Download 478 15.2 Oxidation-Reduction (Redox) Reactions

Section 15.2 Oxidation-Reduction (Redox) Reactions.
15.2 Oxidation-Reduction (Redox) Reactions
Electrochemistry is
driven by
electron
transfer
In the early development of chemical knowledge, an important application was the
refinement of pure metals such as iron, copper, aluminum and tin. These elements exist
naturally as mineral compounds such as aluminum oxide (Al2O3). Early chemists talked
about “reducing” compounds into pure metals long before people understood the
chemistry of what was happening. Today we know that “reducing” aluminum oxide to
elemental aluminum and oxygen involves transferring electrons from oxygen to
aluminum. The atoms that lose electrons are oxidized and the atoms that gain electrons
are reduced. For example, Al3+ gains 3 electrons and is reduced to aluminum metal.
Oxygen gains the electrons and O2- becomes oxidized to molecular oxygen.
Oxidation: Loss of electrons. Element charge becomes more positive
Reduction: Gain of electrons. Element charge becomes more negative
redox=short
for reductionoxidation
Oxidation-Reduction reactions or redox reactions are among the most common
and most important chemical reactions in everyday life. These reactions always happen
in pairs. Oxidation does not happen without reduction and reduction does not happen
without oxidation. The process of oxidation-reduction together with the concept of
charge conservation are the foundation of electrochemistry.
Even though we can’t
easily observe the actual
transfer of electrons, we
can easily see the results. If
you dip a zinc (Zn) nail into
a solution of copper sulfate
(CuSO4) the nail becomes
covered with a brownish
coating of copper metal.
The CuSO4 solution contains Cu2+ and SO42- ions. The copper ions are reduced to pure
copper metal by electrons coming from the zinc. The zinc is oxidized by losing the
electrons to become Zn2+ ions in the solution. The chemical equation that describes the
complete reaction is
Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s)
oxidation - loss of electrons. Element charge increases.
reduction - gain of electrons. Element charge decreases.
redox - abbreviation for oxidation reduction.
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A NATURAL APPROACH TO CHEMISTRY
Oxidation Numbers
Familiar
reactions may
be redox
reactions
Chemical processes such as the burning natural gas (CH4) and the rusting of iron (Fe) are
very familiar. The chemical equations for these reactions are:
4Fe(s) + 3O2(g) → 2Fe2O3(s)
Rusting of iron
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
Burning of methane
How can we determine if these are redox reactions? Which elements are oxidized and
which are reduced?
How to
determine if a
reaction is
redox
In order to determine if a reaction is redox we must find out if there is electron loss and
gain that takes place among the elements involved in the reaction. If we find that there
are some elements that lose electrons and some other elements that gain electrons then
we have the answer. The elements that lose electrons are oxidized, the elements that gain
electrons are reduced and the overall process is a redox reaction.
Hydrochloric acid (HCl) provides a good example that is easier to analyze. Hydrogen and
chlorine combine to form HCl by sharing a pair of electrons. One of these electrons
comes from H and the other from Cl and they form a polar covalent bond.
ne
positive
Since Chlorine is more electronegative than Electrons spend more
hydrogen, the shared electrons stay closer to Cl time closer to Cl
than to H. As a result it appears that chlorine has
gained an electron and hydrogen has lost an
Cl
H
electron. This apparent gain of an electron by the
chlorine atom is indicated by an oxidation
number of -1 which means that Cl is reduced. The
H appears to
oxidation number is the effective unit charge Cl appears to
have gained
have lost
an atom in a compound would have if the electron electrons
electrons
were completely transferred to the more
electronegative atom. In this case, chlorine is more electronegative atom, and if it takes
the electron from hydrogen, its charge is -1. Therefore its oxidation number in the
compound HCl is also -1. In this reaction chlorine is reduced since it gains an electron
and its oxidation number becomes lower (more negative).
gative
Oxidation
number
Since HCl is neutral overall, the corresponding oxidation number of hydrogen should be
+1. Hydrogen is oxidized since it loses an electron and its oxidation number increases
(more positive).
oxidation number - gives the number of electrons that an element has lost or
gained in forming a chemical bond with another element.
A NATURAL APPROACH TO CHEMISTRY
479
Section 15.2 Oxidation-Reduction (Redox) Reactions.
Rules for assigning oxidation numbers
Oxidation
number and
ionic charge
The oxidation number is different from the real charge of
the atom. The assignment of electrons to the atoms and the
associated oxidation number is just a way to account for
the electrons that are associated with the atoms.
The oxidation number is written with a sign followed with
the number like -1, +1,-3, 0, etc. This is different than the
way we write the ionic charge for which the sign follows
the number like 1-, 2-, 3+, etc.
Rules for
assigning
oxidation
numbers
ionic charge
is denoted
2-
sign follows the number
oxidation number
is denoted
-
2
number follows the sign
By looking at the properties of atoms and the way that they bond with other atoms to
form various compounds, scientists have come up with a set of rules that assign an
oxidation number to each atom in a compound. These rules are explicit representations of
the Lewis formulation and give us the number of electrons - the oxidation number - that
is assigned to each atom in a compound
TABLE 15.1. Rules for assigning oxidation numbers
1.
2.
3.
4.
5.
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The oxidation number of an atom in a pure element is 0.
For example, the oxidation number of chlorine in Cl2 or O in O2 is 0
The sum of the oxidation numbers of atoms in a neutral molecule is zero.
For example the sum of the oxidation numbers of C and O in CO2 is 0
Oxidation number of carbon (nC) plus two times the
oxidation number of oxygen = 0: nC + 2nO = 0
The sum of the oxidation numbers of all atoms in an ion is equal to the charge of
the ion.
For example, the oxidation number of Cu in Cu2+ is +2.
For SO42- , the oxidation number of S (nS) plus 4 times the
oxidation number of O (nO) = -2: nS + 4nO = -2
Metals have positive oxidation number according to their group
• Group 1A metals (Na, K, Li) have oxidation number +1
• Group 2A metals (Mg, Ca) have oxidation number +2
The oxidation number of non-metals are as follows:
• Fluorine (F): -1
• Hydrogen (H): +1 except in hydrides such as LiH and NaH
in which the oxidation number is -1.
• Oxygen (O): -2
• Group 7A (Cl, Br, I) : -1
• Group 6A (S, Se, Te) : -2
• Group 5A (N, P, As): -3
A NATURAL APPROACH TO CHEMISTRY
Application of Oxidation Number Rules
Oxidation rules
must be
applied in
order
The oxidation number rules apply in the order they appear on the list. If there is a
conflict, the rule that is higher in the list has priority. For example, consider potassium
peroxide (K2O2).
From Rule 4 we see that K has oxidation number +1 and Rule 5 tells us that oxygen has
oxidation number -2. Obviously there is a conflict since Rule 2 which keeps track of the
overall charge balance is violated. To solve this problem we must give priority to the
order that the rules appear.
Here are the steps to calculate the oxidation numbers of potassium and oxygen in K2O2.
TABLE 15.2.
Step1
From Rule 2 we know that
the sum of the oxidation
numbers of all atoms in
K2O2 must ne zero:
If the oxidation number for
potassium is nK and the
oxidation number for
oxygen is nO then
2(nK) + 2(nO) = 0
Step2
From Rule 4 we know that
the oxidation number of
potassium is +1 (nK= +1).
Since Rule 4 has priority
over Rule 5, the oxidation
number for O (nO) is still
unknown.
But now we have an
equation that we can use to
solve for it.
2(nK) + 2(nO) = 0
Step3
Solving the equation
2(nK) + 2(nO) = 0
we obtain nO = -1
and we are done.
In K2O2:
• the oxidation number of
potassium (K) is +1
• the oxidation number of
oxygen (O) is -1
The oxidation number of elements in a compound can be found by following these steps.
Find the oxidation number for each element in carbon monoxide (CO)
Asked:
Find the oxidation number of O and C in CO.
Relationships: The rules for assigning oxidation numbers
Solve:
The solution procedure is as follows:
• Carbon monoxide (CO) is a neutral compound
• Sum of oxidation numbers of C and O must add to zero (rule 2)
• The oxidation number of oxygen = -2 (rule 5).
• The oxidation number of carbon (nC) = ?
• nC + (-2) = 0 which gives nC = +2
Answer:
The oxidation number for carbon in carbon monoxide (CO) is +2
The oxidation number of oxygen in carbon monoxide is -2
A NATURAL APPROACH TO CHEMISTRY
481
Section 15.2 Oxidation-Reduction (Redox) Reactions.
Finding oxidation numbers
The same
element can
have different
oxidation
numbers
The oxidation number is a way to keep track of the electrons associated with each atom
in a compound. The oxidation number of the same element may be different from one
compound to the other. For example, in carbon dioxide (CO2) the oxidation number of
oxygen is -2. In Potassium peroxide (K2O2) the oxidation number of oxygen is -1.
Find the oxidation number for each element in carbon monoxide (CO2)
Asked:
Find the oxidation number of O and C in CO2.
Relationships: The rules for assigning oxidation numbers
Solve:
The solution procedure is as follows:
• Since carbon dioxide (CO2) is a neutral compound the sum of oxidation numbers of C and O2 must add to zero (rule 2)
• The oxidation number of O is -2 (rule 5)
• The oxidation number of O2 is 2(-2) = -4
• The oxidation number of carbon (nC) = ?
• nC + 2(-2) = 0
which gives
nC = +4
Answer:
The oxidation number for carbon in CO2 is +4 and for oxygen is -2
Find the oxidation number for each element in the nitrite ion NO2-.
Asked:
Find the oxidation number of an ion with overall charge of -1.
Given:
The elemental arrangement of the ion.
Relationships: The rules for assigning oxidation numbers
Solve:
Answer:
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Nitrite (NO2-) is an ion with a charge of -1 and so the oxidation
numbers of the elements must add up to -1.
• The sum of the oxidation numbers of N and O2 must add to -1.
(rule 2)
• The oxidation number of oxygen is -2 (rule 5)
• The oxidation number of nitrogen nN = ?.
• nN + 2(-2) = -1 which gives nN = +3
The oxidation number of oxygen in NO2- is -2.
The oxidation number of nitrogen in NO2- is +3.
+3 + 2(-2) = -1.
A NATURAL APPROACH TO CHEMISTRY
Fractional oxidation numbers
Oxidation
numbers may
also be
fractions
The oxidation number of an atom in a substances is usually a whole number such as +1, 1, +2, -2 etc. However it is also possible for the oxidation number of atoms to be
fractions such as +1/2, -1/2, -2/3 etc. This is often the case for the oxidation number of
carbon in various compounds with hydrogen (hydrocarbons).
Let’s consider methane (C3H8) and go through the various steps for calculating the
oxidation numbers of C and H in this molecule.
TABLE 15.3.
Step1
• From rule 2 we know
the sum of the oxidation
numbers of all atoms in
C3H8 must be zero:
Step2
• From rule 5 we know
that the oxidation
number of hydrogen is
+1 (nH = +1).
Step3
• Solving the equation
3(nC) + 8(+1) = 0
we obtain nC = -8/3
• If the oxidation number
• The oxidation number
for carbon is nC and the
oxidation number for
hydrogen is nH then the
equation that relates
these oxidation numbers
is 3(nC) + 8( nH) = 0
for C (nC) is still
unknown but now we
have an equation that
we can solve for it.
3(nC) + 8(+1)= 0
oxidation number can
be a fraction
• Check the answer by
verifying that the sum of
oxidation numbers is
zero.
3(-8/3) + 8 = 0
• So we see that the
Find the oxidation number for each element in the compound C3H4.
Asked:
Find the oxidation number of each element in methylacetylyne
(C3H4)
Relationships: The rules for assigning oxidation numbers
Solve:
C3H4 is a neutral compound:
The sum of the oxidation numbers of H4 and C3 must add to zero.
(rule 2)
• The oxidation number of hydrogen (nH) is +1. nH = +1 (rule 5)
• The oxidation number of carbon nC = ?.
• 3(nC)+ 4(+1) = 0 which gives nC = -4/3
Answer:
The oxidation number of C in C3H4 is -4/3.
The oxidation number of H in C3H4 is +1.
A NATURAL APPROACH TO CHEMISTRY
483
Section 15.2 Oxidation-Reduction (Redox) Reactions.
Identifying redox reactions
oxidation
numbers in
redox reactions
Determining the oxidation numbers is the first step to identifying a redox reaction. The
next step is to see if the oxidation numbers for any elements have changed in the
reaction. If we find that any oxidation numbers have changed from the left side
(reactants) to the right side (products) then we know that we have a redox reaction. If the
oxidation numbers of the elements in the reactants and the products do not change the
reaction is not redox. The best way to learn this is with an example.
Find the element that is oxidized and the element that is reduced in the
reaction of iron with oxygen resulting in rust (iron oxide Fe2O3).
4Fe(s) + 3O2(g) → 2Fe2O3(s).
Asked:
Given:
Find the element that is oxidized and the element that is reduced
The balanced chemical reaction of Fe and O2 and the rules for
assigning oxidation numbers.
Relationships: The rules for assigning oxidation numbers to each element in compounds.
Solve:
We look at both the left and right sides of the reaction:
• Let’s start with the reactants.
Since both Fe and O2 are free elements the oxidation number of
the Fe and O atoms on the left side of the equation is zero.
• Next let’s look at the product (Fe2O3) of the reaction.
The sum of the oxidation numbers in Fe2O3 must add to zero.
The oxidation number of oxygen is -2.
Now if we add the oxidation number of oxygen and the unknown
oxidation number of iron (nFe) we have the equation:
2(nFe) + 3(-2) = 0
Solving this equation for nFe we find that the oxidation number of
iron in Fe2O3 is +3.
Answer:
The oxidation number of Fe has increased; it went from 0 to +3. Iron
has lost electrons and so it is oxidized.
The oxidation number of oxygen has decreased, it went from 0 to -2.
Oxygen has gained electrons and it is reduced.
3O 2 (g ) + 4Fe(s ) → 2Fe2 O3 ( s )
0
0
+3 -2
Oxidation
oxidation
numbers
Reduction
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A NATURAL APPROACH TO CHEMISTRY
Analyzing a redox reaction
Oxidized - oxidation number increases (lose electrons)
Reduced - oxidation number decreases (gain electrons)
Charge balance
is maintained
The number of electrons that are received by the reduced element is equal to the number
of electrons that have been given up by the oxidized element. The total number of
electrons does not change – overall charge neutrality must be maintained.
In the iron - oxygen reaction we saw that the
iron is oxidized and the oxygen is reduced. We
can also say that “iron is oxidized by oxygen”
which is another way of saying that oxygen
facilitates the oxidation of iron. We may also say
that “oxygen is reduced by iron” oxygen is
reduced and iron facilitates this reduction.
Oxidizing
agent
Reducing
agent
Oxidized
Iron (Fe)
Reducing
agent
Reduction
reaction
Oxidation
reaction
Reduced
The element that oxidizes another element is
Oxygen (O)
called the oxidizing agent. In our example
Oxidizing
agent
oxygen is the oxidizing agent. The element that
reduces another element is called the reducing
agent. In our example iron is the reducing 4Fe(s ) + 3O 2 (g ) → 2Fe 2 O3 (s )
agent.
In the reaction 4Fe(s) + 3O2(g) → 2Fe2O3(s):
Fe is oxidized
O is reduced
Fe is the reducing agent
O is the oxidizing agent
Find the reducing and the oxidizing agents in: 2Mg(s) + O2(g) → 2MgO(s).
Relationships: The rules for assigning oxidation numbers and the chemical reaction
Solve:
O 2 (g ) + 2Mg(s ) → 2Mg O(s )
0
0
Reduction
Oxidation
+2 -2
oxidation numbers
The oxidation number of oxygen decreases from 0 to -2
so oxygen is reduced. Therefore, oxygen is the oxidizing agent.
The oxidation number of magnesium increases from 0 to +2 so magnesium is oxidized. Therefore, magnesium is the reducing agent.
oxidizing agent - the element that oxidizes another element.
reducing agent - the element that reduces another element.
A NATURAL APPROACH TO CHEMISTRY
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