redox reaction好難呀................

First you must remember the following rules.
(1)For any free elements, their oxidation state is zero.
(2)For group I metals, their oxidation state is +1
(3)For group II metals, their oxidation state is +2
(4)For halogen, the oxidation state of florine is -1
(5)For hydrogen, its oxidation state is +1 except hydride (NaH), its oxidation state is -1 [ Na + 1 = 0]
*(6)The oxygen ,its oxidation state is -2 except peroxide O22-, its oxidation state is -1. For superoxide O2- ,its oxidation state is -0.5
(7)When oxidation number of an interested one is increased, oxidation. But when oxidation number of an interested one is decreased, reduction.
example1 : CuO + C -> Cu + CO2
Cu + (-2) = 0 Cu = +2 ; for free element Cu, = 0 [+2 -> 0] oxidation number decreases, Cu is reduced.
For C, it is a free element, C = 0. For CO2 [ C+(-2)x2 = 0]C = +4. For carbon, its oxidation number increases [ 0 -> +4], C is oxidized.
example 2 : Zn + Cu(NO3)2 -> Zn(NO3)2 + Cu
For Zn, it is a free element. Zn = 0, for Zn(NO3)2 ,its (Zn) charge is 2+ so is its oxidation number. Zn2+ = +2 [0 -> +2]. Zn oxidation number increases,
Zn is oxidized. For Cu you can find out by yourself. [Cu is reduced] [Cu 由+2 -> 0]
example 3: 2CrO4 2- + 2 H+ -> Cr2O72- + H2O
Hence for CrO4 2-its oxidation number is Cr+(-2)x4 = -2 ==> Cr+(-8) = -2 ==>Cr = -2 +8 = +6.
while for Cr2O72- its oxidation number is 2Cr+(-2)x7 = -2 ==>2Cr+(-14)=-2 ==>2Cr -14 = -2 ==> 2Cr = -2+14 ===> Cr = 12/2 = +6
[Cr2表示有兩個Cr]Cr 由+6去番+6,表明這個不是redox reaction.
*(6) O22-, its oxidation state is 2O = -2 ==> O = -2/2 = -1 while superoxide O2- ,its oxidation state is 2O = -1 ==> O = -1/2 = -0.5
cpd = compound
I hope you may find my answer helpful.