·¢²¼Ê±¼ä : ÐÇÆÚËÄ ÎÄÕ´óѧÎïÀí»¯Ñ§ºËÐĽ̵̳ڶþ°æ(ÉòÎÄϼ)¿Îºó²Î¿¼´ð°¸µÚ8Õ¸üÐÂÍê±Ï¿ªÊ¼ÔĶÁ
?1.36 V???8.314?298?ln(0.66?1.0?10?7)?V?1.78 V
?1?96500?¸ù¾ÝTafel¹«Ê½£¬Îö³öH2(g)µÄ³¬µçÊÆΪ
?H2?0.73 V?0.11 V?lg0.10?0.62 V ËùÒÔ£¬Êµ¼Ê·Ö½âµçѹΪ
E·Ö½â?E¿ÉÄæ??H2??Cl2
?1.78 V?0.62 V?0?2.40 V
29£®ÔÚ298 KºÍ±ê׼ѹÁ¦Ï£¬ÓÃAg(s)µç¼«µç½âŨ¶ÈΪ0.01 mol?kgµÄNaOHÈÜÒº£¬¿ØÖƵçÁ÷ÃܶÈj?0.1 A?cm¡£ÎÊÕâʱÔÚÁ½¸öÒøµç¼«ÉÏÊ×ÏÈ·¢Éúʲô·´Ó¦£¿´ËʱÍâ¼ÓµçѹΪ¶àÉÙ£¿ÒÑÖªÔڸõçÁ÷ÃܶÈʱ£¬H2(g)ºÍO2(g)ÔÚAg(s)µç¼«Éϵij¬µçÊÆ·Ö±ðΪ0.87 VºÍ0.98 V¡£ÒÑÖªEOH?|AgO|Ag?0.344 V£¬EOH?|H??0.828 V£¬EO|OH??0.401 V£¬
222?1?2ENa?|Na??2.71 V¡£Éè»î¶ÈÒò×Ó¾ùΪ1¡£
½â£º½«ÔÚÒõ¼«ÉÏÓпÉÄÜ·¢Éú»¹Ô·´Ó¦µÄµç¼«µçÊƼÆËã³öÀ´£¬»¹Ôµç¼«µçÊÆ×î´óµÄ½«Ê×ÏÈ
ÔÚÒõ¼«·¢Éú·´Ó¦£¬
£¨1£© Na(aNa??0.01)?e???Na(s) ENa?|Na?ENa?|Na???RT1 lnFaNa?RT1ln??2.83 V F0.01 ??2.71 V? £¨2£© H2O(l)?e????1H2(p)?OH?(aOH??0.01) 2RT EOH?|H?EOH?|H?lnaOH???H2
22zFRTln0.01?0.87 V??1.58 V ??0.828 V?FÕâ¸öµç¼«µçÊÆÒ²¿ÉÒÔÕâÑù¼ÆË㣬Á½¸ö½á¹ûÊÇÒ»ÑùµÄ¡£
2H(aH??10??12)?2e????H2(p) RT1ln??H2 FaH? EH?|H?EH?|H?22?RTln10?12?0.87V??1.58 V F
ÔÚÒõ¼«ÉÏ£¬»¹Ôµç¼«µçÊÆ´óµÄH2(g)ÏÈ»¹ÔÎö³ö¡£Í¨³£Çé¿öÏ£¬Na(s)µÄÎö³öÒ»°ãÊDz»¿¼Âǵģ¬ÕâÀï½öÊÇÓÃÀ´ËµÃ÷¿¼Âǵķ½·¨¡£
Ñô¼«ÉÏ¿ÉÄÜ·¢ÉúµÄ·´Ó¦³ýÒõÀë×ÓÍ⣬»¹Òª¿¼ÂÇÒøµç¼«±¾ÉíÒ²ÓпÉÄÜ·¢ÉúÑõ»¯
11H2O(l)?e?
42RT EO|OH??EO|OH??lnaOH???Ñô
22FRT?0.401 V?ln0.01?0.98 V?1.50 V
F11?£¨2£© Ag(s)?OH(aOH??0.01)???Ag2O(s)?H2O(l)?e?
22RT EOH?|AgO|Ag?EOH?|AgO|Ag?lnaOH?
22zFRT?0.344 V?ln0.01?0.46 V
F£¨1£© OH(aOH??0.01)???O2(p)??ÔÚÑô¼«ÉÏ£¬»¹Ôµç¼«µçÊÆ×îСµÄÊ×ÏÈ·¢Éú·´Ó¦£¬ËùÒÔÒøµç¼«ÏÈÑõ»¯³ÉAg2O(s)¡£ÕâʱÍâ¼ÓµÄ×îСµçѹÊÇ
E·Ö½â?EÑô?EÒõ?0.46 V?(?1.58 V)?2.04 V
30£®ÔÚ298 KºÍ±ê׼ѹÁ¦Ê±£¬µç½âÒ»º¬Zn2+ÈÜÒº£¬Ï£Íûµ±Zn2+Ũ¶È½µÖÁ
1?10?4 mol?kg?1ʱ£¬ÈÔ²»»áÓÐH2(g)Îö³ö£¬ÊÔÎÊÈÜÒºµÄpHÓ¦¿ØÖÆÔÚ¶àÉÙΪºÃ£¿ÒÑÖª
EZn2?|Zn??0.763 V£¬H2(g)ÔÚZn(s)Éϵij¬µçÊÆΪ0.72 V£¬²¢Éè´ËÖµÓëŨ¶ÈÎ޹ء£
½â£ºÒªÊ¹H2(g)²»Îö³ö£¬H2(g)µÄʵ¼ÊÎö³öµçÊÆӦСÓÚZnÁгöÁ½¸öµç¼«µçÊƵļÆËãʽ
EZn2?|Zn?EZn2?|Zn?2+µÄÎö³öµçÊÆ¡£Ê×ÏÈ·Ö±ð
RT1 ln2FaZn2?RT1ln??0.881 V 2F1?10?4??0.763 V? EH?|H?EH?|H?22RT1ln??H2 FaH? ?(?0.05916?pH?0.72) V (?0.05916?pH?0.72) V0.881 V ½âµÃ pH>2.72
ÈÜÒºµÄpHÓ¦¿ØÖÆÔÚ2.72ÒÔÉÏ£¬H2(g)²Å²»»áÓëZn(s)ͬʱÎö³ö¡£
31£®ÔÚ298 KºÍ±ê׼ѹÁ¦Ï£¬ÓÃÌúFe(s)ΪÒõ¼«£¬C£¨Ê¯Ä«£©ÎªÑô¼«£¬µç½â6.0 mol?kg?1µÄNaClË®ÈÜÒº¡£ÈôH2(g)ÔÚÌúÒõ¼«Éϵij¬µçÊÆΪ0.20 V£¬O2(g)ÔÚʯīÑô¼«Éϵij¬µçÊÆΪ0.60 V£¬Cl2(g)µÄ³¬µçÊƿɺöÂÔ²»¼Æ¡£ÊÔÓüÆËã˵Ã÷Á½¼«ÉÏÊ×ÏÈ·¢ÉúµÄ·´Ó¦£¬²¢¼ÆËãʹµç½â³Ø·¢Éú·´Ó¦ÖÁÉÙËùÐè¼ÓµÄÍâ¼Óµçѹ¡£ÒÑÖªECl?Cl??1.36 V£¬EO|OH??0.401 V£¬
22ENa?|Na??2.71 V£¬Éè»î¶ÈÒò×Ó¾ùΪ1¡£
½â£ºÓÉÓÚNa»¹Ô³ÉNa(s)µÄÎö³öµçÊƺÜС£¬¶øÇÒÔÚË®ÈÜÒºÖв»¿ÉÄÜÓÐNa(s)Îö³ö£¬ËùÒÔ²»¿¼ÂÇNa(s)µÄÎö³ö£¬ÔÚÌúÒõ¼«ÉÏÎö³öµÄÊÇH2(g)£¬ÆäÎö³öµçÊÆΪ
EH?|H?EH?|H?22?RT1ln??H2 FaH? ?RTln10?7?0.20 V??0.613 V F?Ñô¼«ÉÏ¿ÉÄÜ·¢ÉúÑõ»¯µÄÀë×ÓÓÐCl?ºÍOH¡£ËüÃǵÄÎö³öµçÊÆ·Ö±ðΪ EO|OH??EO|OH??2 ECl?|Cl2RTlnaOH???O2
2FRT?0.401 V?ln10?7?0.6 V?1.415 V
FRT?ECl?|Cl?lnaCl?
2FRT?1.360 V?ln6.0?1.314 V
F?ËùÒÔÑô¼«ÉÏ·¢ÉúµÄ·´Ó¦ÊÇClÑõ»¯ÎªCl2(g)¡£×îСµÄÍâ¼Ó·Ö½âµçѹΪ E·Ö½â?EÑô?EÒõ?1.314 V?(?0.613 V)?1.927 V
ÕâÀïÀûÓÃC£¨Ê¯Ä«£©ÎªÑô¼«£¬¾ÍÊÇÒòΪÑõÆøÔÚʯīÑô¼«ÉÏÎö³öÓ㬵çÊÆ£¬¶øÂÈÆøûÓУ¬´Ó¶øµç½âNaClŨµÄË®ÈÜÒºÔÚʯīÑô¼«ÉÏ»ñµÃÂÈÆø×÷Ϊ»¯¹¤ÔÁÏ¡£
32£®298 K ʱ£¬Ä³¸ÖÌúÈÝÆ÷ÄÚÊ¢ pH = 4.0 µÄÈÜÒº£¬ÊÔÓüÆËã˵Ã÷£¬´Ëʱ¸ÖÌúÈÝÆ÷ÊÇ·ñ»á±»¸¯Ê´£¿¼Ù¶¨ÈÝÆ÷ÄÚ Fe2+µÄŨ¶È³¬¹ý 10-6 mol¡¤dm-3 ʱ£¬ÔòÈÏΪÈÝÆ÷Òѱ»¸¯Ê´¡£
ÒÑÖª£ºEFe2?|Fe??0.440 V£¬H2£¨g£©ÔÚFe£¨s£©ÉÏÎö³öʱµÄ³¬µçÊÆΪ 0.40 V¡£ ½â£ºÖ»ÒªËã³öÇâµç¼«ºÍÌúµç¼«µÄµç¼«µçÊÆ£¬±È½ÏÆä´óС£¬¾Í¿ÉÒÔÅжÏÈÝÆ÷ÊǷǻᱻ¸¯Ê´¡£
EH?|H?EH?|H?22RT?1ln?F??aH???? ??H2? ?0?RT1ln(?4)?0.40V??0.6365 V F10RT?1ln??a2?F?Fe? ??? EFe2?|Fe?EFe2?|Fe? ??0.4402 V?RT?1?ln????0.6176 V 2F?10?6?ÒòΪ EFe2?|Fe>EH?|H£¬µçÊƸߵÄ×öÕý¼«£¬ÊÇÔµç³ØµÄÒõ¼«£¬¹Ê¸ÖÌúÈÝÆ÷²»»á±»¸¯Ê´¡£
2