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 polyatomic ion species.)
le 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,
B.
and following Rule #6, O has an OS of -2, which
means Fe has an OS of +3.
C.
The OS of Fe corresponds to its charge; therefore,
the OS is +2.
D.
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: Assigni
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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. |
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. |
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 disproportionation 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).
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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