1. Determine the oxidation number of each element.

Reactants:

Products:

2. Write and balance the half reactions.

Oxidation (loss of electrons):

Reduction (gain of electrons): 

3. Balance oxygen by adding water (from the solution) into the half reactions.

Oxidation: 

Reduction: 

4. Balance hydrogen by adding  ions from the acid into the half reactions.

Oxidation: 

Reduction: 

5. Multiply each half reaction times an integer such that the electrons cancel when the equations are added. The oxidation reaction will be multiplied by five, and the reduction reaction will be multiplied by two.

Oxidation times 5: 

Reduction times 2: 

6. Add the half reactions together. The electrons will cancel from both the reactants and products.

Reduction potentials are typically provided in table form. Since the oxidation potential of an element is the negative of the reduction potential, it can be determined by viewing a reduction table.

Since the reduction potential for nickel is given as –0.23V, we can conclude that the oxidation potential for nickel is 0.23V.

A galvanic cell is spontaneous, meaning that there will be a positive cell voltage. In order for this to take place, the more negative reduction potential must be flipped so that the metal is oxidized.

Adding these values together gives a standard cell voltage of .

Since there is no overall charge on the compound the oxidation states must cancel out. Although the oxidation state of hydrides are +1, there are two in the compound that must be accounted for. The same is true for oxygen; although the oxidation number of oxygen is -2, there are four oxygens present accounting for a total of -8. The +2 state contribution from the hydrides and -8 from the oxygens results in a -6 charge. The oxidation state on sulfur must be +6 for the molecule to be neutral.

Since the calcium ion has a charge of +2, the dichromate anion must have a charge of -2. Oxygen always has an oxidation number of -2, so the seven oxygens in the molecule have a total number of -14. That leaves a charge of +12 that the two chromiums must balance out (-2 - (-14) = 12). Each chromium must therefore have an oxidation number of +6.

When applying oxidation numbers there are certain hierarchical rules that must be followed. 

1. The sum of oxidation states of all the elements in a molecule must add up to the overall charge. 

2. Group 1 and Group 2 elements have +1 and +2 oxidation states, respectively.

3. Fluorine has an oxidation state of –1.

4. Hydrogen has an oxidation state of +1 (except in metal hydrides).

5. Oxygen has an oxidation state of –2.

6. Elements of the same group (excluding transition metals) generally have the same oxidation state.

When applying the oxidation state to HNO₃, hydrogen has a +1 oxidation state, and each of the oxygen molecules has a –2 oxidation state. Since there is only one hydrogen molecule and three oxygen molecules, the oxidation state of nitrogen must balance out the charge of the hydrogen molecule and oxygen molecules combined.

Nitrogen must have a +5 oxidation state.

When applying oxidation numbers there are certain hierarchical rules that must be followed. 

1. The sum of oxidation states of all the elements in a molecule must add up to the overall charge. 

2. Group 1 and Group 2 elements have +1 and +2 oxidation states, respectively.

3. Fluorine has an oxidation state of –1.

4. Hydrogen has an oxidation state of +1 (except in metal hydrides).

5. Oxygen has an oxidation state of –2.

6. Elements of the same group (excluding transition metals) generally have the same oxidation state.

Following the oxidation rules, group 1 elements must have an oxidation number of +1.

To balance out the overall charge of zero, the hydrogen in this compound must have an oxidation state of –1. Note that NaH is a metal hydride; therefore, the oxidation state of hydrogen is not necessarily +1.

The sum of oxidation numbers of each element in this compound must add to the total charge of -2. Hydrogen always has an oxidation number of +1, and oxygen always has an oxidation number of -2. 

So, in total for oxygen we have (4) * (-2) = -8, and for hydrogen we have (4) * (+1) = +4.

Combining the oxygen and hydrogen, (+4) + (-8) = -4, so we need an additional +2 to achieve the total charge of -2. Zn must provide this balancing charge, and have an oxidation number of +2.

Oxidation number is a concept that you will also encounter in electrochemistry lectures. It is used to track electrons in an oxidation-reduction reaction. Overall, the sum of the oxidation states of all of the atoms in the molecule must equal the overall charge on the molecule. It is important to remember some of the commonly accepted oxidation states for atoms, so that you can predict the oxidation numbers on other atoms. Here are some key oxidation numbers useful to know for the MCAT:

1. Atoms in elemental form have an oxidation state of 0.

2. Halogens are commonly given an oxidation state of –1.

3. Hydrogen and alkali metals are commonly given an oxidation state of +1.

4. Oxygen is commonly given an oxidation number of –2. 

Keeping these rules in mind, we can assume that the hydrogens and oxygens in the two compounds maintain the same oxidation state. Chlorine, however, is attached to one less oxygen in HClO₃. As a result, the oxidation state of Cl has decreased from +7 to +5, when comparing HClO₄ to HClO₃.

We can calculate the oxidation number in each answer to find the greatest value.

In , oxygen will have a value of and the molecule is neutral. Sulfur must have an oxidation number of to balance the three oxygens.

In , oxygen will have a value of and the molecule is neutral. Carbon must have an oxidation number of  to balance the two oxygens.

In , chlorine will have a value of  and the molecule is neutral. Aluminum must have an oxidation number of  to balance the three chlorines.

In , calcium will have a value of  and the molecule is neutral. Sulfur must have an oxidation number of  to balance the calcium.

is a molecule comprised of a single element; each atom will have an oxidation number of .