Transcript 20.2 Oxidation Numbers

20.2 Oxidation Numbers >
Chapter 20
Oxidation-Reduction
Reactions
20.1 The Meaning of Oxidation
and Reduction
20.2 Oxidation Numbers
20.3 Describing Redox
Equations
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20.2 Oxidation Numbers >
CHEMISTRY
& YOU
Why does a sparkler have such a
bright light?
If you have ever
seen or held a
sparkler, then you
know that sparklers
give off very bright
light. They are like
handheld fireworks.
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20.2 Oxidation Numbers > Assigning Oxidation Numbers
Assigning Oxidation Numbers
What is the general rule for assigning
oxidation numbers?
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20.2 Oxidation Numbers > Assigning Oxidation Numbers
An oxidation number is a positive or
negative number assigned to an atom to
indicate its degree of oxidation or
reduction.
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20.2 Oxidation Numbers > Assigning Oxidation Numbers
As a general rule, a bonded atom’s
oxidation number is the charge that it
would have if the electrons in the
bond were assigned to the atom of
the more electronegative element.
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20.2 Oxidation Numbers > Assigning Oxidation Numbers
In binary ionic compounds, the oxidation
numbers of the atoms equal their ionic
charges.
• The compound sodium chloride is composed
of sodium ions (Na1+) and chloride ions (Cl1–).
• The oxidation number of sodium is +1.
• That of chlorine is –1.
– Notice that the sign is put before the oxidation
number.
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20.2 Oxidation Numbers > Assigning Oxidation Numbers
Because water is a molecular compound,
no ionic charges are associated with its
atoms.
• Oxygen is reduced in the formation of water.
• Oxygen is more electronegative than hydrogen.
• The two shared electrons in the H–O bond are
shifted toward oxygen and away from hydrogen.
– The oxidation number of oxygen is –2.
– The oxidation number of each hydrogen is +1.
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20.2 Oxidation Numbers > Assigning Oxidation Numbers
Oxidation numbers are often written above
the chemical symbols in a formula.
+1 –2
H 2O
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20.2 Oxidation Numbers > Assigning Oxidation Numbers
Rules for Assigning Oxidation Numbers
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1.
The oxidation number of a monatomic ion is equal in
magnitude and sign to its ionic charge. For example, the
oxidation number of the bromide ion (Br1–) is –1; that of
the Fe3+ ion is +3.
2.
The oxidation number of hydrogen in a compound is +1,
except in metal hydrides, such as NaH, where it is –1.
3.
The oxidation number of oxygen in a compound is –2,
except in peroxides, such as H2O2, where it is –1, and in
compounds with the more electronegative fluorine,
where it is positive.
4.
The oxidation number of an atom in uncombined
(elemental) form is 0. For example, the oxidation number
of the potassium atoms in potassium metal (K) or of the
nitrogen atoms in nitrogen gas (N2) is 0.
5.
For any neutral compound, the sum of the oxidation
numbers of the atoms in the compound must equal 0.
6.
For a polyatomic ion, the sum of the oxidation numbers
must equal the ionic charge of the ion.
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20.2 Oxidation Numbers >
Sample Problem 20.2
Assigning Oxidation Numbers to Atoms
What is the oxidation number of each kind of
atom in the following ions and compounds?
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a. SO2
c. Na2SO4
b. CO32–
d. (NH4)2S
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20.2 Oxidation Numbers >
Sample Problem 20.2
1 Analyze Identify the relevant concepts.
Use the set of rules you just learned to
assign and calculate oxidation numbers.
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20.2 Oxidation Numbers >
Sample Problem 20.2
2 Solve Apply concepts to this situation.
a. There are two oxygen atoms, and the
oxidation number of each oxygen is –2
(Rule 3).
The sum of the oxidation numbers for the
neutral compound must be 0 (Rule 5).
Therefore, the oxidation number of sulfur is
+4, because +4 + (2 × (–2)) = 0.
+4 –2
SO2
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20.2 Oxidation Numbers >
Sample Problem 20.2
2 Solve Apply concepts to this situation.
b. The oxidation number of each oxygen is –2
(Rule 3).
? –2
CO32–
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20.2 Oxidation Numbers >
Sample Problem 20.2
2 Solve Apply concepts to this situation.
b. The sum of the oxidation numbers of the
carbon and oxygen atoms must equal the
ionic charge, –2 (Rule 6).
The oxidation number of carbon must be +4,
because +4 + (3 × (–2)) = –2.
+4 –2
CO32–
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20.2 Oxidation Numbers >
Sample Problem 20.2
2 Solve Apply concepts to this situation.
c. The oxidation number of each sodium
ion, Na+, is the same as its ionic charge,
+1 (Rule 1).
The oxidation number of oxygen is –2
(Rule 3).
+1
? –2
Na2SO4
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20.2 Oxidation Numbers >
Sample Problem 20.2
2 Solve Apply concepts to this situation.
c. For the sum of the oxidation numbers in
the compound to be 0 (Rule 5), the
oxidation number of sulfur must be +6,
because (2 × (+1)) + (+6) + (4 × (–2)) =
0.
+1
+6 –2
Na2SO4
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20.2 Oxidation Numbers >
Sample Problem 20.2
2 Solve Apply concepts to this situation.
d. Ammonium ions, NH4+, have an ionic charge
of +1, so the sum of the oxidation numbers of
the atoms in the ammonium ion must be +1.
The oxidation number of hydrogen is +1 in
this ion.
So, the oxidation number of nitrogen must be
–3.
? +1
NH4+
? + 4(+1) = +1
–3 + 4(+1) = +1
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20.2 Oxidation Numbers >
Sample Problem 20.2
2 Solve Apply concepts to this situation.
d. Two ammonium ions have a total charge of
+2.
Since the compound (NH4)2S is neutral,
sulfur must have a balancing oxidation
number of –2.
–3 +1
–2
(NH4)2S
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20.2 Oxidation Numbers >
Sample Problem 20.2
3 Evaluate Do the results make sense?
• The results are consistent with the rules for
determining oxidation numbers.
• Also, addition of the oxidation numbers
correctly gives the final overall charge for
the ion and the three neutral compounds.
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20.2 Oxidation Numbers >
Chromium in its uncombined state is a dull
silvery color. Orange potassium dichromate
(K2Cr2O7) and purple chromium(III) potassium
sulfate (CrK(SO4)2·12H2O) are both compounds
of chromium. What is the oxidation number of
chromium in each compound?
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20.2 Oxidation Numbers >
Chromium in its uncombined state is a dull
silvery color. Orange potassium dichromate
(K2Cr2O7) and purple chromium(III) potassium
sulfate (CrK(SO4)2·12H2O) are both compounds
of chromium. What is the oxidation number of
chromium in each compound?
+1 +6
–2
K2Cr2O7
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+3 +1 +6 –2
CrK(SO4)2·12H2O
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Oxidation-Number Changes in
20.2 Oxidation Numbers > Chemical Reactions
Oxidation-Number Changes in
Chemical Reactions
How are oxidation and reduction
defined in terms of a change in
oxidation number?
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Oxidation-Number Changes in
20.2 Oxidation Numbers > Chemical Reactions
The figure below shows what happens when
copper wire is placed in a solution of silver nitrate.
• The oxidation number of silver decreases from +1 to
0 as each silver ion (Ag1+) gains an electron and is
reduced to silver metal (Ag0).
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Oxidation-Number Changes in
20.2 Oxidation Numbers > Chemical Reactions
The figure below shows what happens when
copper wire is placed in a solution of silver nitrate.
• Copper’s oxidation number increases from 0 to +2
as each atom of copper metal (Cu0) loses two
electrons and is oxidized to copper(II) ion (Cu2+).
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Oxidation-Number Changes in
20.2 Oxidation Numbers > Chemical Reactions
The figure below shows what happens when
copper wire is placed in a solution of silver nitrate.
+1 +5 –2
0
+2
+5 –2
0
2AgNO3(aq) + Cu(s) → Cu(NO3)2(aq) + 2Ag(s)
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Oxidation-Number Changes in
20.2 Oxidation Numbers > Chemical Reactions
This figure illustrates a redox reaction that shows
what occurs when a shiny iron nail is dipped into a
solution of copper(II) sulfate.
• The iron reduces
Cu2+ ions in
solution and is
simultaneously
oxidized to Fe2+.
• The iron becomes
coated with
metallic copper.
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Oxidation-Number Changes in
20.2 Oxidation Numbers > Chemical Reactions
You can define oxidation and reduction in
terms of a change in oxidation number.
An increase in the oxidation number of
an atom or ion indicates oxidation.
A decrease in the oxidation number of an
atom or ion indicates reduction.
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20.2 Oxidation Numbers >
CHEMISTRY
& YOU
What happens to the oxidation
numbers of metals as they burn in a
sparkler?
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20.2 Oxidation Numbers >
CHEMISTRY
& YOU
What happens to the oxidation
numbers of metals as they burn in a
sparkler?
As the metals burn, they gain
oxygen or undergo oxidation. A
substance that undergoes
oxidation has an increase in
oxidation number. Therefore,
as metals burn, their oxidation
numbers increase.
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20.2 Oxidation Numbers >
Sample Problem 20.3
Identifying Oxidized and Reduced Atoms
Use changes in oxidation number to
identify which atoms are oxidized and
which are reduced in the following
reactions. Also identify the oxidizing
agent and the reducing agent.
a. Cl2(g) + 2HBr(aq) → 2HCl(aq) + Br2(l)
b. C(s) + O2(g) → CO2(g)
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20.2 Oxidation Numbers >
Sample Problem 20.3
1 Analyze Identify the relevant concepts.
• An increase in oxidation number
indicates oxidation.
• A decrease in oxidation number
indicates reduction.
• The substance that is oxidized in a redox
reaction is the reducing agent.
• The substance that is reduced is the
oxidizing agent.
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20.2 Oxidation Numbers >
Sample Problem 20.3
2 Solve Apply concepts to this situation.
a. Use the rules to assign oxidation numbers
to each atom in the equation.
0
+1 –1
+1 –1
0
Cl2(g) + 2HBr(aq) → 2HCl(aq) + Br2(l)
The oxidation number of
each chlorine in Cl2 is 0
because of Rule 4.
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20.2 Oxidation Numbers >
Sample Problem 20.3
2 Solve Apply concepts to this situation.
a. Then use the changes in oxidation numbers
to identify which atoms are oxidized and
which are reduced.
0
+1 –1
+1 –1
0
Cl2(g) + 2HBr(aq) → 2HCl(aq) + Br2(l)
• The element chlorine is reduced because its
oxidation number decreases (0 to –1).
• The bromide ion from HBr(aq) is oxidized
because its oxidation number increases (–1 to 0).
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20.2 Oxidation Numbers >
Sample Problem 20.3
2 Solve Apply concepts to this situation.
a. Finally, identify the oxidizing and reducing
agents.
0
+1 –1
+1 –1
0
Cl2(g) + 2HBr(aq) → 2HCl(aq) + Br2(l)
• Chlorine is reduced, so Cl2 is the oxidizing agent.
• The bromide ion from HBr(aq) is oxidized, so Br–
is the reducing agent.
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20.2 Oxidation Numbers >
Sample Problem 20.3
2 Solve Apply concepts to this situation.
b. Use the rules to assign oxidation numbers
to each atom in the equation.
0
0
+4 –2
C(s) + O2(g) → CO2(g)
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20.2 Oxidation Numbers >
Sample Problem 20.3
2 Solve Apply concepts to this situation.
b. Then use the changes in oxidation numbers
to identify which atoms are oxidized and
which are reduced.
0
0
+4 –2
C(s) + O2(g) → CO2(g)
• The element carbon is oxidized because its
oxidation number increases (0 to +4).
• The element oxygen is reduced because its
oxidation number decreases (0 to –2).
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20.2 Oxidation Numbers >
Sample Problem 20.3
2 Solve Apply concepts to this situation.
b. Finally, identify the oxidizing and reducing
agents.
0
0
+4 –2
C(s) + O2(g) → CO2(g)
• Carbon is oxidized, so C is the reducing agent.
• Oxygen is reduced, so O2 is the oxidizing agent.
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20.2 Oxidation Numbers >
Sample Problem 20.3
3 Evaluate Do the results make sense?
• It makes sense that what is oxidized in a
chemical reaction is the reducing agent
because it loses electrons—it becomes the
agent by which the atom that is reduced gains
electrons.
• Conversely, it makes sense that what is
reduced in a chemical reaction is the oxidizing
agent because it gains electrons—it is the
agent by which the atom that is oxidized loses
electrons.
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20.2 Oxidation Numbers >
Sample Problem 20.4
Identifying Oxidized and Reduced
Atoms
Use changes in oxidation number to
identify which atoms are oxidized and
which are reduced in the following
reaction. Also identify the oxidizing agent
and the reducing agent.
Zn(s) + 2MnO2(s) + 2NH4Cl(aq) → ZnCl2(aq) + Mn2O3(s) + 2NH3(g) + H2O(l)
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20.2 Oxidation Numbers >
Sample Problem 20.4
1 Analyze Identify the relevant concepts.
• An increase in oxidation number
indicates oxidation.
• A decrease in oxidation number
indicates reduction.
• The substance that is oxidized in a redox
reaction is the reducing agent.
• The substance that is reduced is the
oxidizing agent.
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20.2 Oxidation Numbers >
Sample Problem 20.4
2 Solve Apply concepts to this situation.
Use the rules to assign oxidation numbers to
each atom in the equation.
0
+4 –2
–3 +1 –1
+2 –1
+3
–2
–3 +1
+1 –2
Zn(s) + 2MnO2(s) + 2NH4Cl(aq) → ZnCl2(aq) + Mn2O3(s) + 2NH3(g) + H2O(l)
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20.2 Oxidation Numbers >
Sample Problem 20.4
2 Solve Apply concepts to this situation.
Then use the changes in oxidation numbers to
identify which atoms are oxidized and which
are reduced.
0
+4 –2
–3 +1 –1
+2 –1
+3
–2
–3 +1
+1 –2
Zn(s) + 2MnO2(s) + 2NH4Cl(aq) → ZnCl2(aq) + Mn2O3(s) + 2NH3(g) + H2O(l)
• The element zinc is oxidized because its oxidation
number increases (0 to +2).
• The manganese ion is reduced because its
oxidation number decreases (+4 to +3).
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20.2 Oxidation Numbers >
Sample Problem 20.4
2 Solve Apply concepts to this situation.
Finally, identify the oxidizing and reducing
agents.
0
+4 –2
–3 +1 –1
+2 –1
+3
–2
–3 +1
+1 –2
Zn(s) + 2MnO2(s) + 2NH4Cl(aq) → ZnCl2(aq) + Mn2O3(s) + 2NH3(g) + H2O(l)
• Zinc is oxidized, so Zn is the reducing agent.
• Manganese (in MnO2) is reduced, so Mn4+ is the
oxidizing agent.
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20.2 Oxidation Numbers >
Use changes in oxidation number to identify
which atoms are oxidized and which are
reduced in the following reaction.
2HNO3(aq) + 3H2S(g) → 2NO(g) + 4H2O(l) + 3S(s)
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20.2 Oxidation Numbers >
Use changes in oxidation number to identify
which atoms are oxidized and which are
reduced in the following reaction.
2HNO3(aq) + 3H2S(g) → 2NO(g) + 4H2O(l) + 3S(s)
+1 +5 –2
+1 –2
+2 –2
+1 –2
0
2HNO3(aq) + 3H2S(g) → 2NO(g) + 4H2O(l) + 3S(s)
Sulfur is oxidized because its oxidation number
increases (–2 to 0). Nitrogen is reduced because its
oxidation number decreases (+5 to +2).
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20.2 Oxidation Numbers > Key Concepts
As a general rule, a bonded atom’s
oxidation number is the charge that it
would have if the electrons in the bond
were assigned to the atom of the more
electronegative element.
An increase in the oxidation number of
an atom or ion indicates oxidation. A
decrease in the oxidation number of an
atom or ion indicates reduction.
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20.2 Oxidation Numbers > Glossary Term
oxidation number: a positive or
negative number assigned to an atom to
indicate its degree of oxidation or
reduction; the oxidation number of an
uncombined element is zero
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20.2 Oxidation Numbers >
BIG IDEA
Reactions
Redox reactions are identified by changes
in oxidation number.
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20.2 Oxidation Numbers >
END OF 20.2
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