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Guide to Redox Reactions and Oxidation Numbers

Updated on April 25, 2017
Picture: the oxidation of Iron is known as rust
Picture: the oxidation of Iron is known as rust

Introduction to Oxidation/Reduction Reactions

Oxidation-Reduction reactions, also known as redox reactions, are reactions in which electrons are transferred.

Consider the reaction:

2Fe3+(aq) + 2I-(aq) → 2Fe2+(aq) + I2(s)

In this reaction, the Iron (III) (Fe3+) molecules each gain an electron from the Iodine- molecules, creating Iron (II) and Iodine. The reactions with respect to electron transfer would look like this:

2I- → I2 + 2e-

2Fe3+ + 2e- → 2Fe2+

Each of these reactions is known as a half-reaction, which are used to show the electron transfer between species in a reaction, showing either the oxidation of a species, or the reduction of a species.

Oxidation - refers to a loss of electrons, or a raising of the oxidation number
Reduction - refers to a gain of electrons, or a lowering of the oxidation number

These half reactions are added together, and so must be equal in terms of electrons lost by the oxidation reaction and electrons gained by the reduction reaction.

Oxidation number refers to the charge an atom would have if it were to transfer its electrons completely. See Oxidation Number Rules below for assigning oxidation numbers to species.

So in the problem above, I- was oxidized, because it lost an electron and raised its oxidation number, and Fe3+ was reduced, because it gained an electron and lowered its oxidation number.

Oxidation Number Rules

1. Free elements ALWAYS have an oxidation number of 0 unless indicated otherwise by a sign.

For example: K P Mg F O2 O P3 P4 all have oxidation numbers of 0. If, however they have a sign: K- would have an oxidation number of -1 and Mg2+ would have an oxidation number of +2, even though they are freestanding elements.

2. All Alkali metals have an oxidation number of +1, and all alkaline earth metals have an oxidation number of +2 in their compounds. Aluminum, however, has an oxidation number of +3 in its compounds.

3. Hydrogen (H) has an oxidation number of +1, except when it is bonded to any metal in a binary compound, where it takes on the oxidation number of -1. (So LiH, NaH, CaH2 etc)

4. the oxidation number of oxygen in most compounds is -2. However, in compounds with Hydrogen, the oxidation number of H takes precedence and therefore dictates the oxidation number of O.

For example: H2O2, Oxygen is -1, because the total charge is 0 and there are two Hydrogens with oxidation numbers of +1 (as stated in rule 3). The total charge left, then, would have to be split up by the number of Oxygen molecules, so 0 - 2 = -2. Charge of -2 split between the 2 Oxygen molecules = charge of -1 per Oxygen.

5. Fluorine (F) always has an oxidation number of -1, in all of its compounds.

6. In any given molecule/compound, the sum of the oxidation numbers of the individuals must equal the charge of the compound as a whole.

For example: in NH3, the total charge of the atom is 0, therefore the sum of the charges must equal 0. Since we know that Hydrogen has a charge of +1 in its non-metal compounds, we have +1 for every Hydrogen (3 total) and we have a charge of +3. Since the net charge is 0, we would have a charge of -3 for Nitrogen.
Another example, Cr2O72-, has a total charge of -2. So the sum of the oxidation numbers must equal -2. We know that in most compounds, Oxygen has an oxidation number of -2, so as have 7 oxygens with charges of -2, which gives us a charge of -14. Since the total charge of the compound is -2, we can take away -2 from -14, to give us -12 or 12 electrons that the remaining Cr atoms must make up. Since there are 2 Cr atoms, the -12 is divided by 2, to get 6 electrons needing to be shared for each Cr atom. So each Cr has a charge of +6.
To check this, we can use the equation [Cr2yO7x]-2 where y and x are the oxidation numbers of Cr and O. So 2(y) + 7(x) should equal -2. Let's test this: 2(+6) + 7(-2) = 12 + -14 = -2, Perfect!

7. One last rule, though not often used, is that oxidation numbers do not have to be integers. (Example is the superoxide ion, O2-, where Oxygen has an oxidation number of -1/2)

Chain showing the effects of oxidation
Chain showing the effects of oxidation | Source

Example Problems

These are some problems one might be asked in a General Chemistry course.

1. Determine the oxidation numbers of all the elements in the following compounds:

a) HNO3
b) HCl
c) Li2O
d) HSO4-

Solution:

a) HNO3 is a compound with a neutral charge, so the oxidation numbers must add up to 0. We know that Hydrogen atoms will have a charge of +1, and we know that Oxygen atoms will have a charge of -2, so we have 3 Oxygen molecules at -2, which gives us -6, plus 1 Hydrogen molecule at +1, which will put us at -5. In order to get to 0 charge, Nitrogen must have a charge of +5.

b) HCl is another neutral charge compound, so the oxidation numbers will add to 0. Hydrogen atoms are +1, and there is just one Hydrogen molecule so the Chlorine molecule must have an oxidation number of -1 in order to get back to 0.

c) Li2O has no charge, so the oxidation numbers will add to 0. Oxygen has a charge of -2 according to our rules, therefore the remaining charge must be split between the 2 Lithium molecules. -2/2 = -1 charge to make up, so each Lithium will have a charge of +1 to get back to 0.

d) HSO4- has a charge of -1, so the oxidation numbers must add up to -1. Hydrogen has a charge of +1 and Oxygen has a charge of -2. 1 Hydrogen (+1) + 4 Oxygens (-2) = -7. We know that our charge is -1, so we can subtract -1 from our remaining charge to get -6, which is the charge the Sulfur must now compensate for, giving Sulfur a charge of +6.

2. Determine the half-reactions of the following compounds, and label the oxidation and reduction reaction:

a) H2(g) + F2(g) → 2HF(g)
b) 2N2O(g) → 2N2(g) + O2(g)

For problems like this, the first step is to identify which element is oxidized, and which is reduced. Do this by determining the oxidation numbers of the elements and seeing which increase/decrease across the reaction.
Then, identify the number of electrons gained/lost and place them on the side of the equation that they are being gained/lost. Viola!

a) H2(g) + F2(g) → 2HF(g)
Oxidation numbers: H2 = 0 F2 = 0 → H = +1 F = -1
So Hydrogen is being oxidized since its oxidation number increases, and Florine is being reduced.
We know that there are 2 Hydrogen and 2 Florine in the reaction, so 2 electrons must be transferred per element in the reaction, giving us the half reactions:

H2 → 2H+ + 2e-
F2 + 2e- → 2F-

Overall: H2 + F2 + 2e- → 2H+ + 2F- + 2e-
The electrons cancel, and the H+ and F+ ions combine to form HF, giving us:
H2 + F2 → 2HF

Remember, the electrons must balance when recombined so that they are not in the overall equation. If they do not, find the LCM (least common multiple) of the two reactions and multiply all species in the half reactions by the LCM before recombining, so that the overall reaction has the same number of electrons on each side (causing them to cancel).

b) 2N2O(g) → 2N2(g) + O2(g)

Oxidation numbers: N2 = +1 O = -2 → N2 = 0 O2 = 0

Oxygen is being oxidized and Nitrogen is being reduced, so:

2O2- → O2 + 4e-
2N2+ + 4e- → 2N2
Overall reaction: 2N2+ + 2O2- + 4e- → O2 + 2N2 + 4e-
The electrons cancel and N2 and O combine to form N2O:
2N2O(g) → O2(g) + 2N2(g)


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