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Oxidation-Reduction Reactions
Oxidation-reduction reactions (or redox) reactions, are a type of reaction in
aqueous solutions 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 (within plants), respiration
(within animals), combustion, and corrosion or rusting. To understand these
types of reactions, you must first understand oxidation numbers or states.
1. Rules for Assigning Oxidation States
1.1. Sample Problems: Determine the oxidation states:
1.2. Solutions
1.2.1.1. Answers to Example 1:
1.2.1.2. Answers to Example 2:
2. Oxidizing and Reducing Agents
3. Oxidation-Reduction Reactions
3.1.1.1. Answer:
3.1.1.2. Answer:
4. Half Reactions
5. Balancing Oxidation-Reduction Reactions
5.1. Half-Equation Method
5.2. Balancing in Basic and Acidic Solution
5.2.1. Example 5:
5.2.2. Example 6:
6. Types of Redox Reactions:
6.1. Combination
6.1.1. Sample 1.
6.2. Decomposition
6.2.1. Sample 2.
6.3. Displacement Reactions
6.3.1. Sample 3.
6.3.2. Sample 4.
6.4. Combustion
6.5. Disproportionation
7. True or False
8. Solutions
9. Summary
10. References
11. Contributors
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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.)
Sample Problems: 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 1: Determine the oxidation state of the bold element in each of the
following:
1. Na3PO3
2. H2PO4-
2
Answers to Example 1:
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 2: 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
Answers to Example 2:
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).
Oxidizing and Reducing Agents
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- =
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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 thermite reaction 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 3. Using the equations from the previous examples determine
what is oxidized?
1. Zn + 2H+ → Zn2+ + H2
Answer:
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 4. What is reduced?
1. Zn + 2H+ → Zn2+ + H2
Answer:
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.
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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 halfequations 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
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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 5:
Balance the redox reaction in acidic solution: MnO4- + I- --> Mn2+ + I2(s)

Write and balance the half reactions:
MnO4- + I- --> Mn2+ + I2(s)
O.S: +7 -2





-1
+2
Rx:2I-(aq)
0
(Mn is reduced and I is oxidized)
-
Oxidation
--> I2(s) + 2e
Reduction Rx: MnO4- + 5e- --> Mn2+
Balance oxygen, O, by adding H2O
Oxidation Rx: 2I-(aq) --> I2(s) + 2eReduction Rx: MnO4- + 5e- --> Mn2+ + 4H2O
Balance hydrogen, H, by adding H+
Oxidation Rx: 2I-(aq) --> I2(s) + 2eReduction Rx: MnO4- + 5e- + 8H+ --> Mn2+ + 4H2O
Balance charge by adding electrons
Oxidation Rx: 2I-(aq) --> I2(s) + 2eReduction 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) + 10eReduction 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
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

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 6:
Balance the above redox reaction in basic 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) + 16OHOn 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) + 16OHCancel 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:
(1) MnO4- → Mn2+
Answers:
(1) a. Because there is one atom of Mn on both sides, no adjustment is
required.
b. Because manganese is reduced from an oxidation number of +7 to +2,
five electrons must be added to the left (MnO4- + 5e- → Mn2+)
c. 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+)
d. To balance the eight H+ ions on the left, add four H2O molecules to the
right. MnO4- + 8H+ + 5e- → Mn2+ + 4H2O
e. 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.
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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
Sample 1.
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
General Equation: AB → A + B
Decomposition reactions are the reverse of combination reactions, meaning
they are the breakdown of a chemical compound into the individual
elements.
Sample 2.
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.
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Single Replacement
General Equation: A + BC → AB + C
A single replacement reaction involves the "replacing" of an element in the
reactants with another element in the products.
Sample 3.
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
General Equation: AB + CD → AD + CB
A double replacement reaction is similar to a double replacement reaction,
but involves "replacing" two elements in the reactants, with two in the
products.
Sample 4.
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
General Equation: 2A → A' + A"
In some redox reactions substances can be both oxidized and reduced.
These are known as disproportionation reactions, which have some
practical significance in everyday life including the reaction of hydrogen
peroxide, H2O2 poured over a cut. This a decomposition reaction of hydrogen
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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






http://www.youtube.com/watch?v=yp60-oVxrT4
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
o Combination and Decomposition
o Displacement Reactions (Single and Double)
o Combustion
o Disproportionation
The oxidizing agent undergoes reduction and the reducing agent
undergoes oxidation.
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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/Redo
x_Chemistry/Balancing_Redox_reactions
Contributors



Christopher Spohrer (UCD)
Christina Breitenbuecher (UCD)
Luvleen Brar (UCD)
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