·¢²¼Ê±¼ä : ÐÇÆÚÁù ÎÄÕÂÎÞ»ú¼°·ÖÎö»¯Ñ§£¨µÚ¶þ°æ£©¸÷ÕÂÒªÇó¼°ÀýÌâ½²Îö5-9Õ¸üÐÂÍê±Ï¿ªÊ¼ÔĶÁ
(4) E+ = E?(H+/H2) = 0
?nE2 ?0.151 5? 0.0592VlgK==0.0592= 5.10 K?=1.3¡Á10
E? = 0V +
E = E+ ? E? = 0V + 0.12V= 0.12V
nE? 2?0??0?0.0592V0.0592lgK= K?=1
0.059220.01Vlg1= ?0.12V
214£®ÒÑÖªÏÂÁбê×¼µç¼«µçÊÆ Cu2++2e? =Cu E? =0.337V Cu2++e?=Cu+ E? =0.153V
(1) ¼ÆËã·´Ó¦ Cu + Cu2+ = 2Cu+µÄƽºâ³£Êý¡£ (2) ÒÑÖªK?sp(CuCl) =1.2¡Á10?6£¬ÊÔ¼ÆËãÏÂÃæ·´Ó¦µÄƽºâ³£Êý¡£ ½â£º(1) Cu 0.153 CuE Cu 0.337
2+2+++
2?E?(Cu/Cu) = 1?E?(Cu/Cu) + 1?E?(Cu/Cu)
2?0.337V = 1?0.153V + 1?E?(Cu+/Cu) E?(Cu+/Cu) = 0.521V
?E (CunE ??
lgK?=0.0592V2+
2?2+
+ ?
Cu + Cu+2Cl
2+?
2CuCl¡ý
/Cu)??E (Cu0.0592V2+
+
??/Cu)?(2) E?+ = E?(Cu/CuCl) = E(Cu/Cu) = E?(Cu/Cu) + 0.0592Vlg[c(Cu)/c(Cu)]
= 0.153V + 0.0592Vlg(1/K?sp,CuCl)
?6
= 0.153V + 0.0592Vlg(1/1.2¡Á10) = 0.504V
+??
E? = E(CuCl/Cu) = E(Cu/Cu) = E?(Cu+/Cu) + 0.0592Vlgc(Cu+)
= 0.521V + 0.0592Vlg K?sp,CuCl = 0.521V + 0.0592Vlg(1.2¡Á10?6) = 0.170
?E (CunE ??
lgK?=0.0592V2?2+
0.153?0.521??6.220.0592K? = 6.03?10?7
+
2+
+
/CuCl)??E (CuCl/Cu)0.504?0.170??5.640.0592V0.0592K? = 4.37?105
15£®ÏÂÁÐÈý¸ö·´Ó¦£º
(1) A + B+ = A+ + B (2) A + B2+ = A2+ + B (3) A + B3+ = A3+ + B
µÄƽºâ³£ÊýÖµÏàͬ£¬ÅжÏÏÂÊöÄÇÒ»ÖÖ˵·¨ÕýÈ·£¿ (a) ·´Ó¦(1)µÄÖµE?×î´ó¶ø·´Ó¦(3)µÄÖµE?×îС£» (b) ·´Ó¦(3)µÄE?Öµ×î´ó£»
(c) ²»Ã÷È·AºÍBÐÔÖʵÄÌõ¼þÏÂÎÞ·¨±È½ÏE?ÖµµÄ´óС£» (d) Èý¸ö·´Ó¦µÄE?ÖµÏàͬ¡£ ½â£º (a)¡£
16£®ÊÔ¸ù¾ÝÏÂÁÐÔªËصçÊÆͼ»Ø´ðCu+, Ag+, Au+, Fe2+µÈÀë×ÓÄÄЩÄÜ·¢ÉúÆ绯·´Ó¦¡£
E?A/V Cu2?????CuAg2?0.153?????Cu
?0.7990.521???Ag0.771Au+
+
2?3?????Ag
1.401.691????Au????Au
2?1.98Fe????Fe½â£º Cu, AuÄÜ·¢ÉúÆ绯¡£
?13+2+
17£®¼ÆËãÔÚ1mol?LHClÈÜÒºÖÐÓÃFeµÎ¶¨SnµÄµçÊÆͻԾ·¶Î§¡£Ôڴ˵ζ¨ÖÐӦѡÓÃʲôָʾ¼Á£¿ÈôÓÃËùѡָʾ¼Á£¬µÎ¶¨ÖÕµãÊÇ·ñºÍ»¯Ñ§¼ÆÁ¿µã·ûºÏ£¿ ½â£ºÒÑÖªE??(Sn4+/ Sn2+) = 0.14V E??( Fe3+/ Fe2+) = 0.70V
????Fe
?0.4400.0592ͻԾ·¶Î§£º(0.14V+3¡Á2?
0.0592V)=0.23V~ (0.70V?3¡Á1V) = 0.52V
1?0.70?2?0.143Ñ¡ÑǼ׻ùÀ¶£»E?In = 0.36V£¬ Ó뻯ѧ¼ÆÁ¿µã Esp==0.33V ·ûºÏ¡£
18£®½«º¬ÓÐBaCl2µÄÊÔÑùÈܽâºó¼ÓÈëK2CrO4ʹ֮Éú³ÉBaCrO4³Áµí£¬¹ýÂËÏ´µÓºó½«³ÁµíÈÜÓÚ
HClÈÜÒº£¬ÔÙ¼ÓÈë¹ýÁ¿µÄKI²¢ÓÃNa2S2O3ÈÜÒºµÎ¶¨Îö³öµÄI2¡£ÈôÊÔÑùΪ0.4392g£¬µÎ¶¨Ê±ºÄÈ¥29.61mL 0.1007 mol?L?1Na2S2O3±ê×¼ÈÜÒº¡£¼ÆËãÊÔÑùÖÐBaCl2µÄÖÊÁ¿·ÖÊý¡£
2+ 2?
½â£ºÓйط´Ó¦ Ba+ CrO4= BaCrO4
2? ? + 3+
2CrO4+ 6I+ 16H= 2Cr+ 8H2O + 3I2
I2 + 2Na2S2O3 = Na2S4O6 + 2NaI
n(Ba) = n(CrO4
2+
2?
21) = 3n(I2) = 3n(S2O32?)
22?1n(S2O3)M(BaClm(BaCl2)n(BaCl2)M(BaCl2)3w(BaCl2)???msmsms)1?0.1007?29.61?103?0.4392?0.4714?3?208.3
19£®ÓÃ30.00mL KMnO4ÈÜҺǡÄÜÑõ»¯Ò»¶¨ÖÊÁ¿µÄKHC2O4?H2O£¬Í¬ÑùÖÊÁ¿µÄKHC2O4?H2OÓÖÇ¡Äܱ»25.20mL0.2000 mol?L?1 KOHÈÜÒºÖк͡£¼ÆËãKMnO4ÈÜÒºµÄŨ¶È¡£ ½â£ºÓйط´Ó¦ 2MnO4? + 5C2O42? + 16H+ ? 2Mn2+ + 10CO2 + 8H2O
? ?2?
HC2O4+ OH ? C2O4 + H2O
n(MnO4?) = 2n(C2O42?)/5 n(KHC2O4?H2O) = n(KOH)
c(MnO?4)??n(MnOV(MnO?4?4))?2n(C2O45V(MnO2??4))?2n(KOH)5V(MnO?4)?12?0.2000?25.20mol?L5?30.00?1
20£®ÓÃKMnO4·¨²â¶¨¹èËáÑÎÑùÆ·ÖеÄCaº¬Á¿¡£³ÆÈ¡ÊÔÑù0.5863g£¬ÔÚÒ»¶¨Ìõ¼þÏ£¬½«¸Æ³ÁµíΪCaC2O4£¬¹ýÂË¡¢Ï´µÓ³Áµí¡£½«Ï´¾»µÄCaC2O4ÈܽâÓÚÏ¡H2SO4ÖУ¬ÓÃ0.05052 mol?L?1 KMnO4±ê×¼ÈÜÒºµÎ¶¨£¬ÏûºÄ25.64mL¡£¼ÆËã¹èËáÑÎÖÐCaµÄÖÊÁ¿·ÖÊý¡£ ½â£º n(Ca) = n(C2O42?) = 5n(MnO4?)/2
?0.06720 mol?L2+
?5n(MnO4)M(Ca)m(Ca)w(Ca) ??2msms?35?0.05052?254?10?40.078?20.5863
21£®´óÇŸÖÁºµÄ³ÄÆáÓú쵤(Pb3O4)×÷ÌîÁÏ£¬³ÆÈ¡0.1000gºìµ¤¼ÓHCl´¦Àí³ÉÈÜÒººóÔÙ¼ÓÈëK2CrO4£¬Ê¹¶¨Á¿³ÁµíΪPbCrO4£º Pb2+ + CrO42? = PbCrO4?
½«³Áµí¹ýÂË¡¢Ï´µÓºóÈÜÓÚËá²¢¼ÓÈë¹ýÁ¿µÄKI£¬Îö³öI2ÒÔµí·Û×÷ָʾ¼ÁÓÃ0.1000 mol?L?1Na2S2O3ÈÜÒºµÎ¶¨ÓÃÈ¥12.00mL£¬ÇóÊÔÑùÖÐPb3O4µÄÖÊÁ¿·ÖÊý¡£
2? ? + 3+
½â£º 2CrO4+ 6I+ 16H= 2Cr+ 8H2O + 3I2
I2 + 2Na2S2O3 = Na2S4O6 + 2NaI
2?2?
n(Pb3O4) = n(Pb)/3 = n(CrO4)/3 = 2n(I2)/9 = n(S2O3)/9
?0.22142?1n(S2O3)M(Pb3O4)m(Pb3O4)n(Pb3O4)M(Pb3O4)9w(Pb3O4)???msms9ms?31?0.1000?12.00?10?685.69?0.1000?0.914122£®¿¹»µÑªËá(Ħ¶ûÖÊÁ¿Îª176.1g?mol)ÊÇÒ»¸ö»¹Ô¼Á£¬ËüµÄ°ë·´Ó¦Îª
+?
C6H6O6 + 2H + 2e === C6H8O6
ËüÄܱ»I2Ñõ»¯¡£Èç¹û10.00mLÄûÃÊË®¹ûÖÑùÆ·ÓÃHAcËữ£¬²¢¼ÓÈë20.00mL0.02500 mol?L?1I2ÈÜÒº£¬´ý·´Ó¦ÍêÈ«ºó£¬¹ýÁ¿µÄI2ÓÃ10.00mL 0.0100 mol?L?1Na2S2O3µÎ¶¨£¬¼ÆËãÿºÁÉýÄûÃÊË®¹ûÖÖп¹»µÑªËáµÄÖÊÁ¿¡£
½â£º I2 + C6H8O6 = 2I? + C6H6O6 + 2H+
n(C6H8O6) = n(I2) = n(Na2S2O3)/2
?(C6H8O6) ?m(C6H8O6)Vs?n(C6H8O6)M(C6H8O6)Vs?31?0.0100?10.00)?10?176.1g210.00mL?1
(0.02500?20.00???0.007925 g?mL?
?
?1
23£®ÎüÈ¡50.00mLº¬ÓеÄIO3ºÍIO4ÊÔÒº£¬ÓÃÅðÉ°µ÷ÈÜÒºpH£¬²¢ÓùýÁ¿KI´¦Àí£¬Ê¹IO4?
??1
ת±äΪIO3£¬Í¬Ê±ÐγɵÄI2ÓÃÈ¥18.40mL 0.1000 mol?LNa2S2O3ÈÜÒº¡£ÁíÈ¡10.00mLÊÔÒº£¬ÓÃÇ¿ËáËữºó£¬¼ÓÈë¹ýÁ¿KI£¬ÐèͬŨ¶ÈµÄNa2S2O3ÈÜÒºÍê³ÉµÎ¶¨£¬ÓÃÈ¥48.70mL¡£¼ÆËãÊÔ
??
ÒºÖÐIO3ºÍIO4µÄŨ¶È¡£
½â£º Óйط´Ó¦ IO4? + 2I? + 2H+ = IO3? + I2 + H2O£¬ I2 + 2S2O32? = S4O62? + 2I?
n(IO4?) = n(I2) = n(S2O32?)/2
Ç¿ËáËữÒÔºó IO4? + 7I? + 8H+ = 4I2 + 4H2O£¬ IO3? + 5I? + 6 H+ = 3I2 + 3H2O
?2??2?
n(IO4) = n(I2)/4 = n(S2O3)/8 £¬ n(IO3) = n(I2)/3 = n(S2O3)/6
c(IO4)??n(IO4V?2?1n(S2O3))?2V1?0.1000?18.40?12?mol?L50.00?0.01840 mol?L??1c(IO4?)
c(IO3)?n(IO3V?2??1[n(S2O3)?8n(IO4)])6?V1?[0.1000?48.70?8?0.01840?10.00]?16?mol?L10.0024£®Calculate the ?rG
for the reaction
Cd(s) + Pb2+(aq) ¡ú Cd2+(aq) + Pb(s)
Solution£º E?(Cd2+/Cd) = ?0.403V E?(Pb2+/Pb) = ?0.126V
E? = E?(Pb2+/Pb) ? E?(Cd2+/Cd) = ?0.126V ?(?0.403V) = 0.277V
?1 ?1 ?1
?rG?m = ?nFE?= ?2¡Á96500¡Á0.277J?mol= ?53461J?mol= ?53.5 kJ?mol
25£®Calculate the potential at 25¡æ for the cell
2+-12+-1
Cd|Cd(2.00mol¡¤L)¡¬Pb(0.0010mol¡¤L)|Pb
Solution£º E+ = E?(Pb2+/Pb) + 0.0592V/2lgc(Pb2+)
= ?0.126V + 0.0592V/2lg0.0010 = ?0.215V
E? = E?(Cd2+/Cd) + 0.0592V/2lgc(Cd2+)
?
m at 25¡æ
?0.05663 mol?L?1
= ?0.403V + 0.0592V/2lg2.00 = ?0.394V E = E+ ? E?
= ?0.215V ?(?0.394V) = 0.179V
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s?1
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= 656?10m = 656nm
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= 6.626?10?34 J?s?3.289?1015s?1(1/12 ?1/?)
?18
=2.179?10J
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