Ò»ÖÖÁ½½ðÊôÐÞÊε缫µÄ·½·¨
¡¾×¨ÀûÕªÒª¡¿±¾·¢Ã÷Éæ¼°Ò»ÖÖÁ½½ðÊôÐÞÊε缫µÄ·½·¨£¬ÖƵõĵ缫ÓÃÓÚ´¼µÄµç´ß»¯Ñõ»¯·´Ó¦£¬°üÀ¨ÒÔϲ½Ö裺ÅäÖÆ½ðǰÌåÈÜÒº¡¢²¬Ç°ÌåÈÜÒº¡¢µç¼«±íÃæÐÞÊΡ¢µç¼«ÓÃÓÚÒÒ´¼µÄµç´ß»¯Ñõ»¯·´Ó¦¡£±¾·¢Ã÷µÄÓÐÒæÐ§¹ûÊÇ£ºµç¼«ÐÞÊη½·¨¼òµ¥£¬µç¼«Îȶ¨£¬Í¨¹ý¼«Ð¡Á¿µÄAuºÍPtÐÞÊÎËùµÃµ½µÄPdµç¼«ÔÚ¼îÐÔÈÜÒºÖжÔÒÒ´¼Ñõ»¯·´Ó¦µÄµç´ß»¯»îÐÔÔ¶¸ßÓÚAu¡¢PtºÍPdµç¼«ÒÔ¼°AuÐÞÊεÄPdµç¼«ºÍPtÐÞÊεÄPdµç¼«£»ÓÉÓÚAuºÍPtÐÞÊÎÁ¿¼«ÉÙ£¬½ÚÊ¡ÁËAuºÍPtµÄʹÓÃÁ¿£»ËùÖÆ±¸µÄµç¼«¶Ô¼îÐÔ½éÖÊÖеÄÒÒ´¼Ñõ»¯·´Ó¦¾ßÓкܸߵĴ߻¯»îÐÔ£¬ÐÞÊÎЧ¹û·Ç³£Ã÷ÏÔ¡£
¡¾×¨Àû˵Ã÷¡¿¡ªÖÖÁ½½ðÊôÐÞÊε缫µÄ·½·¨
¡¾¼¼ÊõÁìÓò¡¿
[0001]±¾·¢Ã÷ÊôÓڵ绯ѧµç¼«²ÄÁÏ¡¾¼¼ÊõÁìÓò¡¿£¬Éæ¼°Ò»ÖÖÁ½½ðÊôÐÞÊε缫µÄ·½·¨¡£
¡¾±³¾°¼¼Êõ¡¿
[0002]´¼µÄµç´ß»¯Ñõ»¯·´Ó¦Êǵ绯ѧÑо¿µÄÖ÷ÒªÁìÓòÖ®Ò»¡£³¤ÆÚÒÔÀ´ÈËÃǶԼ״¼Ö±½ÓȼÁÏµç³Ø½øÐÐÁË´óÁ¿Ñо¿£¬µ«ÊǼ״¼×÷ΪȼÁÏµç³ØµÄȼÁÏÒ²ÓÐһЩ²»×㣬Èç¼×´¼ÓÐÒ»¶¨¶¾ÐÔ£¬·Ðµã½ÏµÍÒ×»Ó·¢£¬·Ö×ÓСÈÝÒ×´©¹ýȼÁÏµç³ØÖеķÖÀëĤµÈ¡£Òò´Ë£¬ÈËÃÇÒ²»ý¼«¿ªÕ¹Ñ¡ÓÃÆäËû´¼×÷ΪȼÁÏµç³ØÔÁϵÄÑо¿£¬ÓÈÆäÊÇÒÒ´¼¡¢ÒÒ¶þ´¼µÈµÄµç´ß»¯Ñõ»¯·´Ó¦Êܵ½¹Ø×¢¡£ÁíÍâ£¬Ëæ×ÅÒõÀë×Ó½»»»Ä¤ÑÐÖÆµÄѸËÙ·¢Õ¹£¬ÀûÓüîÐÔ½éÖÊÌõ¼þÏ´¼µç´ß»¯Ñõ»¯·´Ó¦µÄ¼îÐÔȼÁÏµç³ØÑо¿Ò²Ô½À´Ô½Êܵ½ÖõÄ¿¡£±ÈÆðËáÐÔ½éÖÊ£¬¼îÐÔ½éÖʵÄÒ»¸öÏÔÖøµÄÓŵãÊDz»ÂÛ´¼Ñõ»¯·´Ó¦»¹ÊÇÑõÆø»¹Ô·´Ó¦¶¼ÈÝÒ×½øÐУ¬Ëٶȿ졣
[0003]ÒÒ´¼µÄµç´ß»¯Ñõ»¯·´Ó¦Ñо¿£¬ÔÚ¼îÐÔ½éÖÊÌõ¼þÏÂÖ÷ҪʹÓÃîâ»ùºÍîÙ»ùµÄ¶þ×é·Öµç¼«£¬ÔÚËáÐÔ½éÖÊÌõ¼þÏÂÒÔPtSn»ùºÍPtRu»ùµÄ¶þ×é·ÖºÍÈý×é·Öµç¼«ÎªÖ÷¡£¾¡¹ÜÔÚµç¼«ÖÆ±¸¡¢´ß»¯·´Ó¦»úÀíµÈ·½ÃæÈ¡µÃÁ˺ܴó½øÕ¹£¬µ«ÈÔÓÐÐí¶à¹¤×÷Òª×ö£¬ÆäÖÐÊ×ÒªµÄÎÊÌ⻹Êǵ缫µÄ´ß»¯»îÐԵ͡£ÔÚ¼îÐÔÌõ¼þϳ£ÓõÄPt¡¢PdºÍAuµç¼«ÉÏÒÒ´¼µç´ß»¯Ñõ»¯·´Ó¦ÖÐ(¼û¸½Í¼1)£¬Ptµç¼«ºÍPdµç¼«ÉÏÑõ»¯·´Ó¦Æðʼµçλ±È½Ï½Ó½ü£¬¶øPdµç¼«ÉϵķåµçÁ÷ÂÔ´óÓÚPtµç¼«ÉϵķåµçÁ÷£¬µ«ÈÔÊÇÒ»¸ö½ÏµÍµÄÖµ£»Auµç¼«Éϵķ´Ó¦µçλ±È½ÏÕý£¬²»ÀûÓÚȼÁÏµç³ØÉϵÄÀûÓ㬷åµçÁ÷ҲûÓÐÃ÷ÏÔÓÅÊÆ¡£±È½ÏÕâÈý¸öµç¼«µÄ»îÐÔ¿ÉÖª£¬îÙ»ùµç¼«¶ÔÒÒ´¼Ñõ»¯·´Ó¦¸üÓÐÓÅÊÆ£¬¶øÇÒPdµÄ¼Û¸ñ±ÈPtºÍAuµÍ¡£ÎªÁ˸Ľøµ¥Ò»½ðÊôµç¼«´ß»¯»îÐԵ͵ÄÎÊÌ⣬һ°ã²ÉÓÃÁ½½ðÊôÄÉÃ×Á£×ÓÐÎʽÀ´Ìá¸ßÆä´ß»¯»îÐÔ¡£µç¼«±íÃæÐÞÊÎÒ²ÊÇÒ»¸öÓÐЧµÄ·½·¨£¬¼´°Ñ»ùµ×µç¼«µÄ²¿·Ö±íÃæÓÃÆäËû½ðÊô½øÐÐÐÞÊΣ¬Í¨¹ýÎü¸½Ô×Ӻͻùµ×Ô×ÓÖ®¼äµÄÏ໥×÷Óã¬Ìá¸ßµç¼«»îÐԺͿ¹Öж¾ÄÜÁ¦¡£µ«µ½Ä¿Ç°ÎªÖ¹£¬ÓйØÔÚ½ðÊô»ùµ×±íÃæÓÃÒì½ðÊôÐÞÊεĵ缫ÉÏ´¼Ñõ»¯·´Ó¦µÄ±¨µÀ±È½ÏÉÙ¡£ÔÚ³£Óõļ¸ÖÖ±íÃæÐÞÊη¨ÖУ¬µç»¯Ñ§·¨ÔÚ¿ØÖÆÒì½ðÊôµÄ³Á»ýÁ¿ºÍ³Áµí½á¹¹·½Ãæ¾ßÓкܴóµÄÓÅÊÆ¡£Òò´Ë£¬ÎÒÃÇÑ¡Óõ绯ѧ·¨¶ÔPdµç¼«½øÐÐÁË AuÐÞÊκÍPtÐÞÊΣ¬ÖƱ¸ÁË AuÐÞÊεÄPdµç¼«ºÍPtÐÞÊεÄPdµç¼«(·Ö±ðÒÔAu(x)/PdºÍPtw/Pd±íʾ£¬ÆäÖÐXºÍy´ú±í³ÁµíÁ¿£¬µ¥Î»yg*cm_2£¬Ãæ»ýÒÔPd»ùµ×¼¸ºÎÃæ»ýΪ»ù×¼)£¬²¢ÓÃÓÚÒÒ´¼µÄµç´ß»¯Ñõ»¯·´Ó¦(¼û¸½Í¼2)¡£½á¹û±íÃ÷£¬Pt(Q.4Q)/Pdµç¼«ÉϵķåÓëPdµç¼«Éϵķå±È½Ï½Ó½ü£¬ËµÃ÷PtÐÞÊζÔPdµç¼«´ß»¯×÷ÓõÄÓ°Ïì±È½ÏС£»Au(a52)/Pdµç¼«ÉϵķåÓëPdµç¼«ÉϵķåÏà±È£¬·´Ó¦µçλ½Ó½ü£¬¶ø·åµçÁ÷ֵΪPdµç¼«·åµçÁ÷ÖµµÄ1.8±¶£¬ËµÃ÷AuÐÞÊÎÌá¸ßÁË Pdµç¼«´ß»¯»îÐÔ£¬µ«»îÐÔÔö´óµÄ·ù¶È±È½ÏÓÐÏÞ¡£ÎªÁ˽øÒ»²½Ìá¸ßµç¼«»îÐÔ£¬±¾·¢Ã÷ÓÃÁ½ÖÖÒì½ðÊôÐÞÊλùµ×½ðÊô£¬ÖƱ¸ÐÂÀàÐ͵ĵ缫£¬¼ìÑé¶ÔÒÒ´¼µÄµç´ß»¯»îÐÔ¡£µ½Ä¿Ç°ÎªÖ¹£¬ÓÃÁ½ÖÖÒì½ðÊôÐÞÊλùµ×½ðÊôÒÔ¼°ÕâÀàµç¼«ÔÚµç´ß»¯·´Ó¦ÖÐÓ¦ÓõÄÑо¿±¨µÀºÜÉÙ¡£
¡¾·¢Ã÷ÄÚÈÝ¡¿
[0004]±¾·¢Ã÷Òª½â¾öµÄ¼¼ÊõÎÊÌâÊÇ:Au¡¢PtºÍPdµç¼«¶ÔÒÒ´¼µç´ß»¯Ñõ»¯·´Ó¦»îÐÔµÍÒÔ¼°PtÐÞÊζÔPdµç¼«´ß»¯»îÐԸĽøÐ§¹û²î£¬AuÐÞÊζÔPdµç¼«»îÐԵĸĽø·ù¶ÈºÜÓÐÏÞµÄÎÊÌâ¡£
[0005]±¾·¢Ã÷½â¾öÆä¼¼ÊõÎÊÌâËù²ÉÓõÄÒ»¸ö¼¼Êõ·½°¸ÊÇ:ÌṩһÖÖÁ½½ðÊôÐÞÊε缫µÄ·½·¨£¬ÖƵõĵ缫ÓÃÓÚÒÒ´¼µÄµç´ß»¯Ñõ»¯·´Ó¦¡£°üÀ¨ÒÔϲ½Öè:
[0006]a¡¢ÅäÖÆ½ðǰÌåÈÜÒº¡¢îâǰÌåÈÜÒº:
[0007]½ðǰÌåÈÜÒº:ÓÃÂȽðËá¡¢ÁòËáºÍ¶þ´ÎÕôÁóË®ÅäÖÆ£»
[0008]îâǰÌåÈÜÒº:ÓÃÂÈîâËá¡¢ÁòËáºÍ¶þ´ÎÕôÁóË®ÅäÖÆ¡£
[0009]b¡¢µç¼«±íÃæÐÞÊÎ:
[0010]ÒÔPtƬΪ¶Ôµç¼«£¬¸Ê¹¯µç¼«Îª²Î±Èµç¼«£¬ÔÚºãµçλÌõ¼þÏ´ӽðǰÌåÈÜÒº»¹Ô³ÁµíAuµ½Pdµç¼«±íÃæ,ͨ¹ý¿ØÖƳÁµíʱ¼äµÃµ½¾ßÓв»Í¬AuÐÞÊÎÁ¿µÄPdµç¼«£¬¼´Au(x)/Pdµç¼«£¬ÆäÖÐX´ú±í³ÁµíÁ¿£¬XֵΪ0.26?2.45£¬µ¥Î»¦Ì g.cm_2£¬Ãæ»ýÒÔPd»ùµ×¼¸ºÎÃæ»ýΪ»ù×¼£»
[0011]ÒÔPtƬΪ¶Ôµç¼«£¬¸Ê¹¯µç¼«Îª²Î±Èµç¼«£¬ÔÚºãµçλÌõ¼þÏ´ÓîâǰÌåÈÜÒº»¹Ô³ÁµíPtµ½Au(x)/Pdµç¼«±íÃæ£¬Í¨¹ý¿ØÖƳÁµíʱ¼äµÃµ½¾ßÓв»Í¬PtÐÞÊÎÁ¿µÄAu(x)/Pdµç¼«£¬¼´Pt(y)-Au(x)/Pdµç¼«£¬ÆäÖÐXºÍy´ú±í³ÁµíÁ¿£¬XֵΪ0.26?2.45£¬yֵΪ0.20?1.25£¬µ¥Î»¦Ì g.cm_2£¬Ãæ»ýÒÔPd»ùµ×¼¸ºÎÃæ»ýΪ»ù×¼£»
[0012]C¡¢µç¼«ÓÃÓÚ´¼µÄµç´ß»¯Ñõ»¯·´Ó¦:
[0013]ÒÔ²½ÖèbÖÆ±¸µÄPt(y)_Au(x)/Pdµç¼«×÷Ϊ¹¤×÷µç¼«£¬PtƬΪ¶Ôµç¼«£¬¸Ê¹¯µç¼«Îª²Î±Èµç¼«£¬ÔÚº¬ÓÐÒÒ´¼µÄÇâÑõ»¯ÄÆÈÜÒºÖнøÐÐÑ»··ü°²²â¶¨¡£
[0014]½øÒ»²½µØ£¬²½ÖèaÖÐÂȽðËáµÄŨ¶ÈΪ0.2?1.0mmol.dm_3,ÁòËáµÄŨ¶ÈΪ
0.05mol.dm_3,ÂÈîâËáµÄŨ¶ÈΪ 0.2 ?1.0mmol.dm_3¡£
[0015]½øÒ»²½µØ£¬²½ÖèbÖÐAu³ÁµíºÍPt³ÁµíʱµÄºãµçλÔÚ0.15?0.25V vs.SCE·¶Î§ÄÚȡһ¶¨Öµ£¬ÆäÖÐAu³Áµíʱ¼äΪI?10Ã룬Pt³Áµíʱ¼äΪI?5Ãë¡£
[0016]½øÒ»²½µØ£¬²½ÖècÖÐÒÒ´¼Å¨¶ÈΪ0.1?0.5mol.dm_3£¬ÇâÑõ»¯ÄÆÅ¨¶ÈΪ0.1?
1.0mol.dm 3£¬Ñ»··ü°²É¨ÃèËÙ¶ÈΪ20?50mV.s 1O
[0017]±¾·¢Ã÷µÄÓÐÒæÐ§¹ûÊÇ:±¾·¢Ã÷Öеĵ缫ÐÞÊη½·¨¼òµ¥£¬µç¼«Îȶ¨£¬Í¨¹ý¼«Ð¡Á¿µÄAuºÍPtÐÞÊÎËùµÃµ½µÄPdµç¼«ÔÚ¼îÐÔÈÜÒºÖжÔÒÒ´¼Ñõ»¯·´Ó¦µÄµç´ß»¯»îÐÔÔ¶¸ßÓÚAu¡¢PtºÍPdµç¼«ÒÔ¼°AuÐÞÊεÄPdµç¼«ºÍPtÐÞÊεÄPdµç¼«£¬½â¾öÁË Au¡¢PtºÍPdµç¼«¶ÔÒÒ´¼µç´ß»¯Ñõ»¯·´Ó¦»îÐԵ͵ÄÎÊÌâÒÔ¼°PtÐÞÊζÔPdµç¼«´ß»¯»îÐԸĽøÐ§¹û²î£¬AuÐÞÊζÔPdµç¼«»îÐԵĸĽø·ù¶ÈºÜÓÐÏÞµÄÎÊÌ⣻ÓÉÓÚAuºÍPtÐÞÊÎÁ¿¼«ÉÙ£¬½ÚÊ¡ÁË AuºÍPtµÄʹÓÃÁ¿£»ËùÖÆ±¸µÄµç¼«¶Ô¼îÐÔ½éÖÊÖеÄÒÒ´¼Ñõ»¯·´Ó¦¾ßÓкܸߵĴ߻¯»îÐÔ£¬ÐÞÊÎЧ¹û·Ç³£Ã÷ÏÔ¡£
¡¾×¨Àû¸½Í¼¡¿
¡¾¸½Í¼ËµÃ÷¡¿
[0018]ÏÂÃæ½áºÏ¸½Í¼¶Ô±¾·¢Ã÷½øÒ»²½ËµÃ÷¡£
[0019]ͼ1Ϊº¬0.1mol.dnf3ÒÒ´¼µÄ0.5mol.dnf 3NaOHÈÜÒºÖÐAu, PtºÍPdµç¼«µÄÑ»··ü°²ÇúÏߣ¬É¨ÃèËÙ¶È:50mV.s ¡ª 1 ;
[0020]ͼ2 Ϊº¬ 0.1mol.dm 3 ÒÒ´¼µÄ 0.5mol.dm 3NaOH ÈÜÒºÖÐ Pt((l.4(I)/Pd ºÍ Au((l.52)/Pdµç¼«µÄÑ»··ü°²ÇúÏߣ¬É¨ÃèËÙ¶È:50mV.s ¡ª 1 ;
[0021]ͼ3 Ϊº¬ 0.1mol.dm 3 ÒÒ´¼µÄ 0.5mol.dm 3NaOH ÈÜÒºÖÐ Pt(0.20)_Au(0.ffi)/Pd ºÍPt(0.20)-Au(0.52)/Pdµç¼«µÄÑ»··ü°²ÇúÏߣ¬É¨ÃèËÙ¶È:50mV.s ¡ª 1 ;
[0022]ͼ4ΪPt(Q.2Q)-Auftl52)/Pdµç¼«±íÃæµÄɨÃèµç¾µÍ¼¡£
¡¾¾ßÌåʵʩ·½Ê½¡¿
[0023]ÏÖÔÚ½áºÏ¾ßÌåʵʩÀý¶Ô±¾·¢Ã÷×÷½øÒ»²½ËµÃ÷£¬ÒÔÏÂʵʩÀýÖ¼ÔÚ˵Ã÷±¾·¢Ã÷¶ø²»ÊǶԱ¾·¢Ã÷µÄ½øÒ»²½ÏÞ¶¨¡£
[0024]ʵʩÀýÒ»:
[0025](I)ÏȺóÈ¡2 OmL ¶þ´ÎÕô‘ÖË®¡¢200 ¦Ì L0.1mol.dnf3ÂȽðËáºÍ53.2 ¦Ì LŨÁòËá(´¿¶È98.0% )¼ÓÈëµ½Ò»¸ö¸ÉÔï½à¾»µÄ50mLÉÕ±ÖÐÅäÖÆ½ðǰÌåÈÜÒº,ÆäÖк¬Immol.?!¦Ð¦£3ÂȽðËáºÍ0.05mol.dm 3ÁòËá¡£
[0026](2)ÏȺóÈ¡20mL ¶þ´ÎÕôÁóË®¡¢200 ¦Ì L0.1mol.dnf3ÂÈîâËáºÍ53.2 ¦Ì LŨÁòËá(´¿¶È98.0% )¼ÓÈëµ½Ò»¸ö¸ÉÔï½à¾»µÄ50mLÉÕ±ÖÐÅäÖÆîâǰÌåÈÜÒº,ÆäÖк¬lmmol *dm-3ÂÈîâËáºÍ0.05mol.dm 3ÁòËá¡£
[0027](3)ÏȺóÈ¡20mL ¶þ´ÎÕôÁóË®¡¢118 ¦Ì LÒÒ´¼(´¿¶Èáê99.7% )ºÍ0.4gÇâÑõ»¯ÄÆ(´¿¶Èáê96.0 % )¼ÓÈëµ½Ò»¸ö¸ÉÔï½à¾»µÄ50mLÉÕ±ÖÐÅäÖÆ´¼ÈÜÒº£¬ÆäÖк¬0.1mol.dm_3ÒÒ´¼ºÍ0.5mol.dm 3ÇâÑõ»¯ÄÆ¡£
[0028](4)½«ÉÏÊöÖÆ±¸µÄÈýÖÖÈÜÒºÖÐͨÈëµªÆø£¬ÅųöÆäÖÐÈܽâµÄÑõÆø¡£
[0029](5)ÏȺóÓÃÁ£¾¶0.35 ¦Ì mºÍ0.03 ¦Ì mÑõ»¯ÂÁ·ÛÄ©Äཬ´òÄ¥Pdµç¼«±íÃæ£¬²¢Óöþ´ÎÕôÁóË®³¬Éù²¨Ï´µÓ¡£
[0030](6)½«Pdµç¼«ÖÃÓÚ½ðǰÌåÈÜÒºÖУ¬ÒÔPtƬΪ¶Ôµç¼«£¬SCEΪ²Î±Èµç¼«£¬ÔÚµçλ+0.2V vs.SCEϽøÐкãµçλ»¹Ô³Áµí½ð,³Áµíʱ¼äIÃë¡£Au³ÁµíÁ¿Îª0.26 ¦Ì g.¦Á¦É¦É_2,µÃµ½Au (¦Ï.26)/Pd µç¼«¡£
[0031](7)½«ÖƱ¸ºÃµÄAu(Q.26)/Pdµç¼«Óöþ´ÎÕôÁóË®ÇåÏ´±íÃæ£¬ÖÃÓÚîâǰÌåÈÜÒºÖУ¬ÒÔPtƬΪ¶Ôµç¼«£¬SCEΪ²Î±Èµç¼«£¬ÔÚµçλ+0.2V vs.SCEϽøÐкãµçλ»¹Ô³Áµíî⣬³Áµíʱ¼äIÃë¡£Pt ³ÁµíÁ¿Îª 0.20¦Ì g.cnT2,µÃµ½ Pt(Q.2Q)-Auftl26)/Pd µç¼«¡£
[0032](8)½«ÖƱ¸ºÃµÄPt((l.2CI)-Au(a26)/Pdµç¼«Óöþ´ÎÕôÁóË®ÇåÏ´±íÃæ£¬È»ºóÖÃÓÚº¬ÒÒ´¼µÄÇâÑõ»¯ÄÆÈÜÒºÖнøÐÐÑ»··ü°²·¨²â¶¨£¬½á¹ûÈçͼ3Ëùʾ¡£Pt((l.2CI)-Au(a26)/Pdµç¼«ÉÏÒÒ´¼Ñõ»¯·´Ó¦µÄ·åµçÁ÷ֵΪPdµç¼«ÉϵķåµçÁ÷ÖµµÄ3.7±¶£¬¶ø·´Ó¦µçλ(Ñõ»¯ÆðʼµçλºÍ·åµçλ)ÓëPdµç¼«Éϵķ´Ó¦µçλ»ù±¾Ò»Ö¡£Õâ¸ö½á¹û±íÃ÷£¬Ptftl 2c0-Auftl 26)/Pdµç¼«µÄ´ß»¯»îÐÔÔ¶¸ßÓÚAu¡¢PtºÍPdµç¼«µÄ´ß»¯»îÐÔ¡£
[0033]ʵʩÀý¶þ:
[0034](I)ÏȺóÈ¡20mL ¶þ´ÎÕôÁóË®¡¢200 ¦Ì L0.1mol.dm_3ÂȽðËáºÍ53.2 ¦Ì LŨÁòËá(´¿¶È98.0% )¼ÓÈëµ½Ò»¸ö¸ÉÔï½à¾»µÄ50mLÉÕ±ÖÐÅäÖÆ½ðǰÌåÈÜÒº,ÆäÖк¬Immol.?!¦Ð¦£3ÂȽðËáºÍ0.05mol.dm 3ÁòËá¡£
[0035](2)ÏȺóÈ¡20mL ¶þ´ÎÕôÁóË®¡¢200 ¦Ì L0.1mol.dnf3ÂÈîâËáºÍ53.2 ¦Ì LŨÁòËá(´¿¶È98.0% )¼ÓÈëµ½Ò»¸ö¸ÉÔï½à¾»µÄ50mLÉÕ±ÖÐÅäÖÆîâǰÌåÈÜÒº,ÆäÖк¬lmmol *dm-3ÂÈîâËáºÍ0.05mol.dm 3ÁòËá¡£
[0036](3)ÏȺóÈ¡20mL ¶þ´ÎÕôÁóË®¡¢118 ¦Ì LÒÒ´¼(´¿¶Èáê99.7% )ºÍ0.4gÇâÑõ»¯ÄÆ(´¿¶Èáê96.0 % )¼ÓÈëµ½Ò»¸ö¸ÉÔï½à¾»µÄ50mLÉÕ±ÖÐÅäÖÆ´¼ÈÜÒº£¬ÆäÖк¬0.1mol.dm_3ÒÒ´¼ºÍ0.5mol.dm 3ÇâÑõ»¯ÄÆ¡£
[0037](4)½«ÉÏÊöÖÆ±¸µÄÈýÖÖÈÜÒºÖÐͨÈëµªÆø£¬ÅųöÆäÖÐÈܽâµÄÑõÆø¡£
[0038](5)ÏȺóÓÃÁ£¾¶0.35 ¦Ì mºÍ0.03 ¦Ì mÑõ»¯ÂÁ·ÛÄ©Äཬ´òÄ¥Pdµç¼«±íÃæ£¬²¢Óöþ´ÎÕôÁóË®³¬Éù²¨Ï´µÓ¡£
[0039](6)½«Pdµç¼«ÖÃÓÚ½ðǰÌåÈÜÒºÖУ¬ÒÔPtƬΪ¶Ôµç¼«£¬SCEΪ²Î±Èµç¼«£¬ÔÚµçλ+0.2V vs.SCEϽøÐкãµçλ»¹Ô³Áµí½ð,³Áµíʱ¼ä3Ãë¡£Au³ÁµíÁ¿Îª0.56 ¦Ì g.¦Á¦É¦É_2,µÃµ½Au (¦Ï.52)/Pd µç¼«¡£
[0040](7)½«ÖƱ¸ºÃµÄAu(a52)/Pdµç¼«Óöþ´ÎÕôÁóË®ÇåÏ´±íÃæ£¬ÖÃÓÚîâǰÌåÈÜÒºÖУ¬ÒÔPtƬΪ¶Ôµç¼«£¬SCEΪ²Î±Èµç¼«£¬ÔÚµçλ+0.2V vs.SCEϽøÐкãµçλ»¹Ô³Áµíî⣬³Áµíʱ¼äIÃë¡£Pt³ÁµíÁ¿Îª0.20¦Ì g.¦Á¦É¦£2,µÃµ½Pt((l.2Q)-Au(a52)/Pdµç¼«¡£ÓÉͼ4¿´µ½ÔÚPdµç¼«±íÃæÉÏÉú³ÉÁ˲»Í¬´óСµÄ½ðÊôÄÉÃ×Á£×Ó£¬ÆäÁ£¾¶´óС·Ö²¼±È½Ï¾ùÔÈ¡£ÓÉÓÚ½ðÊô³ÁµíÁ¿ºÜС£¬´ó²¿·ÖµÄPd»ùµ×±íÃæÃ»Óб»³ÁµíµÄ½ðÊôËù¸²¸Ç¡£ÁíÍ⣬Pdµç¼«±íÃæÓÐЩ´Ö²Ú²»Æ½£¬ÕâÊÇÒòΪPdµç¼«±íÃæÔÚÐÞÊÎǰ¾¹ýÁËÑõ»¯ÂÁ·ÛÄ©Äཬ´òÄ¥¡£
[0041](8)½«ÖƱ¸ºÃµÄPt((l.2CI)-Au(a52)/Pdµç¼«Óöþ´ÎÕôÁóË®ÇåÏ´±íÃæ£¬È»ºóÖÃÓÚº¬ÒÒ´¼µÄÇâÑõ»¯ÄÆÈÜÒºÖнøÐÐÑ»··ü°²·¨²â¶¨£¬½á¹ûÈçͼ3Ëùʾ¡£Pt((l.2CI)-Au(a52)/Pdµç¼«ÉÏÒÒ´¼Ñõ»¯·´Ó¦µÄ·åµçÁ÷ֵΪPdµç¼«ÉϵķåµçÁ÷ÖµµÄ4.5±¶£¬¶ø·´Ó¦µçλ(Ñõ»¯ÆðʼµçλºÍ·åµçλ)ÓëPdµç¼«Éϵķ´Ó¦µçλ»ù±¾Ò»Ö¡£Í¼3»¹±íÃ÷£¬Ptftl2c0-Auftl.52)/Pdµç¼«µÄ»îÐÔÂÔ¸ßÓÚPt(0.20)-Au(0.26)/Pd µç¼«µÄ»îÐÔ¡£
[0042]ÒÔÉÏÊöÒÀ¾Ý±¾·¢Ã÷µÄÀíÏëʵʩÀýΪÆôʾ£¬Í¨¹ýÉÏÊöµÄ˵Ã÷ÄÚÈÝ£¬Ïà¹Ø¹¤×÷ÈËÔ±ÍêÈ«¿ÉÒÔÔÚ²»Æ«Àë±¾Ïî·¢Ã÷¼¼Êõ˼ÏëµÄ·¶Î§ÄÚ£¬½øÐжàÑùµÄ±ä¸üÒÔ¼°Ð޸ġ£±¾Ïî·¢Ã÷µÄ¼¼ÊõÐÔ·¶Î§²¢²»¾ÖÏÞÓÚ˵Ã÷ÊéÉϵÄÄÚÈÝ£¬±ØÐëÒª¸ù¾ÝȨÀûÒªÇó·¶Î§À´È·¶¨Æä¼¼ÊõÐÔ·¶Î§¡£
¡¾È¨ÀûÒªÇó¡¿
1.Ò»ÖÖÁ½½ðÊôÐÞÊε缫µÄ·½·¨£¬ÖƵõĵ缫ÓÃÓÚ´¼µÄµç´ß»¯Ñõ»¯·´Ó¦£¬ÆäÌØÕ÷ÊÇ:°üÀ¨ÒÔϲ½Öè: a¡¢ÅäÖÆ½ðǰÌåÈÜÒº¡¢îâǰÌåÈÜÒº: ½ðǰÌåÈÜÒº:ÓÃÂȽðËá¡¢ÁòËáºÍ¶þ´ÎÕôÁóË®ÅäÖÆ£» îâǰÌåÈÜÒº:ÓÃÂÈîâËá¡¢ÁòËáºÍ¶þ´ÎÕôÁóË®ÅäÖÆ£» b¡¢µç¼«±íÃæÐÞÊÎ: ÒÔPtƬΪ¶Ôµç¼«£¬¸Ê¹¯µç¼«Îª²Î±Èµç¼«£¬ÔÚºãµçλÌõ¼þÏ´ӽðǰÌåÈÜÒº»¹Ô³ÁµíAuµ½Pdµç¼«±íÃæ£¬Í¨¹ý¿ØÖƳÁµíʱ¼äµÃµ½¾ßÓв»Í¬AuÐÞÊÎÁ¿µÄPdµç¼«£¬¼´Au(x)/Pdµç¼«£¬ÆäÖÐX´ú±í³ÁµíÁ¿£¬XֵΪ0.26?2.45£¬µ¥Î»¦Ì g.cm-2,Ãæ»ýÒÔPd»ùµ×¼¸ºÎÃæ»ýΪ»ù×¼£» ÒÔPtƬΪ¶Ôµç¼«£¬¸Ê¹¯µç¼«Îª²Î±Èµç¼«£¬ÔÚºãµçλÌõ¼þÏ´ÓîâǰÌåÈÜÒº»¹Ô³ÁµíPtµ½Au(x)/Pdµç¼«±íÃæ,ͨ¹ý¿ØÖƳÁµíʱ¼äµÃµ½¾ßÓв»Í¬PtÐÞÊÎÁ¿µÄAu(x)/Pdµç¼«£¬¼´Pt(y)-Au(x)/Pdµç¼«£¬ÆäÖÐXºÍy´ú±í³ÁµíÁ¿£¬XֵΪ0.26?2.45£¬yֵΪ0.20?1.25£¬µ¥Î»¦Ì g.cm_2£¬Ãæ»ýÒÔPd»ùµ×¼¸ºÎÃæ»ýΪ»ù×¼£» C¡¢µç¼«ÓÃÓÚ´¼µÄµç´ß»¯Ñõ»¯·´Ó¦: ÒÔ²½ÖèbÖÆ±¸µÄPt(y)-Au(x)/Pdµç¼«×÷Ϊ¹¤×÷µç¼«£¬PtƬΪ¶Ôµç¼«£¬¸Ê¹¯µç¼«Îª²Î±Èµç¼«£¬ÔÚº¬ÓÐÒÒ´¼µÄÇâÑõ»¯ÄÆÈÜÒºÖнøÐÐÑ»··ü°²²â¶¨¡£
2.¸ù¾ÝȨÀûÒªÇó1ËùÊöµÄÒ»ÖÖÁ½½ðÊôÐÞÊε缫µÄ·½·¨£¬ÆäÌØÕ÷ÊÇ:ËùÊöµÄ²½ÖèaÖÐÂȽðËáµÄŨ¶ÈΪ0.2?1.0mmol.dnf3£¬ÂÈîâËáµÄŨ¶ÈΪ0.2?1.0mmol.dnf3£¬ÁòËáµÄŨ¶ÈΪ 0.05mol.dm 3¡£
3.¸ù¾ÝȨÀûÒªÇó1ËùÊöµÄÒ»ÖÖÁ½½ðÊôÐÞÊε缫µÄ·½·¨£¬ÆäÌØÕ÷ÊÇ:ËùÊöµÄ²½ÖèbÖÐAu³ÁµíʱµÄºãµçλÔÚ0.15?0.25V vs.SCE·¶Î§ÄÚȡһ¶¨Öµ£¬³Áµíʱ¼äΪI?10Ã룬Pt³ÁµíʱµÄºãµçλÔÚ0.15?0.25V vs.SCE·¶Î§ÄÚȡһ¶¨Öµ£¬³Áµíʱ¼äΪI?5Ãë¡£
4.¸ù¾ÝȨÀûÒªÇó1ËùÊöµÄÒ»ÖÖÁ½½ðÊôÐÞÊε缫µÄ·½·¨£¬ÆäÌØÕ÷ÊÇ:ËùÊöµÄ²½ÖècÖÐÒÒ´¼Å¨¶ÈΪ0.1?0.5mol.dnf3,ÇâÑõ»¯ÄÆÅ¨¶ÈΪ0.1?1.0mol.dnf3,Ñ»··ü°²É¨ÃèËÙ¶ÈΪ.20 ?50mV.s¡¢
¡¾Îĵµ±àºÅ¡¿G01N27/48GK104132977SQ201410329197
¡¾¹«¿ªÈÕ¡¿2014Äê11ÔÂ5ÈÕ ÉêÇëÈÕÆÚ:2014Äê7ÔÂ10ÈÕ ÓÅÏÈȨÈÕ:2014Äê7ÔÂ10ÈÕ
¡¾·¢Ã÷Õß¡¿½ð³¤´º, ÍõÖÒÓî, ¶ÈçÁÖ ÉêÇëÈË:³£ÖÝ´óѧ