原帖由 xodfg1997 於 2014-10-11 19:43 發表
cr2o72----->cr3+
點可以變成
cr2o72-+14h++6e=2cr3++7h2o ?
吾係左的3+>2- 咩?完全吾識
我你個方法係簡單D,冇咁難balance
最正確既做法,而都係老師教個個方法係咁既:
check the oxidation no.(檢查氧化數)
note that always check element involve in both side with oxidation no. change(檢查牽涉氧化數變化的元素,如不知道,檢查所有元素,做得多就會明白是哪個)
so from the above case you should check Cr not Cr2O7^2-(這個例子,檢查Cr)Cr2O7^2- ---->Cr^3+
first balance the element(首先,平衡元素)
Cr2O7^2- ----->2Cr^3+
than use the information of oxidation no.(利用氧化數)
Cr2O7^2- ----->2Cr^3+
+6 +3
(If you don't even know how to check oxidation no. , you can ask me.)
Add electron(e^-), if there are x element(normally 1) that involve ,electrons must multiple x(如果有多少個元素,便乘多少個相差電子)
Cr2O7^2- +((6-3)*2)e^- ------>2Cr^3+
Cr2O7^2- +6e^- -------->2Cr^3+
Than base on acidic media(+H+) or alkaline media(+OH-) to balance the electric charge in both side
酸性用H+平衡,鹼性用OH-平衡電荷
Cr2O7^2- +6e^- -------->2Cr^3+
-2+(-6)=-8 +6
acidic media add H+ to lower charge side
alkaline media addOH- to higher charge side
for this case, it is acidic media
Cr2O7^2- +6e^- +14H+ -------->2Cr^3+
than add H2O to balance O(加水平衡O)
Cr2O7^2- +6e^- +14H+ -------->2Cr^3+ + 7H2O
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本帖最後由 輝輝 於 2014-10-11 21:00 編輯 ]