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Oxidation-Reduction Reactions
Fe2O3(s) + 2 Al(s)  2 Fe(s) + Al2O3(s)
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Chapter 6:
Oxidation-Reduction Reactions
Reactions that involve the transfer of electrons are
called oxidation-reduction or redox reactions
OIL RIG or LEO says GER
• Oxidation is the loss of electrons by a reactant
• Reduction is the gain of electrons by a reactant
• oxidizing agent
• reduction agent
Oxidation numbers provide a way to
keep track of electron transfers :
1) The oxidation number of any free element is zero.
2) The oxidation number of any simple,
monoatomic ion is equal to the charge on the ion.
3) The sum of all oxidation numbers of the atoms in
a molecule or polyatomic ion must equal the
charge on the particle.
4) In its compounds, fluorine has an oxidation
number of –1.
5) In its compounds, hydrogen has an oxidation
number of +1.
6) In its compounds, oxygen has an oxidation
number of –2.
p.218
Problem Solving
Determine the oxidation numbers for the elements
in the following compounds and reactions
MoO2
BrO3-
KN3
HPO4Zr(SrO4)2
Mg2TiO4
Fe2O3
Question:
What is the oxidation number of P in
H3PO4?
a: -5
b: -3
c: 0
d: +3
e: +5
Problem Solving
Determine the oxidation numbers for the elements in the
following compounds and reactions
Cu2+ + Zn  Zn2+ + Cu
Br2 + 2 NaBrO2  2BrO2 + 2 NaCl
NaClO3 + 3HNO2  NaCl + 3 HNO3
Question:
The oxidation number of an atom
increases during a reaction. This
means that the atom
was____________ and the substance
containing it was the____________
a: oxidized; reducing agent
b: reduced; oxidizing agent
c: oxidized; oxidizing agent
d: reduced; reducing agent
Question:
In a redox titration, MnO4- reacts to
Mn2+. In this reaction, is Mn
reduced or oxidized?
a: reduced
b: oxidized
Question:
What is the reducing agent in this
reaction?
Sn + Cl2 SnCl2
a: Cl2
b: Sn
c: SnCl2
2
3
Al( s)  Cu (aq)  Al (aq)  Cu( s)
• The oxidation and
reduction are divided into
equations called halfreactions
• The half-reactions are
balanced separately, then
combined into the fully
balanced net ionic
equation
• Both mass and charge
must be balanced
• Charge is balanced by
adding electrons to the
side of the equation that is
more positive or less
negative
Al( s )  Cu 2 (aq)  Al 3 (aq)  Cu( s )
ANALYSIS : This is a redox reaction.
SOLUTION :
Oxidation : Al( s )  Al 3  3e Reduction : Cu 2 (aq)  2e -  Cu( s)
The least common factor is 6, combining
2 Al( s )  3Cu 2 (aq)  2Al 3 (aq)  3Cu( s )
Ion-electron method
Fe (s) + Cu+2 (aq)  Cu (s) + Fe+3 (aq)
V (s) + Al+3(aq)  V+5 (aq) + Al (s)
Question:
What is the coefficient of Ce3+ when
the equation is balanced with the
ion-electron method?
3+
Ce
+
a: 1
b: 2
c: 3
d: 4
e: 5
5+
V

2+
V
+
4+
Ce
The Ion-Electron Method in Acidic Solution:
1)
2)
3)
4)
5)
6)
Divide the equation into two half-reactions.
Balance atoms other than H and O.
Balance O by adding water.
Balance H by adding hydrogen ion.
Balance net charge by adding electrons.
Make electron gain and loss equal: add halfreactions.
7) Cancel anything that’s the same on both sides
of the equation.
p. 227
Balance in acidic solution
3+
Cr
+ BiO3  Cr2O7 +
-
2-
3+
Bi
Balance in acidic solution
HNO2 + MnO4-  Mn2+ + NO3-
Question:
What must always cancel in writing
the net balanced redox equation?
a: H+
b: H2O
c: OHd: e-
Additional Steps for Basic Solutions
8) Add to both sides of the equation t he same number
of OH- ions as there are H  .
9) Combine H  and OH- to form H 2O.
10) Cancel any H 2 O that you can.
Balance the following in basic solution:
MnO  C 2 O  MnO 2  CO
4
24
23
MnO -4  C 2 O 24-  MnO 2  CO32ANALYSIS : Balance as if in acidic solution t hen " convert".
SOLUTION :
C 2 O 24-  2H 2 O  2CO32-  4H   2e MnO -4  4H   3e -  MnO 2  2H 2 O
Net ionic : 3C 2 O 24-  2H 2 O  2MnO -4  6CO32  4H   2MnO 2
Add OH3C 2 O 24-  2H 2 O  4OH-  2MnO -4  6CO32  4(H   OH  )  2MnO 2
Form H 2 O
3C 2 O 24-  2H 2 O  4OH-  2MnO -4  6CO32  4H 2 O  2MnO 2
Simplify
3C 2 O 24-  4OH-  2MnO -4  6CO32  2H 2 O  2MnO 2
Balance in basic solution
Fe(OH)2 + O2  Fe(OH)3 +
OH
Balance in basic solution
Au +
CN
+ O2  Au(CN)4 +
-
OH
What is the coefficient indicated when the
following is balanced?
MnO4-(aq) + Cr (s) → Cr2O72-(aq) + ? MnO2(s)
A. 1
B. 2
C. 3
D. 4
E. none of these
23
What is the coefficient indicated when
the following is balanced?
? PbSO4(s) →Pb(s) + PbO2(s) + H2SO4(aq)
A. 1
B. 2
C. 3
D. 4
E. none of these
5.2 The ion–electron method creates balanced net ionic equations for redox reactions
24
Reactions of metals with acids
Zn + H2SO4  ZnSO4 + H2
Cu + H2SO4  no reaction
Zinc is more active than hydrogen
Copper is less active than hydrogen
The more active will be
Nonoxidizi ng acids : HCl( aq ), cold dilute H 2SO 4 (aq ),
H 3 PO 4 (aq ), and most organic acids.
Ox. Acid
HNO 3
Reduction Reaction
(conc.) NO3-  2H   e -  NO2 ( g )  H 2 O
(dilute) NO3-  4H   3e -  NO(g )  4H 2 O
(very dilute, with strong reducing agent)


4
NO  10H  8e  NH  3H 2 O
3
H 2SO 4
-
(hot, conc.) SO 24-  4H   3e -  SO 2 ( g )  2H 2 O
(hot conc., with strong reducing agent)
SO -4  10H   8e -  H 2S( g )  4H 2 O
Table 6.1 p231
Nitric acid acts upon copper
• Ira Remsen (1846-1927) founded the chemistry department at Johns
Hopkins University, and founded one of the first centers for chemical
research in the United States; saccharin was discovered in his research
lab in 1879.
Ira Remsen (1846-1927) founded the chemistry department at Johns Hopkins University, and founded one of the first
centers for chemical research in the United States; saccharin was discovered in his research lab in 1879. Like
many chemists, he had a vivid "learning experience," which led to a heightened interest in laboratory work:
While reading a textbook of chemistry I came upon the statement, "nitric acid acts upon
copper." I was getting tired of reading such absurd stuff and I was determined to see
what this meant. Copper was more or less familiar to me, for copper cents were then in
use. I had seen a bottle marked nitric acid on a table in the doctor's office where I was
then "doing time." I did not know its peculiarities, but the spirit of adventure was upon
me. Having nitric acid and copper, I had only to learn what the words "act upon"
meant. The statement "nitric acid acts upon copper" would be something more than
mere words. All was still. In the interest of knowledge I was even willing to sacrifice
one of the few copper cents then in my possession. I put one of them on the table,
opened the bottle marked nitric acid, poured some of the liquid on the copper and
prepared to make an observation. But what was this wonderful thing which I beheld?
The cent was already changed and it was no small change either. A green-blue liquid
foamed and fumed over the cent and over the table. The air in the neighborhood of the
performance became colored dark red. A great colored cloud arose. This was
disagreeable and suffocating. How should I stop this? I tried to get rid of the
objectionable mess by picking it up and throwing it out of the window. I learned
another fact. Nitric acid not only acts upon copper, but it acts upon fingers. The pain
led to another unpremeditated experiment. I drew my fingers across my trousers and
another fact was discovered. Nitric acid acts upon trousers. Taking everything into
consideration, that was the most impressive experiment and relatively probably the
most costly experiment I have ever performed. . . . It was a revelation to me. It resulted
in a desire on my part to learn more about that remarkable kind of action. Plainly, the
only way to learn about it was to see its results, to experiment, to work in a laboratory.
from F. H. Getman, "The Life of Ira Remsen"; Journal of Chemical Education: Easton, Pennsylvania, 1940; pp 9-10; quoted in Richard W. Ramette,
"Exocharmic Reactions" in Bassam Z. Shakhashiri, Chemical Demonstrations: A Handbook for Teachers of Chemistry, Volume 1. Madison: The University
of Wisconsin Press, 1983, p. xiv:
An Active Metal
One metal can displace another from solution
Zn (s) + CuSO4 (aq)  Cu (s) + ZnSO4 (aq)
Zinc is “more active” than copper
Zinc is more active than copper
• The more active metal will displace the
lesser active metal from solution
An activity series
arranges metals
according to their
ease of oxidation
• They can be used
to predict
reactions
Copper is less active than Zinc
Solid copper placed into zinc nitrate solution
Activity Series for Some Metals and Hydrogen
Element Oxidation Product
Least active Gold
Au 3
Silver
Ag 
Copper
Cu 2
Hydrogen H 
Most active
Sodium
Na 
Cesium
Cs 
A given element will be displaced from its
compounds by any element below it in the table
(See Table 6.2
for a more
extensive list.)
Reactivity with acid
Using the following observations, rank these
metals from most reactive to least reactive
• Cu(s) + HCl(aq) → no reaction
• Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)
• Mg(s) + ZnCl2(aq) → MgCl2(aq) + Zn(s)
a. Mg > Zn > H > Cu
b. Mg > H > Cu > Zn
c. Cu > H > Zn > Mg
d. H > Cu > Mg > Zn
35
Activity
Series
5.4 A more active metal will displace a less active one from its compounds
36
Predict the Products of the Following:
• Zn + CuSO4→
• Cu + ZnSO4 →
• AgNO3(aq) + Cu(s) →
• Solid magnesium and aqueous iron(III)
chloride
• Solid nickel and aqueous sodium chloride
37
Predict the Products of the Following:
• Zn + CuSO4→
Cu(s) + ZnSO4(aq)
• Cu + ZnSO4 →
NR
• AgNO3(aq) + Cu(s) →
Ag(s) + Cu(NO3)2 (aq)
• Solid magnesium and aqueous iron(III) chloride
Mg(s) + FeCl3(aq)→ MgCl2(aq) + Fe(s)
• Solid nickel and aqueous sodium chloride
Nis) + NaCl(aq)→
NR
5.4 A more active metal will displace a less active one from its compounds
38
Oxygen reacts with many substances
The products depend, in part, on how much oxygen
is available
Combustion of hydrocarbons
O 2 plentiful :
CH 4  2O 2  CO2  2H 2O
O 2 limited :
2CH 4  3O 2  2CO  4H 2 O
O 2 very limited : CH 4  O 2  C  2H 2O
Organic compounds containing O also produce carbon dioxide
and water
C2 H5OH  3O2  2CO2  3H 2O
Problem Solving:
A 0.3000 g sample of tin ore was dissolved in
acid solution converting all the tin to tin(II).
In a titration, 8.08 mL of 0.0500 M KMnO4
was required to oxidize the tin(II) to tin(IV).
What was the percentage tin in the original
sample? The reaction also produced MnO2
3Sn 2  2MnO -4  8H   3Sn 4  2MnO 2  4H 2O
Problem Solving
When 250.00 mL of 0.200M of sodium sulfite is reacted
with 3.50 g of potassium dichromate (K2Cr2O7
FW=294.18) in an acidic solution, the products of the
reaction include sulfate ion and chromium(III) ion.
What is the concentration of chromium(III) ion if the
total volume is now 251.25 mL?
How would the problem be different with 5.50 g of
potassium dichromate?
Problem Solving
A solution of sodium thiosulfate (Na2S2O3) reacts
with chlorine gas in an acidic solution to give
sulfate ion and chloride ion. How many mL of
0.100 M sodium thiosulfate would be needed to
react with 4.25 g of chlorine gas?
Problem Solving
All the iron in a 2.000 g sample of iron ore is
dissolved in an acidic solution and
converted to iron(II) ion. When titrated
with 0.100 M potassium permanganate the
iron was oxidized to iron(III) and the
permanganate to manganese dioxide. The
titration required 27.45 ml of permanganate
solution to reach the endpoint. What was
the percent by mass iron in the ore?
Balance in acidic solution
I2 + OCl-  IO3- + ClH3AsO3 + Cr2O72-  H3AsO4 + Cr3+
Balance in acidic solution
• VO2+ + Sn2+  VO2+ + Sn4+
Balance in basic solution
• NiO2 + Mn(OH)2  Mn2O3 + Ni(OH)2
• O2 + N2H2  H2O2 + N2
In an acidic solution, Manganese(II) is oxidized
to MnO4- by bismuthate ion (BiO3- ). In the
reaction, BiO3- is reduced to Bi3+
Write a balanced net ionic equation for the
reaction
How many grams of NaBiO3 are needed to
oxidize 25.0 mL of 0.200 M MnSO4
• Aluminum metal and oxygen gas forms
aluminum oxide solid.
• Solid sulfur (S8) burns in oxygen gas to make
gaseous sulfur trioxide
• Copper metal is heated in oxygen to form
black copper(II) oxide solid.
5.5 Molecular oxygen is a powerful oxidizing agent
51
What is the coefficient on O2 when octane,
C8H18 is combusted with scant oxygen?
A. 1
B. 2
C. 3
D. 4
E. none of these
5.5 Molecular oxygen is a powerful oxidizing agent
52
What is the coefficient on O2 when iron
combusts with plentiful oxygen
available ?
A. 1
B. 2
C. 3
D. 4
E. none of these
5.5 Molecular oxygen is a powerful oxidizing agent
53
Ore Analysis
A 0.3000 g sample of tin ore was dissolved in acid
solution converting all the tin to tin(II). In a
titration, 8.08 mL of 0.0500 M KMnO4 was
required to oxidize the tin(II) to tin(IV). What was
the percentage tin in the original sample?
2

4
3Sn  2MnO  8H  3Sn  2MnO 2  4H 2O
4
5.6 Redox reactions follow the same stoichiometric principles as other reactions
54
A 25.0 g sample of granite contains a vein of
copper. What is the % of Cu present if 25.00
mL of concentrated 15 M HNO3 are reacted?
Cu(s) + 2NO3-(aq) + 4H+(aq) →2NO2(g) + 2H2O(l) +
Cu2+
A. 12
B. 38
C. 48
D. 95
E. none of these
5.6 Redox reactions follow the same stoichiometric principles as other reactions
55