Rules
for Assigning Oxidation States
The oxidation state (OS) of an element corresponds to the
number of electrons, e-, that an atom
loses, gains, or appears to use when joining with other atoms in compounds. In
determining the OS of an atom, there are seven guidelines to follow:
1.
The OS of an individual atom is 0.
2.
The total OS 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 OS of +1 and Group 2 an OS of +2
4.
The OS of fluorine is -1 in compounds
5.
Hydrogen generally has an OS of +1 in compounds
6.
Oxygen generally has an OS of -2 in compounds
7.
In binary metal compounds, Group 17 elements have an OS of -1,
Group 16 of -2, and Group 15 of -3.
(Note: The sum of the OSs is equal to zero
for neutral compounds and equal to the charge for poly atomic ion species.)
Example 1: Assigning OSs
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Determine the OSs
of the elements in the following reactions:
a.
Fe(s)+O2(g)→Fe2O3(g)
b.
Fe2+
c.
Ag(s)+H2S→Ag2S(g)+H2(g)
SOLUTIONS
A.
Fe and O2 are
free elements; therefore, they each have an OS of 0 according to
Rule #1. The product has a total OS equal to 0, and following Rule #6,
O has an OS of -2, which means Fe has an OS of +3.
B.
The OS of Fe
corresponds to its charge; therefore, the OS is +2.
C.
Ag has an OS of
0, H has an OS of +1 according to Rule #5, S has an OS of -2 according
to Rule #7, and hence Ag in Ag2S has an OS of
+1.
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Example 2: Assigning OSs
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Determine the OS of the bold
element in each of the following:
A.
Na3PO3
B.
H2PO4-
SOLUTIONS
A.
The oxidation
numbers of Na and O are +1 and -2. Because sodium phosphite is neutral,
the sum of
B.
C.
the oxidation
numbers must be zero. Letting x be
the oxidation number of phosphorus, 0= 3(+1) + x + 3(-2).
D.
x=oxidation number of P= +3.
E.
Hydrogen and
oxygen have oxidation numbers of +1 and -2. The ion has a charge of -1, so
the sum
F.
of the
oxidation numbers must be -1. Letting y be the oxidation number of phosphorus,
-1= y +
2(+1) +4(-2), y= oxidation
G.
number of P=
+5.
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Example 3: Identifying Reduced and
Oxidized Elements
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Determine which element is
oxidized and which element is reduced in the following reactions (be sure to
include the OS of each):
A.
Zn + 2H+ →
Zn2+ +
H2
B.
2Al + 3Cu2+→2Al3+ +3Cu
C.
CO32- + 2H+→ CO2 + H2O
SOLUTIONS
A. Zn is oxidized (Oxidation number: 0 → +2); H+ is
reduced (Oxidation number: +1 → 0)
B. Al is oxidized (Oxidation number: 0 → +3); Cu2+ is
reduced (+2 → 0)
C. This is not a redox reaction because each
element has the same oxidation number in both reactants and products:
D. O= -2, H= +1,
C= +4.
(For further discussion, see
the article on oxidation numbers).
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An atom is oxidized if its oxidation number
increases, the reducing agent, and an atom is reduced if its oxidation number
decreases, the oxidizing agent. The atom that is oxidized is the reducing
agent, and the atom that 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 loses
electrons and the oxidation number increases. Simple ways to remember this
include the mnemonic devices OIL RIG, meaning "oxidation is loss"
and "reduction is gain,"
and 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 up 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 causes the oxidation of
another species. Conversely, the species that donates electrons is called the reducing
agent; when the 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, as in disproportionate reactions).
A good example of a redox reaction is the thermite reaction, in which iron atoms in 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 4: Identifying Oxidized Elements
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Using the equations from the
previous examples, determine what is oxidized in the following reaction.
Zn+2H+→Zn2++H2
SOLUTION
The OS of H
changes from +1 to 0, and the OS of Zn changes from 0 to +2. Hence, Zn is
oxidized and acts as the reducing agent.
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Example 5: Identifying Reduced Elements
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What is reduced species in
this reaction?
Zn+2H+→Zn2++H2
SOLUTION
The OS of H changes from +1 to 0, and the OS of Zn
changes from 0 to +2. Hence, H+ ion is reduced and acts as the oxidizing
agent.
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Combination Reactions
Combination reactions
are among the simplest redox reactions and, as the name suggests,
involves "combining" elements to form a chemical compound. As usual,
oxidation and reduction occur together. The general equation for a combination
reaction is given below:
A+B→AB
Example 6: Combination Reaction
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Equation: H2 + O2 → H2O
Calculation: 0 + 0 → (2)(+1) + (-2) = 0
Explanation: In this equation both H2 and O2 are free elements; following Rule #1, their OSs are 0. The product is
H2O,
which has a total OS of
0. According to Rule #6, the OS of oxygen is usually -2.
Therefore, the OS of H in H2O must be +1.
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Decomposition Reactions
A decomposition reaction
is the reverse of a combination reaction, the breakdown of a chemical
compound into individual elements:
AB→A+B
Example 7: Decomposition Reaction
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Consider the decomposition of
water:
H2O→H2+O2
Calculation: (2)(+1) + (-2) = 0 → 0 + 0
Explanation: In this reaction, water is "decomposed" into
hydrogen and oxygen. As in the previous example the
H2O has a total OS of 0; thus, according to
Rule #6 the OS of oxygen is usually -2, so the OS of hydrogen in H2O must be +1.
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Single Replacement Reactions
A single replacement reaction involves the
"replacing" of an element in the reactants with another element in
the products:
A+BC→AB+C
Example 8: Single Replacement Reaction
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Equation:
Cl2+NaBr−−−→NaCl−−+Br2
Calculation: (0) + ((+1) +
(-1) = 0) -> ((+1) + (-1) = 0) + 0
Explanation: In this equation, Br is replaced with Cl, and
the Cl atoms in Cl2 are
reduced, while the Br ion in NaBr is oxidized.
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Double Replacement Reactions
A double replacement reaction is similar to a double
replacement reaction, but involves "replacing" two elements in the
reactants, with two in the products:
AB+CD→AD+CB
Example 9: Double Replacement Reaction
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Equation: Fe2O3 + HCl → FeCl3 + H2O
Explanation: In this equation, Fe and H trade places, and oxygen and chlorine
trade places.
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Combustion Reactions
Combustion reactions
almost always involve oxygen in the form of O2, and
are almost always exothermic, meaning they produce heat. Chemical reactions
that give off light and heat and light are colloquially referred to as
"burning."
CxHy+O2→CO2+H2O
Although combustion reactions typically involve redox
reactions with a chemical being oxidized by oxygen, many chemicals
"burn" in other environments. For example, both titanium and
magnesium burn in nitrogen as well:
2Ti(s)+N2(g)→2TiN(s)
3Mg(s)+N2(g)→Mg3N2(s)
Moreover, chemicals can be oxidized by other chemicals than
oxygen, such as Cl2 or F2; these processes are also considered
combustion reactions
Disproportionate Reactions
Disproportionate Reactions: In some redox reactions a single
substance can be both oxidized and reduced. These are known as disproportionate reactions, with the
following general equation:
2A→A+n+A−n
Where n is the number of electrons
transferred. Disproportionate reactions do not need begin with neutral
molecules, and can involve more than two species with differing oxidation
states (but rarely).
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Disproportionate 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. The reaction is as follows:
2H2O2(aq)→2H2O(l)+O2(g)
Explanation: On the reactant
side, H has an OS of +1 and O has an OS of -1, which changes
to -2
for the product H2O (oxygen is reduced), and 0 in the
product O2 (oxygen is oxidized).
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