d) 2Al(s) + Fe2O3(s)  2Fe(l) + Al2O3(s)

Oxidation
2Al(s)  2Al+ + 2e-
Reduction
Fe23+ + 3e-  2Fe(l)

Oxidation numbers for chlorine
HCl = -1
MnO2(s)= +4
KMnO4(s)= +7

Oxidation numbers for manganese
Mn(s)= ?
MnCl2(s)= +7
Mn2+(aq)= ?
MnO4- =?

And then SO4 is so confusing...
H2SO4 =?
PbSO4 = ?

1 answer

d) 2Al(s) + Fe2O3(s) „³ 2Fe(l) + Al2O3(s)
Oxidation
2Al(s) „³ 2Al+ + 2e-
Reduction
Fe23+ + 3e- „³ 2Fe(l)


2Al(s) + Fe2O3 ==> 2Fe(l) + Al2O3
did you mean Fe(l) or Fe(s). It doesn't matter for the purpose of balancing Or for oxidation state. This is the "thermite" reaction, in which Fe gets so hot it melts, so you may have intended liquid.
Al ==>Al^+3 + 3e
Fe^+3 + 3e ==> Fe

Oxidation numbers for manganese
Mn(s)= ? All elements in the free state (uncombined with something) = 0
MnCl2(s)= +7 Cl is -1 so Mn must be +2 in order for MnCl2 to be zero
Mn2+(aq)= ? The oxidation state of elements in the ion form is the charge on the ion. Here Mn is +2
MnO4- =? Oxygen is -2 each for a total of -8, so Mn must be +7 in order to leave a -1 charge on the ion.

And then SO4 is so confusing...
H2SO4 =? H is +1 each for total of +2. O is -2 each for a total of -8. Therefore, S must be +6 to make H2SO4 zero.
PbSO4 = ? Pb is +2, S is +6, O is -2, for the same reasons in H2SO4.
S is one of those elements with multiple oxidation states. +2 in SO, zero in S(s), -2 in H2S, +4 in SO2 and SO3^-2, and +6 in SO3 and SO4^-2.
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