£¨s£©=CO£¨g£©+Cu2O£¨s£©¡÷H=+35.5kJ?mol-1£®
£¨2£©¢Ù¼ÓÈëÑõ»¯¼ÁµÄÄ¿µÄÊǰÑÑÇÌúÀë×ÓÑõ»¯ÎªÌúÀë×Ó£¬¹ýÑõ»¯ÇâÑõ»¯ÑÇÌúÀë×ÓµÄÀë×Ó·½³ÌʽΪ£ºH2O2+2Fe2++2H+¨T2Fe3++2H2O£¬ÓÉÓÚ²úÎïÊÇË®£¬²»ÒýÈëеÄÔÓÖÊ£¬Òò´ËÑõ»¯¼ÁÑ¡Ôñ¹ýÑõ»¯Ç⣬
¹Ê´ð°¸Îª£ºH2O2ºÃ£¬²»ÒýÈëÔÓÖÊÀë×Ó£»H2O2+2Fe2++2H+¨T2Fe3++2H2O£»
¢ÚÈÜÒºIIÖгýCu2+Í⻹º¬ÓÐH2O2+2Fe2++2H+¨T2Fe3++2H2OÉú³ÉµÄFe3+£¬È¡ÊÔ¹ÜÉÙÁ¿ÈÜÒº¢ò£¬µÎ¼ÓKSCNÈÜÒº£¬ÈÜÒºÈô±äºìÉ«£¬Ö¤Ã÷ÈÜÒº¢òº¬ÓÐFe3+£¬ÈôÈÜÒº²»±äºìÉ«£¬ÔòÖ¤Ã÷²»º¬Fe3+£¬
¹Ê´ð°¸Îª£ºFe3+£»È¡ÉÙÁ¿ÈÜÒºÓÚÊÔ¹ÜÖУ¬µÎ¼ÓKSCNÈÜÒº£¬ÈôÈÜÒº±äºìÉ«£¬Ö¤Ã÷º¬Fe3+£®ÈôÈÜÒº²»±äºìÉ«£¬ÔòÖ¤Ã÷²»º¬Fe3+£»
¢Û¼ÓÈëµÄÎïÖÊY¿Éµ÷½Ú£¬Ê¹Fe3+È«²¿³Áµí£¬Í¬Ê±²»Òý½øÐÂÔÓÖÊ£¬ËùÒÔ¿ÉÒÔÓÃCu£¨OH£©2¡¢CuCO3¡¢CuO¡¢Cu2£¨OH£©2CO3µÈ£¬ÈôÓÃCaO¡¢NaOH¾ùÒýÈëеÄÔÓÖÊÀë×Ó£¬
¹Ê´ð°¸Îª£ºef£»
¢Ü̼Ëá¸ÆÓëÈÜÒº¢òÖеÄÇâÀë×Ó·´Ó¦Éú³É¶þÑõ»¯Ì¼ÆøÌ壬Òò´Ë»á²úÉúÆøÅÝ£¬Í¬Ê±·´Ó¦Ôö´óÁËÈÜÒºµÄpHֵʹµÃFe3+ÐγɺìºÖÉ«µÄÇâÑõ»¯Ìú³Áµí£¬
¹Ê´ð°¸Îª£ºÌ¼Ëá¸ÆÈܽ⣬²úÉúÆøÅݺͺìºÖÉ«³Áµí£®
¢Ú2C£¨s£©+O2£¨g£©=2CO£¨g£©¡÷H=-221kJ?mol-1
¸ù¾Ý¸Ç˹¶¨ÂÉ£¬¢Ú-¢ÙµÃ4CuO£¨s£©+2C£¨s£©=2CO£¨g£©+2Cu2O£¨s£©£¬¡÷H=+71kJ?mol-1
¼´2CuO£¨s£©+C£¨s£©=CO£¨g£©+Cu2O£¨s£©¡÷H=+35.5kJ?mol-1
¹Ê´ð°¸Îª£º2CuO£¨s£©+C£¨s£©=CO£¨g£©+Cu2O£¨s£©¡÷H=+35.5kJ?mol-1£®
£¨2£©¢Ù¼ÓÈëÑõ»¯¼ÁµÄÄ¿µÄÊǰÑÑÇÌúÀë×ÓÑõ»¯ÎªÌúÀë×Ó£¬¹ýÑõ»¯ÇâÑõ»¯ÑÇÌúÀë×ÓµÄÀë×Ó·½³ÌʽΪ£ºH2O2+2Fe2++2H+¨T2Fe3++2H2O£¬ÓÉÓÚ²úÎïÊÇË®£¬²»ÒýÈëеÄÔÓÖÊ£¬Òò´ËÑõ»¯¼ÁÑ¡Ôñ¹ýÑõ»¯Ç⣬
¹Ê´ð°¸Îª£ºH2O2ºÃ£¬²»ÒýÈëÔÓÖÊÀë×Ó£»H2O2+2Fe2++2H+¨T2Fe3++2H2O£»
¢ÚÈÜÒºIIÖгýCu2+Í⻹º¬ÓÐH2O2+2Fe2++2H+¨T2Fe3++2H2OÉú³ÉµÄFe3+£¬È¡ÊÔ¹ÜÉÙÁ¿ÈÜÒº¢ò£¬µÎ¼ÓKSCNÈÜÒº£¬ÈÜÒºÈô±äºìÉ«£¬Ö¤Ã÷ÈÜÒº¢òº¬ÓÐFe3+£¬ÈôÈÜÒº²»±äºìÉ«£¬ÔòÖ¤Ã÷²»º¬Fe3+£¬
¹Ê´ð°¸Îª£ºFe3+£»È¡ÉÙÁ¿ÈÜÒºÓÚÊÔ¹ÜÖУ¬µÎ¼ÓKSCNÈÜÒº£¬ÈôÈÜÒº±äºìÉ«£¬Ö¤Ã÷º¬Fe3+£®ÈôÈÜÒº²»±äºìÉ«£¬ÔòÖ¤Ã÷²»º¬Fe3+£»
¢Û¼ÓÈëµÄÎïÖÊY¿Éµ÷½Ú£¬Ê¹Fe3+È«²¿³Áµí£¬Í¬Ê±²»Òý½øÐÂÔÓÖÊ£¬ËùÒÔ¿ÉÒÔÓÃCu£¨OH£©2¡¢CuCO3¡¢CuO¡¢Cu2£¨OH£©2CO3µÈ£¬ÈôÓÃCaO¡¢NaOH¾ùÒýÈëеÄÔÓÖÊÀë×Ó£¬
¹Ê´ð°¸Îª£ºef£»
¢Ü̼Ëá¸ÆÓëÈÜÒº¢òÖеÄÇâÀë×Ó·´Ó¦Éú³É¶þÑõ»¯Ì¼ÆøÌ壬Òò´Ë»á²úÉúÆøÅÝ£¬Í¬Ê±·´Ó¦Ôö´óÁËÈÜÒºµÄpHֵʹµÃFe3+ÐγɺìºÖÉ«µÄÇâÑõ»¯Ìú³Áµí£¬
¹Ê´ð°¸Îª£ºÌ¼Ëá¸ÆÈܽ⣬²úÉúÆøÅݺͺìºÖÉ«³Áµí£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿