# Oxidation-Reduction Reactions

Oxidation-reduction reactions (or redox) reactions, are a type of chemical reaction that involves a transfer of electrons between two species. An oxidation-reduction reaction is any chemical reaction in which the oxidation number of a molecule, atom, or ion changes by gaining or losing an e-. They occur every day and are vital to some of the basic functions of life. Some examples include photosynthesis, respiration, combustion, and corrosion or rusting. To understand these types of reactions, you must first understand oxidation numbers or states.

### Rules for Assigning Oxidation States

The oxidation number corresponds to the number of electrons, e-, that an atom loses, gains, or appears to use when joining with other atoms in compounds. When determining the Oxidation State of an atom there are seven guidelines to follow:

1. The Oxidation State of an individual atom is 0.
2. The total Oxidation State of all atoms in: a neutral species is 0 and in an ion is equal to the ion charge.
3. Group 1 metals have an Oxidation State of +1 and group 2 an Oxidation State of +2
4. The Oxidation State of fluorine is -1, when in compounds
5. Hydrogen generally has an Oxidation State of +1 in compounds
6. Oxygen generally has an Oxidation State of -2 in compounds
7. In binary metal compounds, group 17 elements have an Oxidation State of -1, group 16 of -2, and group 15 of -3.

(Note: The sum of the oxidation states is equal to zero for neutral compounds and equal to the charge for polyatomic ion species.)

Example: Assign Oxidation States

Determine the oxidation states:

1. Fe(s) + O2(g) → Fe2O3(g)
2. Fe2+
3. Ag(s) + H2S → Ag2S(g) + H2(g)

Solutions

1. Fe and O2 are free elements, therefore they have an O.S. of "0" according to Rule #1. The product has a total O.S. equal to "0" and following Rule #6, O3 has an O.S. of -2, which means Fe2 has an O.S. of +2.
2. The O.S. of Fe corresponds to its charge, therefore the O.S. is +2.
3. Ag has an O.S. of 0, H2 has an O.S. of +1 according to Rule #5 and S has an O.S. of -2 according to Rule #7.
Example

Determine the oxidation state of the bold element in each of the following:

1. Na3PO3
2. H2PO4-

Solutions

1. The oxidation numbers of Na and O are +1 and -2. Since sodium phosphite is neutral, the sum of the oxidation numbers must be zero.. Letting x be the oxidation number of phosphorus then, 0= 3(+1) + x + 3(-2). x=oxidation number of P= +3.
2. Hydrogen and oxygen have oxidation numbers of +1 and -2. The ion has a charge of -1, so the sum of the oxidation numbers must be -1. Letting y be the oxidation number of phosphorus, -1= y + 2(+1) +4(-2), y= oxidation number of P= +5.
Example 1

Determine which element is oxidized and which element is reduced in the following reactions (be sure to include the oxidation state of each):

1. Zn + 2H+ → Zn2+ + H2
2. 2Al + 3Cu2+→2Al3+ +3Cu
3. CO32+ 2H+→ CO2 + H2O

Solutions

1. Zn is oxidized (Oxidation number: 0 → +2); H+ is reduced (Oxidation number: +1 → 0)
2. Al is oxidized (Oxidation number: 0 → +3); Cu2+ is reduced (+2 → 0)
3. This is not a redox type because each element has the same oxidation number in both reactants and products: O= -2, H= +1, C= +4.

(For a more in depth look see oxidation numbers).

An atom is oxidized when it oxidation number increases, the reducing agent, and an atom is reduced when its oxidation number decreases, the oxidizing agent. In other words, what is oxidized is the reducing agent and what is reduced is the oxidizing agent. (Note: the oxidizing and reducing agents can be the same element or compound).

### Oxidation-Reduction Reactions

Redox reactions are comprised of two parts, a reduced half and an oxidized half, that always occur together. The reduced half gains electrons and the oxidation number decreases, while the oxidized half losses electrons and the oxidation number increases. Simple ways to remember this are the mnemonic devices OIL RIG meaning "oxidation is loss" and "reduction is gain" or LEO says GER meaning "loss of e- = oxidation" and "gain of e- = reduced." There is no net change in the number of electrons in a redox reaction. Those given off in the oxidation half reaction are taken on by another species in the reduction half reaction.

The two species that exchange electrons in a redox reaction are given special names. The ion or molecule that accepts electrons is called the oxidizing agent; by accepting electrons it brings about the oxidation of another species. Conversely, the species that donates electrons is called the reducing agent; when reaction occurs it reduces the other species. In other words, what is oxidized is the reducing agent and what is reduced is the oxidizing agent. (Note: the oxidizing and reducing agents can be the same element or compound This will be further discussed under Types of Redox Reactions: Disproportionation).

A good example of a redox reaction is the in which iron atoms of ferric oxide lose (or give up) O atoms to Al atoms, producing Al2O3.

Fe2O3(s) + 2Al(s) → Al2O3(s) + 2Fe(l)

Another example of the redox reaction is the reaction between Zinc and Copper sulfate.

Example

Using the equations from the previous examples determine what is oxidized?

1. Zn + 2H+ → Zn2+ + H2

Solution

The O.S. of H goes from +1 to 0 and the O.S. of Zn goes from 0 to 2+. Hence, Zn is oxidized and acts as the reducing agent.

Example

What is reduced species in this reaction?

1. Zn + 2H+ → Zn2+ + H2

Solution

The O.S. of H goes from +1 to 0 and the O.S. of Zn goes from 0 to 2+. Hence, H+ ion is reduced and acts as the oxidizing agent.

### Half Reactions

Before one can balance an overall redox equation, one has to be able to balance two half-equations, one for oxidation (electron loss) and one for reduction (electron gain). Collectively, oxidation and reduction are known as redox, or an electron transfer reaction. After balancing the two half-equations one can determine the total net reaction.

Each equation is balanced by adjusting coefficients and adding H2O, H+, and e- in this order:

1. Balance the number of atoms of each element.
2. Balance the number of electrons transferred.
3. Balance the total charge on reactants and products

(Note: If #1 and #2 are done correctly, #3 will follow. Thus, it serves as a means of checking your work).

### Balancing Oxidation-Reduction Reactions

To solve redox reactions accurately, you must first understand how to balance chemical equations. Though this process is more difficult than normal balancing it is a required step in the process of redox reactions. One of the most accepted methods of balancing a redox reaction is known as the half-equation method, however it can become more complex when involving basic or acidic solutions. In this module, a brief introduction to this different method will be explored. For an in depth explanation see: Balancing Oxidation-Reduction Reactions.

#### Half-Equation Method

The half-equation method (for neutral reactions) involves three basic steps which are as follows:

• Write and balance the half reactions.
• Adjust coefficients in both equations so that the same number of electrons appears in each half.
• Add together both halves, canceling out electrons, to obtain the overall equation.

#### Balancing in Basic and Acidic Solution

Balancing in acidic solution is similar to balancing in neutral solutions however, instead of three steps to follow, there are six. These rules are:

• Write and balance the half reactions.
• Balance oxygen, O, by adding with H2O
• Balance hydrogen, H, by adding H+ (acidic)
• Balance charge by adding electrons (you should be adding the same number of electrons as H+ ions)
• Multiply both half reactions by some integer to cancel out electrons
• Add the half reactions together and cancel out what appears on both sides

Example

Balance the redox reaction in acidic solution:

$MnO_4^- + I^- \rightarrow Mn^{2+} + I_2 (s)$

Solution

• Write and balance the half reactions:
MnO4- + I- --> Mn2+ + I2(s)
O.S:  +7 -2       -1          +2          0        (Mn is reduced and I- is oxidized)

Oxidation Rx:2I-(aq) --> I2(s) + 2e-
Reduction Rx: MnO4- + 5e- --> Mn2+
• Balance oxygen, O, by adding H2O
Oxidation Rx: 2I-(aq) --> I2(s) + 2e-
Reduction Rx: MnO4- + 5e- --> Mn2+ + 4H2O
• Balance hydrogen, H, by adding H+
Oxidation Rx: 2I-(aq) --> I2(s) + 2e-
Reduction Rx: MnO4+ 5e- + 8H--> Mn2+ + 4H2O
• Balance charge by adding electrons
Oxidation Rx: 2I-(aq) --> I2(s) + 2e-
Reduction Rx: MnO4- + 5e- + 8H+ --> Mn2+ + 4H2O
• Multiply both half reactions by some integer to cancel out electrons
(Oxidation Rx: 2I-(aq) --> I2(s) + 2e-) * 5
(Reduction Rx: MnO4- + 5e- + 8H--> Mn2+ + 4H2O) *2
Oxidation Rx: 10I-(aq) --> 5I2(s) + 10e-
Reduction Rx: 2MnO4- + 10e- + 16H+ --> 2Mn2+ + 8H2O
• Add the half reactions together and cancel out what appears on both sides:
10I-(aq) + 2MnO4-(aq) + 16H+(aq) --> 2Mn2+(aq) + 5I2(s) + 8H2O(l)
(Note: Don't forget the states of matter! Generally, anything with a charge is (aq) and H2O is (l))

Balancing in basic solution follows balancing in acidic solutions in three steps:

• Balance the reaction in acidic solution
• Add the same amount of OH- ions as H+ ions to both sides of the equation. On one side, the OH- and H+ will react to form water (H2O) in a 1:1 ratio.
• Cancel out water molecules appearing on both sides
Example

Balance the above redox reaction in basic solution

Solution

• Balance the reaction in acidic solution
10I-(aq) + 2MnO4-(aq) + 16H+(aq) --> 2Mn2+(aq) + 5I2(s) + 8H2O(l)
• Add the same amount of OH- ions as H+ ions to both sides of the equation.
10I-(aq) + 2MnO4-(aq) + 16H+(aq) + 16OH--> 2Mn2+(aq) + 5I2(s) + 8H2O(l) + 16OH-
On one side, the OH- and H+ will react to form water (H2O) in a 1:1 ratio.
10I-(aq) + 2MnO4-(aq) + 16H2O --> 2Mn2+(aq) + 5I2(s) + 8H2O(l) 16OH-
• Cancel out the water molecules appearing on both sides
10I-(aq) + 2MnO4-(aq) + 8H2O(l) --> 2Mn2+(aq) + 5I2(s) 16OH-(aq)
Example

Balance the following half-equation:

MnO4- → Mn2+

Solution

1. Because there is one atom of Mn on both sides, no adjustment is required.
2. Because manganese is reduced from an oxidation number of +7 to +2, five electrons must be added to the left (MnO4- + 5e- → Mn2+)
3. There is a total charge of -6 on the left versus +2 on the right. To balance, add eight H+ to the left to give a charge of +2 on both sides. (MnO4- + 8H+ 5e- → Mn2+)
4. To balance the eight H+ ions on the left, add four H2O molecules to the right. MnO4- + 8H+ + 5e- Mn2+ + 4H2O
5. Note that there are the same number of oxygen atoms, four, on both sides, as there should be. The equation shown in green is the correctly balanced reduction half-equation.

### Types of Redox Reactions

#### Combination

Combination reactions are some of the simplest redox reactions and as the name suggests involves the "combining" of elements to form a chemical compound. As usual, oxidation and reduction occur together. General Equation:

A + B → AB

Example

Equation: H2 + O2 → H2O
Calculation: 0 + 0 → (2)(+1) + (-2) = 0
Explanation: In this equation both H2 and O2 are free elements and following Rule #1, their oxidation state is "0." The product is H2O, which has a total oxidation state of "0." According to Rule #6, the O.S. of oxygen is usually -2. So, the O.S. of H2 must be +1.

#### Decomposition

Decomposition reactions are the reverse of combination reactions, meaning they are the breakdown of a chemical compound into the individual elements. General Equation:

AB → A + B

Example

Equation: H2O → H2 + O2
Calculation: (2)(+1) + (-2) = 0 → 0 + 0
Explanation: In this equation the water is "decomposed" into a Hydrogen and Oxygen. Similar to the previous sample the H2O has a total oxidation state of "0," thus according to Rule #6 the O.S. of oxygen is usually -2 so the O.S. of H2 must be +1.

#### Displacement Reactions

Displacement reactions, also known as replacement reactions, involve compounds and the "replacing" of elements. They occur as single replacement and double replacement reactions.

##### Single Replacement

A single replacement reaction involves the "replacing" of an element in the reactants with another element in the products.General Equation:

A + BC → AB + C

Example

Equation:

Cl2 + NaBr → NaCl + Br2

Calculation: (0) + ((+1) + (-1) = 0) -> ((+1) + (-1) = 0) + 0
Explanation: In this equation Br is replaced with Cl and Cl is reduced, while Br is oxidized.

##### Double Replacement

A double replacement reaction is similar to a double replacement reaction, but involves "replacing" two elements in the reactants, with two in the products. General Equation:

AB + CD → AD + CB

Example

Equation: Fe2O3 + HCl → FeCl3 + H2O
Explanation: In this equation Fe and H trade places and oxygen and chlorine trade places.

#### Combustion

Combustion reactions always involve oxygen, in the form of O2 and are almost always exothermic, meaning they produce heat.

General Equation: CxHy + O2 → CO2 + H2O

#### Disproportionation

In some redox reactions substances can be both oxidized and reduced. These are known as disproportionation reactions with a general equation:

2A → A' + A"

Disproportionation reactions have some practical significance in everyday life including the reaction of hydrogen peroxide, H2O2 poured over a cut. This a decomposition reaction of hydrogen peroxide, which produces oxygen and water. Oxygen is present in all parts of the chemical equation and as a result it is both oxidized and reduced. Reaction:

2H2O2(aq) → 2H2O(l) + O2(g)

Explanation: In the reactants H has an O has an O.S. of -1, which changes to -2 for the product, H2O (reduced) and 0 for the product, O2 (oxidized).

### True or False

1. The oxidation state of an individual atom is always +1
2. The oxidation agent has to be reduced
3. Combustion reactions always involve oxygen
4. Electrons and Hydrogen are almost never on the same side

### Solutions

1. False: is always 0
2. True
3. True
4. False: electrons and hydrogen are almost always on the same side

### Summary

• Remember the 7 Rules of Oxidation States (these are vital to undertanding redox reactions)
• Oxidation signifies a loss of electron and reduction signifies a gain of electrons.
• Balancing redox reactions is an important step that changes in neutral, basic, and acidic solutions.
• Remember the various types of redox reactions
• Combination and Decomposition
• Displacement Reactions (Single and Double)
• Combustion
• Disproportionation
• The oxidizing agent undergoes reduction and the reducing agent undergoes oxidation.

### References

1. Petrucci, et al. General Chemistry: Principles & Modern Applications. 9th ed. Upper Saddle River, New Jersey: Pearson/Prentice Hall, 2007.
2. Sadava, et al. Life: The Science of Biology. 8th ed. New York, NY. W.H. Freeman and Company, 2007
3. "oxidation–reduction reaction." Encyclopædia Britannica. 2008. Encyclopædia Britannica Online. 19 May 2008.
4. http://chemwiki.ucdavis.edu/Analytical_Chemistry/Electrochemistry/Redox_Chemistry/Balancing_Redox_reactions

### Contributors

• Christopher Spohrer (UCD), Christina Breitenbuecher (UCD), Luvleen Brar (UCD)

19:06, 13 Dec 2013