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.  

Potassium always has an oxidation state of , while oxygen always has an oxidation state of . Since we have four oxygens, there is a total charge of  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 .

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 .

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

This compound has four oxygens, which always have a oxidation state.

Chlorine usually has an oxidation state of , but in this case it must balance out the oxygens.

Chlorine must be .

On the left side of the equation, 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, , 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: .

(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 , we first must realize that  has 3 oxygen atoms, each with a charge of , for a total charge contribution of . Furthermore, since  has no net charge, the  atoms must contribute a total charge of  to balance out the  charge coming from the oxygens. And since there are two  atoms, then each must have a charge of .

On the product side of the reaction, notice that  is all by itself without any charge. The oxidation state of any individual atom is .

We have an oxidizing   and a reducing   agent, then is a red-ox reaction and the two semi reactions are:

  (oxidation,  losses electrons) 

  (reduction,  gains electrons)

In total 32 electrons are transferred from the  to .

In order to find the potential of a galvanic cell in which copper is reduced by iron, we need to combine the two half reactions for the metals in question.

Since copper is reduced in the net ionic equation, we can use the reduction potential seen in the table.

Iron must be oxidized in order to reduce the copper. As a result, we must use the reverse reaction in the table. Remember that the table only gives reduction potentials; you will frequently need to charge the reaction to find an oxidation potential.

Notice how the potential for the half reaction has been switched as well. Combining these two reactions gives the following balanced reaction, once the electron transfer is balanced.

Reduction potentials are intensive properties, so we do not multiply the potential of the copper half reaction by two. The total potential is simply the addition of both half reaction potentials.

The reducing agent is the metal that will be oxidized in the reaction. Since all of the half reactions shown are reduction potentials, we need to switch the half reactions in order to determine which results in the greatest cell potential when the metal is oxidized. Iron will result in a cell potential of 0.44 volts when oxidized, which makes it the strongest reducing agent on the list.

You do not multiply the coefficents that you need to balance to the Eo cell; you just have to see that the copper is being oxidized, so the sign changes (0.34 → –0.34) and the Cr is being reduced so the sign doesn't change. Then, just add the Eo cells: –0.74 + (–0.34) = –1.08.