[急]balancing redox equation!!

最傳統o既方法...
首先balance左兩條half equation先
寫左d main change出離先

C ---->CO2

H2SO4--->SO2

之後再加番d H+/ OH-/H2O離balance番d oxygen同hydrogen

C ---->CO2
step 1. 2H2O + C ------> CO2
step 2. 2H2O + C ------> CO2+4H^+

H2SO4 ---> SO2
step 1. H2SO4 ------> SO2+2H2O
step 2. 2H^+ + H2SO4 ------>SO2+ 2H2O

之後就加番d electrons上去離balance番d charge

2H2O + C ------> CO2+4H^+
2H2O + C ------> CO2 + 4H^+ +4e^- --------------(1)

2H^+ + H2SO4 ------>SO2+ 2H2O
2H^+ + H2SO4 + 2e^- ------>SO2+ 2H2O --------------(2)

之後再搵番兩條equation o既electrons數目o既LCM(最小公倍數)
equation(1)有4粒electrons
equation(2)有2粒electrons
佢地o既LCM係4
咁我地就將equation (2)乘2
令到佢有4粒electrons

2H^+ + H2SO4 + 2e^- ------>SO2+ 2H2O
4H^+ + 2H2SO4 + 4e^- ------>2SO2+ 4H2O-------(3)

之後我地就將(3)同(1)兩條equation加埋佢
再整走晒d兩邊都係o既野

4H^+ + 2H2SO4 + 4e^- ------>2SO2+ 4H2O----(3)
2H2O + C ------> CO2 + 4H^+ +4e^- ----(1)
(3)+(2):
step 1:
4H^+ + 2H2SO4 + 4e^- + 2H2O + C ------>2SO2+ 4H2O +CO2 + 4H^+
+ 4e^-
step 2:
4H^+ + 2H2SO4 + 2H2O + C ------>2SO2+ 4H2O +CO2 + 4H^+

step 2.
2H2SO4 + 2H2O + C ------>2SO2+ 4H2O +CO2

step 3.
2H2SO4 + C ------>2SO2+ 2H2O +CO2

就係咁就完成啦....

C變成CO2
係因為H2SO4有oxidizing property
H2SO4 oxidize左C
所以有CO2出

H2SO4變成SO2
係因為H2SO4俾C reduce左做SO2
S o既oxidation state由+6變成+4
所以就出左SO2啦

2007-11-29 18:30:17 補充：
4Al 3O2 - 4Al^(3 ) 6O^(2-)依條絕對唔係一條岩o既 equation正確o既應該係咁4Al 3O2 ---- 2Al2O3(s)Al2O3冇任何mobile ion留意番只有mobile ion (in aq / l)先可以將佢就咁寫做一粒ion所以根本就冇話com唔combine埋一齊P.S.sorry...加埋(3)同(2)o個度打左兩個step 2.出離第二個step 2. 應該係step 3.先岩之後step 3.應該改番做step 4.

The balance equation is shown below:
C(s) + 2 H2SO4(l) ---> CO2(g) + 2 SO2(g) + 2H2O(l)
Carbon will change to CO2. It is because carbon will oxidize to form CO or CO2 but CO will oxidize to form the most stable oxide -- CO2.
The H2SO4 will reduce to form the most stable oxide - SO2. It is because SO is not the stable sulphur oxide. It will oxide again to form SO2 gas. The oxidation number of SO3 gas is the same as H2SO4. SO3 is not the oxide which is reduced from H2SO4.