燃料电池的电极方程式应如何书写?负极反应与正极反应如何相加??回答后加悬赏!

2025-03-15 19:03:47
推荐回答(1个)
回答(1):

燃料电池的电极反应式的书写
一、首先分清原电池的正、负极均为惰性电极,电极均不参与反应.
二、正极发生还原反应,通入的气体一般是氧气,氧气得到电子首先变为氧离子,根据电解质的不同,其负极电极反应式书写分以下几种情况:
(1)在酸性溶液中生成的氧离子与氢离子结合生成水,其电极反应式为:
O2 + 4e^- + H^+== 4H2O
(2)在碱性溶液中,氧离子与氢氧根离子不能结合,只能与水结合生成氢氧根离子,其电极反应式为:O2+ 4e^- + 2H2O= 4OH^-
(3)在熔融碳酸盐中,氧离子与碳酸根离子不能结合,只能与二氧化碳结合生成碳酸根离子,其电极反应式为:O2+2CO2+4e^-=2 CO3^2-
(4)在熔融氧化物介质中,氧气得到电子转化为氧离子,其电极反应式为:O2+ 4e^-= 2O^2-
三、负极发生氧化反应,负极生成的离子一般与正极产场结合:
若负极通入的气体为CH4、CH3OH、C2H5OH等,碳元素均转化为正四价碳的化合物、在酸性溶液中生成二氧化物气体、在碱性溶液中生成碳酸根离子,熔融碳酸盐中生成二氧化碳,熔融氧化物中生成碳酸根离子.含有氢元素的化合物最终都有水生成.
如CH3OH燃料电池:
酸性溶液负极:CH3OH - 6e^-+ H2O == CO2↑ + 6H^+
碱性溶浚中负极:CH3OH - 8e- + 10OH^- == CO3^2-+ 7H2O
甲醇燃料电池
甲醇燃料电池以铂为两极:
1、碱性电解质(KOH溶液为例)
总反应:2CH3OH + 3O2 +4KOH = 2K2CO3 + 6H2O
 正极:3O2+12e^-+ 6H20=12OH-
 负极:CH3OH -6e^-+8OH^-  = CO3^2- + 6H2O
2、 酸性电解质(H2SO4溶液为例)
总反应:2CH3OH + 3O2 = 2CO2 + 4H2O
正极:3O2+12e^-+12H^+= 6H2O
负极:2CH3OH-12e^-+2H2O = 12H^++2CO2
3、中性电解质(NaCl溶液为例)
总反应:2CH3OH + 3O2 = 2CO2 + 4H2O
负极:2CH3OH + 2H2O - 12e^- = 2CO2 + 12H^+
     正极:3O2 + 12H^+ + 12e^- = 6H2O
说明:乙醇燃料电池与甲醇燃料电池原理基本相同

甲烷燃料电池
 甲烷燃料电池以多孔镍板为两极,
 1、碱性电解质(KOH溶液为例)
 总反应为:CH4 + 2KOH+ 2O2= K2CO3 + 3H2O.
    负极:CH4 + 10 OH^-– 8e^-=  CO3^2- + 7H2O    
正极:O2 + 2H2O + 4e^-= 4OH^-
2、酸性电解质( H2SO4溶液为例)
总反应:CH4 + 2O2 = CO2 + 2H2O  
正极:2O2 +8e^- +8H^+ =4H2O
  负极:CH4 +2H2O-8e^-=CO2 +8H^+
3、中性电解质(NaCl溶液为例)
总反应:CH4 + 2O2 = CO2 + 2H2O  
   正极:2O2 +8e^- +4H2O=8OH^-    负极:CH4 +2H2O-8e^-=CO2 +8H^+
说明:掌握了甲烷燃料电池的电极反应式,就掌握了其它气态烃燃料电池的电极反应式

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