Ó¦Óõ绯ѧ£¬Ñî»Ô¬ÎÄÇì È«Êé˼¿¼ÌâºÍϰÌâ µÚÒ»ÕÂϰÌâ½â´ð£º
1ÊÔÍÆµ¼ÏÂÁи÷µç¼«·´Ó¦µÄÀàÐͼ°µç¼«·´Ó¦µÄ¹ý³Ì¡£ (1)Ce4??2e?Ce2?
½â£ºÊôÓÚ¼òµ¥Àë×ÓµçÇ¨ÒÆ·´Ó¦£¬Ö¸µç¼«/ÈÜÒº½çÃæµÄÈÜÒºÒ»²àµÄÑõ»¯Ì¬ÎïÖÖCe4?½èÖúÓڵ缫µÃµ½µç×Ó£¬Éú³É»¹Ô̬µÄÎïÖÖCe2?¶øÈܽâÓÚÈÜÒº
ÖУ¬¶øµç¼«ÔÚ¾ÀúÑõ»¯-»¹ÔºóÆäÎïÀí»¯Ñ§ÐÔÖʺͱíÃæ×´Ì¬µÈ²¢Î´·¢Éú±ä»¯£¬ (2)
O2?2H2O?4e?4OH?
½â£º¶à¿×ÆøÌåÀ©É¢µç¼«ÖÐµÄÆøÌ廹Է´Ó¦¡£ÆøÏàÖÐµÄÆøÌåO2ÈܽâÓÚÈÜÒººó£¬ÔÙÀ©É¢µ½µç¼«±íÃæ£¬È»ºó½èÖúÓÚÆøÌåÀ©É¢µç¼«µÃµ½µç×Ó£¬ÆøÌåÀ©É¢µç¼«µÄʹÓÃÌá¸ßÁ˵缫¹ý³ÌµÄµçÁ÷ЧÂÊ¡£ (3)
Ni2??2e?Ni
2?½â£º½ðÊô³Á»ý·´Ó¦¡£ÈÜÒºÖеĽðÊôÀë×ÓNi´Óµç¼«Éϵõ½µç×Ó»¹
ÔΪ½ðÊôNi£¬¸½×ÅÓڵ缫±íÃæ£¬´Ëʱµç¼«±íÃæ×´Ì¬Óë³Á»ýǰÏà±È·¢ÉúÁ˱仯¡£ (4)
MnO2(s)?e?H2O?MnOOH(s)?OH?
½â£º±íÃæÄ¤µÄ×ªÒÆ·´Ó¦¡£¸²¸ÇÓڵ缫±íÃæµÄÎïÖÖ(µç¼«Ò»²à)¾¹ýÑõ»¯-»¹ÔÐγÉÁíÒ»ÖÖ¸½×ÅÓڵ缫±íÃæµÄÎïÖÖ£¬ËüÃÇ¿ÉÄÜÊÇÑõ»¯Îï¡¢ÇâÑõ»¯Îï¡¢ÁòËáÑεȡ£
(5)
Zn?2OH??2e?Zn(OH)2£»
Zn(OH)2?2OH??[Zn(OH)4]2?
½â£º¸¯Ê´·´Ó¦£ºÒ༴½ðÊôµÄÈܽⷴӦ£¬µç¼«µÄÖØÁ¿²»¶Ï¼õÇá¡£¼´½ðÊôпÔÚ¼îÐÔ½éÖÊÖз¢ÉúÈܽâÐγɶþôÇ»ùºÏ¶þ¼ÛпÂçºÏÎËùÐγɵĶþôÇ»ùºÏ¶þ¼ÛпÂçºÏÎïÓÖºÍôÇ»ù½øÒ»²½ÐγÉËÄôÇ»ùºÏ¶þ¼ÛпÂçºÏÎï¡£
2£®ÊÔ˵Ã÷²Î±Èµç¼«Ó¦¾ßÓеÄÐÔÄܺÍÓÃ;¡£
²Î±Èµç¼«(reference electrode£¬¼ò³ÆRE)£ºÊÇÖ¸Ò»¸öÒÑÖªµçÊÆµÄ½Ó½üÓÚÀíÏë²»¼«»¯µÄµç¼«£¬²Î±Èµç¼«ÉÏ»ù±¾Ã»ÓеçÁ÷ͨ¹ý£¬ÓÃÓڲⶨÑо¿µç¼«(Ïà¶ÔÓڲαȵ缫)µÄµç¼«µçÊÆ¡£
¼ÈÈ»²Î±Èµç¼«ÊÇÀíÏë²»¼«»¯µç¼«£¬ËüÓ¦¾ß±¸ÏÂÁÐÐÔÄÜ£ºÓ¦ÊÇ¿ÉÄæµç¼«£¬Æäµç¼«µçÊÆ·ûºÏNernst·½³Ì£»²Î±Èµç¼«·´Ó¦Ó¦ÓнϴóµÄ½»»»µçÁ÷Ãܶȣ¬Á÷¹ý΢СµÄµçÁ÷ʱµç¼«µçÊÆÄÜѸËÙ»Ö¸´Ô×´£»Ó¦¾ßÓÐÁ¼ºÃµÄµçÊÆÎȶ¨ÐÔºÍÖØÏÖÐԵȡ£
1
²»Í¬Ñо¿Ìåϵ¿ÉÒÔÑ¡Ôñ²»Í¬µÄ²Î±Èµç¼«£¬Ë®ÈÜÒºÌåϵÖг£¼ûµÄ²Î±Èµç¼«ÓУº±¥ºÍ¸Ê¹¯µç¼«(SCE)¡¢Ag/AgClµç¼«¡¢±ê»´Çâµç¼«(SHE»òNHE)µÈ¡£Ðí¶àÓлúµç»¯Ñ§²âÁ¿ÊÇÔÚ·ÇË®ÈܼÁÖнøÐе쬾¡¹ÜË®ÈÜÒº²Î±Èµç¼«Ò²¿ÉÒÔʹÓ㬵«²»¿É±ÜÃâµØ»á¸øÌåϵ´øÈëË®·Ö£¬Ó°ÏìÑо¿Ð§¹û£¬Òò´Ë£¬½¨Òé×îºÃʹÓ÷ÇË®²Î±ÈÌåϵ¡£³£ÓõķÇË®²Î±ÈÌåϵΪAg/Ag+(ÒÒëæ)¡£¹¤ÒµÉϳ£Ó¦ÓüòÒײαȵ缫£¬»òÓø¨Öúµç¼«¼æ×ö²Î±Èµç¼«¡£ÔÚ²âÁ¿¹¤×÷µç¼«µÄµçÊÆÊ±£¬²Î±Èµç¼«ÄÚµÄÈÜÒººÍ±»Ñо¿ÌåϵµÄÈÜÒº×é³ÉÍùÍù²»¡ªÑù£¬Îª½µµÍ»òÏû³ýÒº½ÓµçÊÆ£¬³£Ñ¡ÓÃÑÎÇÅ£»Îª¼õСĩ²¹³¥µÄÈÜÒºµç×裬³£Ê¹Óó½ðëϸ¹Ü¡£
3£®ÊÔÃèÊöË«µç²ãÀíÂ۵ĸÅÒª¡£
½â£ºµç¼«/ÈÜÒº½çÃæÇøµÄ×îÔçÄ£ÐÍÊÇ19ÊÀ¼ÍÄ©HelmholtzÌá³öµÄƽ°åµçÈÝÆ÷Ä£ÐÍ(Ò²³Æ½ôÃܲãÄ£ÐÍ)£¬ËûÈÏΪ½ðÊô±íÃæ¹ýÊ£µÄµçºÉ±ØÐë±»ÈÜÒºÏàÖп¿½üµç¼«±íÃæµÄ´øÏà·´µçºÉµÄÀë×Ó²ãËùÖкͣ¬Á½¸öµçºÉ²ã¼äµÄ¾àÀëÔ¼µÈÓÚÀë×Ó°ë¾¶£¬Èçͬһ¸öƽ°åµçÈÝÆ÷¡£ÕâÖÖÓÉ·ûºÅÏà·´µÄÁ½¸öµçºÉ²ã¹¹³ÉµÄ½çÃæÇøµÄ¸ÅÄ±ãÊÇ¡°Ë«µç²ã¡±Ò»´ÊµÄÆðÔ´¡£
¼ÌHelmholtzÖ®ºó£¬GouyºÍChapmanÔÚ1913Ä겻ı¶øºÏµØÌá³öÁËÀ©É¢Ë«µç²ãÄ£ÐÍ¡£ËûÃÇ¿¼Âǵ½½çÃæÈÜÒº²àµÄÀë×Ó²»½öÊܽðÊôÉϵçºÉµÄ¾²µç×÷Ó㬶øÇÒÊÜÈÈÔ˶¯µÄÓ°Ï죬Òò´Ë£¬µç¼«±íÃæ¸½½üÈÜÒº²ãÖеÄÀë×ÓŨ¶ÈÊÇÑØ×ÅÔ¶Àëµç¼«µÄ·½ÏòÖ𽥱仯µÄ£¬Ö±µ½×îºóÓëÈÜÒº±¾Ìå³Ê¾ùÔÈ·Ö²¼¡£¸ÃÄ£ÐÍÈÏΪÔÚÈÜÒºÖÐÓëµç¼«±íÃæÀë×ÓµçºÉÏà·´µÄÀë×ÓÖ»ÓÐÒ»²¿·Ö½ôÃܵØÅÅÁÐÔڵ缫/ÈÜÒº½çÃæµÄÈÜÒºÒ»²à(³Æ½ôÃܲ㣬²ã¼ä¾àÀëԼΪһ¡¢¶þ¸öÀë×ӵĺñ¶È)£¬ÁíÒ»²¿·ÖÀë×ÓÓëµç¼«±íÃæµÄ¾àÀëÔò¿ÉÒÔ´Ó½ôÃܲãÒ»Ö±·ÖÉ¢µ½±¾ÌåÈÜÒºÖÐ(³ÆÀ©É¢²ã)£¬ÔÚÀ©É¢²ãÖÐÀë×ӵķֲ¼¿ÉÓò£¶û×ÈÂü·Ö²¼¹«Ê½±íʾ¡£Gouy-ChapmanÄ£Ð͵ÄȱµãÊǺöÂÔÁËÀë×ӵijߴ磬°ÑÀë×ÓÊÓΪµãµçºÉ£¬Ö»ÄÜ˵Ã÷¼«Ï¡µç½âÖÊÈÜÒºµÄʵÑé½á¹û¡£
ÓÉÓÚº¥Ä·»ô´ÄÄ£Ðͺ͹ÅÒÀ¡ª²éÆÕÂüÄ£ÐͶ¼Óв»×ãÖ®´¦£¬1924Ä꣬SternÎüÈ¡ÁËHelmholtzÄ£ÐͺÍGouy-ChapmanÄ£Ð͵ĺÏÀíÒòËØ£¬Ìá³öÕû¸öË«µç²ãÊdzö½ôÃܲãºÍÀ©É¢²ã×é³ÉµÄ£¬´Ó¶øÊ¹ÀíÂÛ¸ü¼ÓÇкÏʵ¼Ê¡£Stern»¹Ö¸³öÀë×ÓÌØÐÔÎü¸½µÄ¿ÉÄÜÐÔ£¬¿ÉÊÇûÓп¼ÂÇËü¶ÔË«µç²ã½á¹¹µÄÓ°Ïì¡£
ĿǰÆÕ±é¹«ÈϵÄÊÇÔÚGCSÄ£ÐÍ»ù´¡ÉÏ·¢Õ¹ÆðÀ´µÄBDM
2
£¨Bockris-Davanathan-muller£©Ä£ÐÍ×î¾ßÓдú±íÐÔ£¬ÆäÒªµãÈçÏ¡£
µç¼«/ÈÜÒº½çÃæµÄË«µç²ãµÄÈÜÒºÒ»²à±»ÈÏΪÊÇÓÉÈô¸É¡±²ã¡±×é³ÉµÄ¡£×î¿¿½üµç¼«µÄÒ»²ãΪÄڲ㣬Ëü°üº¬ÓÐÈܼÁ·Ö×ÓºÍËùνµÄÌØÐÔÎü¸½µÄÎïÖÊ(Àë×ӳɷÖ×Ó)£¬ÕâÖÖÄÚ²ãÒ²³ÆÎª½ôÃܲ㡢Helmholtz²ã»òStern²ã¡£
4£®¸ù¾Ýµç¼«·´Ó¦Ox?ne£½10-7cm¡¤s-1£¬?£½03£¬nRed£¬ÒÑÖª£ºcR£½cOx£½1mmol¡¤Ll£¬k¡ª
**??1£»(1)¼ÆËã½»»»µçÁ÷ÃܶÈj0?I0/A(ÒÔ
?A?cm?2±íʾ)£»(2)ÊÔ»³öÑô¼«µçÁ÷ºÍÒõ¼«µçÁ÷ÔÚ600?A?cm?2·¶Î§
ÄÚµÄTafelÇúÏß(lg½â£º£¨1£©
I??)¡£T=298K£¬ºöÂÔÎïÖÊ´«µÝµÄÓ°Ïì¡£
*(1??)*?j0?I0/A?zFk0cOXCR
-1
-7
-1
¡ªl
0.7
£½1¡Á96500 £Ã¡¤mol¡Á10 cm¡¤s¡Á(1 mmol¡¤L)¡Á(1 mmol¡¤L)
¡ªl0.3
-7
-2
-1
-6
-9
-2
-1
=1¡Á96500 £Ã¡Á10 cm¡¤s¡Á1¡Á10£½9.65¡Á10c ¡¤cm¡¤s £½9.65¡Á10A¡¤cm=9.65¡Á10¦ÌA¡¤cm £¨£±£Ã£½£±A¡¤s£© Ò²¿ÉÒÔµÈÓÚ9.6£´£¸£´¡Á10¦ÌA¡¤cm ¡£
(2)
¶ÔÓÚÒõ¼«£º
-3
-2
-9
-2
-3
-2
?£½RTRTlni0?lni?zF?zF0.059120.05912 ?lgi0?lgi?0.1971?(lgi0/i)0.30.3¶Ô
ÓÚ
Ñô
¼«
¼«
£º
£¨1£©
?£½RTRTlni0?lni?0.08446lg(i0/i) £¨2£© ?zF?zF
-2µçÁ÷/¦ÌA¡¤cm lgi Òõ¼«³¬µçÊÆ/V Ñô¼«³¬µçÊÆ/V
100 2.000 -0.7914 -0.3391 200 2.301 -0.8508 0.3646 300 2.477 -0.8855 0.3794 400 2.602 -0.9101 0.3900 52.-0.90.35£®ÊÔÍÆµ¼¸ßÕý³¬µçÊÆÊ±?¡«I¹ØÏµÊ½£¬²¢Í¬Tafel·½³Ì±È½Ï¡£ ½â£º
3
¸ß³¬µçÊÆÊ±£¬·½³Ì(1.43)ÓÒʽÁ½ÏîÖеÄÒ»Ïî¿ÉÒÔºöÂÔ¡£µ±µç¼«ÉÏ·¢ÉúÒõ¼«»¹Ô·´Ó¦£¬ÇÒ?ºÜ´óʱ(´Ëʱ£¬µç¼«µçÊÆ·Ç³£¸º£¬Ñô¼«Ñõ»¯·´Ó¦ÊÇ¿ÉÒÔºöÂÔµÄ)£¬
¶ÔÓÚÒ»¶¨Ìõ¼þÏÂÔÚÖ¸¶¨µç¼«ÉÏ·¢ÉúµÄÌØ¶¨·´Ó¦£¬
(RT/?zF)lni0ºÍ?RT/?zFΪһȷ¶¨µÄÖµ£¬¼´·½³Ì(1.47)¿ÉÒÔ¼ò
»¯Îª£º??a?blgi¡£Òò´Ë£¬ÔÚÇ¿¼«»¯µÄÌõ¼þÏ£¬ÓÉ
Butler-Volmer
·½³Ì¿ÉÒÔÍÆµ¼³öTafel¾Ñé·½³Ì¡£Tafel¾Ñé·½³ÌÖеÄa,b¿ÉÒÔÈ·¶¨Îª£º
6.¸ù¾ÝÎÄÏ×ÌṩµÄÊý¾Ý£¬
Pt|
??1Fe(CN)36(20mmol?L)£¬
4?Fe(CN)6(20mmol?L?1)£¬NaCl(1.0mmol?L?1)ÔÚ
25¡æÊ±µÄ0.50£¬¼Æ
j0?I0/A£½2.0mA?cm?2£¬Õâ¸öÌåϵµÄµç×Ó´«µÝϵÊýΪ
Ë㣺(1)
k?µÄÖµ£»(2)ÈÜÒºÖÐÁ½ÖÖÂçºÏÎïŨ¶È¶¼Îª1mol?L?1ʱ£¬µÄ½»
j0£»(3)µç¼«Ãæ»ýΪ
0.1cm£¬ÈÜÒºÖÐÁ½ÖÖÂçºÏÎïŨ¶ÈΪ
2»»µçÁ÷ÃܶÈ
10?4mol?L?1ʱµÄµçºÉ´«µÝµç×è¡£
½â£º(1)
k?µÄÖµ:
i0*(1??)*?AzFcOXcRk??2.0mA?cm?2?1?96500C?mol?1?(20mmol?L?1)0.5(20mmol?L?1)0.52.0mA?cm?23?3?1 =?10?1.036?10cm?s1?96500C?mol?1?20mmol?L?1
**£¨2£©cOX?cR?1mol?L?1
4
*(1??)*?0j0?iA?k?zFcOXcR
0.50.5?2 ?1.036?10?1?96500?1?1?99.974mA?cm£¨3£©
*(1??)*?i0?zFAk?cOXcR?3 ?1?96500C?mol?1?0.1cm2?1.036?10?3cm?s?1?(10?4mol?L?1)0.5?(10?4mol?L?1)0.5 ?9.9974?10?7C?cm?2?9.9974?10?7A?s?cm?2
Rct?RT8.314?298.15??25693.92? ?7zFi01?96500?9.9974?107£®¸ù¾ÝÎÄÏ×J Am£®Chem. Soc. £¬77£¬6488(1955)±¨µÀ£¬Ñо¿µç¼«·´Ó¦£ºµ±cCd(Hg)?0.40Cd2??Hg?2e?Cd(Hg)£¬ÊµÑéÊý¾Ý£º
mol?L?1ʱ£¬µÃµ½ÈçÏÂ
[Cd2?]/mmol?L?1 J0/mA?cm?2 ÊÔ¼ÆËã?ºÍkµÄÖµ¡£ ½â£º
?1.0 30.0 0.50 17.3 0.25 10.1 *(1??)*?*(1??)*?j0?zFk?cOXcR?zFk?cCd2?cCd(Hg)
ÓɱêÖÐÊý¾Ý¿ÉµÃ£º
3011???0.4??1??1?????17.30.5?0.4??0.5?17.30.51???0.4??0.5?????10.10.251???0.4??0.25?10.10.251???0.4??0.25?????4.940.101???0.4??0.10?1?? £¨1£©
1?? £¨2£©
1?? £¨3£©
¶Ô·½³Ì£¨1£©È¡¶ÔÊý£ºln30?(1??)ln2 µÃ£º1-¦Á£½0.7942£¬¦Á£½0.2058
17.317.3¶Ô·½³Ì£¨2£©È¡¶ÔÊý£º1-¦Á£½0.7764£¬¦Á£½0.2236 ln?(1??)ln2 µÃ£º
10.1¶Ô·½³Ì£¨3£©È¡¶ÔÊý£ºln10.10?(1??)ln2.5 µÃ£º1-¦Á£½0.7805£¬¦Á£½
4.940.2195
ËùÒÔ£º¦Á=£¨0.2058+0.2236+0.2195£©/3=0.2163
k??j0*(1??)*?zFcCd2?cCd(Hg)30?10?3A?cm?2?2?96500C?mol?1?(1.0?10?3mol?L?1)1?0.2163?(0.4mol?L?1)0.783730?10?3A?cm?2 ?2?96500A?s?mol?1?(1.0?10?6mol?cm?3)0.7837?(0.4?10?3mol?cm?3)0.783730?10?3A?cm ==0.15544cm?s?1=1..5544?10-4m?s?1 ?62?96500A?s?9.681?105
¡¡¡¡¡¡¡¡¡¡¡
8. ¶ÔÓÚÒ»¸öÐýתԲÅ̵缫£¬Ó¦ÓÃÎÈ̬ÎïÖÊ´«µÝ¿ØÖƵ缫·´Ó¦µÄ´¦Àí£¬ÎïÖÊ´«µÝϵÊým0£½0.62D02/3?1/2??/6£¬Ê½ÖУ¬D0ΪÀ©É¢ÏµÊý(cm?s?1)£¬?ΪԲÅ̵ĽÇËÙ¶È
£¬
(
s?1)£¨??2?f?1fΪÐýתƵÂÊr?s?1)£¬?ÊǶ¯Á¦Ç¿¶È£¬Ë®ÈÜÒºÖÐ
?Ϊ
0.010(cm2?s)¡£Ê¹ÓÃ0.30 cm2µÄÔ²Å̵缫£¬ÔÚ1
mol?L?1H2SO4ÖÐʹ
0.010mol?L?1
Fe3?»¹ÔΪFe2?¡£ÒÑÖªFe3?µÄD0Ϊ5.2¡Á10-6 cm2?s?1£¬
?s?1ʱµÄ»¹Ô¼«ÏÞµçÁ÷¡£
¼ÆËãÒòÅ̵缫תËÙΪ10r½â£ºm0£½0.62D02/3?1/2??/6
?1/6 =0.62?(5.2?10?6cm2?s?1)2/3?(2?10r?s?1)1/2??0.010cm2?s?1? =0.0252cm¡¤s-1
*id?zFAm0cOX =1?96500A?s?mol-1?0.30cm2?0.0252cm?s?1?0.01mol?cm?3?10?3 =7.2954?10?3A
9£®ÏÖÓÃ70A¡¤m-2µÄµçÁ÷Ãܶȵç½âÎö³öÍ£¬¼Ù¶¨ÈÜÒºÖÐCu2+µÄ»î¶ÈΪ1£¬ÊµÑé²âµÃÆäTafelÇúÏßбÂÊΪ(0.06V)-1£¬½»»»µçÁ÷ÃܶÈj0Ϊ1A¡¤m-2£¬ÊÔÎʵç½âÎö³öÍʱÒõ¼«µçλӦΪ¶àÉÙ?
½â£º???eq???0.06lgi0?0.06lgi?0.06(lg??Cu,Îö³ö??eq????Cu1)??0.1107V 702?£/CuRTaCuln?0.1107VzFaCu2?
=0.34£
RT1ln+0.1107V=0.4507VzF1µÚÈýÕÂϰÌâ½â´ð
10.ÊÔд³öÏÂÁÐµç³ØµÄµç¼«·´Ó¦¡¢³ÉÁ÷·´Ó¦ÒÔ¼°µç½âÒººÍ¼¯µçÆ÷Ãû³Æ£º (1)¼îÐÔп£ÃÌÔµç³Ø (2)ﮡª¶þÑõ»¯ÃÌÔµç³Ø (3)п¡ªÑõ»¯¹¯µç³Ø (4)¼îÐÔÄø£ïÓµç³Ø (5)ÇâÄøÐîµç³Ø (6)Ç⡪ÑõȼÁÏµç³Ø ½â£º(1) ¼îÐÔп£ÃÌÔµç³Ø
6
µç½âÒº£¬Å¨KOHµç½âÖÊÈÜÒº.¼îÐÔÔ²ÖùÐÎп£ÃÌµç³ØµÄ¸Ö¿ÇΪΪÕý¼«µÄ¼¯Á÷Æ÷£¬¶øÖÐÐĵĸּ¯Á÷Æ÷Óëп½ôÃܽӴ¥£¬²¢Á¬½Óµç³Øµ×ÁÚ³ÉΪ¸º¼«¶Ë¡£ (2)ﮡª¶þÑõ»¯ÃÌÔµç³Ø
½â£ºï®¡ª¶þÑõ»¯ÃÌµç³Ø±í´ïʽΪ£º
(-)Li | LiClO4+PC+DME| MnO2,C£¨+£©
¸º¼«·´Ó¦£º Li?e?Li?
Õý¼«·´Ó¦£º MnO2?Li??e?LiMnO2 µç³Ø·´Ó¦£º MnO2?Li?LiMnO2
µç½âÒºÊÇ£¬LiClO4+PC+DME£¨µç½âÖÊLiClO4ÈܽâÓÚPCºÍ1,2-DME»ìºÏÈܼÁÖУ©¡£
µç³ØÒÔ²»Ðâ¸ÖÍâ¿ÇºÍʯī·Ö±ð×÷Ϊ¸º¼«ºÍÕý¼«µÄ¼¯µçÆ÷¡£
(3)п¡ªÑõ»¯¹¯µç³Ø
½â£ºÐ¿¡ªÑõ»¯¹¯µç³ØµÄ±í´ïʽΪ£º(-)Zn|ŨKOH | HgO£¬C(+)£¬µç³ØµÄ¸º¼«·´Ó¦Óë¼îÐÔµç½âÒºµÄпÃÌµç³ØÏàͬ£¬
Õý¼«·´Ó¦£º HgO?H2O?2e?Hg?2OH?
µç³Ø·´Ó¦£º Zn?HgO?2OH??H2O?Hg?[Zn(OH)4]2?
¸Ãµç³Ø²ÉÓÃŨµÄKOHÈÜÒº×÷µç½âÒº£¬¼¯µçÆ÷·Ö±ðΪZnºÍʯī¡£
(4)¼îÐÔÄø£ïÓµç³Ø ½â£º
¶ÔÓÚ¼îÐÔNi/Cdµç³ØµÄ³ÉÁ÷·´Ó¦£¬µç³Ø·Åµçʱ¸º¼«ïÓ±»Ñõ»¯Éú³ÉÇâÑõ»¯ïÓ£»ÔÚÕý¼«ÉÏôÇ»ùÇâÑõ»¯Äø½ÓÊÜÁËÓɸº¼«¾Íâµç·Á÷¹ýÀ´µÄµç×Ó£¬±»»¹ÔΪÇâÑõ»¯Äø¡£¼¯µçÆ÷·Ö±ðΪCdºÍôÇ»ùÇâÑõ»¯Äø¡£µç½âҺΪÏà¶ÔÃܶÈΪ1.25¡«1.28µÄKOHÈÜÒº¡£ (5)ÇâÄøÐîµç³Ø
½â£ºÇâÄøµç³ØµÄ¸º¼«¿É²ÉÓûìºÏÏ¡ÍÁÖüÇâºÏ½ð(ÈçLaNiHx£¬x£½6)»òîÑ£ÄøºÏ½ð(MHx)£¬Õý¼«²ÉÓüîÐÔNi/Cdµç³ØÖÐNiµç¼«¼¼Êõ£¬²¢¼ÓÒԸĽø¡£µç³Ø±í´ïʽΪ£º
¸Ãµç³ØÒÔKOHÈÜÒº×÷Ϊµç½âÒº¡£ (6)Ç⡪ÑõȼÁÏµç³Ø
7
µÍμîÐÔÇâÉÏÑõȼÁÏµç³Ø¡£¸º¼«ÊÇÓÃNi·ÛºÍPt¡¢PdÉÕ½á¶ø³É£¬»òÓÃÄøµÄ»¯ÎïNi2BÖÆ×÷£¬Õý¼«ÊÇÓÐÐ§Ãæ»ýºÜ´óµÄÒø£¬¸ßŨ¶ÈKOHΪµç½âÖÊ£¬²ÉÓÃʯÃÞ»òîÑËá¼Ø×÷¸ôĤ¡£Æäµç¼«·´Ó¦Îª ¸º¼«·´Ó¦£ºH2£«2OH
Ò»
2H2O£«2e
¡ª
¡ª
¡ª
Õý¼«·´Ó¦£º1/2O2Ê®H2O£«2e µç³Ø·´Ó¦£ºH2£«1/2O2Ìì·É»úÉÏ¡£
2OH
H2O
µç¶¯ÊÆ1.15V£¬¹¤×÷µçѹ0.95V£¬¹¤×÷ζÈ353¡«363K£¬´Ëµç³ØÒÑÓÃÓÚº½
11.ϱíΪ´Óµç³ØÔÚ²»Í¬·ÅµçµçÁ÷ϵķŵçÊý¾Ý¼Ç¼(µç³ØÖÊÁ¿50g) (1)»æ³öÁ½·ÅµçµçÁ÷ϵķŵçÇúÏߣ¨E/V-Q/mA¡¤h¡¤g-1£©¡£
(2)½âÊÍΪʲôÏàͬ³õÖշŵçµçѹ¶ø²»Í¬·ÅµçµçÁ÷ÏÂµç³ØÈÝÁ¿²»¾¡Ò»Ñù? i=15mA i=40mA E/V t/min E/V t/min 3.24 0 3.24 0 3.11 60 3.12 45 3.10 120 3.08 90
½â£º(1)»æ³öÁ½·ÅµçµçÁ÷ϵķŵçÇúÏߣ¨E/V-Q/mA¡¤h¡¤g-1£©¡£
E/V i=15mA t/min Q//mA¡¤h¡¤g-1 E/V i=40mA t/min Q//mA¡¤h¡¤g-1
3.532.53.08 240 3.00 180 3.06 330 2.80 240 3.00 600 2.72 270 2.85 660 2.50 310 2.75 700 2.60 740 3.24 0 0 3.24 0 0 3.11 60 0.3 3.12 45 0.6 3.10 120 0.6 3.08 90 1.2 3.08 240 1.2 3.00 180 2.4 3.06 330 1.65 2.80 240 3.2 3.00 600 3 2.72 270 3.6 2.85 660 3.3 2.50 310 4.1 2.75 700 3.5 2.60 740 3.7 E/V21.510.500100200300400500Q/mA¡¤h¡¤g-1600700800i=40mAi=15mAͼ1Á½·ÅµçµçÁ÷ÏÂµç³ØµÄ·ÅµçÇúÏß (2) ÓÉÉÏͼ¿É¼û£¬·ÅµçµçÁ÷µÄ´óС¶Ôµç³ØÈÝÁ¿ÓнϴóµÄÓ°Ïì¡£¶ÔÓÚ¸ø¶¨µÄµç³Ø£¬ÓÉÓÚÅ·Ä·ÄÚ×èºÍÓеçÁ÷ͨ¹ýʱ¼«»¯ÄÚ×èµÄ´æÔÚ£¬µç³ØÈÝÁ¿ºÍ·ÅµçµçÑ¹Ëæ·ÅµçµçÁ÷µÄÔö¼Ó¶ø¼õС£¬µç³ØµÄʹÓÃÊÙÃüÒ²Ëæ×żõС¡£¾¡¹ÜÔÚ³õ
8
ÖշŵçµçѹÏàͬÌõ¼þÏ·ŵ磬µ«µç³ØÈÝÁ¿ÈÔÈ»Ëæ·ÅµçµçÁ÷µÄÔö¼Ó¶ø¼õС£¬µç³ØµÄʹÓÃÊÙÃüÒ²Ëæ×żõС¡£µç³Ø·ÅµçµçÁ÷µÄ´óС³£Ó÷ŵ籶Âʱíʾ£¬¼´¶ÔÓÚÒ»¸ö¾ßÓжÈÝÁ¿CµÄµç³Ø£¬°´¹æ¶¨µÄСʱÊý·ÅµçµÄµçÁ÷¡£
·Åµç±¶Êý£½¶î¶¨ÈÝÁ¿£¨A?h)·ÅµçµçÁ÷£¨A)
ÀýÈ磬ijµç³Ø¶î¶¨ÈÝÁ¿Îª20A¡¤h£¬ÈôÒÔ4AµçÁ÷·Åµç£¬Ôò·ÅÍê20A¡¤hµÄ¶î¶¨ÈÝÁ¿ÐèÒª5h£¬¼´ÒÔ5hÂʷŵ磬·Åµç±¶ÂʱíʾΪ¡°C/5¡±(»ò¡°0.2C¡¯¡¯)£»ÈôÒÔ0.5hÂʷŵ磬¶ÔÓڶÈÝÁ¿Îª20A¡¤hµÄµç³Ø£¬¾ÍÊÇÓÃ40AµÄµçÁ÷·Åµç£¬·Åµç±¶ÂÊΪ¡°C/0.5¡±(»ò¡°2C¡±)¡£¸ù¾Ý·Åµç±¶ÂʵĴóС£¬µç³Ø¿ÉÒԷֳɵͱ¶ÂÊ(£¼0.5C)¡¢Öб¶ÂÊ(0.5¡«3.5C)¡¢¸ß±¶ÂÊ(3.5¡«7C)ºÍ³¬¸ß±¶ÂÊ(£¾7C)ËÄÀà¡£·Åµç±¶ÂÊÔ½´ó£¬±íʾ·ÅµçµçÁ÷Ô½´ó£¬µç³ØÈÝÁ¿Òà»á½µµÍ½Ï´ó¡£Õâ¾Í½âÊÍΪʲôÏàͬ³õÖշŵçµçѹ¶ø²»Í¬·ÅµçµçÁ÷ÏÂµç³ØÈÝÁ¿²»¾¡Ò»ÑùµÄÔÒò¡£
12.ϱíΪ¼îÐÔп-ÃÌµç³ØµÄ¿ªÂ·µçѹ(OCV)ºÍ·ÅÖÃʱ¼äµÄ¹ØÏµ£¬·ÅÖÃ10¸öÔºóµç³Ø´æÁ¿Ï½µÁË10£¥£¬(1)ÊÔÖËãÆ½¾ù×ԷŵçËÙÂÊ£¬²¢»æ³öOCV-·ÅÖÃʱ¼äÇúÏߣ»(2)ÊÔ˵Ã÷Òý·¢¸Ãµç³ØA·ÅµçµÄÖ÷ÒªÔÒò¡£
OCV/V t/d 1.52 0 ʾ¡£
ÿÌìÆ½¾ù×ԷŵçËÙÂÊ=10%C/300d=0.033333%C/d ²¢»æ³öOCV-·ÅÖÃʱ¼äÇúÏßÈçÏÂͼ¡£
(2)Òý·¢¸Ãµç³ØA·ÅµçµÄÖ÷ÒªÔÒòΪ£ºÐ¿ÔÚÓë¼îÈÜÒº½Ó´¥Ê±µÄÈÈÁ¦Ñ§²»Îȶ¨ÐÔ£¬µ¼ÖÂп¸º¼«ÔÚ¼îÐÔµç½âÒºÖеÄÈܽ⼰ˮÖеÄHÀë×Ó»¹ÔΪÇâÆø¶øµ¼ÖÂ×Էŵ硣
1.541.521.501.481.461.441.421.401.381.361.34050100150t/dͼ12Ìâ. ¿ªÂ·µçѹ¶Ô·ÅÖÃʱ¼äÇúÏß200250300350+1.44 30 1.42 60 1.40 120 1.38 180 1.36 240 1.36 300 ½â£º(1)µç³Ø×ԷŵçµÄ´óСһ°ãÓõ¥Î»Ê±¼äÄÚµç³ØÈÝÁ¿¼õÉٵİٷÖÊýÀ´±í
OCV/V 13.¶ÔÓÚǶÈë·´Ó¦£º
xLi?V3O8?LixV3O8LiClO4+PC:DME(1:1)]
[µç½âҺΪ1
mol?L?1
(1)ÊÔ½«¸Ã·´Ó¦Éè¼Æ³É¶þ´Îµç³Ø²¢Ð´³öÏàÓ¦µÄµç¼«·´Ó¦¡£ (2)ÊÔ¸ù¾ÝǶÈëµÄï®Àë×ÓµÄÁ¿x£¬¼ÆËãµç³ØµÄÀíÂÛÈÝÁ¿¡£
9
(3) ÊÔ¼òÊöÑо¿¸Ã¶þ´Îµç³ØÐÔÄܵÄÒ»°ã·½·¨¡£ ½â£º(1)
Li/V3O8³É¶þ´Îµç³Ø±í´ïʽΪ
(?)Li1mol?L?1LiClO4?PC:DME(1:1)V3O8(?)
¸º¼«·´Ó¦:: Õý¼«·´Ó¦:
xLi?xLi??xe
xLi??V3O8?xe?LixV3O8
·Åµç³äµçµç³Ø³ä·Åµç·´Ó¦Îª£ºxLi?VO38LixV3O8
(2) Éè
x?1
mzF1000?2?26.8??7722.23 A¡¤h¡¤kg-1»ò£½7722.23 mA¡¤h¡¤g-1 M6.941C?(3) ½üÄêÀ´ï®Àë×Óµç³ØµÄÑо¿¹¤×÷ÖØµãÔÚ̼¸º¼«²ÄÁϵÄÑо¿ÉÏ£¬ÇÒÒѾȡµÃÁËеĽøÕ¹¡£µ«ÊÇï®Àë×Óµç³ØÒª´ïµ½´ó¹æÄ£µÄÓ¦Ó㬶ÔÓÚ̼¸º¼«²ÄÁÏ»¹ÐèÒªÌá¸ß﮵ĿÉÄæ´¢Á¿ºÍ¼õÉÙ²»¿ÉĿĿÈÝÁ¿Ëðʧ£¬´Ó¶øÓÐÀûÓÚ¸º¼«±ÈÈÝÁ¿µÄÌá¸ßºÍµç³Ø±ÈÄÜÁ¿µÄÌá¸ß¡£
Óëï®Àë×Óµç³Ø¸º¼«µÄ·¢Õ¹Ïà±È£¬Õý¼«²ÄÁϵķ¢Õ¹ÉÔÏÔ»ºÂý£¬Ö÷ҪͣÁôÔÚ¶Ôº¬ï®½ðÊôÑõ»¯ÎïµÄÑо¿ÉÏ¡£ÔÒòÔÚÓÚ¾¡¹Ü´ÓÀíÂÛÉÏÄÜÍÑǶ﮵ÄÎïÖʺܶ࣬µ«Òª½«¹²ÖƱ¸³ÉÄÜʵ¼ÊÓ¦ÓõIJÄÁÏÈ´·ÇÒ×Ê£¬ÖƱ¸¹ý³ÌÖеÄ΢С±ä»¯¶¼¿ÉÄܵ¼ÖÂÑùÆ·½á¹¹ºÍÐÔÖʾ޴ó²îÒ죬Òò¶ø¶ÔÏÖÓвÄÁϵĸĽøÈÔÈ»Êǹ¤×÷µÄÖØµã¡£
14. ¶ÔÓÚȼÁÏµç³Ø£º£¨-£©£¨Pt£©,CH3OH | 1mol?LH2SO4 | O2(Pt)£¬(+) (1)ÊÔд³öµç¼«·´Ó¦ºÍµç³Ø·´Ó¦£¬²¢¼ÆËã±ê×¼µç³Øµç¶¯ÊÆ¡£
(2)ÊÔ¸ù¾ÝÈÈÁ¦Ñ§ÖªÊ¶ÍƵ¼Æäµç³ØÄÜÁ¿Ð§ÂÊ((?id??G/?H)¡Ý100£¥ÊÇ¿ÉÄܵġ£
(3)ÊÔÐðÊö¸Ä½ø¸ÃȼÁÏµç³ØÐÔÄܵķ½·¨¡£
½â£º(1)¸º¼«·´Ó¦: CH3OH?H2O?6e?CO2?6H Õý¼«·´Ó¦:
??1
6H??6e?1.5O2?3H2O
3µç³Ø·´Ó¦:CHOH?1.5O2?2H2O?CO2
CH3OH?1.5O2?2H2O?CO2
(2) ¶Ô¿ÉÄæµç³Ø·´Ó¦:
²éÈÈÈÈÁ¦Ñ§Êý¾Ý:
??fGm(CH3OH)=-166.27kJ?mol?1;?G?(HO(l))=-237.129kJfm2?mol?1;
?1??fGm(CO2)?-394.359kJ?mol;
???fHm(CH3OH)=-238.66kJ?mol?1;?fHm(H2O(l))=-285.83kJ?mol?1;
??fHm(CO2)?-393.5kJ?mol?1
10
?1??rGm?1*(-394.359)+2*(-237.129)-( -166.27)= -702.347kJ?mol
??rHm?1*(-393.5)+2*(-285.83)-( -238.66)= -726.5kJ?mol?1
ÀíÂÛת»»Ð§ÂÊ£º ??702.347?100%?96.6754%=96.68%
726.5(3): ÀíÂÛ¼ÆËã½á¹û±íÃ÷£ºÖ±½Ó¼×´¼È¼ÁÏµç³ØµÄÀíÂÛÄÜÁ¿×ª»»Ð§ÂÊΪ96.68£¥¡£¾¡¹ÜDMPEMFC¾ßÓÐÎ޿ɱÈÄâµÄÓŵ㣬µ«Òª´ïµ½Êµ¼ÊÓ¦Óû¹ÓдóÁ¿ÎÊÌâÓдý½øÒ»²½½â¾ö£¬Ä¿Ç°ËüµÄ¼¼Êõ»¹ºÜ²»³ÉÊ죬½ö´¦ÓÚÑÐÖÆ½×¶Î£¬ÐÔÄÜ×îºÃµÄÒ²Ö»ÓÐ0.1W¡¤cm¡£¶øÒª´ïµ½Êµ¼ÊÓ¦Ó㬹¦ÂʱØÐë´ïµ½0.25 W¡¤cmÒÔÉÏ£¬Í¬Ê±»¹ÒªÊ¹µç³ØÂú×ãÐÔÄܸߣ¬´æÃü³¤ºÍ¼Û¸ñµÍÈý¸öÌõ¼þ¡£Ä¿Ç°ÏÞÖÆDMPEMFCʵ¼ÊÓ¦ÓõÄÖ÷ÒªÎÊÌâÊÇÑô¼«´ß»¯¼ÁµÍµÄ»îÐÔ¡¢¸ßµÄ¼Û¸ñ¼°´ß»¯¼ÁµÄ¶¾»¯¡£Òò´Ë±ØÐëÌá¸ßÑô¼«´ß»¯¼ÁµÄ»îÐÔ£¬½µµÍ´ß»¯¼ÁµÄÓÃÁ¿£¬½µµÍ»òÏû³ý´ß»¯¼ÁµÄ¶¾»¯¡£
15ÊÔÒÀ¾ÝÈÈÁ¦Ñ§Êý¾Ý¼ÆËã¼îÐÔп-¿ÕÆøµç³ØµÄÀíÂÛÈÝÁ¿¡¢µç³ØµÄµç¶¯ÊÆ¡£ ½â£ºµç³Ø·´Ó¦:Zn?1O?ZnO
2-2
-2
2²éÈÈÈÈÁ¦Ñ§Êý¾Ý: ?rGm??rG??rGm?RTln????fGm(ZnO)?-318.30kJ?mol?1,
1(0.21p?/p?)1/2 ??318300?298.15?8.34?(?0.5)ln(0.21) =-316365.72J?mol-1??rG316365.72??1.6392V zF2?96500µç¶¯ÊÆ:E?ÀíÂÛÈÝÁ¿:C?mzF1000?2?26.8£½819.46 A¡¤h¡¤kg-1
?M65.409£¨×¢Ò⣺Á½Õß»»Ëã¹ØÏµÎª96500¿âÂØÏ൱ÓÚ26.8°²Ê±£©
16£®ÒÔÐðÊöȼÁÏµç³ØµÄÀàÐͼ°Ìص㡣
½â£ºÈ¼ÁÏµç³Ø¿ÉÒÀ¾ÝÆä¹¤×÷ζȡ¢ËùÓÃȼÁϵÄÖÖÀàºÍµç½âÖÊÀàÐͽøÐзÖÀà¡£°´ÕÕ¹¤×÷ζȣ¬È¼ÁÏµç³Ø¿É·ÖΪ¸ß¡¢ÖС¢µÍÎÂÐÍÈýÀà¡£°´È¼ÁÏÀ´Ô´£¬È¼ÁÏµç³Ø¿É·ÖΪֱ½ÓʽȼÁÏµç³Ø(ÈçÖ±½Óѧ´¼È¼ÁÏµç³Ø)£¬¼ä½ÓʽȼÁÏµç³Ø(¼×´¼Í¨¹ýÖØÕûÆ÷²úÉúÇâÆø£¬È»ºóÒÔÇâÆøÎªÈ¼ÁÏµç³ØµÄȼÁÏ)ºÍÔÙÉúÀàÐͽøÐзÖÀà¡£ÏÖÔÚÒ»°ã¶¼ÒÀ¾Ýµç½âÖÊÀàÐÍÀ´·ÖÀ࣬¿ÉÒÔ·ÖΪÎå´óÀàȼÁÏµç³Ø£¬¼´£¬
Á×ËáÐÍȼÁÏµç³Ø(phosphoric acid fuel cell PAFC)¡¢ ÈÛÈÚ̼ËáÑÎȼÁÏµç³Ø(molten carbonate fuel cell £¬MCFC)¡¢ ¹ÌÌåÑõ»¯ÎïȼÁÏµç³Ø(solid oxide fuel cell,SOFC)ºÍ ¼îÐÔȼÁÏµç³Ø(alkaline fuel cell,AFC)
11
ÖÊ×Ó½»»»Ä¤È¼ÁÏµç³Ø(proton exchange membrane fuel cell, PEMFC).
ȼÁÏµç³ØµÄµÚÒ»¸öÏÔÖøÌØµãÊDz»ÊÜ¿¨ÅµÑ»·µÄÏÞÖÆ£¬ÄÜÁ¿×ª»»Ð§Âʸߡ£ÓÉÓÚȼÁÏµç³ØÖ±½Ó½«»¯Ñ§ÄÜת±äΪµçÄÜ£¬Öмäδ¾È¼ÉÕ¹ý³Ì(Ò༴ȼÁÏµç³Ø²»ÊÇÒ»ÖÖÈÈ»ú)£¬Òò´Ë£¬²»ÊÜ¿¨ÅµÑ»·µÄÏÞÖÆ£¬¿ÉÒÔ»ñµÃ¸ü¸ßµÄת»¯Ð§ÂÊ¡£È¼ÁÏµç³ØµÄÆäËûÓŵãÊÇ£ºµÍµÄ»·¾³ÎÛȾºÍÔëÒôÎÛȾ£¬°²È«¿É¿¿ÐԸߣ»²Ù×÷¼òµ¥£¬Áé»îÐԴ󣬽¨ÉèÖÜÆÚ¶ÌµÈ¡£
17£®Çëд³öǦËáÐîµç³ØµÄµç¼«·´Ó¦ºÍ³ÉÁ÷·´Ó¦£¬²¢ÐðÊöÓ°ÏìÆäÊÙÃüºÍÈÝÁ¿¼õСµÄÔÒò¼°Ç¦ËáÐîµç³ØµÄ¸Ä½ø·½·¨¡£
½â£ºµç³Ø±íʾʽΪ £¨-£©Pb,PbSO4|H2SO4|PbO2£¬Pb£¨+£©
Ó°ÏìÈÝÁ¿ºÍÑ»·ÊÙÃüµÄÖ÷ÒªÔÒòÓУº
(1)¼«°åÕ¤¸¯Ê´£ºPbµç¼«ÔÚÓëPbO2ºÍËá½Ó´¥µÄµØ·½¸¯Ê´ÒÔ¼°Pb°åÕ¤µÄ±©Â¶²¿·Ö³äµçʱ¿ÉÄÜ·¢ÉúµÄÑô¼«Ñõ»¯¶øµ¼Öµĸ¯Ê´¡£ÕâЩ¹ý³ÌµÄÓк¦×÷ÓÃÔÚÓÚÆÆ»µ°åÕ¤Óë»î¼þÎïÁϵĽӴ¥£¬´ËÍ⣬Éú³ÉµÄPbO2¾ßÓбÈPb¸ü´óµÄ±ÈÌå»ý£¬Òò¶øÊ¹¼«°åÕ¤±äÐΡ£
(2)Õý¼«»îÐÔÎïÖʵÄÍÑÂ䣺ÊÇÓÉÓÚ¾§ÌåºÍСÓÚ0.1¦ÌmµÄPbO2¿ÅÁ£Í¬°åÕ¤·ÖÀ룬ÕâÒ»°ãÔÚ³äµç¿ªÊ¼ºÍ½áÊøÊ±·¢Éú¡£ÏÖÈÏΪ·ÅµçʱPbSO4½ôÃܲãµÄÐγÉÊǵ¼ÖÂÕý¼«»îÐÔÎïÖÊÍÑÂäµÄÖ÷ÒªÔÒò£¬Í¬Ê±BaSO4µÄ¼ÓÈëÒ²»á´ÙʹÍÑÂ䡣ΪÁË·ÀÖ¹Õý¼«»îÐÔÎïÖʵÄÍÑÂ䣬µç¼«²ÉÓýôÃÜ×°Å䣬²¢»ìÈë²£Á§ÏËά£¬ÓÐʱҲÔÚ»îÐÔÎïÖÊÖмÓÈëһЩð¤ºÏ¼Á¡£
(3)¸º¼«×Էŵ磺Ö÷ÒªÔÒòÊÇÓÉÓڵ缫ÌåϵºÍµç½âÒºÖдæÔÚµÄÔÓÖÊ(ÈçFe£¬Cu£¬Mn)Ï໥×÷ÓöøÊ¹º£ÃàǦ¸¯Ê´¡£Ç¦µÄ¸¯Ê´ËÙ¶ÈËæÎ¶ÈÉý¸ßºÍÁòËáŨ¶ÈÔö´ó¶øÔö¼Ó¡£Òò´Ë£¬ÎªÁ˽µµÍ×Էŵ磬±ØÐëÓô¿PbÖÆ±¸»îÐÔÎïÁϵĺϽð·ÛÄ©£¬²ÉÓô¿ÁòËáºÍµçµ¼Ë®ÅäÖÆµç½âÒº£¬²¢±£³ÖÊÊÒ˵ÄÔËÐÐÌõ¼þ¡£
(4)¼«°åÕ¤ÁòËữ£º±íÏÖΪÔڵ缫ÉÏÉú³É½ôÃܵİ×É«ÁòËáÑÎÍâÆ¤£¬´Ëʱµç³Ø²»ÄÜÔÙ³äµç£¬ÔÒòÊǵ±Ðîµç³Ø±£´æÔڷŵç״̬ʱÁòËáÑÎÔٽᾧ£¬Òò´ËÐîµç³Ø²»ÄÜÒԷŵç״̬Öü´æ¡£
ΪÁ˿˷þÕâЩȱµã£¬µç»¯Ñ§¹¤×÷Õß¶ÔǦËáµç³Ø½øÐÐÁËһЩ¸Ä½ø:Èç²ÉÓÃÇáÇá²ÄÁÏÖÆ±¸°åÕ¤£¬ÒÔÌá¸ß±ÈÈÝÁ¿£»²ÉÓ÷ÖÉ¢¶È¸ü¸ßµÄµç¼«ÒÔÌá¸ß»îÐÔÎïÖʵÄÀûÓÃÂÊ£»²ÉÓýº×´µç½âÒº(¼ÓSiO2»ò¹è½º)ʹµç³ØÔÚÈκÎÇé¿ö϶¼ÄÜÔËÐУ»²ÉÓÃPb-CaºÏ½ðºÍPb-SbºÏ½ð£¨Ca¡¢Sbº¬Á¿Ô¼0.1£¥£©£¬ÒÔ½µ
12
µÍ×ԷŵçºÍË®µÄ·Ö½â£»ËÜÁϿǵÄÃÜ·âµç³ØÓÐÅÅÆø·§Ãŵȡ£20ÊÀ¼Í80Äê´úºóÆÚ¿ª·¢³öµÄµÍά»¤ºÍÃâά»¤µç³Ø½øÒ»²½Ôö´óÁË¸Ãµç³ØµÄʵÓ÷¶Î§¡£
18. ÊÔÐðÊöÈçºÎʵÏÖÄø-ïÓµç³ØµÄÃÜ·â¡£
½â£ºÔÚ¼îÐÔNi/Cdµç³ØÖУ¬²ÉÓøº¼«ÈÝÁ¿¹ýÁ¿£¬¿ØÖƵç½âÒºµÄÓÃÁ¿£¬Ê¹Óøß΢ÃܶȵÄ΢¿×¸ôĤ£¨ÉøÍ¸ÐÔ¸ôĤ£©£»Õý¼«ÖÐÌí¼ÓÇâÑõ»¯ïÓ£¬²¢¼ÓÃÜ·âȦ»ò½ðÊôÌÕ´É·â½ÓµÈ´ëÊ©£¬ÑÐÖÆ³É¹¦ÁËÃÜ·âʽ¼îÐÔNi/Cdµç³Ø¡£ÈçÔ²ÖùÐÎÃÜ·âʽµç³ØÊÇÓÃ;×î¹ã·ºµÄÀàÐÍ¡£Ô²ÖùÐÎµç³ØµÄÕý¼«Îª¶à¿×ÉÕ½áÄøµç¼«£¬²É¾ù½þ×ÕÈÛÈÚÄøÑΣ¬ÔÙ½þÈë¼îÈÜÒºÖгÁµíÇâÑõ»¯ÄøµÄ·½·¨³äÌî»îÐÔÎïÖÊ¡£¸º¼«µÄÖÆÔìÓм¸ÖÖ·½·¨£ºÓеÄÈçÏòÕý¼«Ò»Ñù²ÉÓÃÉÕ½áÄø»ù¼«£»ÓеÄÓÃÍ¿¸à·¨»òÑ¹ÖÆ·¨½«»îÐÔÎïÖÊÍ¿Èë»ù¼«ÄÚ£»Ò²ÓеIJÉÓÃÁ¬ÐøµÄµç»¯Ñ§³Á»ý·¨»òеķ¢Åݵ缫¼¼ÊõÖÆÔì¡£½«Á¬Ðø¼Ó¹¤³ÉÐ͵ÄÕý¡¢¸º¼«Á¬Í¬ÖмäµÄ¸ôĤһÆðÅÌÐý¾íÈÆ¡£È»ºó×°Èë¶ÆÄøµÄ¸Ö¿ÇÄÚ¡£×îºó½«¸º¼«º¸½ÓÔÚ¿ÇÉÏ£¬Õý¼«º¸½ÓÔÚ¶¥¸ÇÉϽøÐзâ×°¡£
µÚËÄÕÂ
19ÊÔÐðÊöÓ°Ïìµç¶Æ²ãÖÊÁ¿µÄÒòËØ¡£
½â£ºÓ°Ïìµç¶Æ²ãÖÊÁ¿µÄÖ÷ÒªÒòËØÓжÆÒºµÄÐÔÄÜ¡¢µç¶Æ¹¤ÒÕÌõ¼þ¡¢Ñô¼«¡£ £¨1£©¶ÆÒºµÄÐÔÄÜ¿ÉÒÔÓ°Ïì¶Æ²ãµÄÖÊÁ¿£¬¶ø¶ÆÒºÅäÖÆÇ§²îÍò±ð£¬µ«Ò»°ã¶¼ÊÇÓÉÖ÷ÑΡ¢µ¼µçÑÎ(ÓÖ³ÆÎªÖ§³Öµç½âÖÊ)¡¢ÂçºÏ¼ÁºÍһЩÌí¼Ó¼ÁµÈ×é³É¡£
Ö÷ÑζԶƲãµÄÓ°ÏìÌåÏÖÔÚ£ºÖ÷ÑÎŨ¶È¸ß£¬¶Æ²ã½Ï×â²Ú£¬µ«ÔÊÐíµÄµçÁ÷Ãܶȴó£»Ö÷ÑÎŨ¶ÈµÍ£¬ÔÊÐíͨ¹ýµÄµçÁ÷ÃܶÈС£¬Ó°Ïì³Á»ýËÙ¶È¡£
µ¼µçÑÎ(Ö§³Öµç½âÖÊ)µÄ×÷ÓÃÊÇÔö¼Óµç¶ÆÒºµÄµ¼µçÄÜÁ¦£¬µ÷½ÚÈÜÒºµÄpHÖµ£¬ÕâÑù²»½ö¿É½µµÍ²Ûѹ¡¢Ìá¸ß¶ÆÒºµÄ·ÖÉ¢ÄÜÁ¦£¬¸üÖØÒªµÄÊÇijЩµ¼µçÑεÄÌí¼ÓÓй¦ÓÚ¸ÄÉÆ¶ÆÒºµÄÎïÀí»¯Ñ§ÐÔÄܺÍÑô¼«ÐÔÄÜ¡£
¶ø¼ÓÈëÂçºÏ¼ÁµÄ¸´Ñεç½âҺʹ½ðÊôÀë×ÓµÄÒõ¼«»¹Ô¼«»¯µÃµ½ÁËÌá¸ß£¬ÓÐÀûÓڵõ½Ï¸Ö¡¢½ôÃÜ¡¢ÖÊÁ¿ºÃµÄ¶Æ²ã£¬µ«³É±¾½Ï¸ß¡£
Ìí¼Ó¼Á¶Ô¶Æ²ãµÄÓ°ÏìÌåÏÖÔÚÌí¼Ó¼ÁÄÜÎü¸½Óڵ缫±íÃæ£¬¿É¸Ä±äµç¼«-ÈÜÒº½çÃæË«µç²ãµÄ½á¹¹£¬´ïµ½Ìá¸ßÒõ¼«»¹Ô¹ý³Ì³¬µçÊÆ¡¢¸Ä±äTafelÇúÏßбÂʵÈÄ¿µÄ¡£Í¬Ê±£¬Ìí¼Ó¼ÁµÄ´æÔÚ¶Ôµç³Á»ý²ãµÄÐÔÄÜÓ°Ï켫´ó£¬Í¨³£Ê¹¶Æ²ãµÄÓ²¶ÈÔö¼Ó£¬¶øÄÚÓ¦Á¦ºÍ´àÐÔÔò¿ÉÄÜÊÇÌá¸ß£¬Ò²¿ÉÄÜÊǽµµÍ£¬¶øÇÒ¼´Ê¹ÊÇͬһÖÖ±íÃæ»îÐÔ¼Á£¬ËæÆäŨ¶ÈµÄ²»Í¬£¬ÆäÓ°ÏìÇé¿öÒ²²»Ò»Ñù¡£
´ËÍ⣬ÈܼÁ¶Ô¶Æ²ãÖÊÁ¿Ò²Ó¦ÓÐÒ»¶¨Ó°Ïì¡£µç¶ÆÒºÈܼÁ±ØÐë¾ßÓÐÏÂÁÐÐÔÖÊ£º(1)µç½âÖÊÔÚÆäÖÐÊÇ¿ÉÈܵģ»(2)¾ßÓнϸߵĽéµç³£Êý£¬Ê¹ÈܽâµÄµç½âÖÊÍêÈ«»ò´ó²¿·ÖµçÀë³ÉÀë×Ó¡£
£¨2£©µç¶Æ¹¤ÒÕÌõ¼þµÄÓ°Ï죺µçÁ÷ÃܶȶԶƲãµÄÓ°ÏìÖ÷ÒªÌåÏÖÔÚ£ºµçÁ÷Ãܶȴ󣬶ÆÍ¬Ñùºñ¶ÈµÄ¶Æ²ãËùÐèʱ¼ä¶Ì£¬¿ÉÌá¸ßÉú²úЧÂÊ£¬Í¬Ê±£¬µçÁ÷Ãܶȴó£¬Ðγɵľ§ºËÊýÔö¼Ó£¬¶Æ²ã½á¾§Ï¸¶ø½ôÃÜ£¬´Ó¶øÔö¼Ó¶Æ²ãµÄÓ²¶È¡¢ÄÚÓ¦Á¦ºÍ´àÐÔ£¬µ«µçÁ÷ÃܶÈÌ«´ó»á³öÏÖ֦״¾§ÌåºÍÕë¿×µÈ¡£¶ÔÓÚµç¶Æ¹ý³Ì£¬µçÁ÷ÃܶȴæÔÚÒ»¸ö×îÊÊÒË·¶Î§¡£
µç½âҺζȶԶƲãµÄÓ°ÏìÌåÏÖÔÚ£ºÌá¸ß¶ÆÒºÎ¶ÈÓÐÀûÓÚÉú³É½Ï´óµÄ¾§Á££¬Òò¶ø¶Æ²ãµÄÓ²¶È¡¢ÄÚÓ¦Á¦ºÍ´àÐÔÒÔ¼°¿¹ÀÇ¿¶È½µµÍ¡£Í¬Ê±Î¶ÈÌá¸ß£¬ÄÜÌá¸ßÒõ¼«ºÍÑô¼«µçÁ÷ЧÂÊ£¬Ïû³ýÑô¼«¶Û»¯£¬Ôö¼ÓÑεÄÈܽâ¶ÈºÍÈÜÒºµ¼µçÄÜÁ¦£¬½µµÍŨ²î¼«»¯ºÍµç»¯Ñ§¼«»¯£¬µ«Î¶ÈÌ«¸ß£¬½á¾§Éú³¤µÄËٶȳ¬¹ýÁËÐγɽᾧ»îÐÔµÄÉú³¤µã£¬Òò¶øµ¼ÖÂÐγɴ־§ºÍ¿×϶½Ï¶àµÄ¶Æ²ã¡£
µç½âÒºµÄ½Á°èÓÐÀûÓÚ¼õÉÙŨ²î¼«»¯£¬ÀûÓڵõ½ÖÂÃܵĶƲ㣬¼õÉÙÇâ
13
ϰÌâ½â´ð
´à¡£Í¬Ê±£¬µç½âÒºµÄpHÖµ¡¢³å»÷µçÁ÷ºÍ»»ÏòµçÁ÷µÈµÄʹÓöԶƲãÖÊÁ¿ÒàÓÐÒ»¶¨Ó°Ïì¡£
£¨3£©Ñô¼«µç¶ÆÊ±Ñô¼«¶Ô¶Æ²ãÖÊÁ¿ÒàÓÐÓ°Ïì¡£Ñô¼«Ñõ»¯Ò»°ã¾Àú»î»¯Çø(¼´½ðÊôÈܽâÇø)¡¢¶Û»¯Çø(±íÃæÉú³É¶Û»¯Ä¤)ºÍ¹ý¶Û»¯Çø(±íÃæ²úÉú¸ß¼Û½ðÊôÀë×Ó»òÎö³öÑõÆø)Èý¸ö²½Ö裬µç¶ÆÖÐÑô¼«µÄÑ¡ÔñÓ¦ÊÇÓëÒõ¼«³Á»ýÎïÖÖÏàͬ£¬¶ÆÒºÖеĵç½âÖÊӦѡÔñ²»Ê¹Ñô¼«·¢Éú¶Û»¯µÄÎïÖÊ£¬µç¶Æ¹ý³ÌÖпɵ÷½ÚµçÁ÷Ãܶȱ£³ÖÑô¼«ÔÚ»î»¯ÇøÓò¡£Èç¹ûijЩÑô¼«(ÈçCr)ÄÜ·¢Éú¾çÁÒ¶Û»¯£¬Ôò¿ÉÓöèÐÔÑô¼«¡£
20£®ÊÔÐðÊöµç¶Æ¹ý³ÌÖÐÌí¼Ó¼ÁµÄ×÷ÓÃÔÀí¡£
½â£ºµç¶ÆÉú²úÖÐÀûÓþßÓбíÃæ»îÐÔµÄÌí¼ÓÀûÀ´¿ØÖƺ͵÷½Ú½ðÊôµç³Á»ý¹ý³Ì£¬ÒÔ´ïµ½¸ÄÉÆ¶ÆÒºµÄ·ÖÉ¢ÄÜÁ¦£¬»ñµÃ½á¾§Ï¸Ö¡¢½ôÃܵĶƲ㣻¸ÄÉÆÎ¢¹ÛµçÁ÷·Ö²¼£¬ÒԵõ½Æ½ÕûºÍ¹âÁÁµÄ¶Æ²ã±íÃæ£»ÒÔ¼°¶Ô¶Æ²ãÎïÀíÐÔÄܵÄÓ°ÏìµÈ¡£µç¶ÆÌí¼Ó¼ÁÖ÷ÒªÓÐÕûƽ¼Á¼°¹âÁÁ¼ÁµÈ¡£
Õûƽ¼Á×÷ÓûúÀí¿ÉÒÔ±íÊöΪ£º(1)ÔÚÕû¸ö»ùµ×±íÃæÉÏ£¬½ðÊôµç³Á»ý¹ý³ÌÊÇÊܵ绯ѧ»î»¯¿ØÖÆ(¼´µç×Ó´«µÝ²½ÖèÊÇËÙ¶È¿ØÖƲ½Öè)µÄ£»(2)Õûƽ¼ÁÄÜÔÚ»ùµ×µç¼«±íÃæ·¢ÉúÎü¸½£¬²¢¶Ôµç³Á»ý¹ý³ÌÆð×軯×÷Óã»(3)ÔÚÕûƽ¹ý³ÌÖУ¬Îü¸½ÔÚ±íÃæÉϵÄÕûƽ¼Á·Ö×ÓÊDz»¶ÏÏûºÄµÄ£¬¼´Õûƽ¼ÁÔÚ±íÃæµÄ¸²¸Ç¶È²»ÊÇ´¦ÓÚÆ½ºâ״̬£¬Õûƽ¼ÁÔÚ»ùµ×ÉϵÄÎü¸½¹ý³ÌÊÜÆä±¾Éí´Ó±¾ÌåÈÜÒºÏòµç¼«±íÃæÀ©É¢²½Öè¿ØÖÆ¡£ÕâÑùÕûƽ×÷ÓÿÉÒÔ½èÖúÓÚ΢¹Û±íÃæÉÏÕûƽ¼Á¹©Ó¦µÄ¾Ö²¿²îÒìÀ´ËµÃ÷¡£ÓÉÓÚ΢¹Û±íÃæÉÏ΢·åºÍ΢¹ÈµÄ´æÔÚ£¬Õûƽ¼ÁÔÚµç³Á»ý¹ý³ÌÖÐÏò¡°Î¢·å¡±À©É¢µÄÁ÷Á¿Òª´óÓÚÏò¡°Î¢¹È¡±À©É¢µÄÁ÷Á¿£¬ËùÒÔ¡°Î¢·å¡±´¦»ñµÃµÄÕûƽ¼ÁµÄÁ¿Òª½Ï¡°Î¢¹È¡±´¦µÄ¶à£¬Í¬Ê±ÓÉÓÚ»¹Ô·´Ó¦²»ÄÜ·¢ÉúÔÚÕûƽ¼Á·Ö×ÓËù¸²¸ÇµÄλÖÃÉÏ£¬ÓÚÊÇ£¬¡°Î¢·å¡±´¦Êܵ½µÄ×軯×÷ÓÃÒª½Ï¡°Î¢¹È¡±´¦µÄ´ó£¬Ê¹µÃ½ðÊôÔڵ缫±íÃæ¡°Î¢·å¡±´¦µç³Á»ýµÄËÙ¶ÈҪСÓÚ¡°Î¢¹È¡±´¦µÄËÙ¶È£¬×îÖÕµ¼Ö±íÃæµÄ¡°Î¢·å¡±ºÍ¡°Î¢¹È¡±´ïµ½Æ½Õû¡£
¹âÁÁ¼Á×÷ÓûúÀí¿ÉÒÔ±íÊöΪ£ºÒ»ÖÖ¿´·¨ÈÏΪ¹âÁÁ×÷ÓÃÊÇÒ»¸ö·Ç³£ÓÐЧµÄÕûƽ×÷Ó㬿ÉÒÔÓÃÇ°ÃæÌáµ½µÄÀ©É¢¿ØÖÆ×軯»úÀíÀ´ËµÃ÷Ôö¹â×÷Óã»ÁíÒ»ÖÖ½âÊÍÊǹâÁÁ¼Á¾ßÓÐʹ²»Í¬¾§ÃæµÄÉú³¤ËÙ¶ÈÇ÷ÓÚÒ»ÖµÄÄÜÁ¦¡£µÚÒ»ÖÖ»úÀí¼ÙÉè¹âÁÁ¼ÁÔڶƼþ±íÃæÉÏÐγÉÁ˼¸ºõÍêÕûµÄÎü¸½µ¥²ã£¬Îü¸½²ãÉÏ´æÔÚÁ¬ÐøÐγÉÓëÏûʧµÄ΢¿×£¬¶ø½ðÊôÖ»ÔÚ΢¿×´¦½øÐгÁ»ý¡£ÓÉÓÚÕâЩ΢¿×ÊÇÎÞÐò·Ö²¼µÄ£¬¹Ê½ðÊô³Á»ýÊÇÍêÈ«¾ùÔȵ쬲»»áµ¼ÖÂС¾§ÃæµÄÐγɡ£Óë´Ëͬʱ£¬½èÖú¼¸ºÎƽÕû×÷Óã¬ÔÏÈ´æÔÚµÄС¾§ÃæÖð½¥±»Ïû³ý£¬×îÖյõ½¹âÁÁµÄ¶Æ²ã¡£µÚ¶þÖÖ»úÀí¼ÙÉè¹âÁÁ¼Á·Ö×ÓÄÜÓÅÏÈÎü¸½ÔÚ½ðÊôµç½á¾§Éú³¤½Ï¿ìµÄ¾§ÃæÉÏ£®¿ÉÄܶԵç³Á»ýÆð×軯×÷Óã¬Òò¶øµ¼Ö¶Ƽþ±íÃæ²»Í¬Î»ÖõÄÉú³¤ËÙ¶ÈÇ÷ÓÚÒ»Ö£¬¼ÓÉϼ¸ºÎÆ½ÒÆ×÷Óã¬ÖÕÓڵõ½¹âÁÁµÄ¶Æ²ã¡£
21£®ÊÔÐðÊöµç¶ÆÉú²úµÄÖ÷Òª¹¤ÒնԶƲãÖÊÁ¿µÄÓ°Ïì¡£
½â£ºµç¶ÆÉú²ú¹¤ÒÕÁ÷³ÌÒ»°ã°üÀ¨¶ÆÇ°´¦Àí¡¢µç¶ÆºÍ¶Æºó´¦ÀíÈý´ó²½¡£ ¶ÆÇ°´¦Àí£ºÊÇ»ñµÃÁ¼ºÃ¶Æ²ãµÄǰÌá¡£¶ÆÇ°´¦ÀíÒ»°ã°üÀ¨»úе¼Ó¹¤¡¢ËáÏ´¡¢³ýÓ͵Ȳ½Öè¡£»úе¼Ó¹¤ÊÇÖ¸ÓûúеµÄ·½·¨£¬³ýÈ¥¶Æ¼þ±íÃæµÄë´Ì¡¢Ñõ»¯Îï²ãºÍÆäËû»úеÔÓÖÊ£¬Ê¹¶Æ¼þ±íÃæ¹â½àƽÕû£¬ÕâÑù¿Éʹ¶Æ²ãÓëÕûÌå
14
½áºÏÁ¼ºÃ£¬·Àֹë´ÌµÄ·¢Éú¡£ÓÐʱ¶ÔÓÚ¸´ºÏ¶Æ²ã£¬Ã¿¶ÆÒ»ÖÖ½ðÊô¾ùÐëÏȽøÐиô¦Àí¡£³ý»úе¼Ó¹¤Å×¹âÍ⣬»¹¿ÉÓõç½âÅ×¹âʹ¶Æ¼þ±íÃæ¹â½àƽÕû¡£µç½âÅ×¹âÊǽ«½ðÊô¶Æ¼þ·ÅÈ븯ʴǿ¶ÈÖеȡ¢Å¨¶È½Ï¸ßµÄµç½âÒºÖÐÔڽϸßζÈÏÂÒԽϴóµÄµçÁ÷ÃܶÈʹ½ðÊôÔÚÑô¼«Èܽ⣬ÕâÑù¿É³ýÈ¥¶Æ¼þȱÏÝ£¬µÃµ½Ò»¸ö½à¾»Æ½ÕûµÄ±íÃæ£¬ÇҶƲãÓë»ùÌåÓнϺõĽáºÏÁ¦£¬¼õÉÙÂé¿ÓºÍ¿Õ϶£¬Ê¹¶Æ²ãÄÍÊ´ÐÔÌáÉУ¬µ«µç½âÅ׹ⲻÄÜ´úÌæ»úеÅ׹⡣
ËáÏ´µÄÄ¿µÄÊÇΪÁ˳ýÈ¥¶Æ¼þ±íÃæÑõ»¯²ã»òÆäËû¸¯Ê´Îï¡£³ýÓ͵ÄÄ¿µÄÊÇÏû³ý»ùÌå±íÃæÉϵÄÓÍÖ¬¡£
ÐèҪ˵Ã÷µÄÊÇÔÚ¶ÆÇ°´¦ÀíµÄ¸÷²½ÖèÖУ¬ÓÉÒ»µÀ¹¤ÐòתÈëÁíÒ»µÀ¹¤Ðò¾ùÐè¾¹ýˮϴ²½Öè¡£
µç¶Æ£º¶Æ¼þ¾¶ÆÇ°´¦Àí£¬¼´¿É½øÈëµç¶Æ¹¤Ðò¡£ÔÚ½øÐÐµç¶ÆÊ±»¹±ØÐë×¢Òâµç¶ÆÒºµÄÅä·½£¬µçÁ÷ÃܶȵÄÑ¡ÔñÒÔ¼°Î¶ȡ¢pHµÈµÄµ÷½Ú¡£ÐèҪ˵Ã÷µÄÊÇ£¬µ¥Ñεç½âÒºÊÊÓÃÓÚÐÎ×´¼òµ¥¡¢Íâ¹ÛÒªÇóÓÖ²»¸ßµÄ¶Æ²ã£¬ÂçÑεç½âÒº·ÖÉ¢ÄÜÁ¦¸ß£¬µç¶ÆÊ±µçÁ÷ÃܶȺÍЧÂʵͣ¬Ö÷ÒªÊÊÓÃÓÚ±íÃæÐÎ×´½Ï¸´ÔӵĶƲ㡣
¶Æºó´¦Àí£º¶Æ¼þ¾µç¶Æºó±íÃæ³£Îü¸½×ŶÆÒº£¬Èô²»¾´¦Àí¿ÉÄܸ¯Ê´¶Æ²ã¡£Ë®Ï´ºÍºæ¸ÉÊÇ×î¼òµ¥µÄ¶Æºó´¦Àí¡£ÊӶƲãʹÓõÄÄ¿µÄ£¬¶Æ²ã¿ÉÄÜ»¹ÐèÒª½øÐÐÒ»Ð©ÌØÊâµÄ¶Æºó´¦Àí¡£Èç¶ÆZn£¬CdµÄ¶Û»¯´¦ÀíºÍ¶ÆAgºóµÄ·À±äÉ«´¦ÀíµÈ¡£
22£®Ð´³öËáÐÔ¹âÁÁ¶ÆÍ¡¢¶ÆÁÁÄøºÍ¶Æ¸õµÄµç¼«·´Ó¦£¬²¢¼òҪ˵Ã÷¶ÆÒºÖи÷³É·ÖµÄ×÷Óá£
½â£º£¨1£©ËáÐÔ¹âÁÁ¶ÆÍÒõ¼«·´Ó¦ÓÉÁ½²½Á¬Ðø·Åµç·´Ó¦ºÍÒ»¸ö½á¾§¹ý³Ì×é³É£º
Cu2+?e?Cu? Cu+?e?Cu(Îü¸½£©±íÃæÀ©É¢Cu(Îü¸½£©?????Cu(¾§¸ñ)
£¨1£©
£¨2£©
£¨3£©
ËáÐÔ¹âÁÁ¶ÆÍÑô¼«·´Ó¦£º
Õý³£ÈܽⷴӦΪ£ºCu-2e?Cu2?
µ±Ñô¼«²»ÍêÈ«Ñõ»¯Ê±¿ÉÄܲúÉúCu£ºCu-e?Cu? Ñô¼«¸½½üCu»ýÀۻᵼÖÂÆç»¯·´Ó¦£º2Cu?++Cu?Cu2?
Ö÷ÑÎÁòËáͺ¬Á¿Ô½¸ß£¬ÔÊÐíµÄÒõ¼«µçÁ÷ÃܶÈÉÏÏÞÔ½´ó£¬µ«¶ÆÒºµÄ·ÖÉ¢ÄÜÁ¦ÔòËæÖ®Ï½µ£¬ËùÒÔ£¬Ò»Ï¸ÖµÄ×÷Óá£
£¨2£©¶ÆÁÁÄøµÄÒõ¼«·´Ó¦
ÁòËá¿ÉÌá¸ß¶ÆÒºµÄµ¼µçÐԺ͵çÁ÷ЧÂÊ£»ÊÊÁ¿ÂÈÀë×ÓÓÐÀûÓڵõ½¹âÁÁ¶Æ²ã£¬·ÀÖ¹Ñô¼«¼«»¯£»Ìí
Ni2+?2e?Ni
2H+?2e?H2
¶ÆÁÁÄøµÄÑô¼«·´Ó¦Ö÷ÒªÊÇ£º Ni?2e?Ni2?
15
ÄøÑô¼«¾ßÓнϸߵĶۻ¯ÄÜÁ¦£¬ÈôÑô¼«¼«»¯½Ï´ó£¬Ôò¿ÉÄܵ¼ÖÂÎö³öÑõÆø»òÂÈÆø£º
2H2O?4e?4H??O2?
»ò2Cl??2e?Cl2?
2+
ÁòËáÄøÊÇÖ÷ÑΣ¬²úÉúNiÀë×Ó£»¶øÂÈ»¯ÄøÒ²ÊÇÖ÷ÑΣ¬µ«·ÖÉ¢ÄÜÁ¦±ÈÁòËáÄøºÃ£¬Í¬Ê±ÂÈÀë×ÓÒ²
Ëá×÷Ϊ»º³å¼Á£¬²¢ÄܸÄÉÆÄø²ãºÍ»ùµ×½ðÊôµÄ½áºÏÁ¦£¨×¢Ò⣺pH=2-3ʱÒ×Îö³öÇâÆø£»ÌǾ«ÊÇ
¾§Á£³ß´ç£¬Ê¹¶Æ²ã¹âÁÁ£¬Í¬Ê±Ê¹Äø²ãÑÓÕ¹ÐԺã»1£¬4-¶¡È²¶þ´¼ÊǴμ¶¹âÁÁ¼Á£¬¾ßÓÐÔö¹âºÍ
ÁÁ¼Á£¬¾ßÓÐÔö¹âºÍÕûƽ×÷Óã¬ÇÒÕûƽ×÷ÓñÈ1£¬4-¶¡È²¶þ´¼¸üÏÔÖø£»Ê®¶þÍé»ù»ÇËáÄÆÊÇÈóʪ
£¨3£©¶Æ¸õµÄµç¼«·´Ó¦
Òõ¼«·´Ó¦Îª£ºCrO4+8H+6e¡úCr+4H2O
Ñô¼«·´Ó¦Îª£ºCr¡úCr+3eºÍÎöÑõ·´Ó¦2H2O?4e?4H3+
6+
2-+
Á¦£¬¼ÓËÙÇâÆøÅݴӶƲã±íÃæÊÍ·Å£¬·ÀÖ¹Õë¿×ºÍÂéµã²úÉú¡£´ËÍ⣬Òõ¼«ÐèÒªÒÆ¶¯£¬ÐèÒª½Á°è¡£
??O2?¡£
-1
¶ÆÒº×é³É£º×÷ΪÖ÷ÑεÄCr2O3(¸õôû)Ũ¶ÈΪ360¡«380g¡¤L£¬H2SO4Ũ¶ÈΪ2¡«2.5g¡¤L
-1
ºÍÖÂÃÜÐԺ㬵«µçÁ÷ЧÂʺͷÖÉ¢ÄÜÁ¦Ï½µ£»ÁòËáŨ¶ÈµÍʱ£¬·ÖÉ¢ÄÜÁ¦Ìá¸ß£¬µ«¹âÔó¶ÈÏÔÖøÏÂ
±ÈÒ»°ãΪ0.55¡«0.70£º100£¬ÕâÑùÒõ¼«ÉϲÅÄÜ»¹Ô³ö¸õ¡£ÁòËá¼ÓÈë¹ýÁ¿Ê±³£Í¨¹ýBaCO3ÖкͺÍÖØ¸õËáÐÎʽ´æÔÚ£¬×ª»¯¹ØÏµÎª£º
23£®ÊÔ˵Ã÷Ñô¼«ºÍÒõ¼«µçӾͿװµÄÖ÷ÒªÔÀíºÍͿӾͿװµÄÓŵ㡣 ½â£ºÑô¼«µçӾͿװ(anodic electrophoretic coating£¬¼ò³ÆAEC)ÊÇÒÔ±»¶Æ½ðÊô»ùµ××÷ΪÑô¼«£¬´øµçµÄÒõÀë×ÓÊ÷Ö¬Ôڵ糡×÷ÓÃϽøÐж¨ÏòÒÆ¶¯£¬´Ó˳ÔÚ½ðÊô±íÃæÊµÏÖµç³Á»ýµÄ·½·¨¡£Ñô¼«µçӾͿװµÄÒõÀë×ÓÊ÷Ö¬Ö÷ҪΪ±ûÏ©ËáϵÁС£¶ÔÓÚÑô¼«µçÓ¾¹ý³Ì£¬Ñô¼«·´¸®¿ÉÄܰéËæºÏÇâÆøµÄÎö³öºÍ½ðÊôµÄÑõ»¯£¬µçÓ¾¹ý³Ì±íʾΪ£º
Òõ¼«µçӾͿװ(cathodic electrophoretic coating£¬¼ò³ÆCEC)ÊÇÒÔË®ÈÜÐÔÑôÀë×ÓÊ÷֬Ϊ³ÉĤ»ùÁÏ£¬ÒÔ¹¤¼þ×÷ΪÒõ¼«£¬´Ó¶øÔÚ½ðÊô±íÏòʵÏÖµç³Á»ýµÄÒ»ÖÖµç¶Æ·½·¨¡£Òõ¼«µçӾͿװµÄÑôÀë×ÓÊ÷Ö¬µçÓ¾ÆáÖ÷ҪΪ»·ÑõÊ÷֬ϵÁкÍÒìÇèËáÖ¬µÄ»ìºÏÎï¡£¶ÔÓÚÒõ¼«µçÓ¾¹ý³Ì£º
16
µçӾͿװ(electrophoretic coating£¬¼ò³ÆEC)¼¼ÊõÓë½ðÊôµç¶Æ²»Í¬µÄÊÇ£¬µçÓ¾ÆáÈÜÒºÖдý¶ÆµÄÒõ¡¢ÑôÀë×ÓÊÇÓлúÊ÷Ö¬£¬¶ø²»ÊǽðÊôÀë×Ó¡£µçӾͿװÒÔˮΪÈܼÁ£¬¼ÛÁ®Ò׵ã»ÓлúÈܼÁº¬Á¿ÉÙ£¬¼õÉÙÁË»·¾³ÎÛȾºÍ»ðÔÖ£»µÃµ½µÄÆáĤÖÊÁ¿ºÃÇÒºñ¶ÈÒ׿ØÖÆ£®Ã»Óкñ±ß¡¢Á÷¹ÒµÈ±×²¡£¬Í¬Ê±Í¿ÁÏÀûÓÃÂʸߣ¬Ò×ÓÚ×Ô¶¯»¯Éú²ú£¬ÕýÊÇÓÉÓÚÕâЩÓŵ㣬µçӾͿװÒѹ㷺ӦÓÃÓÚÆû³µ¡¢×ÔÐгµ¡¢µç·çÉȵȽðÊô±íÃæµÄ¾«ÊΡ£
24£®298Kʱ£¬Óû»ØÊÕµç¶ÆÒºÖеÄÒø£¬·ÏÒºÖÐAgNO3Ũ¶ÈΪ10
2+
-6
mol?L?1£¬
»¹º¬ÓÐÉÙÁ¿µÄCu´æÔÚ£¬ÈôÒÔʯīΪÒõ¼«£¬ÒøÓÃÒõ¼«µç½â·¨»ØÊÕ£¬ÒªÇó»ØÊÕ99£¥µÄÒø£®¼ÆÆäÖÐÁî»î¶ÈϵÊý?1£½1¡£ÊÔ¼ÆË㣺(1) CuµÄŨ¶ÈÓ¦µÍÓÚ
2+
¶àÉÙ£¿(2)Òõ¼«µçλӦ¿ØÖÆÔÚʲô·¶Î§ÄÚ£¿(3)ʲôÇé¿öÏÂCuÓëAgͬʱÎö³ö?
½â£º(1)ÉèÒøÓëÍÔÚʯīµç¼«Îö³öµÄ³¬µçλΪÁ㣬ҪÇó»ØÊÕ99£¥µÄÒø,Ò²¾ÍÊÇҪʹÈÜÒºÖеÄAgNO3Ũ¶ÈϽµµ½Îª10
??Ag-8
mol?L?1£¬ÓÚÊÇ
+/Ag?RTaAg8.314?298.151ln£½0.7994-ln?8£½0.3271V zFaAg?1?96500102+??Ag£¬Îö³ö£½?Cu,Îö³ö£½?Cu?/CuRTRTlnaCu2?£½0.3400?lncCu2??0.3271VzFzFcCu2??0.3654mol?L?1
(2) ·ÏÒºÖÐAgNO3Ũ¶ÈΪ10
-6
mol?L?1£¬¿ªÊ¼Îö³öµÄµçλΪ
??Ag+/Ag?RTaAg8.314?298.151ln£½0.7994-ln?6£½0.4451V zFaAg?1?9650010ËùÒÔ£¬Òõ¼«µÄµçλӦ¿ØÖÆÔÚ0.3271V-0.4451V·¶Î§ÄÚ¡£
(3)Ö»ÓÐÔÚ?Ag£¬Îö³ö£½?Cu,Îö³öÇé¿öÏÂCuÓëAgͬʱÎö³ö£¬ÎªÊ¹Á½ÕßÎö³öµçλÏàµÈ£¬¿ÉÒÔͨ¹ýÌá¸ßCuÀë×ÓŨ¶È»ò¼ÓÈëÂçºÏ¼Á¡£
25£®ÒÔ¶èÐԵ缫µç½âSnCl2µÄË®ÈÜÒº½øÐÐÒõ¼«¶ÆSn£¬Ñô¼«ÖвúÉúO2£¬ÒÑÖª£º
2+
aSn2??0.10£¬aH?£½0.010£¬Ñô¼«ÉÏ?O2
-0.140V£»?H?,H2O/O2£½0.50V£¬ÒѼӣº?Sn2?/Sn£½
£½1.23V£¬£¨1£©ÊÔд³öµç¼«·´Ó¦£¬²¢¼ÆËãʵ¼Ê·Ö½â
µçѹ£¬(2)Èô?H2,Òõ£½0.40V£¬ÊÔÎÊҪʹaSn2?½µµÍµ½ºÎֵʱ²Å¿ªÊ¼Îö³öÇâ?
17
½â£º£¨1£©Ñô¼«·´Ó¦£º2H2O?4e?4HÒõ¼«·´Ó¦£º2Sn2???O2?
?4e?2Sn
?O£½?H2??,H2O/O24?aH/p?)?(0.21pRT?ln4?965001 ?1.23???Sn£½?Sn2?8.314?298.15ln(0.014?0.21)£½1.1017 V
4?96500RT2 lnaSn2?4?96500?/Sn ??0.140?ʵ¼Ê·Ö½âµçѹ£½
8.314?298.15ln(0.102)£½-0.1695 V
4?9650022?O+?O-?Sn£½1.1.017+0.50-£¨-0.1695£©£½1.7707
(2) 4H??4e?2H2
??Sn??H??Sn2?/H2
RTRT2?lnaSn?lnc4??H22????H/H2 4?96500zF8.314?298.158.314?298.152?0.14?lnaSnln0.014?0.402??0?4?965004?96500?/Sn½âµÃ£º
aSn2??1.536?10?13
26£®ÔÚÂÁµÄÑô¼«Ñõ»¯ÖзֱðÒªµÃµ½ºñ²ãºÍ±¡²ãÑõ»¯ÎïĤ¶ÔÓ¦µÄµç½âÖÊÌõ¼þ¡¢Î¶ÈÌõ¼þºÍµçÁ÷ÃܶÈÌõ¼þ¸÷ÊÇʲô?²¢ËµÃ÷ÔÒò¡£
½â£ºÂÁ¼°ÆäºÏ½ðµÄÑô¼«Ñõ»¯Ò»°ãÊÇÔÚÈܽâÄÜÁ¦¸ßµÄÁòËá¡¢¸õËá¡¢²ÝËá»òÁ×ËáµÈµç½âÒºÖÐʵÏֵģ¬ÔÚÕâЩµç½âÒºÖÐÂÁ¼°ÆäºÏ½ðµÄÑô¼«Ñõ»¯ºñ¶ÈÓÐʱ¿É´ïµ½500¦Ìm¡£ÀýÈç¶ÔÓÚ10£¥¡«20£¥µÄÁòËáµç½âÒºÖÐÂÁ¼°ÆäºÏ½ðµÄÑô¼«Ñõ»¯£¬Ò»°ãµç½âÌõ¼þΪ£ºµçÁ÷ÃܶÈ190¡«250A¡¤dm£¬¹¤×÷ζÈΪ15¡«25¡æ£¬µç½âʱ¼äΪ20¡«60min¡£Ñô¼«Ñõ»¯Ê±Ò»°ã²ÉÓÃֱͨÁ÷µçµÄ·½Ê½£¬Ê©¼ÓµçѹÊÓµç½âÒºµÄµ¼µçÐÔ¡¢Î¶ÈÒÔ¼°ÈܽâÓÚÆäÖеÄÂÁµÄº¬Á¿¶ø¶¨£¬´óÖÂΪ12¡«28V¡£ÔÚÂÁ¼°ÆäºÏ½ðµÄÑô¼«»¯¹ý³ÌµÄËùÓÐÓ°ÏìÒòËØÖУ¬ÆðÖ÷µ¼×÷ÓõÄÊÇÁòËáµÄŨ¶ÈºÍ¹¤×÷ζȡ£µ±Ñô¼«¼«»¯Ôڽϵ͵ÄÁòËáŨ¶ÈºÍζÈϽøÐÐʱ£¬¿ÉÒԵõ½½ÏºñºÍ½ÏÓ²µÄĤ²ã¡£ÈôͬʱÌá¸ßµçÁ÷Ãܶȣ¬ÔòĤ²ãÓ²¶ÈËä¿É½øÒ»²½µÃµ½Ìá¸ß£¬µ«Ä¤²ãÒײúÉúȱÏÝ£¬µ¼ÖÂÑõ»¯ÎïĤ·À»¤¼þÄÜϽµ¡£µ±µç½âÒºµÄŨ¶ÈºÍζÈÒ»¶¨Ê±£¬Ñõ»¯Ä¤µÄºñ¶ÈÈ¡¾öÓÚËùÓõĵçÁ÷ÃܶȺÍÑõ»¯Ê±¼ä£¬¼´Ñõ»¯µçÁ¿£¬ËùÒÔ³£ÓÃͨ¹ýµçÁ¿À´¿ØÖÆÄ¤²ãµÄºñ¶È¡£
µÚÎåÕ ÎÞ»úÎïµÄµç½â¹¤Òµ
27£®ÊÔд³öÏÂÁÐÎÞ»úµçºÏ³ÉµÄµç¼«·´Ó¦ºÍ×Ü·´Ó¦¡£
(1)Ë®ÒøÊ½µç½â²ÛÖÆ±¸C12ºÍNaOH (2)ÂÈËá¼ØµÄµçºÏ³É (3)¶þÑõ»¯Ã̵ĵçºÏ³É (4)¼îÐÔ½éÖÊÖеç½âË® £¨5£©¸ßÂÈËáÄÆµÄµçºÏ³É £¨6£©ÓÉMnO2µÄÃÌ¿óµç½â
18
-2
ºÏ³ÉKMnO4
½â£º(1) Ë®ÒøÊ½µç½â²ÛÖÆ±¸C12ºÍNaOH£º ¹¯²ÛÑô¼«·´Ó¦Îª£ºCl
?1?Cl2?e
2
(2)ÂÈËá¼ØµÄµçºÏ³É£º ÔÚÖÐÐÔKClÈÜÒºÖÐͨ¹ýÖ±Á÷µçʱ£¬ÔÚÒõ¼«·ÅµçµÄÊÇÎö³öµçλµÍµÄH
Ê®
Àë×Ó£¬¼´
H2
2HÊ®2eÔÚÑô¼«£¬ÔòÊÇCl·Åµç£º2ClÒ»
??Cl2?2e
Ò»
ÓÉÓÚµç½âÎÞ¸ôĤ£¬C12ÓëOH·´Ó¦ Cl2Ê®2OH £¨A£© ClO
Ò»
Ò»
ClOʮCl
Ò»Ò»
£« H2O
¼ÌÐøÔÚÑô¼«ÉϷŵçÉú³ÉÂÈËáÑΡ£
? 6ClO(B)
3?3H2O?6e??2ClO3??6H??4Cl??O2
2(3) ¶þÑõ»¯Ã̵ĵçºÏ³É
(4)¼îÐÔ½éÖÊÖеç½âË®:
×Ü·´Ó¦Îª£ºH2O?1/2O2?H2
£¨5£©¸ßÂÈËáÄÆµÄµçºÏ³É£º
19
ÕâÒ»µçºÏ³ÉµÄ·´Ó¦Îª£ºNaClO3+H2O-2e £¨6£©ÓÉMnO2µÄÃÌ¿óµç½âºÏ³ÉKMnO4£º
µç½âÃÌËá¼ØÈÜÒº¿ÉÖÆµÃKMnO4¡£ÃÌËá¼ØÓû¯Ñ§·½·¨ÖƱ¸£¬ÔÁÏΪÈíÃ̿󣨴óÔ¼º¬60£¥MnO2£©£¬½þÈë50£¥¡«80£¥µÃKOHÈÜÒº¼ÓÈÈÖÁ200¡«700¡æ²¢ÓÉ¿ÕÆøÑõ»¯ÎªK2MnO4¡£
NaClO4 + 2H
£«
28.ij¹¤³§Äâ²ÉÓùýÁòËá·¨£¬Éè¼ÆÄê²ú10000 ¶Ö30£¥H2O2µÄ³µ¼ä£¬ÊÔ¼ÆËãÆäµçÄÜÏûºÄÏàÉ豸µÈ²ÎÊý¡£ÉèË®½âЧÂÊΪ80£¥£¬µçÁ÷ЧÂÊΪ85£¥¡£ ½â£º£¨1£©Äê²ú´¿Ë«ÑõË®Á¿£½10000*0.30£½3000¶Ö£½3000*10/34£½88235294 mol
£¨2£©ÀíÂÛÉÏ1 molË«ÑõË®¿ÉÓÉ1 mol¹ýÁòËá¼Óˮˮ½â¶ø³É£¬µ«Ë®½âЧÂÊΪ80£¥£¬Òò´Ëʵ¼ÊÐèÒªÉú²úµÄ¹ýÁòËáµÄÊýÁ¿Îª£ºnH/0.80£½11029412 mol
£¨3£©ÀíÂÛÉÏͨÈë2F£¨1F£½26.8 A¡¤h£©µçÁ¿£¬¾Í¿ÉÉú²ú1 mol¹ýÁòËᣬÔòÄê²ú11029412 molÐèÒªµçÁ¿Îª
Äê²úÀíÂÛµçÁ¿£½11029412*2*26.8£½591176483 A¡¤h µ«µçÁ÷ЧÂÊΪ85£¥£¬
Äê²úʵ¼ÊÓõçÁ¿£½11029412*2*26.8/0.85£½695501745 A¡¤h
£¨4£©Ã¿Äê°´310ÌìÉú²ú°²ÅÅ£¬Ã¿Ìì°´24Сʱ¼ÆË㣬Ôòµç½â²ÛÿСʱͨ¹ýµÄµçÁ÷Á¿Îª£º
I=695501745/(310*24)= 93481.417 A µç½â²ÛÿÃëͨ¹ýµÄµçÁ÷Á¿£½93481.417/3600£½25.96706 A
(5) ¸ù¾ÝÉÏÊö¼ÆËã½á¹û£¬Ã¿Ãëͨ¹ýµÄµçÁ÷Ϊ25.96706 A£¬Ôò¿ÉÑ¡ÔñWeissenstein¹¤ÒÕÀ´Éú²úË«ÑõË®¡£ÆäËû²ÎÊý²Î¼ûp175±í5.6¡££¨Éè¼Æ»¹Ó¦²Î¿¼ÓйØ×¨Öø£©¡£
29ÊԱȽÏÂȼҵÖÐÈýÖÖ²»Í¬µç½â²ÛµÄÓÅȱµã¡£
½â£º1£©²ÉÓÃÐÎÎÈÑô¼«(DSA)µÄ¸ôĤÔãµç½â·¨µÄÓŵãÊÇ£ºÐÎÎÈÑô¼«(DSA)ÊDz»Êܸ¯Ê´£¬³ß´çÎȶ¨£¬ÊÙÃü³¤£¬Âȳ¬µçÊÆºÜµÍ£¬¶øÑõ³¬µçÊÆÈ´¸ß£¬Òò¶øËùµÃC12ºÜ´¿£¬¶øÇÒ²ÛµçѹҲ½ÏµÍ£¬½µµÍµçÄÜÏûºÄ´ï10£¥£¬Ìá¸ßÉ豸Éú²úÄÜÁ¦´ï50£¥¡£
²»×ãÖ÷Òª±íÏÖÔÚ£º(1)ËùµÃ¼îҺϡ£¬Ô¼10£¥×óÓÒ£¬ÐèŨËõÖÁ50£¥²Å¾Í³öÊÛ£»(2)¼îÒºº¬ÔÓÖÊCl£¬¾Å¨ËõºóÔ¼ÖÁ1£¥×óÓÒ£»(3)µç½â²Ûµç×è¸ß£¬µçÁ÷Ãܶȵͣ¬Ô¼0.2A¡¤cm£»(4)ʯÃÞ¸ôĤÊÙÃü¶Ì£¬³£Ö»Óм¸¸öÔÂÖÁÒ»Äê×óÓÒ£¬Òò´Ë³£Ðè¸ü»»¡£
20
-2-6
2S2O2?88235294
2£©¹¯²Û·¨µÄÖ÷ÒªÓŵãÊÇËùµÃ¼îÒºµÄŨ¶È¸ß£¬½Ó½ü50£¥£¬¿ÉÖ±½Ó×÷ΪÉÌÆ·³öÊÛ£¬¶øÇÒ´¿¶È¸ß£¬¼¸ºõ²»º¬Cl¡£ÆäÖ±Á÷µçÄÜÏûºÄËä¸ß£¬µ«Ëü²»ÐèÒªÕô·¢Å¨Ëõ¼îÒºµÄºó´¦Àí²Ù×÷£¬¹Êÿ¶Ö¼îµÄ×ÜÄܺÄÈÔºÍÆäËû¶þ·¨Ïà·Â£¬¶øÇÒ¶ÔÑÎË®µÄ¾»»¯ÒªÇó²»Ïñ¸ôĤ²ÛÄÇÑù¸ß¡£´ÓÉú²úÄÜÁ¦ÉÏ¿´£¬¹¯²ÛµÄÓÅÔ½ÐÔÊÇËùÓõçÁ÷Ãܶȴ󣬶øÇÒ¿É´ó·ù¶ÈµØ±ä»¯£¬¿É±Ü¿ª³ÇÊÐÓõç¸ß·å£¬ËæÊ±µ÷ÕûµçÁ÷Ãܶȡ£
¹¯²ÛµÄÖ÷ҪȱµãÊÇÓй¯¶¾ÎÛȾ»·¾³£¬±ØÐëÑϸñ¿ØÖÆÅÅ·ÅÎÛË®Öй¯µÄº¬Á¿£¬°´Ê±¼ì²é²Ù×÷¹¤È˽¡¿µ×´¿ö£¬¼ÓÇ¿ÀͶ¯±£»¤´ëÊ©¡£¹¯µÄ¼Û¸ñ¹ó£¬Í¶×Ê´óÊÇËüµÄÁíһȱµã¡£
3£©Àë×ÓĤ²Ûµç½âµÄÓŵãÊÇûÓй¯ºÍʯÃ޵Ĺ«º¦£»ËùµÃNaOH²»º¬C1£¬ºÜ´¿£¬ÆäŨ¶È¿É´ï10£¥¡«40£¥£¬¹ÊÕô·¢Å¨ËõµÄºó´¦Àí·ÑÓÃÒªÉٵöࣻµçÁ÷ÃܶȿɱȸôĤ²ÛËùʹÓõĴóÒ»±¶¶øÈÔ±£³Ö3.1¡«3.8VµÄ²Ûµçѹ£¬¡°×ÜÄܺġ±(°üÀ¨µç½âÓõ硢¶¯Á¦ÓõçºÍÕôÆøÏûºÄ)Ïà¶Ô½ÏµÍ£¬Ò»°ã½Ï¸ôĤ·¨ºÍ¹¯·¨µÍ25£¥ÒÔÉÏ£¬Ã¿Éú²úÒ»¶ÖÉÕ¼î¿ÉÊ¡½Ó½ü1000kW¡¤hµç¡£³£°Ñ¶àÖ»µç½â³Ø»ã¼¯×é×°³ÉѹÂË»úʽµÄµç½â²Û£¬µ¥²Û²úÂʿɴï100¶ÖNaOH/Ä꣬¶ø²ÛÌå»ý±ÈǰÁ½·¨µÄµç½â²ÛСµÃ¶à£¬ÌرðÊÊÓÃÓÚС¹æÄ£Éú²ú£¬²úÁ¿¿É°´Êг¡ÐèÒªµ÷½Ú¡£ÆäȱµãÊÇÀë×ÓĤµÄʹÓÃÊÙÃü¼°Àë×ÓĤµÄÐÔÄÜ»¹ÐèÒª¸Ä½ø¡£ÀýÈ磬ΪÁËÑÓ³¤Àë×ÓĤµÄʹÓÃÊÙÃü£¬ÒªÇó°ÑÑÎË®¸öµÄ¸Æ¡¢Ã¾Àë×Óº¬Á¿½µµÍÁÐ10ÊýÁ¿¼¶¡£Ä¿Ç°£¬Àë×ÓĤµÄÐÔÄÜ»¹ÔÚ²»¶Ï¸Ä½ø£¬´Ó·¢Õ¹Ç÷ÊÆ¿´£¬Àë×ÓÖ¬²Û×îÓÐÉúÃüÁ¦£¬Ô¤¼ÆÔÙ¹ýÈô¸ÉÄ꽫ȫ²¿È¡´úÆäËûµÄµç½â²Û¡£
30. ijÂȼÓÃË®Òø²Ûµç½âʳÑÎË®£¬Í¨¹ýµç½â²ÛµÄµçÁ÷Ϊ100 kA¡£ÎÊÀíÂÛÉÏÿÌìÉú²ú¶àÉÙC12¡¢H2ºÍNaOH£¿ÉèÑô¼«µÄµçÁ÷ЧÂÊΪ97£¥£¬ÎÊʵ¼ÊÉÏÿÌìÉú²ú¶àÉÙC12£¿(ÒÔÿÌì24h¼Æ)
½â£º£¨1£©ÉèµçÁ÷ЧÂÊΪ100£¥£¬ÂÈÆø°´ÏÂʽ·´Ó¦Ê½¼ÆË㣺
-6
-1
-
2Cl??2e?Cl2(g)
ÿÌìµçÁ¿Q=It=100000¡Á24¡Á60¡Á60£½8640000000 C ÿÌìÉú²úµÄC12£ºN(ÂÈÆø)£½Q/£¨2¡Á96485£©£½44773.80 mol
ÿÌìÉú²úµÄNaOH£ºN(NaHgm)= Q/£¨2¡Á96485£©=89547.60 mol= N(NaOH) ÿÌìÉú²úµÄH2£¨Æø£©£ºN£¨ÇâÆø£©= N(NaHgm)/2=44773.80 mol £¨2£©ÉèÑô¼«µÄµçÁ÷ЧÂÊΪ97£¥£¬Ã¿ÌìÉú²úµÄC12£º
N(ÂÈÆø)£½¦ÇQ/£¨2¡Á96485£©£½0.97¡Á44773.80 mol£½43430.59 mol
31£®ÂȼҵÖÐÔÚÒõ¼«µÄ¸Ä½ø·½Ã棬ÈËÃÇÊÔÓÃÑõ»¹Ô·´Ó¦´úÌæHµÄÎö³ö
+
1/2O2·´Ó¦£¬¼´ÓÃÒõ¼«·´Ó¦£º
´úÌæÎöÇâ·´Ó¦¡£
?H2O?2e£½2OH?£¬???0.410V£¬
(1)ÊÔд³ö¸Ä½øÇ°µÄµç¼«·´Ó¦ºÍ×Ü·´Ó¦£»(2)Òõ¼«·´Ó¦¸Ä½øºó£¬ÀíÂÛ·Ö½âµçѹ½µµÍ¶àÉÙ£»(3)ÀíÂÛÉÏ£¬Ã¿Éú²ú1¶ÖNaOH½«½ÚÔ¼¶àÉÙµçÄÜ?
½â£º(1)¸Ä½øÇ°£ºÑô¼«·´Ó¦Îª£º
2Cl??Cl2?2e ???1.36V
21
Òõ¼«·´Ó¦Îª£º
2H2O?2e?H2?2OH? ???-0.828V
×Ü·´Ó¦Îª£º
2NaCl?2H2O?2e?H2?2NaOH?Cl2
E??2.188V
(1)¸Ä½øºó£º Ñô¼«·´Ó¦Îª£º
2Cl??Cl2?2e ???1.36V
1O?H2O?2e?2OH? ???0.401V 21 ×Ü·´Ó¦Îª£º2NaCl?O2?H2O?2e?2NaOH?Cl2
2E??0.959 V
Òõ¼«·´Ó¦Îª£º
ÀíÂÛ·Ö½âѹ½µµÍ£½2.188-0.959£½1.229 V (3)ÀíÂÛµçÄܼÆË㹫ʽΪ£ºW?Ee?(m/M)zF
ÀíÂÛÉÏÿÉú²ú1¶ÖNaOH½«½ÚÔ¼µçÄÜ
£½£¨2.188-0.959£©*£¨1000*1000/40£©*2*26.8£½1646860 V¡¤A¡¤h £½1646860 W¡¤h=1646.860 kW¡¤h
µÚ°ËÕ µç»¯Ñ§¸¯Ê´Óë·À»¤
43£®¼ò»¯µÄ?¡ªpHͼÒÔ¿ÉÈÜÐÔÀë×ÓŨ¶È×ܺÍСÓÚ10 mol¡¤L×÷Ϊ½ðÊô
-6
-1
¸¯Ê´Óë·ñµÄ½çÏÞ£¬ºÏÀíÂð?ʲô½Ð¸¯Ê´Çø£¬¶Û»¯ÇøºÍÎȶ¨Çø? ´ð£ººÏÀí¡£
44.
?(?Cu?2?/Cu)±È
?(?H??/H2,Pt)¸ß£¬ÎªÊ²Ã´CuÔÚ³±Êª¿ÕÆøÖÐÒ²¸¯
Ê´??(Ti2?/Ti)±È??(H?/H2,Pt)µÍµÃ¶à£¬µ«ÎªÊ²Ã´TiÓÖÊÇÁ¼ºÃµÄÄ͸¯Ê´½ðÊô?
?(?Zn2?/Zn)±È?(H?/H2,Pt)µÍ£¬µ«ÓÃZnͰʢˮ²»¸¯Ê´£¬ÎªÊ²Ã´?
?(H?/H2,Pt)½â£º(1)¾¡¹Ü??(Cu2?/Cu)±È?¸ß£¬ CuÔÚ³±Êª¿ÕÆøÖлᱻ¸¯Ê´£¬
Ò²¾ÍÊÇÍÔÚº¬ÓÐÈܽâÑõµÄËá»òË®½éÖÊÖÐÔò¿ÉÄܱ»¸¯Ê´£¬ÒòΪ͵ıê×¼µç¼«µçλ±ÈÑõµÄÁ½¸ö·´Ó¦µÄ±ê×¼µç¼«µçλ¾ù¸ü¸ºÒ»Ð©¡£ £¨2£©¾¡¹Ü??(Ti2?/Ti)±È??(H?/H2,Pt)µÍµÃ¶à£¬ÒòΪîÑÔÚ¿ÕÆøÖÐÉú³ÉÁËÒ»²ã
Ñõ»¯Ä¤TiO2£¬Õâ²ãÑõ»¯Ä¤¸ô¾ø³±Êª¿ÕÆø£¬½«±£»¤½ðÊôîÑÃâÔⸯʴ¡£ £¨3£©¾¡¹Ü??(Zn2?/Zn)±È??(H?/H2,Pt)µÍ£¬µ«Ð¿ÔÚÔÚ³±Êª¿ÕÆøÖÐÈÝÒ×Éú³ÉÒ»
22
²ãÖ÷ÒªÓɼîʽ̼Ëáп×é³ÉµÄ±¡Ä¤£¨3Zn(OH)2?ZnCO3£©£¬Õâ²ã±¡Ä¤ÓÐÒ»¶¨µÄÄ͸¯Ê´ÄÜÁ¦£¬Òò´ËÓÃZnͰʢˮ²»¸¯Ê´¡£
45£®¾Ö²¿¸¯Ê´Ö»ÊǾֲ¿ÎÊÌ⣬Ϊʲô˵±È¾ùÔȸ¯Ê´¸üΪÑÏÖØ?ÊÔ¾ÙÀý˵Ã÷¿×Ê´£¬¾§¼ä¸¯Ê´£¬Ó¦Á¦¸¯Ê´ÆÆÁѺͳåÊ´¡£
½â£ºÕâÖÖ½ðÊô/ÈÜÒº½çÃæµÄ²»¾ùÒ»ÐÔÊDzúÉú¾Ö²¿¸¯Ê´µÄÔÒò¡£¾Ö²¿¸¯Ê´µÄΣº¦±È¾ùÔȸ¯Ê´ÒªÑÏÖØµÃ¶à£¬ÒòΪ½ðÊô¸¯Ê´µÄÑô¼«·´Ó¦ºÍ¹²¶óÒõ¼«·´Ó¦£¬ÓÉÓÚ½ðÊôÈÜ/Òº½çÃæµÄ²»¾ùÒ»¶ø²úÉúÁ˿ռä·ÖÀ룬Ñô¼«·´Ó¦ÍùÍùÔÚ¼«Ð¡µÄ¾Ö²¿·¶Î§ÄÚ·¢Éú£¬´Ëʱ×ܵÄÑô¼«ÈܽâËÙÂÊËäÈ»ÈԾɵÈÓÚ×ܵĹ²¶óÒõ¼«·´Ó¦ËÙÂÊ£¬µ«ÊÇÒõ¼«µçÁ÷ÃܶÈ(µ¥Î»Ãæ»ýÄڵķ´Ó¦ËÙÂÊ)È´´ó´óÔö¼ÓÁË£¬¼´¾Ö²¿µÄ¸¯Ê´Ç¿¶È´ó´ó¼Ó¾çÁË¡£ÀýÈçÒ»¸ù¾ùÔȸ¯Ê´µÄÌú¹Ü¿ÉÒÔÁ¬ÐøÊ¹Óúܳ¤Ê±¼ä¶øÎ޴󰣬µ«Èç¾Ö²¿¸¯´©¾ÍÖ»Äܱ¨·Ï¡£
¢Ù¿×Ê´ÊÇÔÚ½ðÊô²ÄÁϱíÃæÉÏÐγÉÖ±¾¶Ð¡ÓÚ1mm£¨²¢Ïò°åºñ·½Ïò·¢Õ¹µÄ£©Ð¡¿×µÄ¾Ö²¿¸¯Ê´¡£ÀýÈ磬½ðÊô²ÛÄÚ½éÖÊ·¢Éúй©£¬´ó¶àÊÇ¿×Ê´Ôì³ÉµÄ£¬¶øÇÒËüµÄ·¢Õ¹ËÙ¶ÈÒ²ÊǺܿìµÄ£¬´ó¶àΪÿÄêÊýºÁÃס£
¢Ú¾§¼ä¸¯Ê´ÊÇÓɾ§½çµÄÔÓÖÊ»ò¾§½çÇøÄ³Ò»ºÏ½ðÔªËØÔö¶à»ò¼õÉÙÒýÆðµÄ£¬ÑØ×ŽðÊô²ÄÁϵľ§½ç²úÉúµÄÑ¡ÔñÐÔ¸¯Ê´£¬¾¡¹Ü¾§Á£¼¸ºõ²»·¢Éú¸¯Ê´£¬µ«ÈÔÈ»µ¼Ö²ÄÁÏÆÆ»µ¡£ÀýÈ磬²»Ðâ¸Öƶ¸õÇø²úÉúµÄ¾§¼ä¸¯Ê´£¬ÊÇÓÉCr23C6µÈ̼»¯ÎïÔÚ¾§½çÎö³ö£¬Ê¹¾§½ç½üÅԵĸõº¬Á¿½µµ½°Ù·ÖÖ®¼¸ÒÔÏ£¬¹ÊÕⲿ·ÖÄÍÊ´ÐÔ½µµÍ¡£¢ÛÓ¦Á¦¸¯Ê´ÆÆÁÑÊÇÖ¸Ó¦Á¦ºÍÒ»ÖÖÔÚÌØ¶¨¸¯Ê´½éÖʹ²Í¬´æÔÚ¶øÒýÆðµÄÆÆÁÑ¡£ÆÆÁÑÓÐÑØ¾§(¾§½çÆÆÁÑ)ºÍ´©¾§(¾§Á£ÆÆÁÑ)Á½ÖÖ¡£ËüÃǶÔÓÚÊÜÓ¦Á¦µÄÆ÷еΣº¦×î´ó£¬Èç¸ßѹ¹øÂ¯¡¢·É»úÉϲàÃæ±¡±Ú¡¢¸ÖË÷¡¢»úÆ÷µÄÖáµÈ£¬Èç¹û·¢ÉúÕâÀฯʴ¾Í¿ÉÄÜͻȻ±ÀÁѶøÄð³Éʹʡ£
¢Ü³åÊ´ÊÇÔÚ³å»÷µÄ»úе×÷ÓÃÏ£¬²ÄÁϱíÃæ·¢ÉúÄ¥ËðµÄͬʱÓÖ¼ÓÈ븯ʴ×÷Óã¬Á½ÕßÏ໥´Ù½ø£¬²úÉúÑÏÖØµÄÇÖÊ´¡£ÆøÏàÁ÷ÌåÖеÄÒºµÎ¡¢ÒºÏàÁ÷ÌåСµÄ¹ÌÌå·Ûľ¡¢ÒºÌåÖÐÐýÎвúÉúµÄ¿ÕѨ¡¢Íä¹ÜµÈ²¿Î»·¢ÉúµÄÎÐÁ÷µÈ£¬¶¼ÄÜÆÆ»µ±íÃæÄ¤£¬¼ÓËÙ¸¯Ê´¡£±©Â¶ÔÚÔ˶¯Á÷ÌåÖеÄËùÓÐÀàÐ͵ÄÉ豸¶¼»áÔâµ½³åÊ´¡£Èç¹ÜµÀϵͳµÄÍäÍ·¡¢·§ÃÅ¡¢±ÃµÈ¡£
46.ʲô½Ð¾Ö²¿µç³ØºÍ¾Ö²¿µçÁ÷£¬ËüÃÇÓë¾Ö²¿¸¯Ê´¹ØÏµÈçºÎ£¿ÊÔÁбí¶Ô¾ùÔȸ¯Ê´ºÍ¾Ö²¿¸¯Ê´µÄÌØµã¼ÓÒÔ˵Ã÷¡£
½â£ºµ±½ðÊô±íÃæº¬ÓÐһЩÔÓÖÊʱ£¬ÓÉÓÚ½ðÊôµÄµçÊÆºÍÔÓÖʵĵçÊÆ²»¾¡Ïàͬ£¬¿É¹¹³ÉÒÔ½ðÊôºÍÔÓÖÊΪµç¼«µÄÐí¶à΢СµÄ¶Ì·µç³Ø£¬³ÆÎªÎ¢µç³Ø(»ò¾Ö²¿µç³Ø)£¬´Ó¶øÒýÆð¸¯Ê´²úÉú¾Ö²¿µçÁ÷¡£¶ÔÓÚ¾Ö²¿µç³Ø»ò¾Ö²¿¸¯Ê´¶øÑÔ£¬ÓÉÓÚ½ðÊôÈÜ/Òº½çÃæµÄ²»¾ùÒ»¶ø²úÉúÁ˿ռä·ÖÀ룬Ñô¼«·´Ó¦ÍùÍùÔÚ¼«Ð¡µÄ¾Ö²¿·¶Î§ÄÚ·¢Éú£¬´Ëʱ×ܵÄÑô¼«ÈܽâËÙÂÊËäÈ»ÈԾɵÈÓÚ×ܵĹ²¶óÒõ¼«·´Ó¦ËÙÂÊ£¬µ«ÊÇÒõ¼«µçÁ÷ÃܶÈ(µ¥Î»Ãæ»ýÄڵķ´Ó¦ËÙÂÊ)È´´ó´óÔö¼ÓÁË£¬¼´¾Ö²¿µÄ¸¯Ê´Ç¿¶È´ó´ó¼Ó¾çÁË¡£ÀýÈçÒ»¸ù¾ùÔȸ¯Ê´µÄÌú¹Ü¿ÉÒÔÁ¬ÐøÊ¹ÓúÜ
23
³¤Ê±¼ä¶øÎ޴󰣬µ«Èç¾Ö²¿¸¯´©¾ÍÖ»Äܱ¨·Ï¡£¶ø¾ùÔȸ¯Ê´¶¼ÊÇÑô¼«ºÍÒõ¼«·´Ó¦ÊÇÔÚ½ðÊô±íÃæÏàͬµÄλÖ÷¢ÉúµÄ£¬ÕâÑùÒýÆðµÄ½ðÊô¸¯Ê´ÊǾùÔȵ쬳ÆÎª¾ùÔȸ¯Ê´»òÈ«Ãæ¸¯Ê´¡£ÀýÈçÒ»¿éп»ò¸Ö½þÔÚÏ¡ÁòËáÖУ¬Í¨³£ÔÚÈ«²¿±íÃæÉÏÒÔ¾ùÔÈËÙ¶ÈÈܽ⡣ 47.»³öÒ»ÖÖµäÐ͵ĽðÊô¶Û»¯I??ÇúÏߣ¬²¢ËµÃ÷¸÷¶ÎµÄÒâÒå¡£ÈçºÎÄÜ˵
Ã÷½ðÊô¶Û»¯µÄÔÒòÊǽðÊô±íÃæÐγÉÖÂÃÜÑõ»¯ÎﱻĤµÄ½á¹û¡£
½â£º47ÌâͼÊÇÒÔFe/lmolH2SO4½ðÊôµç¼«Îª»ù´¡£¬Ãè»æÒ»¸ö·´Ó³¶Û̬½ðÊôÑô¼«¼«»¯Ò»°ãÌØÕ÷µÄ¼«»¯ÇúÏßʾÒâͼ£¬Õû¸öÇúÏ߿ɷֳÉËĸöÇø£º
47Ìâͼ£º ÌúÔÚlmolH2SO4ÈÜÒºÖеÄÎÈ̬Ñô¼«¼«»¯ÇúÏß
(1)A-KÇø: ½ðÊô»îÐÔÈܽâÇø¡£´Ó¸¯Ê´½ðÊôµÄ×Ô¸¯Ê´µçλ¦Åcµ½¶Û»¯µçλ¦Å
p
£¬½ðÊôµç¼«µÄÑô¼«µçÁ÷ÃܶÈËæ×ŵçλ¦ÅµÄÉý¸ß¶ø²»¶ÏÔö´ó¡£µ±¦Å£½¦Åpʱ
½ðÊôµç¼«µÄÑô¼«µçÁ÷Ãܶȴﵽ×î´óÖµiÁÙ¡£iÁÙ³ÆÎªÁÙ½ç¶Û»¯µçÁ÷Ãܶȡ£ÀýÈ磬ÔÚÕâ¸öµçÎ»Çø¼äÖÐ,½ðÊôµÄÌúµç¼«ÔÚlmolH2SO4Öв»¶ÏÒÔFeÐÎʽ»îÐÔÈܽ⣬ÈܽâÏÂÀ´µÄFe»ý¾ÛÔڵ缫Óëµç½âÖʽçÃæ£¬µ±ÆäŨ¶È´ïµ½FeÓëSO4
¡ª
2+
2+
2
2+
µÄŨ¶È»ýʱ£¬FeSO4¿ªÊ¼ÒÔÑÎĤÐÎʽ³Á»ýÔÚ½ðÊô±íÃæ¡£
p
(2)K-IÇø : ͨ³£³ÆÎª»î»¯Ò»¶Û»¯¹ý¶ÉÇø¡£ ´Ó¦ÅÖÁ¦Å
F
ÊÇÇúÏߵĵڶþ²¿
·Ý¡£¦ÅF½Ð×ö»î»¯µçλ¡£ÔÚÕâÒ»µçÎ»Çø¼ä£¬½ðÊô±íÃæ×´Ì¬·¢Éú¼±¾çµÄ±ä»¯£¬½ðÊô±íÃæ´¦ÓÚ²»Îȶ¨µÄ״̬£¬ÔÚÓÐЩÇé¿öÏ£¨ÈçÉÏÊöµÄFe£¯lmolH2SO4Ìåϵ£©£¬µçÁ÷ÃܶÈÓдó·ù¶ÈµÄÕñµ´¡£
(3)I-EÇø: ³ÆÖ®Îª¶Û»¯Çø»òÎȶ¨¶ÛÌ¬Çø¡£ µ±µçλ¸ßÓÚ¦Å
F
ʱ£¬½øÈëÇúÏß
µÄµÚÈý²¿·Ý¡£´Ëʱ£¬½ðÊô±íÃæÓÐijÖÖÖÂÃܵġ¢Èܽ⻺ÂýµÄ±¡Ä¤Éú³É£¬½ðÊôµÄÑô¼«µçÁ÷ÃܶȽ«ÏÔÖøÏ½µ£¬½ðÊô±»ÈÏΪ´¦ÓÚ¶Û»¯×´Ì¬¡£ÔÚÒ»¸öÏ൱¿í¹ãµÄµçλ·¶Î§ÄÚ£¬½ðÊôµÄÑô¼«ÈܽâµçÁ÷²»±ä²¢±£³ÖÔÚÒ»¸öºÜµÍµÄˮƽ¡£ÓÐЩ½ðÊôÔÚÕâÒ»µçÎ»Çø¼äµÄÑô¼«Èܽâ²úÎïÒ²²»Í¬ÓÚ»îÐÔÑô¼«ÈܽâʱµÄ²úÎï¡£ÀýÈ磬ÌúÔÚlmolH2SO4ÈÜÒºÖУ¬»îÐÔÑô¼«ÈܽâµÄ²úÎïÊǶþ¼ÛµÄÑÇÌúÀë×Ó£¬¶øÔÚÕâÒ»µçÎ»Çø¼ä£¬Ñô¼«Èܽâ²úÎïÊÇÈý¼ÛµÄÕýÌúÀë×Ó¡£ (4)E-GÇø ÕâÒ»²¿·Ý½Ð×ö¹ý¶Û»¯Çø¡£¦Å
TP
½Ð×ö¹ý¶Û»¯µçλ¡£ÕâÒ»²¿·ÝµÄ
ÇúÏßµÄÌØÕ÷ÊÇ£¬Ñô¼«µçÁ÷ÃܶÈÔÙ´ÎËæµçλµÄÉý¸ß¶øÔö´ó¡£ÕâÊÇÒòΪÔÚ¹ý¶Û»¯Çø£¬½ðÊôµç¼«ÉÏ·¢Éúеĵ缫·´Ó¦£¨»ò¶Û»¯Ä¤±»ÆÆ»µÁË£©¡£´ËʱµÄÑô
24
¼«µçÁ÷ÃܶȱíÕ÷ÕâÖÖеĵ缫·´Ó¦ËÙ¶È¡£ÓÐЩ½ðÊôÔÚÕâµçÎ»Çø¼ä±»Ñõ»¯³É¸ß¼ÛµÄ»¯ºÏÎï¡£ÀýÈ磬½ðÊô¸õÔÚ¶Û»¯ÇøÊÇÒÔCrÐÎʽÈܽ⣬¶øÔÚ¹ý¶Û»¯Çø£¬½ðÊô¸õµÄÑô¼«Èܽâ²úÎïÊÇÖØ¸õËáÑÎÒõÀë×ÓCr2O7»ò¸õËáÑÎÒõÀë×ÓCrO4¡£¶øÓÐЩ½ðÊô£¬ÀýÈçÌú£¬ÔÚ¹ý¶Û»¯Çø³ýÁ˼ÌÐøÉú³ÉÈý¼ÛµÄÌúÀë×ÓÍ⣬»¹·¢ÉúÎö³öÑõÆøµÄµç¼«·´Ó¦¡£
48.ÊÔ¸ù¾Ý½ðÊô¸¯Ê´µÄµç»¯Ñ§»úÀí£¬ËµÃ÷¿ÉÄܵķÀ¸¯·½·¨¡£ËµÃ÷Òõ¼«±£»¤ºÍÑô¼«±£»¤µÄÒìͬ¡£
½â£º½ðÊôºÍºÏ½ðÔÚ¸¯Ê´½éÖÊÖеĻ-¶Û»¯¼«»¯ÇúÏß¿ÉÒÔÖ¸Ã÷Íâ¼ÓµçÁ÷µÄÒõ¡¢Ñô¼«±£»¤·¨¡£Èçͼ8.5.2Ëùʾ¡£´Ëͼչʾ³öÁ½¸ö¿ÉÒԳɹ¦±£»¤Òõ¡¢Ñô¼«µÄÁ½¸öµçÎ»Çø¼ä¡£ÆäÒ»ÔÚ¶Û»¯Çø£¬³ÆÖ®ÎªÑô¼«±£»¤·¨£»Æä¶þÔÚ»î»¯Çø£¬³ÆÖ®ÎªÒõ¼«±£»¤·¨¡£
2¡ª
2¡ª
3+
ͼ8.5.2
Èç¹û°Ñ½þÔÚ½éÖÊÖеĽðÊô¹¹¼þºÍÁíÒ»¸¨Öúµç¼«×é³Éµç³Ø£¬ÓúãµçλÒǰѽðÊô¹¹¼þµÄµçÊÆ¿ØÖÆÔÚͼ8.5.2µÄI-E¶ÎÄÚ£¬Ôò¿ÉÒ԰ѽðÊôÔÚ½éÖÊÖеĸ¯Ê´½µµÍµ½×îСµÄÏÞ¶È¡£ÕâÖÖÓÃÑô¼«¼«»¯Ê¹½ðÊôµÀµÀ±£»¤µÄ·½·¨½ÏÍâ¼ÓµçÁ÷Ñô¼«±£»¤·¨¡£¶ÔÓÚÍâ¼ÓµçÁ÷µÄÑô¼«±£»¤£¬Èçͼ8.5.3Ëùʾ¡£
ͼ8.5.4 ×°ÁòËáµÄ¸ÖÖü²ÛµÄÑô¼«±£»¤
¶ÔÓÚÍâ¼ÓµçÁ÷µÄÒõ¼«±£»¤£¬Èçͼ8.5.5Ëùʾ¡£ÍⲿµÄÖ±Á÷µçÔ´µÄ¸º¼«Í¬±»±£»¤µÄµØÏÂÖü´æ²ÛÏàÁ¬½Ó£¬µçÔ´µÄÕý¼«Á¬½ÓÔÚ¶èÐÔÑô¼«£¨×÷Ϊ¶Ôµç¼«£¬Óֽи¨Öúµç¼«£©ÉÏ¡£ÒªÊ¹½ðÊô»ñµÃÀíÏëµÄ±£»¤£¬¹Ø¼üÔÚÑ¡ÔñºÃ±£»¤µçλºÍ±£»¤µçÁ÷¡£
25
ͼ8.5.5 µØÏ²ÛÓÃÍâ¼ÓµçÁ÷Òõ¼«±£»¤
ÎþÉüÑô¼«·¨µÄÔÀíºÍÒõ¼«±£»¤ÊÇÒ»ÑùµÄ£¬ËüÊÇÀûÓÃÑô¼«Óë±»±£»¤µÄ½ðÊôÖ®¼äµÄµçλ²îËù²úÉúµçÁ÷À´´ïµ½±£»¤µÄÄ¿µÄ¡£¸¨Öú½ðÊôµÄÈܽ⽫µ¼ÖÂÆ½ºâµçÎ»ÒÆÏò¸ü¸ºµÄµçλֵ£¬µ±½ðÊô¾ßÓм¸ºõÏàͬµÄµç루Á½ÕßÖ®¼äÓÐÒ»¸öIR½µ£©Ê±£¬¸¨Öú½ðÊô³ä×÷Ñô¼«£¬¶ø±»±£»¤µÄ½ðÊô±íÃæ³ÉΪÒõ¼«¡£ÏÔÈ»ÊǽðÊôËðʧµÄ×ÜËٶȽ«Ôö´ó£¬µ«ÕýÊǸ¨Öú½ðÊô³ä×÷Ñô¼«²»¶ÏÈܽ⣬¶ø±»±£»¤µÄ½ðÊôËðʧËٶȽ«¼õС¡£ÕâÖÖͨ¹ý´óÁ¿µÄ¸¯Ê´¾ÖÏÞÓÚ¸¨Öú½ðÊôÑô¼«£¬´Ó¶ø±£»¤Á˽ðÊôÉ豸Ó빤¼þµÄ·½·¨³ÆÖ®ÎªÎþÉüÑô¼«·¨¡£Èç¹ûÔÚµç½âÒºÖеı»±£»¤µÄ½ðÊôÉÏÓõ¼ÏßÁ¬½ÓÒ»¿éÒ×ÓÚÑõ»¯µÄ¸¨Öú½ðÊô£¬Ôò¿É¼õÇá±»±£»¤µÄ½ðÊôµÄ¸¯Ê´¡£
ͼ8.5.6ÎþÉüÑô¼«±£»¤Ë®ÖиÖ×®(a)Óë µØÏ¹ܵÀ(b)
49£®Ê²Ã´½Ð»ºÊ´¼Á?Æä·ÀÊ´»úÀíÈçºÎ?ÊÔ˵Ã÷Ö®¡£
½â£ºÄ³Ð©¸¯Ê´½ðÊôϵͳÎãÐëÍâ¼ÓµçÁ÷£¬Ö»ÒªÌí¼ÓһЩÄÜ´Ùʹ¶Û»¯¡¢Ðγɶۻ¯Ä¤»òÐγÉÎü¸½²ãµÄÎïÖÊ£¬¾ÍÄܽµµÍ¸¯Ê´ËÙ¶È¡£ÕâЩÎïÖʳÆÎª»ºÊ´¼Á¡£
Ñô¼«ÐÍ»ºÊ´¼Á:ÕâÀ໺ʴ¼ÁÍùÍù´Ùʹ¸¯Ê´µçλÕýÒÆ£¨Èçͼ8.5.6aÖЦÅc¡ú¦Åc£§£©£¬²¢Ê¹½ðÊô±íÃæÐγɳÁµíĤ£¬Ò²¿ÉÒÔ´ß»¯¼ÓËÙ¶Û»¯Ä¤µÄÐγɡ£¶ÔÓÚǰÕß³£³ÆÎª·ÇÑõ»¯ÐÍ»ºÊ´¼Á£¬ÈçNaOH¡¢NaCO3¡¢Na3PO4¡¢C6H5COONaµÈ£»¶øºóÕß½ÐÑõ»¯ÐÍ»ºÊ´¼Á£¬Èç¸õËáÑΡ¢ÖظõËáÑΡ¢ÏõËáÄÆ¡¢ÑÇÏõËáÑεȡ£¶ÔÓÚ·ÇÑõ»¯Ð͵ĻºÊ´¼Á£¬Í¨³£Ö»ÓÐÔÚÈܽâÑõ´æÔÚʱ²ÅÄÜÆðÒÖÖÆ¸¯Ê´µÄ×÷Óá£Ñô¼«ÐÍ»ºÊ´¼ÁÔÚʵ¼ÊÖеõ½¹ã·ºÓ¦Óá£ÈçÓÃÁ¿²»×㣬²»Äܳä·ÖʹÑô¼«±íÃæÐγÉÍêÕûµÄ¸²¸ÇĤʱ£¬ÓÉÓÚ±©Â¶ÔÚ½éÖÊÖеÄÑô¼«Ãæ»ýԶСÓÚÒõ¼«Ãæ»ý£¬ÐγÉÁËСÑô¼«´óÒõ¼«µÄ¸¯Ê´µç³Ø£¬·´¶ø»á¼Ó¾ç½ðÊôµÄ¾Ö²¿¸¯Ê´¡£Òò´ËÕâÀ໺ʴ¼ÁÓÖÓС°Î£ÏÕ»ºÊ´¼Á¡±Ö®³Æ¡£
Òõ¼«ÐÍ»ºÊ´¼Á:Æä×÷ÓÃÖ÷ÒªÔÚÓÚ¿ØÖÆÒõ¼«¹ý³ÌµÄ½øÐУ¬Ôö´óÒõ¼«¼«»¯£¬Ê¹µçλ¸ºÒÆ£¬Ëü²¢²»¸Ä±äÑô¼«µÄÃæ»ý£¬Òò´Ë£¬¼´Ê¹Ìí¼Ó»ºÊ´¼ÁµÄŨ¶È²»¹»£¬Ò²²»»á·¢Éú¾Ö²¿¸¯Ê´µÄΣÏÕÐÔ£¬Òò¶øÓС°°²È«»ºÊ´¼Á¡±Ö®³Æ¡£ÀýÈ磺ÔÚÖÐÐÔÈÜÒºÖÐÓ¦ÓõĻºÊ´¼ÁÓÐпÑΡ¢Ca£¨HCO3£©2µÈ¡£ËüÃÇÄÜÓëÒõ¼«¸½½üËùÐγɵÄOHÀë×Ó½áºÏÉú³ÉÄÑÈÜÐÔÇâÑõ»¯Îï»ò̼ËáÑγÁµíĤ£¬¸´¸ÇÔÚÒõ¼«±íÃæÉÏ¡£
»ìºÏÐÍ»ºÊ´¼Á£ºËüÃǵÄ×÷ÓÃÖ÷ÒªÊÇͬʱÒÖÖÆÁËÒõ¼«¹ý³ÌºÍÑô°å¹ý³Ì¡£ÕâÖÖ»ºÊ´¼ÁÖ÷ÒªÊÇÓлú»¯ºÏÎÈç°·Àà¡¢Óлú°·µÄÑÇÏõËáÑΣ¬ÁòÃÑ¡¢Áò´¼£¬»·×´º¬Áò»¯ºÏÎÁòëå¼°ÆäÑÜÉúÎïµÈµÈ£¬¶ÔÓÚÕâÀ໺ʴ¼ÁµÄ»ºÊ´×÷
26
¡ª
ÓûúÀí£¬ÏÖÔÚÈÏΪÖ÷ÒªÊÇÓÉÓÚËü±»Îü¸½ÔÚ½ðÊôÒõ¼«±íÃæÉ϶øÔö¼ÓÁËÇⳬµçλ¡¢·Á°ÁËH·Åµç¹ý³ÌµÄ½øÐУ¬´Ó¶øÊ¹µÃ½ðÊôÈܽâËٶȼõÂý£»Ò²ÓеÄÎü¸½ÔÚÑô¼«±íÃæ£¬Ê¹Ñô¼«¹ý³ÌÊܵ½ÒÖÖÆ¡£ÕâÀ໺ʴ¼Áͨ³£ÓÃÔÚËáÐÔ½éÖÊÖС£
Ê®
27