We typically think of Hydrogen as having an oxidation number of +1.  However when it is bonded to a less electronegative element such as Na it is actually assigned an oxidation number of -1.  

\(\displaystyle KMnO_4\)

Potassium always has an oxidation state of \(\displaystyle +1\), while oxygen always has an oxidation state of \(\displaystyle -2\). Since we have four oxygens, there is a total charge of \(\displaystyle -8\) from them.

The most important rule of oxidation numbers is that their sum must equal the charge on the molecule. In this compound, we have \(\displaystyle (+1)+4(-2)+Mn=0\).

\(\displaystyle (+1)-8+Mn=0\)

\(\displaystyle -7+Mn=0\)

\(\displaystyle Mn=+7\)

Manganese needs to cancel out the charge from potassium and oxygen in order to give us a neutral compound.

Perchlorate is a complex ion with the formula \(\displaystyle ClO_4^-\).

The most important rule for oxidation number is that the sum of the oxidation states of the atoms must equal the overall molecular charge.

\(\displaystyle (Cl)+4(O)=-1\)

This compound has four oxygens, which always have a \(\displaystyle -2\) oxidation state.

\(\displaystyle (Cl)+4(-2)=-1\)

\(\displaystyle (Cl)-8=-1\)

Chlorine usually has an oxidation state of \(\displaystyle -1\), but in this case it must balance out the oxygens.

\(\displaystyle Cl=+7\)

Chlorine must be \(\displaystyle +7\).

\(\displaystyle 2Al_{(s)}+3H_2SO_{4\hspace{1 mm}(aq)}\rightarrow Al_2(SO_4)_{3\hspace{1 mm}(aq)}+\hspace{1 mm}H_{2\hspace{1 mm}(g)}\)

On the left side of the equation, \(\displaystyle Al_{(s)}\) is a solid, so its oxidation state is zero, but on the right side it is in a salt, so it is not in its zero state.

Sulfate, \(\displaystyle SO_4^{-2}\), is an anionic salt, and there are three sulfate ions in each complex, yielding a net charge of -6. The two aluminum ions must have a net charge of +6, which, divided over two aluminum ions, yields an oxidation state of +3 for each aluminum ion.

The difference comes from simple subtraction: \(\displaystyle 3-0=3\).

(a) Since O has a –2 oxidation number and K has a +1 oxidation number (1 valence
electron it gives up), that means that Cr must have an oxidation number of +6. (b) Since H
has a +1 oxidation number and O has a –2 oxidation number, S has a +6 oxidation number.
(c) Fe has an oxidation number of +3 in order for it to have a net 0 oxidation state.

To determine the initial oxidation state of \(\displaystyle Fe\), we first must realize that \(\displaystyle Fe_{2}O_{3}\) has 3 oxygen atoms, each with a charge of \(\displaystyle -2\), for a total charge contribution of \(\displaystyle -6\). Furthermore, since \(\displaystyle Fe_{2}O_{3}\) has no net charge, the \(\displaystyle Fe\) atoms must contribute a total charge of \(\displaystyle +6\) to balance out the \(\displaystyle -6\) charge coming from the oxygens. And since there are two \(\displaystyle Fe\) atoms, then each must have a charge of \(\displaystyle +3\).

On the product side of the reaction, notice that \(\displaystyle Fe\) is all by itself without any charge. The oxidation state of any individual atom is \(\displaystyle 0\).

We have an oxidizing \(\displaystyle \left ( SO_{2}\right )\)  and a reducing \(\displaystyle \left ( H_{2}S\right )\)  agent, then is a red-ox reaction and the two semi reactions are:

\(\displaystyle 2\left ( H_{2}S\; \rightarrow \; S_{8}+16H^{+}+16H_{2}O+16e^{-}\right )\)  (oxidation, \(\displaystyle S^{-2}\) losses electrons) 

 \(\displaystyle 32e^{-}+32H^{+}+8SO_{2}\: \rightarrow \:S_{8}\, +\,16H_{2}O\) (reduction, \(\displaystyle S^{+4}\) gains electrons)

In total 32 electrons are transferred from the \(\displaystyle S^{-2}\) to \(\displaystyle S^{+4}\).