Chapter 14 – Oxidation-Reduction Reactions
14.1
(a) electrolytic cell; (b) cathode; (c) oxidation; (d) half-reaction; (e) electrochemistry; (f) electrode; (g)
oxidizing agent; (h) oxidation-reduction reaction; (i) corrosion; (j) oxidation number
14.2
(a) voltaic (galvanic)cell; (b) anode; (c) reduction; (d) half-cell; (e) battery; (f) salt bridge; (g) reducing
agent; (h) electrolysis; (i) spontaneous
14.3
An oxidation-reduction reaction (redox reaction) is a reaction where electrons are transferred from one
reactant to another.
14.4
Oxidation is always coupled with reduction because the electrons released in the oxidation process must be
absorbed by another substance, resulting in its reduction.
14.5
A substance is oxidized when its oxidation number increases. For monatomic ions, the charge becomes
more positive as a result of losing electrons (i.e. Zn → Zn2+ + 2e−; the charge on Zn changes from zero
to 2+).
14.6
A substance is reduced when its oxidation number decreases. For monatomic ions, the charge becomes
more negative as a result of gaining electrons (i.e. Cl2 + 2e− → 2Cl−; the charge on Cl changes from zero
to 2−).
14.7
(a) Mg(s) + Cu(NO3)2(aq) → Cu(s) + Mg(NO3)2(aq)
(b) Magnesium is in its elemental form as a reactant, so we know that its charge is zero. However, as a
product, the charge on magnesium is 2+ (because the charge of each nitrate ion is 1−). The charge on
magnesium increases by two when it transfers two electrons to copper (Mg → Mg2+ + 2e−).
2+ 1–
2+ 1–
0
0
Mg(s) + Cu(NO3)2(aq)
Cu(s) + Mg(NO3)2(aq)
(c) As a reactant, the charge on copper is 2+ (because the charge of each nitrate ion is 1−). As a product,
copper is in its elemental form so its charge is zero. The charge on copper decreases by 2 when it
gains 2 electrons from magnesium (Cu2+ + 2e− → Cu).
14.8
(a) Ni(s) + Pb(NO3)2(aq) → Pb(s) + Ni(NO3)2(aq)
(b) As a reactant, nickel is in its elemental form so we know that its charge is zero. As a product, the
charge on nickel is 2+ (because the charge of each nitrate ion is 1−). The charge on nickel increases by
2 when it transfers 2 electrons to lead (Ni → Ni2+ + 2e−).
2+ 1–
2+ 1–
0
0
Pb(s) + Ni(NO3)2(aq)
Ni(s) + Pb(NO3)2(aq)
(c) As a reactant, lead has a charge of 2+ (because the charge of each nitrate ion is 1−). As a product, the
charge on lead is zero because it is in its elemental form. The charge on lead decreases by two when it
gains two electrons from nickel (Pb2+ + 2e− → Pb).
14.9
To begin, we determine the charge on each reactant and product. The species whose charge increases is
oxidized, and the species whose charge decreases is reduced. The species that contains the element being
reduced is the oxidizing agent, and the species that contains the element being oxidized is the reducing
agent.
14−1
0
(a)
(b)
(c)
(d)
(e)
2+ 2–
2+ 2–
0
Mg(s) + SnSO4(aq)
MgSO4(aq) + Sn(s)
The oxidation number of magnesium increases (0 → 2+), so Mg is oxidized in this reaction.
The oxidation number of tin decreases (2+ → 0), so Sn2+ is reduced in this reaction.
SnO4 is the oxidizing agent because it accepts electrons from Mg.
Mg is the reducing agent because it gives two electrons to Sn2+.
Sulfate is a spectator ion. Tin ions (red) make contact with the surface of the magnesium metal.
Magnesium atoms (gray) transfer two electrons to the tin ions. The tin ions deposit as tin metal.
Mg2+
Mg
2e–
14.10
S n2+
To begin, we determine the charge on each reactant and product. The species whose charge increases is
oxidized, and the species whose charge decreases is reduced. The species that contains the element being
reduced is the oxidizing agent, and the species that contains the element being oxidized is the reducing
agent.
0
2+ 2–
2+ 2–
0
Mn(s) + CdSO4(aq)
(a)
(b)
(c)
(d)
MnSO4(aq) + Cd(s)
The oxidation number of manganese increases (0 → 2+), so Mn is oxidized in this reaction.
The oxidation number of cadmium decreases (2+ → 0), so Cd2+ is reduced in this reaction.
Cd2+ is the oxidizing agent because it accepts electrons from Mn.
Mn is the reducing agent because it gives two electrons to Cd2+.
(e) Sulfate is a spectator ion. Cadmium ions (red) make contact with the surface of the manganese metal.
Manganese atoms (grey) transfer two electrons to the cadmium ions. The cadmium ions deposit as
cadmium metal.
14−2
Mn2+
Mn
2e–
Cd2+
14.11
An oxidation number is part of a bookkeeping method for tracking where electrons go during a chemical
reaction. We assign oxidation numbers to atoms in covalent compounds or polyatomic ions as if they were
ions, giving them charges established by a set of rules. Then we can determine which substance contains
an atom(s) that loses electrons, and which contains an atom(s) that gains electrons. Substances with atoms
that lose electrons are oxidized, and the oxidation number of the oxidized atom(s) increases. Substances
with atoms that gain electrons are reduced, and the oxidation number of the oxidized atom(s) decreases.
14.12
The oxidation number of the atom that is oxidized increases. The oxidation number of the atom that is
reduced decreases.
14.13
The oxidation number of atoms in their elemental form is zero. The charges of monatomic ions are their
oxidation numbers.
(a) Cr, 0; (b) Br2, 0; (c) Cr3+, 3+; (d) Br−, 1−
14.14
The oxidation number of atoms in their elemental form is zero. The charges of monatomic ions are their
oxidation numbers.
(a) Ca, 0; (b) S8, 0; (c) Ca2+, 2+; (d) S2−, 2−
14.15
In compounds and ions, we generally assign to the most electronegative element in a substance the
oxidation number we expect from its position in the periodic table. For example, oxygen is more
electronegative than sulfur, so the oxidation number of oxygen is 2− (because an oxygen atom needs two
more electrons to fill its valence shell, Table 14.1 Rule 6). We determine the oxidation number of sulfur
from the chemical formula and net charge (if any) on the species.
For SO3 we can write:

 

 total positive   total negative 
Net charge 
=
+

 oxidation numbers   oxidation numbers 
SO3
1× S
3× O

 

SO3 does not have a charge, so its net charge is zero. From Table 14.1 (Rule 6), assign oxygen an
oxidation number of oxygen of 2−. We assign an oxidation number to sulfur that gives SO3 a net charge of
zero.
Net charge = 0
Total negative oxidation numbers = 3 × (−2) = −6.
14−3
This gives us:


 total positive 
=
0 
 + ( –6 )
 oxidation numbers 
S


The oxidation number of sulfur in SO3 is 6+. S = 6+, O = 2−
14.16
In compounds and ions, we generally assign to the most electronegative element in a substance the
oxidation number we expect from its position in the periodic table. For example, oxygen is more
electronegative than nitrogen, so the oxidation number of oxygen is 2− (because an oxygen atom needs two
more electrons to fill its valence shell, Table 14.1 Rule 6). We determine the oxidation number of nitrogen
from the chemical formula of the species and its net charge (if any).
For NO2 we can write:

 

 total positive   total negative 
Net charge 
=
+

 oxidation numbers   oxidation numbers 
NO 2
1× N
2×O

 

NO2 does not have a charge, so its net charge is zero. From Table 14.1 (Rule 6), we assign oxygen an
oxidation number of oxygen of 2−. We assign an oxidation number to nitrogen that gives NO2 a net charge
of zero.
Net charge = 0
Total negative oxidation numbers = 2 × (−2) = −4.
This gives us:


 total positive 
=
0 
 + ( –4 )
 oxidation numbers 
N


The oxidation number of nitrogen in NO2 is 4+. N = 4+, O = 2−
14.17
In compounds and ions, we assign to the most electronegative element in a substance the oxidation number
we expect from its position in the periodic table. For example, oxygen is more electronegative than
phosphorus, so the oxidation number we assign to oxygen is 2− (because an oxygen atom needs two more
electrons to fill its valence shell, Table 14.1 Rule 6). We assign an oxidation number to phosphorus based
on the chemical formula and net charge (if any) of the substance.
(a) P4O10

 

 total positive   total negative 
Net charge 
=
+

 oxidation numbers   oxidation numbers 
P4 O10
4× N
10 × O

 

Net charge = 0
Total negative oxidation numbers = 10 × (−2) = −20
This gives us:


 total positive 
0= 
 + ( –20 )
 oxidation numbers 
4×P


14−4
P=
20
=5
4
Each phosphorus atom in P4O10 has an oxidation number of 5+.
(b) P4O6

 

 total positive   total negative 
Net charge 
=
+

 oxidation numbers   oxidation numbers 
P4 O6
4× N
6×O

 

Net charge = 0
Total negative oxidation numbers = 6 × (−2) = −12
This gives us:


 total positive 
0= 
 + ( –12 )
 oxidation numbers 
4×P


P=
12
=3
4
Each phosphorus atom in P4O6 has an oxidation number of 3+.
(c) P4O8

 

 total positive   total negative 
Net charge 
=
+

 oxidation numbers   oxidation numbers 
P4 O8
4× N
8×O

 

Net charge = 0
Total negative oxidation numbers = 8 × (−2) = −16
This gives us:


 total positive 
0= 
 + ( –16 )
 oxidation numbers 
4×P


P=
16
=4
4
Each phosphorus atom in P4O8 has an oxidation number of 4+.
14.18
In compounds and ions, the most electronegative element in a substance generally takes the oxidation
number we expect from its position in the periodic table. For example, oxygen is more electronegative than
chlorine, so the oxidation number of oxygen is 2− (because an oxygen atom needs two more electrons to
fill its valence shell, Rule 6). We assign oxidation numbers to chlorine based on the chemical formulas and
net charges of the substances.
(a) Cl2O
14−5

 

 total positive   total negative 
Net charge 
=
+

 oxidation numbers   oxidation numbers 
Cl2 O
2 × Cl
1× O

 

Net charge = 0
Total negative oxidation numbers = 1 × (−2) = −2
This gives us:


 total positive 
0= 
 + ( –2 )
 oxidation numbers 
2 × Cl


Cl =
2
=1
2
Each chorine atom in Cl2O has an oxidation number of 1+.
(b) ClO3

 

 total positive   total negative 
Net charge 
=
+

 oxidation numbers   oxidation numbers 
ClO3
1 × Cl
3× O

 

Net charge = 0
Total negative oxidation numbers = 3 × (−2) = −6
This gives us:


 total positive 
0= 
 + ( –6 )
 oxidation numbers 
Cl


Cl = 6
Each chlorine atom in ClO3 has an oxidation number of 6+.
(c) Cl2O3

 

 total positive   total negative 
Net charge 
=
+

 oxidation numbers   oxidation numbers 
Cl2 O3
2 × Cl
3× O

 

Net charge = 0
Total negative oxidation numbers = 3 × (−2) = −6
This gives us:


 total positive 
0= 
 + ( –6 )
 oxidation numbers 
2 × Cl


14−6
Cl =
6
=3
2
Each chlorine atom in Cl2O3 has an oxidation number of 3+.
(d) ClO2

 

 total positive   total negative 
Net charge 
=
+

 oxidation numbers   oxidation numbers 
ClO 2
1 × Cl
2×O

 

Net charge = 0
Total negative oxidation numbers = 2 × (−2) = −4
This gives us:


 total positive 
0= 
 + ( –4 )
 oxidation numbers 
Cl


Cl = 4
Each chlorine atom in ClO2 has an oxidation number of 4+.
(e) Cl2O5

 

 total positive   total negative 
Net charge 
=
+

oxidation
numbers

  oxidation numbers 
Cl2 O5
2 × Cl
5×O

 

Net charge = 0
Total negative oxidation numbers = 5 × (−2) = −10
This give us:


 total positive 
0= 
 + ( –10 )
 oxidation numbers 
2 × Cl


Cl =
10
=5
2
Each chlorine atom in Cl2O5 has an oxidation number of 5+.
14.19
In compounds and ions, the most electronegative element in a substance generally takes the oxidation
number we expect from its position in the periodic table. When a substance contains hydrogen, we must
remember (Table 14.1 Rule 4) that the oxidation number of hydrogen is 1+, unless it is combined with a
metal (where it is 1−).
(a) AlPO4 – Aluminum’s oxidation number is 3+ (predicted from the periodic table).


 
 total positive   total negative 
Net charge 
=

+
 oxidation numbers   oxidation numbers 
AlPO 4
4×O
 1 × Al + 1 × P  

14−7
Net charge = 0
Total negative oxidation numbers = 4 × (−2) = −8
This gives us:


 total positive 
0= 
 + ( –8 )
 oxidation numbers 
3+P


0=3+P−8
P=5
The oxidation number of P in AlPO4 is 5+.
(b) PF5 – Fluorine’s oxidation number is 1− (predicted from Table 14.1 Rule 3 and the periodic table).


 
 total positive   total negative 
Net charge 
=

+
 oxidation numbers   oxidation numbers 
PF5
P
5× F

 

Net charge = 0
Total negative oxidation numbers = 5 × (−1) = −5
This gives:


 total positive 
0= 
 + ( –5 )
 oxidation numbers 
P


P=5
The oxidation number of P in PF5 is 5+.
(c) H3PO4 − The oxidation number of hydrogen is 1+ (predicted from Table 14.1 Rule 4 and the periodic
table). Because we analyzed the phosphate ion in part (a), we can predict that the oxidation number of
phosphorus is 5+.

 

 total positive   total negative 
Net charge 
=
+

 oxidation numbers   oxidation numbers 
H3 PO 4
4×O
 3 × H + 1× P  

Net charge = 0
Total negative oxidation numbers = 4 × (−2) = −8
This gives us:


 total positive 
0= 
 + ( –8 )
 oxidation numbers 
3+P


0=3+P−8
P=5
The oxidation number of P in H3PO4 is 5+.
14−8
(d) H3PO2 − The oxidation number of hydrogen is 1+ (predicted from Table 14.1 Rule 4 and the periodic
table).

 

 total positive   total negative 
Net charge 
=
+

 oxidation numbers   oxidation numbers 
H3 PO 2
2×O
 3 × H + 1× P  

Net charge = 0
Total negative oxidation numbers = 2 × (−2) = −4
This gives:


 total positive 
0= 
 + ( –4 )
 oxidation numbers 
3+P


0=3+P−4
P=1
The oxidation number of P in H3PO2 is 1+.
(e) PH3 − The oxidation number of hydrogen is 1+ (predicted from Table 14.1 Rule 4 and the periodic
table). In this example, hydrogen has a positive oxidation number, so phosphorus has a negative
oxidation number.

 

 total positive   total negative 
Net charge 
=
+

 oxidation numbers   oxidation numbers 
PH3
3× H
P


 
Net charge = 0
Total positive oxidation numbers = 3 × (+1) = −3
This gives us:
 total positive 


0 =  oxidation numbers  + ( P )


3


0=3+P
P = 3−
The oxidation number of P in PH3 is 3−.
(f) H3PO3 − The oxidation number of hydrogen is 1+ (predicted from Table 14.1 Rule 4 and the periodic
table).
 total positive 
 total negative 




=
Net charge  oxidation numbers  +  oxidation numbers 
 3 × H + 1× P 


H3 PO3
3× O




Net charge = 0
Total negative oxidation numbers = 3 × (−2) = −6
14−9
This gives us:


 total positive 
0= 
 + ( –6 )
 oxidation numbers 
3+P


0=3+P−6
P=3
The oxidation number of P in H3PO3 is 3+.
14.20
In compounds and ions, the most electronegative element in a substance generally takes the oxidation
number we expect from its position in the periodic table.
(a) NaI – NaI is an ionic compound, so sodium’s oxidation number is 1+, based on its position in the
periodic table. Because the overall charge on NaI is zero, iodine must have a 1− oxidation number.
The oxidation number of I in NaI is 1−.
(b) I2 – Any element in its pure form (i.e. O3, P4, S8, Fe, Na) has an oxidation number of zero.
The oxidation number of I in I2 is 0.
(c) IO2


 
 total positive   total negative 
Net charge 
=

+
 oxidation numbers   oxidation numbers 
IO 2
1× I
2×O

 

Net charge = 0
Total negative oxidation numbers = 2 × (−2) = −4
This gives us:


 total positive 
0= 
 + ( –4 )
 oxidation numbers 
I


I=4
The oxidation number of I in IO2 is 4+.
(d) I2O7


 
 total positive   total negative 
Net charge 
=

+
 oxidation numbers   oxidation numbers 
I 2 O7
2×I
7×O

 

Net charge = 0
Total negative oxidation numbers = 7 × (−2) = −14
14−10
This gives us:


 total positive 
0= 
 + ( –14 )
 oxidation numbers 
2×I


I=
14
2
I=7
The oxidation number of I in I2O7 is 7+.
(e) KIO4 − Potassium’s oxidation number is 1+ (predicted from Table 14.1 Rule 4 and the periodic table).


 
 total positive   total negative 
Net charge 
=

+
 oxidation numbers   oxidation numbers 
KIO 4
4×O
 1× K + 1× I  

Net charge = 0
Total negative oxidation numbers = 4 × (−2) = −8
This give us:


 total positive 
0= 
 + ( –8 )
 oxidation numbers 
1+ I


0=1+I−8
I=7
The oxidation number of I in KIO4 is 7+.
(f) Ca(IO)2 − Calcium’s oxidation number is 2+ (predicted from Table 14.1 Rule 4 and the periodic table).

 

 total positive   total negative 
Net charge 
=
+

 oxidation numbers   oxidation numbers 
Ca(IO) 2
2×O
 1 × Ca + 2 × I  

Net charge = 0
Total negative oxidation numbers = 2 × (−2) = −4
This gives us:


 total positive 
0= 
 + ( –4 )
 oxidation numbers 
2+ 2×I


0=2+ 2×I −4
I=
2
2
14−11
I = 1+
The oxidation number of I in Ca(IO)2 is 1+.
14.21
SO42−

 

 total positive   total negative 
Net charge 
=
+

oxidation numbers   oxidation numbers 
2−

SO 4
S
4×O

 

Net charge = −2
Total negative oxidation numbers = 4 × (−2) = −8
This gives us:


 total positive 
−2 = 
 + ( –8 )
 oxidation numbers 
S


S=6
The oxidation numbers of sulfur and oxygen in SO42− are 6+ and 2−, respectively.
14.22
NO3−

 

 total positive   total negative 
Net charge 
=
+

 oxidation numbers   oxidation numbers 
NO3−
N
3× O

 

Net charge = −1
Total negative oxidation numbers = 3 × (−2) = −6
This gives us:


 total positive 
−1 = 
 + ( –6 )
 oxidation numbers 
N


N=5
The oxidation numbers of nitrogen and oxygen in NO3− are 5+ and 2−, respectively.
14.23
(a) BrF7 − Fluorine’s oxidation number is 1− (predicted from Table 14.1 Rule 3 and the periodic table).


 
 total positive   total negative 
Net charge 
=

+
 oxidation numbers   oxidation numbers 
BrF7
1 × Br
7×F

 

Net charge = 0
Total negative oxidation numbers = 7 × (−1) = −7
This gives us:
14−12


 total positive 
0= 
 + ( –7 )
 oxidation numbers 
Br


Br = 7
The oxidation number of Br in BrF7 is 7+.
(b) BrO3−


 
 total positive   total negative 
Net charge 
=

+
 oxidation numbers   oxidation numbers 
BrO3−
1 × Br
3× O

 

Net charge = −1
Total negative oxidation numbers = 3 × (−2) = −6
This gives us:


 total positive 
−1 = 
 + ( –6 )
 oxidation numbers 
Br


Br = 5
The oxidation number of Br in BrO3− is 5+.
(c) For monatomic ions, the oxidation number and charge are the same.
Br− has an oxidation number of 1−.
(d) BrCl3 − Chloride has an oxidation number of 1− (predicted from Table 14.1 Rule 5c and the periodic
table).


 
 total positive   total negative 
Net charge 
=

+
 oxidation numbers   oxidation numbers 
BrCl3
1 × Br
3 × Cl

 

Net charge = 0
Total negative oxidation numbers = 3 × (−2) = −3
This gives us:


 total positive 
0= 
 + ( –3)
 oxidation numbers 
I


Br = 3
The oxidation number of Br in BrCl3 is 3+.
14−13
(e) BrOCl3 − We predict the oxidation numbers of oxygen (2−) and chlorine (1−) from Table 14.1 Rules 4
and 6, and the periodic table.


 
 total positive   total negative 
Net charge 
=

+
 oxidation numbers   oxidation numbers 
BrOCl3
1 × Br

  1 × O + 3 × Cl 
Net charge = 0
Total negative oxidation numbers = 1 × (2−) + 3 × (−1) = −5
This gives us:


 total positive 
0= 
 + ( –5 )
 oxidation numbers 
Br


Br = 5
The oxidation number of Br in BrOCl3 is 5+.
14.24
(a) CrCl2 – The oxidation number of chlorine is 1− (periodic table). This means that the oxidation number
of chromium in CrCl2 must be 2+.
(b) CrSO4 – The sulfate ion has a 2− charge. This means that the oxidation number of chromium in CrSO4
is 2+.
(c) Cr(NO3)3 – Each nitrate ion has a 1− charge. Because there are three nitrate ions in the chemical
formula, the oxidation number of chromium in Cr(NO3)3 is 3+.
(d) Cr2(SO4)3 – Each sulfate ion has a 2− charge. We can write:



 

 total positive   total negative 
Net charge 
=
+

Cr2 (SO 4 )3  oxidation numbers   oxidation numbers 
−
2
2 × Cr

  3 × SO 4



(
)
Total negative oxidation numbers = 3 × (−2) = −6


 total positive 
0= 
 + ( –6 )
 oxidation numbers 
2 × Cr


Cr =
6
=3
2
The oxidation number of chromium in Cr2(SO4)3 is 3+.
14.25
(a) ClO2

 

 total positive   total negative 
Net charge 
=
+

oxidation
numbers

  oxidation numbers 
ClO 2
1 × Cl
2×O

 

Net charge = 0
Total negative oxidation numbers = 2 × (−2) = −4
14−14
This gives us:


 total positive 
0= 
 + ( –4 )
 oxidation numbers 
Cl


Cl = 4
The oxidation numbers of oxygen and chlorine in ClO2 are 2− and 4+, respectively.
(b) CaF2 − Because fluorine is the most electronegative element, it’s oxidation number is 1−. Because
there are two fluoride ions and only one calcium ion, calcium’s oxidation number is 2+.
(c) H2TeO3 – We predict the oxidation numbers of hydrogen (1+) and oxygen (2−) from Table 14.1 Rules
4 and 6, and the periodic table.

 

 total positive   total negative 
Net charge 
=
+

 oxidation numbers   oxidation numbers 
H 2 TeO3
3× O
 2 × H + 1 × Te  

Net charge = 0
Total negative oxidation numbers = 3 × (−2) = −6
This gives us:


 total positive 
0= 
 + ( –6 )
 oxidation numbers 
2 + Te


Te = 4
The oxidation numbers of H, O, and Te in H2TeO3 are 1+, 2−, and 4+, respectively.
(d) NaH – In compounds with metals, the oxidation number of hydrogen is 1− (Table 14.1 Rule 4). The
oxidation number of sodium is 1+.
14.26
(a) Na2O2 – The oxidation number of sodium ion is always 1+ in compounds. Because each sodium ion
has a 1+ oxidation number, each oxygen ion must have a 1− oxidation number (there is a 1:1 Na:O
ratio).
(b) Fe(NO3)3 – The charge on each nitrate ion is 1−. Because there are three nitrate ions in the chemical
formula, the total negative charge is 3−. The oxidation number of iron is 3+. The oxidation number of
oxygen is 2− (Table 14.1 Rule 6 and the periodic table) so, in NO3−, the oxidation number of nitrogen
is 5+. Because we know the charge of a nitrate ion, we do not need to include the iron in the
calculation.

 

 total positive   total negative 
Net charge 
=
+

oxidation numbers   oxidation numbers 
−

NO3
N
3× O

 

Net charge = −1
Total negative oxidation numbers = 3 × (−2) = −6
14−15
This gives us:


 total positive 
−1 = 
 + ( –6 )
 oxidation numbers 
N


N=5
(c) Sc2O3 – The oxidation number of oxygen is 2− (Table 14.1 Rule 6 and the periodic table). The
oxidation number of scandium is 3+.

 

 total positive   total negative 
Net charge 
=
+

 oxidation numbers   oxidation numbers 
Sc 2 O3
2 × Sc
3× O

 

Net charge = 0
Total negative oxidation numbers = 3 × (−2) = −6
This gives us:


 total positive 
0= 
 + ( –6 )
 oxidation numbers 
2 × Sc


6
=3
2
(d) LiH – According to Table 14.1 Rule 4, in compounds with metals the oxidation number of hydrogen is
1−. The oxidation number of lithium is 1+.
Sc =
14.27
(a) NO2−

 

 total positive   total negative 
Net charge 
=
+

 oxidation numbers   oxidation numbers 
NO 2 −
N
2×O

 

Net charge = −1
Total negative oxidation numbers = 2 × (−2) = −4
This gives us:


 total positive 
−1 = 
 + ( –4 )
 oxidation numbers 
N


N=3
The oxidation numbers of N and O in NO2− are 3+ and 2−, respectively.
14−16
(b) Cr2O72−

 

 total positive   total negative 
Net charge 
=
+

oxidation numbers   oxidation numbers 
2−

Cr2 O7
2 × Cr
7×O

 

Net charge = −2
Total negative oxidation numbers = 7 × (−2) = −14
This gives us:


 total positive 
−2 = 
 + ( –14 )
 oxidation numbers 
2 × Cr


Cr = 6
The oxidation numbers of Cr and O in Cr2O72− are 6+ and 2−, respectively.
(c) AgCl2− − From the periodic table we can predict that chlorine has an oxidation number of 1−. Because
there are two chloride ions in the species, and the overall charge of the species is 1–, we can predict
that silver has an oxidation number of 1+. To verify this, we can calculate the overall charge and see if
it matches the net charge on the ion.

 

 total positive   total negative 
+
Net charge 
=

oxidation numbers   oxidation numbers 
−

AgCl2


Ag
2 × Cl


 
Net charge = −1
Total negative oxidation numbers = 2 × (1−) = −2
Total positive oxidation numbers = +1
Overall charge = 1 − 2 = −1
The overall charge and the charge we calculated from the predicted oxidation numbers are the same.
The oxidation numbers of Ag and Cl in AgCl2− are 1+ and 1−, respectively.
(d) SO32−

 

 total positive   total negative 
Net charge 
=
+

oxidation numbers   oxidation numbers 
2−

SO3
1× S
3× O

 

Net charge = –2
Total negative oxidation numbers = 3 × (−2) = −6.
This gives us:


 total positive 
−2 = 
 + ( –6 )
 oxidation numbers 
S


S=4
The oxidation numbers of S and O in SO32− are 4+ and 2−, respectively.
14−17
(e) CO32−

 

 total positive   total negative 
Net charge 
=
+

oxidation numbers   oxidation numbers 
2−

CO3
1× C
3× O

 

Net charge = –2
Total negative oxidation numbers = 3 × (−2) = −6.
This gives us:


 total positive 
−2 = 
 + ( –6 )
 oxidation numbers 
C


C=4
The oxidation numbers of C and O in CO32− are 4+ and 2−, respectively.
14.28
(a) H2PO4− − We determine the oxidation numbers of hydrogen (1+) and oxygen (2−) from Table 14.1
Rules 4 and 6, and their positions on the periodic table. The oxidation number of phosphorus is 5+ as
shown below:


 
 total positive   total negative 
Net charge 
=

+
oxidation numbers   oxidation numbers 
−

H 2 PO 4
4×O
 2 × H + 1× P  

Net charge = –1
Total negative oxidation numbers = 4 × (−2) = −8
This gives us:


 total positive 
−1 = 
 + ( –8 )
 oxidation numbers 
2+P


−1 = 2 + P − 8
P=5
(b) FeCl63− − We determine the oxidation number of chlorine (1−) from Table 14.1 Rule 5c and its
position on the periodic table. The oxidation number of iron in FeCl63− is 3+, as shown:

 

 total positive   total negative 
Net charge 
=
+

 oxidation numbers   oxidation numbers 
FeCl63 −
Fe
6 × Cl

 

Net charge = −3
Total negative oxidation numbers = 6 × (−1) = −6
This gives us:
14−18


 total positive 
−3 = 
 + ( –6 )
 oxidation numbers 
Fe


Fe = 3
(c) ClO2− − We determine the oxidation number of oxygen (2−) from Table 14.1 Rule 6 and its position on
the periodic table. The oxidation number of chlorine in ClO2− is 3+, as shown:

 

 total positive   total negative 
Net charge 
=
+

oxidation numbers   oxidation numbers 
−

ClO 2
Cl
2×O

 

Net charge = −1
Total negative oxidation numbers = 2 × (−2) = −4
This gives us:


 total positive 
−1 = 
 + ( –4 )
 oxidation numbers 
Cl


Cl = 3
(d) SiF62− − We determine the oxidation number of fluorine (1−) from Table 14.1 Rule 5c and its position
on the periodic table. The oxidation number of silicon in SiF62− is 4+ as shown:

 

 total positive   total negative 
Net charge 
=
+

oxidation numbers   oxidation numbers 
2−

SiF6
Si
6×F

 

Net charge = −2
Total negative oxidation numbers = 6 × (−1) = −6
This gives us:


 total positive 
−2 = 
 + ( –6 )
 oxidation numbers 
Si


Si = 4
(e) AsO43− − We determine the oxidation number of oxygen (2−) from Table 14.1 Rule 6 and its position
on the periodic table. The oxidation number of arsenic in AsO43− is 5+ as shown:

 

 total positive   total negative 
Net charge 
=
+

oxidation numbers   oxidation numbers 
3−

AsO 4
As
4×O

 

Net charge = −3
Total negative oxidation numbers = 4 × (−2) = −8
14−19
This gives us:


 total positive 
−3 = 
 + ( –8 )
 oxidation numbers 
As


As = 5
14.29
To be classified as a redox reaction, the oxidation number of at least one element must change. For each
reaction, we calculate, or predict, the oxidation numbers of each element in each species and determine if a
change occurs. If an element’s oxidation number becomes more positive, that element loses electrons by
reducing another substance and, the substance containing that element is the reducing agent. If an
element’s oxidation number becomes more negative, that element gains electrons by oxidizing another
substance, and the substance containing that element is the oxidizing agent.
(a) This is not a redox reaction because none of the oxidation numbers change.
2+
2+
1+ 6+ 2–
1–
BaCl2(aq) + H2SO4(aq)
6+ 2–
1+
1–
BaSO4(s) + 2HCl(aq)
(b) The oxidation number of N decreases because N atoms gain electrons from H atoms. N2 is the
oxidizing agent. H2 is the reducing agent because the oxidation number of H increases as it supplies
electrons to nitrogen.
0
0
3–
3H2(g) + N2(g)
1+
2NH3(g)
(c) This is not a redox reaction because none of the oxidation numbers change.
1+ 4+
1+
2–
H2CO3(aq)
4+ 2–
2–
H2O(l) + CO2(g)
(d) This is not a redox reaction because none of the oxidation numbers change.
1+
5+
2–
1+
1–
1+
1–
1+ 5+ 2–
AgCl(s) + NaNO3(aq)
AgNO3(aq) + NaCl(aq)
(e) The oxidation number of O decreases because it gains electrons by oxidizing C. O2 is the oxidizing
agent. C2H6 is the reducing agent because the oxidation number of C increases when it supplies
electrons to O.
3–
1+
0
2C2H6(g) + 7O2(g)
14.30
4+
2–
1+
2–
4CO2(g) + 6H2O(g)
To be classified as a redox reaction, the oxidation number of at least one element must change. For each
reaction, we calculate, or predict, the oxidation numbers of each element in each species and determine if a
change occurs. If an element’s oxidation number becomes more positive, that element loses electrons by
reducing another substance and, the substance containing that element is the reducing agent. If an
element’s oxidation number becomes more negative, that element gains electrons by oxidizing another
substance, and the substance containing that element is the oxidizing agent
14−20
(a) The oxidation number of H decreases because it gains electrons by oxidizing Na. H2O is the oxidizing
agent. Na is the reducing agent because the oxidation number of Na increases when Na transfers
electrons to H.
0
1+
2–
1+
2Na(s) + 2H2O(l)
2– 1+
0
2NaOH(aq) + H2(g)
(b) The oxidation number of F decreases because it gains electrons when it oxidizes H. F2 is the oxidizing
agent. H2 is the reducing agent because its oxidation number increases when it provides electrons to
fluorine.
0
0
1+
H2(g) + F2(g)
1–
2HF(g)
(c) The oxidation number of H decreases because it gains electrons when it oxidizes C. H2O is the
oxidizing agent. C is the reducing agent because its oxidation number increases when it supplies
electrons to H.
0
1+
2–
0
2–
2+
CO(g) + H2(g)
C(s) + H2O(g)
(d) This is not a redox reaction because the oxidation numbers of the elements do not change.
2+
5+ 2–
1–
1+
2+
Pb(NO3)2(aq) + 2NaCl(aq)
1–
1+ 5+ 2–
PbCl2(s) + 2NaNO3(aq)
(e) The oxidation number of O in O2 decreases because it gains electrons by oxidizing the C in C2H5OH.
O2 is the oxidizing agent. C2H5OH is the reducing agent because the oxidation number of C increases
as it supplies electrons to O.
2–
1+ 2– 1+
0
4+
C2H5OH(g) + 3O2(g)
14.31
2–
1+
2–
2CO2(g) + 3H2O(g)
(a)
0
0
2+
N2(g) + O2(g)
2–
2NO(g)
2O × 2e/O = 4e–
2N × (–2e/N) = –4e–
(b) When N2 is transformed into NO, the oxidation number of nitrogen increases by 2. Because there are
two nitrogen atoms in N2, a total of four electrons are transferred. Each N atom loses two electrons.
(c) In any redox reaction, the numbers of electrons gained and lost must be equal. When O2 reacts with N2
to produce NO, the oxidation number of oxygen decreases by 2. Because there are two oxygen atoms
in O2, a total of four electrons are transferred from N to O atoms. Each O atom gains two electrons.
14−21
14.32
(a)
0
0
2+
2C(g) + O2(g)
2–
2CO(g)
2O × 2e/O = 4e–
2C × (–2e/C) = –4e–
(b) When C is transformed into CO, the oxidation number of oxygen increases by 2. Because there are
two oxygen atoms in the reactants, a total of four electrons are transferred. Each C loses two electrons.
(c) In any redox reaction, the numbers of electrons gained and lost must be equal. When O2 reacts with C
and produces CO, the oxidation number of oxygen decreases by 2. Because there are two oxygen
atoms in O2, a total of four electrons are transferred from C to O atoms. Each O atom gains two
electrons.
14.33
The oxidation numbers of both vanadium and chromium change in this reaction (see below).
2+
6+
3+
6V2+(aq) + Cr2O72–(aq) + 14H+(aq)
3+
6V3+(aq) + 2Cr3+(aq) + 7H2O(l)
2Cr × (–3e/Cr) = –6e–
6V × (+1e/V) = 6e–
(a)
(b)
(c)
(d)
14.34
V2+ is oxidized from 2+ to 3+.
Cr2O72− is reduced (Cr changes from 6+ to 3+).
Because Cr (in Cr2O72−) gains electrons, Cr2O72−is the oxidizing agent.
The reducing agent transfers electrons to reduce another element. Because vanadium (in V2+) loses
electrons, V2+ is the reducing agent.
The oxidation numbers of both manganese and chlorine change in this reaction (see below).
7+
1–
2+
2MnO4–(aq) + 10Cl–(aq) + 16H+(aq)
0
2Mn2+(aq) + 5Cl2(aq) + 8H2O(l)
10Cl × (–1e/Cl) = –10e–
2Mn × (+7e/Mn) = 14e–
(a)
(b)
(c)
(d)
The oxidation number of chlorine changes from 1− to 0, so Cl− is oxidized.
The oxidation number of manganese changes from 7+ to 2+, so it is reduced.
Because Mn (in MnO4−) gains electrons, MnO4− is the oxidizing agent.
The reducing agent transfers electrons to reduce another element. Because Cl (in Cl−) loses electrons,
Cl− is the reducing agent.
14−22
14.35
(a) In this reaction, I2 is both the oxidizing agent and the reducing agent. The oxidation number of I atoms
in molecular iodine (I2) is zero. The oxidation number of iodine in I− is 1−, indicating a gain of one
electron. In IO−, iodine has an oxidation number of 1+, indicating a loss of one electron. In this
reaction, 1 electron is transferred between I atoms.
0
1–
1+
I2(aq) + 2OH–(aq)
I–(aq) + IO–(aq) + H2O(l)
1I × (–1e/I) = –1e–
1I × (1e/I) = 1e–
(b) Cr = reducing agent, H+ = oxidizing agent; 2 electrons are transferred.
0
1+
2+
Cr(s) + 2H+(aq)
0
Cr2+(aq) + H2(g)
Cr × (–2e/Cr) = –2e–
2H × (+1e/H) = 2e–
(c) Cr2O72− is both the oxidizing and reducing agent, and 12 electrons are transferred. Note that this
reaction is different than the one in 14.35(a) where the same element (iodine) is both oxidized and
reduced. In the case of Cr2O72−, all of the chromium atoms are reduced (oxidation number changes
from 6+ to 3+), while only 6 oxygen atoms are oxidized (oxidation number changes from 2– to 0).
The remaining oxygen atoms contribute to the production of water.
6+
2–
0
3+
2Cr2O72–(aq) + 16H+(aq)
4Cr3+(aq) + 3O2(g) + 8H2O(l)
4Cr × (–3e/Cr) = –12e–
6O × (+2e/O) = 12e–
(d) Fe3+ = oxidizing agent, Al = reducing agent; 3 electrons are transferred.
3+
0
2+
3Fe3+(aq) + Al(s)
3+
3Fe2+(aq) + Al3+(aq)
3Fe × (1e/Fe) = 3e–
1Al × (–3e/Al) = –3e–
14−23
14.36
(a) BrO3− = oxidizing agent, I− = reducing agent; 6 electrons are transferred.
1–
1–
0
5+
6I–(aq) + BrO3–(aq) + 6H+(aq)
3I2(aq) + Br–(aq) + 3H2O(l)
1Br × (6e/Cl) = 6e–
6I × (–1e/I) = –6e–
(b) Br− = reducing agent, BrO3− = oxidizing agent, 5 electrons are transferred.
1–
0
5+
5Br–(aq) + BrO3–(aq) + 6H+(aq)
3Br2(aq) + 3H2O(l)
1Br × (5e/Br) = 5e–
5Br × (–1e/Br) = 5e–
(c) XeF4 = oxidizing agent, H2O = reducing agent; 4 electrons are transferred.
4+
0
0
Xe(g) + 4HF(aq) + O2(g)
XeF4(s) + 2H2O(l)
1Xe × (4e/Xe) = 4e–
2O × (–2e/O) = 4e–
(d) Pb = reducing agent, H+ = oxidizing agent; 2 electrons are transferred
0
1+
0
2+
Pb(s) + 2H+(aq)
Pb2+(aq) + H2(g)
1Pb × (2e/Pb) = 2e–
2H × (–1e/H) = 2e–
14−24
14.37
In an electrochemical cell, oxidation takes place at the anode and reduction takes place at the cathode. Iron
is being oxidized, so the iron electrode in the Fe(NO3)2 solution is the anode half-cell. Nickel is being
reduced, so the nickel electrode in the Ni(NO3)2 solution is the cathode half-cell.
The half-reaction in the anode half-cell is: Fe(s) → Fe2+(aq) + 2e−
The half-reaction in the cathode half-cell is: Ni2+(aq) + 2e− → Ni(s)
Voltmeter
Anode
Cathode
Salt Bridge
Ni
Fe
Ni(NO3)2(aq)
Fe(NO3)2(aq)
14.38
In an electrochemical cell, oxidation takes place at the anode and reduction takes place at the cathode.
Because magnesium is being oxidized, the Mg electrode in the Mg(NO3)2 solution is the anode half-cell.
Tin is being reduced, so the Sn electrode in the Sn(NO3)2 solution is the cathode half-cell.
The half-reaction in the anode half-cell is: Mg(s) → Mg2+(aq) + 2e−
The half-reaction in the cathode half-cell is: Sn2+(aq) + 2e− → Sn(s)
Voltmeter
Anode
Cathode
Salt Bridge
Mg
Sn
Mg(NO3)2(aq)
Sn(NO3)2(aq)
14−25
14.39
As the cell runs, iron oxidizes causing the reduction of Ni2+ ions. The iron electrode loses mass as Fe
atoms on its surface oxidize and become Fe2+(aq). The nickel electrode gains mass as Ni2+ ions from the
solution are reduced to Ni(s) which deposits on the electrode. The nitrate ions are not shown. The
concentration of Fe2+ in the solution in the anode half-cell increases and the concentration of Ni2+ in the
solution in the cathode half-cell decreases.
Fe2+
Fe
Ni2+
Iron Electrode
14.40
Ni
Nickel Electrode
As the cell runs, magnesium oxidizes causing the reduction of Sn2+. The magnesium electrode loses mass
as Mg atoms on its surface oxidize and become Mg2+(aq). The tin electrode gains mass as Sn2+ ions from
the solution are reduced to Sn(s) and deposit on the Sn electrode. The nitrate ions are not shown. The
concentration of Mg2+ in the solution in the anode half-cell increases, and the concentration of Sn2+ in the
solution in the cathode half-cell decreases.
Sn2+
Mg
Sn
Mg2+
Iron Electrode
14.41
Nickel Electrode
To answer this question, we determine the oxidation numbers of the elements in each substance. The
substances whose oxidation numbers change are shown below:
0
2+
4+
Pb(s) + PbO2(s) + 2H2SO4(aq)
2PbSO4(s) + 2H2O(l)
Pb(s) is oxidized, and the Pb in PbO2(s) is reduced. The oxidizing agent is PbO2(s), and the reducing agent
is Pb(s).
14.42
To answer this question, we determine the oxidation numbers of the elements in each substance. The
substances whose oxidation numbers change are shown below:
0
2+
4+
Zn(s) + 2MnO2(s) + 2NH4+(aq)
3+
Zn2+(aq) + Mn2O3(s) + 2NH3(aq) + 2H2O(l)
Zn is oxidized, and the Mn in MnO2 is reduced. The oxidizing agent is MnO2, and the reducing agent is Zn.
14−26
14.43
Note that in the cases of both half-reactions, there is a material balance (the atoms are balanced) but not a
charge balance. To determine which half-reaction represents oxidation and which represents reduction, we
begin by assigning oxidation numbers. In the Cd/Cd(OH)2 half-reaction, the oxidation number of cadmium
changes from 0 to 2+. This is the oxidation half-reaction because cadmium is oxidized.
0
2+
Cd(s) + 2OH–(aq)
Cd(OH)2(s)
By adding two electrons as reaction products, we balance the overall charge in the half-reaction. Both the
product and reactant sides of the equation have a net 2− charge. When a half-reaction is properly balanced,
both the atoms and charge balance.
Cd(s) + 2OH–(aq)
Cd(OH)2(s) + 2e–
2–
2–
In the NiO2/Ni(OH)2 half-reaction, the oxidation number of nickel changes from 4+ to 2+. This is the
reduction half-reaction because nickel is reduced.
4+
2+
NiO2(s) + 2H2O(l)
Ni(OH)2(s) + 2OH–(aq)
By adding two electrons as reactants, we balance the overall charge in the half-reaction. Both the product
and reactant sides of the equation have a net 2− charge. When a half-reaction is properly balanced, both the
atoms and charge balance.
2e– + NiO2(s) + 2H2O(l)
Ni(OH)2(s) + 2OH–(aq)
2–
2–
14.44
Both half-reactions show a material balance, but not a charge balance. In the Zn/Zn(OH)2 half-reaction, the
oxidation number of Zn changes from 0 to 2+. This is the oxidation half-reaction because zinc is oxidized.
0
2+
Zn(s) + 2OH–(aq)
Zn(OH)2(s)
By adding two electrons as products, we balance the overall charge in the half-reaction. Both the product
and reactant sides of the equation have a 2− charge. When a half-reaction is properly balanced, both the
atoms and charge balance.
Zn(s) + 2OH–(aq)
Zn(OH)2(s) + 2e–
2–
2–
In the AgO/Ag2O half-reaction the oxidation number of silver changes from 2+ to 1+. This is the reduction
half-reaction because silver is reduced.
2+
1+
2AgO(s) + 2H2O(l)
Ag2O(s) + 2OH–(aq)
We add two electrons as reactants to balance the charge, because two silver atoms undergo the change.
Both the product and reactant sides of the equation have a 2− charge. When a half-reaction is properly
balanced, both the atoms and charge balance.
14−27
2e– + 2AgO(s) + 2H2O(l)
Ag2O(s) + 2OH–(aq)
2–
14.45
2–
To balance a simple half-reaction, we first balance the atoms and then add electrons to the side with more
positive charge to balance the charge.
(a) Fe3+(aq) → Fe(s)
The atoms are balanced. We balance the charge by adding three electrons to the reactant side of the
equation.
3e– + Fe3+(aq)
Fe(s)
0
0
2+
(b) Zn(s) → Zn (aq)
The atoms are balanced. We balance the charge by adding two electrons to the product side of the
equation.
Zn(s)
Zn2+(aq) + 2e–
0
0
−
(c) Cl (aq) → Cl2(g)
The atoms are not balanced, so we add a coefficient of 2 in front of Cl−.
2Cl−(aq) → Cl2(g)
Then we balance the charge by adding two electrons to the product side of the equation.
2Cl–(aq)
Cl2 (g) + 2e–
2–
2–
(d) Fe (aq) → Fe (aq)
The atoms are balanced. We balance the charge by adding one electron to the product side of the
equation.
Fe2+(aq)
Fe3+(aq) + e–
2+
3+
2+
14.46
2+
(a) Ni(s) → Ni2+(aq)
The atoms are balanced. We balance the charge by adding two electrons to the product side of the
equation.
Ni(s)
Ni2+(aq) + 2e–
0
0
−
(b) Br2(l) → Br (aq)
The atoms are not balanced, so we add a coefficient of 2 in front of Br−.
Br2(l) → 2Br−(aq)
Then we balance the charge by adding two electrons to the reactant side of the equation.
Br2 (l) + 2e–
2Br–(aq)
2–
2–
(c) Mg (aq) → Mg(s)
The atoms are balanced. We balance the charge by adding two electrons to the reactant side of the
equation.
2+
14−28
Mg2+(aq) + 2e–
Mg(s)
0
0
(d) Cr (aq) → Cr (aq)
The atoms are balanced. We balance the charge by adding one electron to the reactant side of the
equation.
Cr3+(aq) + e–
Cr2+(aq)
3+
2+
2+
14.47
2+
(a) Zn(s) + Fe(NO3)3(aq) → Zn(NO3)2(aq) + Fe(s)
Nitrate ions, NO3−, are spectator ions in this reaction so we can eliminate them in the first part of the
balancing process, and add them back into the equation after we have balanced the half-reactions.
Eliminating the nitrate ions gives us the ionic equation:
Zn(s) + Fe3+(aq) → Zn2+(aq) + Fe(s)
skeletal ionic equation
We can write the oxidation half-reaction as:
Zn(s) → Zn2+(aq)
oxidation half-reaction
The atoms are balanced. We can balance the charge by adding two electrons to the right side of the
equation:
Zn(s) → Zn2+(aq) + 2e−
balanced oxidation half-reaction
The reduction half-reaction is:
Fe3+(aq) → Fe(s)
reduction half-reaction
We can balance the charge by adding three electrons to the reactant side of the equation:
3e− + Fe3+(aq) → Fe(s)
balanced reduction half-reaction
Then we equalize the number of electrons lost and gained in the two half-reactions by multiplying each
by an appropriate coefficient:
3 × [Zn(s) → Zn2+(aq) + 2e−]
2 × [3e− + Fe3+(aq) → Fe(s)]
3Zn(s) → 3Zn2+(aq) + 6e−
6e− + 2Fe3+(aq) → 2Fe(s)
Next we add the two half-reactions and cancel the six electrons that appear on both sides of the
equation.
3Zn(s) → 3Zn2+(aq) + 6e –
6e – + 2Fe3+(aq) → 2Fe(s)
3Zn(s) + 2Fe3+(aq) → 3Zn2+(aq) + 2Fe(s)
balanced skeletal equation
Finally, we replace the nitrate ions to complete the balanced equation.
3Zn(s) + 2Fe(NO3)3(aq) → 3Zn(NO3)2(aq) + 2Fe(s)
balanced
(b) Mn(s) + HCl(aq) → MnCl2 (aq) + H2(g)
Chloride ions, Cl−, are spectator ions in this reaction so we can eliminate them in the first part of the
balancing process, and add them back into the equation after we have balanced the half-reactions.
Eliminating the chloride ions gives us the ionic equation:
14−29
Mn(s) + H+(aq) → Mn2+(aq) + H2(g)
skeletal ionic equation
The oxidation half-reaction is:
Mn(s) → Mn2+(aq)
oxidation half-reaction
The atoms are balanced, and we can balance the charge by adding two electrons to the right side of the
equation:
Mn(s) → Mn2+(aq) + 2e−
balanced oxidation half-reaction
The reduction half-reaction is:
H+(aq) → H2(g)
reduction half-reaction
First, we balance the atoms by adding a coefficient of 2 in front of H+. Next we balance the charge by
adding two electrons to the reactant side of the equation.
2e− + 2H+(aq) → H2(g)
balanced reduction half-reaction
The electrons exchanged between the two half-reactions are already balanced, so we add the two halfreactions and cancel the two electrons that appear on both sides of the equation.
Mn(s) → Mn2+(aq) + 2e –
2e – + 2H+(aq) → H2(g)
Mn(s) + 2H+(aq) → Mn2+(aq) + H2(g)
balanced skeletal equation
Finally, we replace the chloride ions to complete the balanced equation.
Mn(s) + 2HCl(aq) → MnCl2 (aq) + H2(g)
14.48
balanced
(a) Al(s) + Fe(NO3)3(aq) → Al(NO3)3(aq) + Fe(NO3)2(aq)
Nitrate ions, NO3−, are spectator ions in this reaction, so we can eliminate them while we balance the
two half-reactions and add them back when we complete the balanced chemical equation. Eliminating
the nitrate ions gives us the following ionic equation:
Al(s) + Fe3+(aq) → Al3+(aq) + Fe2+(aq) skeletal ionic equation
The oxidation half-reaction is:
Al(s) → Al3+(aq)
oxidation half-reaction
The atoms are balanced, so we can balance the charge by adding three electrons to the right side of the
equation:
Al(s) → Al3+(aq) + 3e−
balanced oxidation half-reaction
The reduction half-reaction is:
Fe3+(aq) → Fe2+(aq)
reduction half-reaction
The atoms are balanced. We can balance the charge by adding one electron to the reactant side of the
equation:
e− + Fe3+(aq) → Fe2+(aq)
balanced reduction half-reaction
To equalize the number of electrons lost and gained, we multiply each half-reaction by an appropriate
coefficient:
1 × [Al(s) → Al3+(aq) + 3e−]
3 × [e− + Fe3+(aq) → Fe2+(aq)]
Al(s) → Al3+(aq) + 3e−
14−30
3e− + 3Fe3+(aq) → 3Fe2+(aq)
Then we add the two half-reactions and cancel the three electrons that appear on both sides of the
equation.
Al(s) → Al3+(aq) + 3e –
3e – + 3Fe3+(aq) → 3Fe2+(aq)
Al(s) + 3Fe3+(aq) → Al3+(aq) + 3Fe2+( aq)
balanced skeletal equation
Finally, we replace the nitrate ions to complete the balanced equation.
Al(s) + 3Fe(NO3)3(aq) → Al(NO3)3(aq) + 3Fe(NO3)2(aq)
balanced
(b) Na(s) + HNO3(aq) → NaNO3 (aq) + H2(g)
Nitrate ions, NO3−, are spectator ions in this reaction so we can eliminate them while we balance the
two half-reactions and add them back when we complete the balanced chemical equation. Eliminating
the nitrate ions gives us the following ionic equation:
Na(s) + H+(aq) → Na+(aq) + H2(g)
skeletal ionic equation
The oxidation half-reaction is:
Na(s) → Na+(aq)
oxidation half-reaction
The atoms are balanced; we can balance the charge by adding one electron to the right side of the
equation:
Na(s) → Na+(aq) + e−
balanced oxidation half-reaction
The reduction half-reaction is:
H+(aq) → H2(g)
reduction half-reaction
First, we balance the atoms by adding a coefficient of 2 in front of H+. Next we balance the charge by
adding two electrons to the reactant side of the equation.
2e− + 2H+(aq) → H2(g)
balanced reduction half-reaction
To equalize the number of electrons exchanged by the two half-reactions, we multiply each halfreaction by the appropriate coefficient:
2 × [Na(s) → Na+(aq) + e−]
1 × [2e− + 2H+(aq) → H2(g)]
2Na(s) → 2Na+(aq) + 2e−
2e− + 2H+(aq) → H2(g)
Then we add the two half-reactions and cancel the two electrons that appear on both sides of the
equation.
2Na(s) → 2Na+(aq) + 2e –
2e – + 2H+(aq) → H2(g)
2Na(s) + 2H+(aq) → 2Na+(aq) + H2(g)
balanced skeletal equation
Finally, we replace the nitrate ions to complete the balanced equation.
2Na(s) + 2HNO3(aq) → 2NaNO3(aq) + H2(g)
14−31
balanced
14.49
From the problem we can write:
Fe2(SO4)3(aq) + KI(aq) → FeSO4(aq) + K2SO4(aq) + I2(aq)
Sulfate ions and potassium ions are spectator ions in this reaction, so we can eliminate them when we
balance the half-reactions. The oxidation number of Fe in Fe2(SO4)3 is 3+.
Fe3+(aq) + I−(aq) → Fe2+(aq) + I2(aq)
skeletal ionic equation
The oxidation half-reaction is:
I−(aq) → I2(aq)
oxidation half-reaction
First, we balance the atoms by adding a coefficient of 2 in front of I−. Next, we balance the charge by
adding two electrons to the right side of the equation.
2I−(aq) → I2(aq) + 2e−
balanced oxidation half-reaction
The reduction half-reaction is:
Fe3+(aq) → Fe2+(aq)
reduction half-reaction
The atoms are balanced, so we can balance the charge by adding one electron to the reactant side of the
equation.
e− + Fe3+(aq) → Fe2+(aq)
balanced reduction half-reaction
To balance the electrons between the two half-reactions we multiply each half-reaction by the appropriate
coefficient:
1 × [2I−(aq) → I2(aq) + 2e−]
2 × [e− + Fe3+(aq) → Fe2+(aq)]
2I−(aq) → I2(aq) + 2e−
2e− + 2Fe3+(aq) → 2Fe2+(aq)
The electrons between the two half-reactions are already balanced, so we add the two half-reactions and
cancel the two electrons that appear on both sides of the equation.
2I−(aq) → I2(aq) + 2e –
2e – + 2Fe3+(aq) → 2Fe2+(aq)
2Fe3+(aq) + 2I−(aq) → 2Fe2+(aq) + I2(aq)
balanced skeletal equation
Finally, we replace the sulfate and potassium ions to complete the balanced equation. Because the formula
for Fe2(SO4)3 includes two iron ions, we incorporate both Fe3+ from the balanced skeletal equation into that
chemical formula.
Fe2(SO4)3(aq) + 2KI(aq) → 2FeSO4(aq) + K2SO4(aq) + I2(aq)
14.50
balanced
From the problem we can write:
SO2(g) + HNO3(aq) → H2SO4(aq) + NO2(g)
To begin, we assign an oxidation number to each element in the reaction. The elements whose oxidation
numbers change are shown below.
4+
5+
SO2(g) + HNO3(aq)
6+
4+
H2SO4(aq) + NO2
Sulfur is oxidized, so we write the oxidation half-reaction:
SO2(g) → H2SO4(aq)
oxidation half-reaction
14−32
We add two electrons to the right side of the equation to account for the change in oxidation number.
SO2(g) → H2SO4(aq) + 2e−
add electrons
Then we balance the charge by adding hydrogen ions (H+).
SO2(g) → H2SO4(aq) + 2H+(aq) + 2e− add H+
Finally, we balance the atoms by adding water (H2O).
2H2O(l) + SO2(g) → H2SO4(aq) + 2H+(aq) + 2e−
add H2O
The reduction half-reaction is:
HNO3(aq) → NO2(g)
reduction half-reaction
We add one electron to the left side of the equation to account for the change in oxidation number.
e− + HNO3(aq) → NO2(g)
add electrons
We balance the charge by adding hydrogen ions (H+).
e− + H+(aq) + HNO3(aq) → NO2(g) add H+
Then we balance the atoms by adding water (H2O).
e− + H+(aq) + HNO3(aq) → NO2(g) + H2O(l)
add H2O
Next, we equalize the number of electrons exchanged in the two half-reactions by multiplying each halfreaction by the appropriate coefficient:
1 × [2H2O(l) + SO2(g) → H2SO4(aq) + 2H+(aq) + 2e−]
2 × [e− + H+(aq) + HNO3(aq) → NO2(g) + H2O(l)]
2H2O(l) + SO2(g) → H2SO4(aq) + 2H+(aq) + 2e−
2e− + 2H+(aq) + 2HNO3(aq) → 2NO2(g) + 2H2O(l)
Then we add the two half-reactions and cancel the two electrons that appear on both sides of the equation.
2H2O(l) + SO2(g) → H2SO4(aq) + 2H+(aq) + 2e –
2e – + 2H+(aq) + 2HNO3(aq) → 2NO2(g) + 2H2O(l)
2H2O(l) + 2H+(aq) + SO2(g) + 2HNO3(aq) → H2SO4(aq) + 2NO2(g) + 2H+(aq) + 2H2O(l)
Finally, we complete the balancing process by eliminating substances that appear on both the reactant and
product sides of the equation.
SO2(g) + 2HNO3(aq) → H2SO4(aq) + 2NO2(g)
14.51
Balanced reaction
For simple half-reactions (see part (a) below) we determine the oxidation numbers of the atoms and add the
number of electrons needed to balance the charge. To balance more complicated half-reactions (in acidic
conditions) we follow the steps outlined below. Because these reactions occur in acidic solution, we add
H+ to balance the charge.
(a) Ba(s) → Ba2+(aq)
We balance the charge by adding two electrons to the product side:
Ba(s) → Ba2+(aq) + 2e–
0
0
14−33
(b) HNO2(aq) → NO(g)
2+
3+
HNO2(aq)
→
NO(g)
determine oxidation numbers
e− + HNO2(aq) → NO(g)
add e− to balance oxidation numbers
e− + H+(aq) + HNO2(aq) → NO(g)
add H+ to balance charge
e− + H+(aq) + HNO2(aq) → NO(g) + H2O
add H2O to balance atoms
(c) H2O2(aq) → H2O(l)
1–
2–
H2O2(aq) → H2O(l)
determine oxidation numbers
H2O2(aq) → 2H2O(l)
balance atoms being oxidized or reduced
2e + H2O2(aq) → 2H2O(l)
add e− to balance oxidation numbers
2e− + 2H+(aq) + H2O2(aq) → 2H2O(l)
add H+ to balance charge, equation balanced
−
(d) Cr3+(aq) → Cr2O72−(aq)
3+
6+
Cr3+ → Cr2O72–
determine oxidation numbers
2Cr (aq) → Cr2O7 (aq)
3+
2−
14.52
balance atoms being oxidized or reduced
2Cr (aq) → Cr2O7 (aq) + 6e
add e− to balance oxidation numbers
2Cr3+(aq) → Cr2O72−(aq) + 14H+ + 6e−
add H+ to balance charge
2Cr3+(aq) + 7H2O(l) → Cr2O72−(aq) + 14H+ + 6e−
add H2O to balance atoms
3+
2−
−
For simple half-reactions (see part (a) below) we determine the oxidation numbers of the atoms and add the
number of electrons needed to balance the charge. To balance more complicated half-reactions (in acidic
conditions) we follow the steps outlined below. Because these reactions occur in acidic conditions, we add
H+ to balance the charge.
(a) Br2(aq) → Br−(aq)
0
Br2(aq)
1–
→ Br–(aq)
−
determine oxidation numbers
Br2(aq) → 2Br (aq)
balance atoms being oxidized or reduced
2e− + Br2(aq) → 2Br−(aq)
add e− to balance oxidation numbers
(b) H2O(l) → O2(g)
2–
0
H2O(l)
→ O2(g)
2H2O(l) → O2(g)
determine oxidation numbers
balance atoms being oxidized or reduced
14−34
2H2O(l) → O2(g) + 4e−
add e− to balance oxidation numbers
2H2O(l) → O2(g) + 4e− + 4H+(aq)
add H+ to balance charge, equation balanced
(c) IO3−(aq) →
I2(aq)
0+
5+
→
IO3–(aq)
I2(aq)
determine oxidation numbers
−
2IO3 (aq) → I2(aq)
balance iodine
10e + 2IO3 (aq) → 2I2(aq)
add e− to balance oxidation numbers
10e− + 12H+(aq) + 2IO3−(aq) → I2(aq)
add H+ to balance charge
10e− + 12H+(aq) + 2IO3−(aq) → I2(aq) + 6H2O(l)
add H2O to balance atoms
−
−
(d) MnO2(s) → Mn2+(aq)
4+
2+
MnO2(s)
→ Mn2+(aq)
14.53
determine oxidation numbers
2e + MnO2(s) → Mn (aq)
add e− to balance oxidation numbers
2e− + 4H+(aq) + MnO2(s) → Mn2+(aq)
add H+ to balance charge
2e− + 4H+(aq) + MnO2(s) → Mn2+(aq) + 2H2O(l)
add H2O to balance atoms
−
2+
To balance more complicated half-reactions that occur in basic solutions, we follow the steps outlined
below. Because these reactions occur in basic solutions, we add OH− to balance the charge.
(a) La(s) → La(OH)3(s)
0
3+
La(s)
→ Li(OH)3(s)
determine oxidation numbers
La(s) → La(OH)3(s) + 3e−
add e− to balance oxidation numbers
3OH−(aq) + La(s) → La(OH)3(s) + 3e−
add OH− to balance charge
(b) NO3−(aq) → NO2−(aq)
5+
3+
→ NO2–(aq)
NO3–(aq)
determine oxidation numbers
2e + NO3 (aq) → NO2 (aq)
add e− to balance oxidation numbers
2e− + NO3−(aq) → NO2−(aq) + 2OH−(aq)
add OH− to balance charge
2e− + H2O(l) + NO3−(aq) → NO2−(aq) + 2OH−(aq)
add H2O to balance atoms
−
−
−
(c) H2O2(aq) → O2(g)
1–
H2O2(aq)
0
→ O2(g)
determine oxidation numbers
14−35
H2O2(aq) → O2(g) + 2e−
add e− to balance oxidation numbers
H2O2(aq) + 2OH−(aq) → O2(g) + 2e−
add OH− to balance charge
H2O2(aq) + 2OH−(aq) → O2(g) + 2e− + 2H2O(l)
add H2O to balance atoms
(d) Cl2O7(aq) → 2ClO2−(aq)
7+
3+
Cl2O7(aq)
→ ClO2–(aq)
determine oxidation numbers
Cl2O7(aq) → 2ClO2−(aq)
balance atoms being oxidized or reduced
8e + Cl2O7(aq) → 2ClO2 (aq)
add e− to balance oxidation numbers
8e− + Cl2O7(aq) → 2ClO2−(aq) + 6OH−(aq)
add OH− to balance charge
−
−
8e− + 3H2O(l) + Cl2O7(aq) → 2ClO2−(aq) + 6OH−(aq) add H2O to balance atoms
14.54
To balance more complicated half-reactions that occur in basic solutions, we follow the steps outlined
below. Because these reactions occur in basic solution, we add OH− to balance the charge.
(a) CrO42−(aq) → Cr(OH)3(s)
6+
3+
CrO42–(aq)
→ Cr(OH)3(s)
determine oxidation numbers
3e− + CrO42−(aq) → Cr(OH)3(s)
add e− to balance oxidation numbers
3e− + CrO42−(aq) → Cr(OH)3(s)+ 5OH−(aq)
add OH− to balance charge
3e− + 4H2O(l) + CrO42−(aq) → Cr(OH)3(s) + 5OH−(aq)
add H2O to balance atoms
(b) ClO2 (aq) → ClO−(aq)
4+
1+
ClO2(aq) → ClO–(aq)
determine oxidation numbers
3e + ClO2 (aq) → ClO (aq)
add e− to balance oxidation numbers
3e− + ClO2 (aq) → ClO−(aq) + 2OH−(aq)
add OH− to balance charge
3e− + H2O(l) + ClO2 (aq) → ClO−(aq) + 2OH−(aq)
add H2O to balance atoms
−
−
−
(c) MnO4 (aq) → MnO2(s)
4+
7+
MnO4–(aq) → MnO2(s)
determine oxidation numbers
3e− + MnO4−(aq) → MnO2(s)
add e− to balance oxidation numbers
3e− + MnO4−(aq) → MnO2(s) + 4OH−(aq)
add OH− to balance charge
3e− + 2H2O(l) + MnO4−(aq) → MnO2(s) + 4OH−(aq)
add H2O to balance atoms
−
−
(d) Br (aq) → BrO2 (aq)
14−36
3+
1–
Br–(aq)
→
−
BrO2–(aq)
−
Br (aq) → BrO2 (aq) + 4e
14.55
determine oxidation numbers
add e− to balance oxidation numbers
−
4OH−(aq) + Br−(aq) → BrO2−(aq) + 4e−
add OH− to balance charge
4OH−(aq) + Br−(aq) → BrO2−(aq) + 2H2O(l) + 4e−
add H2O to balance atoms
(a) H2S(aq) + Cr2O72−(aq) → S(s) + Cr3+(aq)
2–
6+
0
→
H2S(aq) + Cr2O72–(aq)
3+
S(s) + Cr3+(aq)
determine oxidation numbers
Balance the oxidation half-reaction:
H2S(aq) → S(s) + 2e−
add e− to balance oxidation numbers
H2S(aq) → S(s) + 2H+(aq) + 2e−
add H+ to balance charge
Balance the reduction half-reaction:
Cr2O72−(aq) → 2Cr3+(aq)
balance atoms being oxidized or reduced
6e + Cr2O7 (aq) → 2Cr (aq)
add e− to balance oxidation numbers
6e− + 14H+(aq) + Cr2O72−(aq) → 2Cr3+(aq)
add H+ to balance charge
6e− + 14H+(aq) + Cr2O72−(aq) → 2Cr3+(aq) + 7H2O(l)
add H2O to balance atoms
−
3+
2−
Equalize the number of electrons exchanged between the two half-reactions:
3 × [H2S(aq) → S(s) + 2H+(aq) + 2e−]
1 × [6e− + 14H+(aq) + Cr2O72−(aq) → 2Cr3+(aq) + 7H2O(l)]
3H2S(aq) → 3S(s) + 6H+(aq) + 6e−
6e− + 14H+(aq) + Cr2O72−(aq) → 2Cr3+(aq) + 7H2O(l)
3H2S(aq) → 3S(s) + 6H+(aq) + 6e –
6e – + 14H+(aq) + Cr2O72−(aq) → 2Cr3+(aq) + 7H2O(l)
14H+(aq) + 3H2S(aq) + Cr2O72−(aq) → 3S(s) + 2Cr3+(aq) + 7H2O(l) + 6H+(aq)
Eliminate duplicate substances (bold type):
8H+(aq) + 3H2S(aq) + Cr2O72−(aq) → 3S(s) + 2Cr3+(aq) + 7H2O(l)
−
balanced equation
(b) V (aq) + MnO4 (aq) → VO2 (aq) + Mn (aq)
2+
2+
+
7+
V2+(aq) + MnO4–(aq)
5+
→
2+
2+
VO2+(aq) + Mn2+(aq)
determine oxidation numbers
Balance the oxidation half-reaction:
V2+(aq) → VO2+(aq) + 3e−
add e− to balance oxidation numbers
V2+(aq) → VO2+(aq) + 4H+ + 3e−
add H+ to balance charge
14−37
2H2O(l) + V2+(aq) → VO2+(aq) + 4H+ + 3e−
add H2O to balance atoms
Balance the reduction half-reaction:
5e− + MnO4−(aq) → Mn2+(aq)
add e− to balance oxidation numbers
5e− + 8H+(aq) + MnO4−(aq) → Mn2+(aq)
add H+ to balance charge
5e− + 8H+(aq) + MnO4−(aq) → Mn2+(aq) + 4H2O(l)
add H2O to balance atoms
Equalize the number of electrons exchanged between the two half-reactions:
5 × [2H2O(l) + V2+(aq) → VO2+(aq) + 4H+ + 3e−]
3 × [5e− + 8H+(aq) + MnO4−(aq) → Mn2+(aq) + 4H2O(l)]
10H2O(l) + 5V2+(aq) → 5VO2+(aq) + 20H+ + 15e−
15e− + 24H+(aq) + 3MnO4−(aq) → 3Mn2+(aq) + 12H2O(l)
10H2O(l) + 5V2+(aq) → 5VO2+(aq) + 20H+(aq) + 15e –
15e – + 24H+(aq) + 3MnO4−(aq) → 3Mn2+(aq) + 12H2O(l)
24H+(aq) + 10H2O(l) + 5V2+(aq) + 3MnO4−(aq) → 5VO2+(aq) + 3Mn2+(aq) + 20H+(aq) + 12H2O(l)
Eliminate duplicate substances (bold type):
4H+(aq) + 5V2+(aq) + 3MnO4−(aq) → 5VO2+(aq) + 3Mn2+(aq) + 2H2O(l)
balanced equation
(c) Fe2+(aq) + ClO3−(aq) → Fe3+(aq) + Cl−(aq)
2+
5+
1–
3+
Fe2+(aq) + ClO3–(aq)
→
Fe3+(aq) + Cl–(aq)
determine oxidation numbers
Balance the oxidation half-reaction:
add e− to balance oxidation numbers
Fe2+(aq) → Fe3+(aq) + e−
Balance the reduction half-reaction:
6e− + ClO3−(aq) → Cl−(aq)
add e− to balance oxidation numbers
6e− + 6H+(aq) + ClO3−(aq) → Cl−(aq)
add H+ to balance charge
6e− + 6H+(aq) + ClO3−(aq) → Cl−(aq) + 3H2O(l)
add H2O to balance atoms
Equalize the number of electrons exchanged between the two half-reactions:
6 × [Fe2+(aq) → Fe3+(aq) + e−]
1 × [6e− + 6H+(aq) + ClO3−(aq) → Cl−(aq) + 3H2O(l)]
6Fe2+(aq) → 6Fe3+(aq) + 6e−
6e− + 6H+(aq) + ClO3−(aq) → Cl−(aq) + 3H2O(l)
6Fe2+(aq) → 6Fe3+(aq) + 6e –
6e – + 6H+(aq) + ClO3−(aq) → Cl−(aq) + 3H2O(l)
6H+(aq) + 6Fe2+(aq) + ClO3−(aq) → 6Fe3+(aq) + Cl−(aq) + 3H2O(l)
14−38
balanced equation
14.56
(a) S2−(aq) + CrO42−(aq) → S(s) + Cr(OH)3(s)
6+
2–
3+
0
S2–(aq) + CrO42–(aq)
→
S(s) + Cr(OH)3(s)
determine oxidation numbers
Balance the oxidation half-reaction:
add e− to balance oxidation numbers
S2−(aq) → S(s) + 2e−
Balance the reduction half-reaction:
3e− + CrO42−(aq) → Cr(OH)3(s)
add e− to balance oxidation numbers
3e− + CrO42−(aq) → Cr(OH)3(s) +5OH−(aq)
add OH− to balance charge
3e− + 4H2O(l) + CrO42−(aq) → Cr(OH)3(s) +5OH−(aq)
add H2O to balance atoms
Equalize the number of electrons exchanged between the two half-reactions:
3 × [S2−(aq) → S(s) + 2e−]
2 × [3e− + 4H2O(l) + CrO42−(aq) → Cr(OH)3(s) +5OH−(aq)]
3S2−(aq) → 3S(s) + 6e−
6e− + 8H2O(l) + 2CrO42−(aq) → 2Cr(OH)3(s) +10OH−(aq)
3S2−(aq) → 3S(s) + 6e –
6e – + 8H2O(l) + 2CrO42−(aq) → 2Cr(OH)3(s) +10OH−(aq)
8H2O(l) + 3S2−(aq) + 2CrO42−(aq) → 3S(s) + 2Cr(OH)3(s) +10OH−(aq)
−
−
balanced equation
−
(b) MnO4 (aq) + CN (aq) → MnO4 (aq) + CNO (aq)
7+
2−
2+
6+
MnO4–(aq) + CN–(aq)
→
4+
MnO42–(aq) + CNO–(aq)
determine oxidation numbers
Balance oxidation half-reaction:
CN–(aq) → CNO–(aq) + 2e−
add e− to balance oxidation numbers
2OH−(aq) + CN–(aq) → CNO–(aq) + 2e−
add OH− to balance charge
2OH−(aq) + CN–(aq) → CNO–(aq) + H2O(l) + 2e−
add H2O to balance atoms
Balance reduction half-reaction:
add e− to balance oxidation numbers
e− + MnO4–(aq) → MnO42–(aq)
Equalize the number of electrons exchanged between the two half-reactions:
1 × [2OH−(aq) + CN–(aq) → CNO–(aq) + H2O(l) + 2e−]
2 × [e− + MnO4–(aq) → MnO42–(aq)]
2OH−(aq) + CN–(aq) → CNO–(aq) + H2O(l) + 2e−
2e− + 2MnO4–(aq) → 2MnO42–(aq)
2OH−(aq) + CN–(aq) → CNO–(aq) + H2O(l) + 2e –
14−39
2e – + 2MnO4–(aq) → 2MnO42–(aq)
2OH−(aq) + 2MnO4–(aq) + CN–(aq) → 2MnO42–(aq) + CNO–(aq) + H2O(l)
−
balanced equation
−
(c) Al(s) + OH (aq) → Al(OH)4 (aq) + H2(g)
0
1+
Al(s) + OH–(aq)
0
3+
→
Al(OH)4–(aq) + H2(g)
determine oxidation numbers
Balance oxidation half-reaction:
Al(s) → Al(OH)4−(aq) + 3e−
add e− to balance oxidation numbers
4OH−(aq) + Al(s) → Al(OH)4−(aq) + 3e−
add OH− to balance charge
Balance reduction half-reaction:
2OH−(aq) → H2(g)
balance atoms being oxidized or reduced
2e− + 2OH−(aq) → H2(g)
add e− to balance oxidation numbers
2e− + 2OH−(aq) → H2(g) + 4OH−(aq)
add OH− to balance charge
2e− + 2H2O(l) + 2OH−(aq) → H2(g) + 4OH−(aq)
add H2O to balance atoms
−
−
2e + 2H2O(l) → H2(g) + 2OH (aq)
eliminate duplicate substances
Equalize the number of electrons exchanged between the two half-reactions:
2 × [4OH−(aq) + Al(s) → Al(OH)4−(aq) + 3e−]
3 × [2e− + 2H2O(l) → H2(g) + 2OH−(aq)]
8OH−(aq) + 2Al(s) → 2Al(OH)4−(aq) + 6e−
6e− + 6H2O(l) → 3H2(g) + 6OH−(aq)
8OH−(aq) + 2Al(s) → 2Al(OH)4−(aq) + 6e –
6e – + 6H2O(l) → 3H2(g) + 6OH−(aq)
8OH−(aq) + 6H2O(l) + 2Al(s) → 2Al(OH)4−(aq) + 3H2(g) + 6OH−(aq)
Eliminate duplicate substances (bold type):
2OH−(aq) + 6H2O(l) + 2Al(s) → 2Al(OH)4−(aq) + 3H2(g)
14.57
balanced equation
(a) NH3(aq) + ClO−(aq) → N2H4(aq) + Cl−(aq)
3–
1+
NH3(aq) + ClO–(aq)
2–
→
1–
N2H4(aq) + Cl–(g)
determine oxidation numbers
Balance oxidation half-reaction:
2NH3(aq) → N2H4(aq)
balance atoms being oxidized or reduced
2NH3(aq) → N2H4(aq) + 2e
add e− to balance oxidation numbers
2OH−(aq) + 2NH3(aq) → N2H4(aq) + 2e−
add OH− to balance charge
2OH−(aq) + 2NH3(aq) → N2H4(aq) + 2H2O(l) + 2e−
add H2O to balance atoms
−
14−40
Balance reduction half-reaction:
2e− + ClO−(aq) → Cl−(aq)
add e− to balance oxidation numbers
2e− + ClO−(aq) → Cl−(aq) + 2OH−(aq)
add OH− to balance charge
2e− + H2O(l) + ClO−(aq) → Cl−(aq) + 2OH−(aq)
add H2O to balance atoms
Equalize the number of electrons exchanged between the two half-reactions:
2OH−(aq) + 2NH3(aq) → N2H4(aq) + 2H2O(l) + 2e –
2e – + H2O(l) + ClO−(aq) → Cl−(aq) + 2OH−(aq)
2OH−(aq) + H2O(l) + 2NH3(aq) + ClO−(aq) → N2H4(aq) + Cl−(aq) + 2OH−(aq) + 2H2O(l)
Eliminate duplicate substances (bold type):
2NH3(aq) + ClO−(aq) → N2H4(aq) + Cl−(aq) + H2O(l)
balanced equation
(b) Cr(OH)4−(aq) + HO2−(aq) → CrO42−(aq) + H2O(l)
3+
1–
6+
→
Cr(OH)4–(aq) + HO2–(aq)
2–
CrO42–(aq) + H2O(g)
determine oxidation numbers
Balance oxidation half-reaction:
Cr(OH)4−(aq) → CrO42−(aq) + 3e−
add e− to balance oxidation numbers
4OH−(aq) + Cr(OH)4−(aq) → CrO42−(aq) + 3e−
add OH− to balance charge
4OH−(aq) + Cr(OH)4−(aq) → CrO42−(aq) + 4H2O(l) + 3e−
add H2O to balance atoms
Balance reduction half-reaction:
HO2−(aq) → 2H2O(l)
balance atoms being oxidized or reduced
2e + HO2 (aq) → 2H2O(l)
add e− to balance oxidation numbers
2e− + HO2−(aq) → 2H2O(l) + 3OH−(aq)
add OH− to balance charge
2e− + 3H2O(l) + HO2−(aq) → 2H2O(l) + 3OH−(aq)
add H2O to balance atoms
−
−
Equalize the number of electrons exchanged between the two half-reactions:
2 × [4OH−(aq) + Cr(OH)4−(aq) → CrO42−(aq) + 4H2O(l) + 3e−]
3 × [2e− + 3H2O(l) + HO2−(aq) → 2H2O(l) + 3OH−(aq)]
8OH−(aq) + 2Cr(OH)4−(aq) → 2CrO42−(aq) + 8H2O(l) + 6e−
6e− + 9H2O(l) + 3HO2−(aq) → 6H2O(l) + 9OH−(aq)
8OH−(aq) + 2Cr(OH)4−(aq) → 2CrO42−(aq) + 8H2O(l) + 6e –
6e – + 9H2O(l) + 3HO2−(aq) → 6H2O(l) + 9OH−(aq)
8OH−(aq) + 9H2O(l) + 2Cr(OH)4−(aq) + 3HO2−(aq) → 2CrO42−(aq) + 14H2O(l) + 9OH−(aq)
Eliminate duplicate substances (bold type):
2Cr(OH)4−(aq) + 3HO2−(aq) → 2CrO42−(aq) + 5H2O(l) + OH−(aq) balanced equation
14−41
(c) Br2(aq) → Br−(aq) + BrO−(aq)
0
1–
Br2(aq)
→
1+
Br–(aq) + BrO–(aq)
determine oxidation numbers
Balance oxidation half-reaction:
Br2(aq) → 2BrO−(aq)
balance atoms being oxidized or reduced
Br2(aq) → 2BrO−(aq) + 2e−
add e− to balance oxidation numbers
4OH−(aq) + Br2(aq) → 2BrO−(aq) + 2e−
add OH− to balance charge
4OH−(aq) + Br2(aq) → 2BrO−(aq) + 2H2O(l) + 2e−
add H2O to balance atoms
Balance reduction half-reaction:
Br2(aq) → 2Br−(aq)
−
balance atoms being oxidized or reduced
add e− to balance oxidation numbers
−
2e + Br2(aq) → 2Br (aq)
Equalize the number of electrons exchanged between the two half-reactions:
4OH−(aq) + Br2(aq) → 2BrO−(aq) + 2H2O(l) + 2e –
2e – + Br2(aq) → 2Br−(aq)
4OH−(aq) + 2Br2(aq) → 2Br−(aq) + 2BrO−(aq) + 2H2O(l)
Divide coefficients by 2 (bold type):
2OH−(aq) + Br2(aq) → Br−(aq) + BrO−(aq) + H2O(l)
14.58
balanced equation
(a) C2O42−(aq) + MnO4−(aq) → CO2(g) + Cr3+(aq)
3+
7+
C2O42–(aq) + MnO4–(aq)
4+
→
2+
CO2(g) + Mn2+(aq)
determine oxidation numbers
Balance oxidation half-reaction:
C2O42−(aq) → 2CO2(g)
C2O4 (aq) → 2CO2(g) + 2e
2−
balance atoms being oxidized or reduced
add e− to balance oxidation numbers
−
Balance reduction half-reaction:
5e− + MnO4−(aq) → Mn2+(aq)
add e− to balance oxidation numbers
5e− + 8H+(aq) + MnO4−(aq) → Mn2+(aq)
add H+ to balance charge
5e− + 8H+(aq) + MnO4−(aq) → Mn2+(aq) + 4H2O(l)
add H2O to balance atoms
Equalize the number of electrons exchanged between the two half-reactions:
5 × [C2O42−(aq) → 2CO2(g) + 2e−]
2 × [5e− + 8H+(aq) + MnO4−(aq) → Mn2+(aq) + 4H2O(l)]
5C2O42−(aq) → 10CO2(g) + 10e−
10e− + 16H+(aq) + 2MnO4−(aq) → 2Mn2+(aq) + 8H2O(l)
5C2O42−(aq) → 10CO2(g) + 10e –
14−42
10e – + 16H+(aq) + 2MnO4−(aq) → 2Mn2+(aq) + 8H2O(l)
16H+(aq) + 5C2O42−(aq) + 2MnO4−(aq) → 10CO2(g) + 2Mn2+(aq) + 8H2O(l) balanced equation
(b) ClO3−(aq) + Cl−(aq) → Cl2(aq)
5+
1–
0
ClO3–(aq) + Cl–(aq)
→
Cl2(aq)
determine oxidation numbers
Balance oxidation half-reaction:
2Cl−(aq) → Cl2(g)
balance atoms being oxidized or reduced
−
2Cl (aq) → Cl2(g) + 2e
add e− to balance oxidation numbers
−
Balance reduction half-reaction:
2ClO3−(aq) → Cl2(g)
balance atoms being oxidized or reduced
10e− + 2ClO3−(aq) → Cl2(g)
add e− to balance oxidation numbers
10e− + 12H+(aq) + 2ClO3−(aq) → Cl2(g)
add H+ to balance charge
10e− + 12H+(aq) + 2ClO3−(aq) → Cl2(g) + 6H2O(l)
add H2O to balance atoms
Equalize the number of electrons exchanged between the two half-reactions:
5 × [2Cl−(aq) → Cl2(g) + 2e−]
1 × [12H+(aq) + 10e− + 2ClO3−(aq) → Cl2(g) + 6H2O(l)]
10Cl−(aq) → 5Cl2(g) + 10e−
10e− + 12H+(aq) + 2ClO3−(aq) → Cl2(g) + 6H2O(l)
10Cl−(aq) → 5Cl2(g) + 10e –
10e – + 12H+(aq) + 2ClO3−(aq) → Cl2(g) + 6H2O(l)
12H+(aq) + 2ClO3−(aq) + 10Cl−(aq) → 6Cl2(g) + 6H2O(l)
Divide coefficients by 2 (bold type):
6H+(aq) + ClO3−(aq) + 5Cl−(aq) → 3Cl2(g) + 3H2O(l)
balanced equation
(c) S2O32-(aq) + I2(aq) → SO42−(aq) + I−
2+
0
1–
6+
S2O32–(aq) + I2(aq)
→
SO42–(aq) + I–(aq)
determine oxidation numbers
Balance oxidation half-reaction:
S2O32-(aq) → 2SO42−(aq)
balance atoms being oxidized or reduced
S2O3 (aq) → 2SO4 (aq) + 8e
add e− to balance oxidation numbers
S2O32-(aq) → 2SO42−(aq) + 10H+ + 8e−
add H+ to balance charge
5H2O(l) + S2O32-(aq) → 2SO42−(aq) + 10H+ + 8e−
add H2O to balance atoms
2-
2−
−
Balance reduction half-reaction:
14−43
I2(aq) → 2I−(aq)
−
balance atoms being oxidized or reduced
add e− to balance oxidation numbers
−
2e + I2(aq) → 2I (aq)
Equalize the number of electrons exchanged between the two half-reactions:
1 × [5H2O(l) + S2O32-(aq) → 2SO42−(aq) + 10H+ + 8e−]
4 × [2e− + I2(aq) → 2I−(aq)]
5H2O(l) + S2O32-(aq) → 2SO42−(aq) + 10H+ + 8e−
8e− + 4I2(aq) → 8I−(aq)
5H2O(l) + S2O32-(aq) → 2SO42−(aq) + 10H+ + 8e –
8e – + 4I2(aq) → 8I−(aq)
5H2O(l) + S2O32− (aq) + 4I2(aq) → 2SO42−(aq) + 8I−(aq) + 10H+ balanced equation
14.59
We begin by calculating the oxidation numbers of each atom in the chemical equation:
0
5+
0
4+
C6H12O6(aq) + NO3–(aq)
→
CO2(g) + N2(g)
determine oxidation numbers
Balance oxidation half-reaction:
C6H12O6(aq) → 6CO2(g)
C6H12O6(aq) → 6CO2(g) + 24e
balance atoms being oxidized or reduced
add e− to balance oxidation numbers
−
C6H12O6(aq) → 6CO2(g) + 24H+(aq) + 24e−
add H+ to balance charge
6H2O(l) + C6H12O6(aq) → 6CO2(g) + 24H+(aq) + 24e−
add H2O to balance atoms
Balance reduction half-reaction:
2NO3−(aq) → N2(g)
balance atoms being oxidized or reduced
10e− + 2NO3−(aq) → N2(g)
add e− to balance oxidation numbers
10e− + 12H+(aq) + 2NO3−(aq) → N2(g)
add H+ to balance charge
10e− + 12H+(aq) + NO3−(aq) → N2(g) + 6H2O(l)
add H2O to balance atoms
Equalize the number of electrons exchanged between the two half-reactions:
Rather than multiplying by 10 and 24 respectively, we can use the factors 5 and 12 to produce smaller
coefficients:
5 × [6H2O(l) + C6H12O6(aq) → 6CO2(g) + 24H+(aq) + 24e−]
12 × [10e− + 12H+(aq) + 2NO3−(aq) → N2(g) + 6H2O(l)]
30H2O(l) + 5C6H12O6(aq) → 30CO2(g) + 120H+(aq) + 120e−
120e− + 144H+(aq) + 24NO3−(aq) → 12N2(g) + 72H2O(l)
30H2O(l) + 5C6H12O6(aq) → 30CO2(g) + 120H+(aq) + 120e –
120e – + 144H+(aq) + 24NO3−(aq) → 12N2(g) + 72H2O(l)
144H+(aq) + 30H2O(l) + 5C6H12O6(aq) + 24NO3−(aq) → 30CO2(g) + 12N2(g) + 72H2O(l) + 120H+(aq)
Eliminate duplicate substances (bold type):
14−44
24H+(aq) + 5C6H12O6(aq) + 24NO3−(aq) → 30CO2(g) + 12N2(g) + 42H2O(l)
14.60
balanced equation
We begin by determining the oxidation numbers of each atom in the chemical equation:
0
5+
5+
I2(s) + NO3–(aq)
→
4+
HIO3(g ) + NO2(g )
determine oxidation numbers
Balance oxidation half-reaction:
I2(s) → 2HIO3(s)
balance atoms being oxidized or reduced
I2(s) → 2HIO3(s) + 10e
add e− to balance oxidation numbers
I2(s) → 2HIO3(s) + 10H+(aq) + 10e−
add H+ to balance charge
6H2O(l) + I2(s) → 2HIO3(s) + 10H+(aq) + 10e−
add H2O to balance atoms
−
Balance reduction half-reaction:
e− + NO3−(aq) → NO2(g)
add e− to balance oxidation numbers
e− + 2H+(aq) + NO3−(aq) → NO2(g)
add H+ to balance charge
e− + 2H+(aq) + NO3−(aq) → NO2(g) + H2O(l)
add H2O to balance atoms
Equalize the number of electrons exchanged between the two half-reactions:
1 × [6H2O(l) + I2(s) → 2HIO3(s) + 10H+(aq) + 10e−]
10 × [e− + 2H+(aq) + NO3−(aq) → NO2(g) + H2O(l)]
6H2O(l) + I2(s) → 2HIO3(s) + 10H+(aq) + 10e−
10e− + 20H+(aq) + 10NO3−(aq) → 10NO2(g) + 10H2O(l)
6H2O(l) + I2(s) → 2HIO3(s) + 10H+(aq) + 10e –
10e – + 20H+(aq) + 10NO3−(aq) → 10NO2(g) + 10H2O(l)
20H+(aq) + 6H2O(l) + I2(s) + 10NO3−(aq) → 2HIO3(s) + 10NO2(g) + 10H+(aq) + 10H2O(l)
Eliminate duplicate substances (bold type):
10H+(aq) + I2(s) + 10NO3−(aq) →2HIO3(s) + 10NO2(g) + 4H2O(l)
balanced equation
14.61
The anode can be any metal that is more active than iron (e.g. aluminum or zinc; see Figure 14.22). The
supporting electrolyte can be any soluble salt that does not precipitate with Fe3+ (e.g. potassium nitrate,
because all nitrates are soluble). The anodic half-cell must also contain a salt that contains the metal used
for the anode (e.g. aluminum nitrate for an aluminum anode, or zinc nitrate for a zinc anode).
14.62
The anode can be any metal that is more active than tin (e.g. zinc or nickel; see Figure 14.22). The
supporting electrolyte can be any soluble salt that does not precipitate with tin. The anodic half-cell must
also contain a salt that contains the metal used for the anode (e.g. zinc nitrate or nickel(II) nitrate for a zinc
or nickel anode, respectively).
14.63
Au < Bi < Ni < Zn < Al < Ca (see Figure 14.22). The activity series orders metals according to reducing
power. The most active reducing metals are listed at the top of the activity series, with potassium being the
most active metal of those listed. Of the metals listed, gold (Au) is the least active.
14.64
K+ < Mg2+ < Zn2+ < Fe2+ < Ni2+ < Ag+ (see Figure 14.22). The activity series orders metals according to
reducing power. Of the metals listed, the most active are listed at the top of the activity series. The order for
14−45
the oxidizing power of the metal ions is the opposite of the reducing power of the metal atoms. The best
reducing agent produces an ion with smallest oxidizing power. Metal ions at the bottom of the activity
series have the greatest oxidizing strength.
14.65
Electrolysis is the process of using electricity (electrical energy) to cause a nonspontaneous redox reaction
to occur.
14.66
During the electrolysis of molten NaCl, sodium ions are reduced at the cathode to produce Na(l), and
chloride ions are oxidized at the anode to produce Cl2(g).
14.67
When the temperature of sodium iodide is above its melting point, NaI is a molten salt. Two types of ions
are present, Na+ and I−. Both are in the molten state (which is designated as (l)). The molten salt is similar
to a liquid, except that the ions are not bound together (like H2O in liquid water) and can move separately
from one another. The sodium ions cannot lose any additional electrons because they already have a noble
gas configuration; they can only be reduced. This means that iodide ions are oxidized (to I2). At the high
temperatures employed with molten salts, molecular iodine is in the gas state.
(a) Reduction takes place at the cathode: sodium metal forms from sodium ions in the molten salt.
(b) Oxidation takes place at the anode: molecular iodine forms from iodide ions in the molten salt.
(c) 2I−(l) → I2(g) + 2e− (anode, oxidation)
Na+(l) + e− → Na(l) (cathode, reduction)
(d) To properly balance the overall equation, we double the reduction half-reaction and add it to the
oxidation half-reaction.
2Na+ + 2I−(l) → 2Na(l) + I2(g)
14.68
When the temperature of aluminum chloride is above its melting point, aluminum chloride is a molten salt.
Two types of ions are present, Al3+ and Cl−. Both are in the molten state (which is designated as (l)). The
molten salt is similar to a liquid, except the ions are not bound together (like H2O in liquid water) and can
move indendently of one another. The aluminum ions cannot lose any additional electrons because they
already have a noble gas configuration; they can only be reduced. This means that chloride is oxidized
(to Cl2). At the high temperatures employed with molten salts, molecular chlorine is in the gas state.
(a) Reduction takes place at the cathode: aluminum metal forms from aluminum ions in the molten salt.
(b) Oxidation takes place at the anode: chlorine gas forms from chloride ions in the molten salt.
(anode, oxidation)
(c) 2Cl−(l) → Cl2(g) + 2e−
(cathode, reduction)
3e− + Al3+(l) → Al(l)
(d) Equalize the number of electrons exchanged between the two half-reactions
3 × [2Cl−(l) → Cl2(g) + 2e−]
2 × [3e− + Al3+(l) → Al(l)]
6Cl−(l) → 3Cl2(g) + 6e−
6e− + 2Al3+(l) → 2Al(l)
6Cl−(l) → 3Cl2(g) + 6e –
6e – + 2Al3+(l) → 2Al(l)
6Cl−(l) + 2Al3+(l) → 2Al(l) + 3Cl2(g)
14−46
14.69
At the anode, the iodine atoms decrease in size as iodide ions in solution lose electrons and combine to
form neutral I2 molecules. At the catholde, the sodium atoms increase in size as the netural metal atoms
gain electrons to become ions in solution.
14.70
Chlorine gas, Cl2(g),
forms at the anode.
Solid aluminum, Al(s),
forms at the cathode.
14.71
Steel is an alloy, or mixture, of metals. The principle component is usually iron. Steel corrodes when the
iron actively reduces oxygen in the presence of water. The metal is oxidized and the oxygen is reduced.
For example, pure iron, Fe, corrodes eventually forming iron(III) oxide (Fe2O3, which is Fe3+ and O2−).
Magnesium reduces oxygen more actively than iron does. When the two metals are connected, the
magnesium will reduce the O2(g), preventing the Fe from being oxidized. When all the Mg(s) is oxidized,
the iron will begin to corrode.
14.72
Corrosion of iron takes place when the iron actively reduces oxygen in the presence of water. The iron is
oxidized and the oxygen is reduced, forming iron(III) oxide (Fe2O3, which is Fe3+ and O2−). Zinc reduces
oxygen more actively than iron does. When the two metals are connected, the zinc will reduce the O2(g),
preventing the Fe from being oxidized. When all of the Zn(s) is oxidized, the iron will begin to corrode.
14.73
Tin is less active than iron, so it does not prevent corrosion in the same way as a coating with a more active
metal would. An alternative method of preventing corrosion is to prevent contact of an easily corroded
metal with oxygen and moisture. The tin coating on a tin can serves as a barrier to moisture and oxygen.
Paints serve the same purpose.
14.74
Aluminum is more active than iron. As the iron nails begin to rust, the aluminum adjacent to the nails
oxidizes preferentially. As a result, the aluminum that is in contact with the iron weakens, and the siding
fails. A similar fate would probably also be observed if nickel nails were used. However, if aluminum
nails are used, the aluminum siding and nails corrode at the same rate. Note: unlike iron oxide, which
flakes off the metal surface, aluminum oxide forms a stable, protective coating over the metallic aluminum
beneath it. Once the coating forms, aluminum does not continue to corrode as iron does.
14.75
When iron is exposed to oxygen and moisture, it will corrode (rust). Chrome plating protects iron in two
ways. By isolating the iron from oxygen and moisture, it can not corrode. Chromium is also a more active
14−47
metal than iron. If the chromium coating is scratched, the chromium oxidizes preferentially because of its
higher activity. As a result, the iron will not rust.
14.76
When iron is exposed to oxygen and moisture, it will corrode (rust). Galvanizing (zinc coating) protects
iron in two ways. By isolating the iron from oxygen and moisture, it can not corrode. Zinc is also a more
active metal than iron. If the zinc coating is scratched, the zinc oxidizes preferentially because of its higher
activity. As a result, the iron will not rust.
14.77
See Table 14.1
(a) NH3

 

 total positive   total negative 
Net charge 
=
+

 oxidation numbers   oxidation numbers 
NH3
3× H
1× N

 

Net charge = 0
Total positive oxidation numbers = 3 × (+1) = 3
This gives us:


 total negative 
0 = ( 3 )+ 

 oxidation numbers 
N


The nitrogen in NH3 has an oxidation number of 3−. This is also the oxidation number you expect for
nitrogen based on its position in the periodic table. The most electronegative element in a compound
usually takes the oxidation number predicted from the periodic table. We base our assignment of the
oxidation number of H on Rule 4 (see Table 14.1).
(b) N2H4


 
 total positive   total negative 
Net charge 
=

+
 oxidation numbers   oxidation numbers 
N2H4
4×H
2× N

 

Net charge = 0
Total positive oxidation numbers = 4 × (+1) = 4
This gives us:


 total negative 
0 = ( 4 )+ 

 oxidation numbers 
2× N


N=
−4
= −2
2
The nitrogen in N2H4 has an oxidation number of 2−.
(c) NF3
14−48

 

 total positive   total negative 
Net charge 
=
+

 oxidation numbers   oxidation numbers 
NF3
N
3× F

 

Net charge = 0
Total negative oxidation numbers = 3 × (−1) = −3
This gives us:


 total positive 
0= 
 + ( –3)
 oxidation numbers 
N


The nitrogen in NF3 has an oxidation number of 3+.
(d) NH2OH

 

 total positive   total negative 
Net charge 
=
+

 oxidation numbers   oxidation numbers 
NH 2 OH
N + 3× H
1× O

 

Net charge = 0
Total negative oxidation numbers = 1 × (−2) = −2
Total positive oxidation numbers = N + 3 × (+1) = N + 3
This gives us:


 total positive 
0= 
 + ( –2 )
 oxidation numbers 
N+3


0=N+1
The nitrogen in NH2OH has an oxidation number of 1−.
Note that even though we assumed at first that the nitrogen had a positive oxidation number, this
assumption did affect the fact that the oxidation number turned out to be 1−.
(e) Fe(NO3)3 – The easiest way to calculate the oxidation number N in iron(III) nitrate is to recognize that
nitrate ion has a chemical formula of NO3−. The oxidation number of nitrogen will be the same in
nitrate as it is in compounds containing nitrate.

 

 total positive   total negative 
Net charge 
=
+

oxidation numbers   oxidation numbers 
−

NO3
N
3× O

 

Net charge = −1
Total negative oxidation numbers = 3 × (−2) = −6
This gives us:
14−49


 total positive 
−1 = 
 + ( –6 )
 oxidation numbers 
N


N=5
The oxidation number of nitrogen in NO3− is 5+.
(f) HNO2

 

 total positive   total negative 
Net charge 
=
+

 oxidation numbers   oxidation numbers 
HNO 2
2×O
 1× N + 1× H  

Net charge = 0
Total negative oxidation numbers = 2 × (−2) = −4
Total positive oxidation numbers = N + 1
This gives us:


 total positive 
0= 
 + ( –4 )
 oxidation numbers 
N +1


N=3
The oxidation number of nitrogen in HNO2 is 3+.
14.78
The most active metal will displace all other metal ions from solution. According to the table, a piece of
manganese metal (listed at the top of the table), will displace all other metal ions from solution. Mn(s) is
the most active metal. Fe(s) displaces all metal ions except Mn2+ (Mn(NO3)2(aq)), so Fe(s) is the next most
active metal. Ni(s) displaces all metal ions except Mn2+ and Fe2+ (Fe(NO3)2(aq)), so Ni(s) is the next most
active. The completed list is: (most active) Mn > Fe > Ni > Ag > Hg (least active).
14.79
(a) Answer: 2H2O(l) + 2OH−(aq) + 2CoCl2(s) + Na2O2(aq) → 2Co(OH)3(s) + 4Cl−(aq) + 2Na+(aq)
The oxidizing agent is Na2O2 (oxygen is being reduced from 1− to 2−).
The reducing agent is CoCl2 (cobalt is being oxidized from 2+ to 3+).
Solution: First, we determine the oxidation numbers of each atom and separate the half-reactions.
Because chloride is part of CoCl2, we include chloride in that half-reaction. Sodium is part of the
peroxide half-reaction.
2+
1–
CoCl2(s) + Na2O2(aq)
3+ 2–
→
Co(OH)3(s) + Cl–(aq) + Na+(aq)
determine oxidation numbers
The oxidizing agent is Na2O2 (oxygen is being reduced from 1− to 2−).
The reducing agent is CoCl2 (cobalt is being oxidized from 2+ to 3+).
The peroxide ion is reduced to hydroxide under basic conditions, so the reduction half-reaction is:
Na2O2 (aq) → OH−(aq) + Na+(aq)
For the oxidation reaction, the reaction is:
CoCl2(s) → Co(OH)3(s) + Cl−(aq)
14−50
Balance oxidation half-reaction:
CoCl2(s) → Co(OH)3(s) + 2Cl−(aq)
balance atoms
CoCl2(s) → Co(OH)3(s) + 2Cl−(aq) + e−
add e− to balance oxidation numbers
3OH−(aq) + CoCl2 (s) → Co(OH)3(s) + 2Cl−(aq) + e−
add OH− to balance charge
Balance reduction half-reaction:
Na2O2 (aq) → 2OH−(aq) + 2Na+(aq)
balance atoms
2e− + Na2O2(aq) → 2OH−(aq) + 2Na+(aq)
add e− to balance oxidation numbers
2e− + Na2O2(aq) → 2OH−(aq) + 2Na+(aq) + 2OH−(aq)
add OH− to balance charge
2e− + 2H2O(l) + Na2O2 (aq) → 4OH−(aq) + 2Na+(aq)
add H2O to balance atoms
Balance the electrons between the two half-reactions:
2 × [3OH−(aq) + CoCl2(s) → Co(OH)3(s) + 2Cl−(aq) + e−]
1 × [2e− + 2H2O(l) + Na2O2 (aq) → 4OH−(aq) + 2Na+(aq)]
6OH−(aq) + 2CoCl2(aq) → 2Co(OH)3(s) + 4Cl−(aq) + 2e−
2e− + 2H2O(l) + Na2O2(aq) → 4OH−(aq) + 2Na+(aq)
6OH−(aq) + 2CoCl2(aq) → 2Co(OH)3(s) + 4Cl−(aq) + 2e –
2e – + 2H2O(l) + Na2O2(aq) → 4OH−(aq) + 2Na+(aq)
6OH−(aq) + 2H2O(l) + 2CoCl2(s) + Na2O2(aq) → 2Co(OH)3(s) + 2Na+(aq) + 4Cl−(aq) + 4OH−(aq)
Eliminate duplicate substances (bold type):
2OH−(aq) + 2H2O(l) + 2CoCl2(s) + Na2O2(aq) → 2Co(OH)3(s) + 2Na+(aq) + 4Cl−(aq)
(b) Answer: 2OH−(aq) + Bi2O3(s) + 2ClO−(aq) → 2BiO3−(aq) + 2Cl−(aq) + H2O(l)
The oxidizing agent is ClO− (chlorine is reduced from 1+ to 1−).
The reducing agent is Bi2O3 (bismuth is oxidized from 3+ to 5+).
Solution:
3+
5+
1+
Bi2O3(s) + ClO–(aq)
→
1–
BiO3–(aq) + Cl–(aq)
determine oxidation numbers
Balance oxidation half-reaction:
Bi2O3(s) → 2BiO3−(aq)
balance atoms
Bi2O3(s) → 2BiO3 (aq) + 4e
add e− to balance oxidation numbers
6OH− + Bi2O3(s) → 2BiO3−(aq) + 4e−
add OH− to balance charge
6OH− + Bi2O3(s) → 2BiO3−(aq) + 3H2O(l) +4e−
add H2O to balance atoms
−
−
Balance reduction half-reaction:
2e− + ClO−(aq) → Cl−(aq)
add e− to balance oxidation numbers
2e− + ClO−(aq) → Cl−(aq) + 2OH−(aq)
add OH− to balance charge
2e− + H2O(l) + ClO−(aq) → Cl−(aq) + 2OH−(aq)
add H2O to balance atoms
14−51
Equalize the number of electrons exchanged between the two half-reactions:
1 × [6OH− + Bi2O3(s) → 2BiO3−(aq) + 3H2O(l) +4e−]
2 × [2e− + H2O(l) + ClO−(aq) → Cl−(aq) + 2OH−(aq)]
6OH− + Bi2O3(s) → 2BiO3−(aq) + 3H2O(l) +4e−
4e− + 2H2O(l) + 2ClO−(aq) → 2Cl−(aq) + 4OH−(aq)
6OH− + Bi2O3(s) → 2BiO3−(aq) + 3H2O(l) + 4e –
4e – + 2H2O(l) + 2ClO−(aq) → 2Cl−(aq) + 4OH−(aq)
2H2O(l) + 6OH−(aq) + Bi2O3(s) + 2ClO−(aq) → 2BiO3−(aq) + 2Cl−(aq) + 3H2O(l) + 4OH−(aq)
Eliminate duplicate substances (bold type):
2OH−(aq) + Bi2O3(s) + 2ClO−(aq) → 2BiO3−(aq) + 2Cl−(aq) + H2O(l)
14.80
Pt(s) + NO3−(aq) + Cl−(aq) → PtCl42−(aq) + NO2(g)
0
5+
2+
Pt(s) + NO3–(aq) + Cl–(aq)
→
4+
PtCl42–(aq) + NO2(g )
determine oxidation numbers
Balance oxidation half-reaction:
Pt(s) + 4Cl−(aq) → PtCl42−(aq)
−
Pt(s) + 4Cl (aq) → PtCl4 (aq) + 2e
2−
balance atoms
add e− to balance oxidation numbers
−
Balance reduction half-reaction:
e− + NO3−(aq) → NO2(g)
add e− to balance oxidation numbers
e− + 2H+(aq) + NO3−(aq) → NO2(g)
add H+ to balance charge
e− + 2H+(aq) + NO3−(aq) → NO2(g) + H2O(l)
add H2O to balance atoms
Equalize the number of electrons exchanged between the two half-reactions
1 × [Pt(s) + 4Cl−(aq) → PtCl42−(aq) + 2e−]
2 × [e− + 2H+(aq) + NO3−(aq) → NO2(g) + H2O(l)]
Pt(s) + 4Cl−(aq) → PtCl42−(aq) + 2e−
2e− + 4H+(aq) + 2NO3−(aq) → 2NO2(g) + 2H2O(l)
Pt(s) + 4Cl−(aq) → PtCl42−(aq) + 2e –
2e – + 4H+(aq) + 2NO3−(aq) → 2NO2(g) + 2H2O(l)
4H+(aq) + Pt(s) + 2NO3−(aq) + 4Cl−(aq) → PtCl42−(aq) + 2NO2(g) + 2H2O(l)
14.81
NH4+(aq) + NO3−(aq) → N2O(g) + 2H2O(l)
The oxidizing agent is NO3− (N is reduced from 5+ to 1+).
The reducing agent is NH4+ (N is oxidized from 3− to 1+).
14−52
5+
3–
1+
→
NH4+(aq) + NO3–(aq)
N2O(g) + H2O(l)
determine oxidation numbers
Balance oxidation half-reaction:
2NH4+(aq) → N2O(g)
balance atoms being oxidized or reduced
2NH4 (aq) → N2O(g) + 8e
+
add e− to balance oxidation numbers
−
2NH4+(aq) → N2O(g) + 10H+(aq) + 8e−
add H+ to balance charge
H2O(l) + 2NH4+(aq) → N2O(g) + 10H+(aq) + 8e−
add H2O to balance atoms
Balance reduction half-reaction:
2NO3−(aq) → N2O(g)
balance atoms being oxidized or reduced
8e− + 2NO3−(aq) → N2O(g)
add e− to balance oxidation numbers
8e− + 10H+(aq) + 2NO3−(aq) → N2O(g)
add H+ to balance charge
8e− + 10H+(aq) + 2NO3−(aq) → N2O(g) + 5H2O(l)
add H2O to balance atoms
Equalize the number of electrons exchanged between the two half-reactions
H2O(l) + 2NH4+(aq) → N2O(g) + 10H+(aq) + 8e –
8e – + 10H+(aq) + 2NO3−(aq) → N2O(g) + 5H2O(l)
H2O(l) + 10H+(aq) + 2NH4+(aq) + 2NO3−(aq)→ 2N2O(g) + 5H2O(l) + 10H+(aq)
Eliminate species appearing on both sides of the equation (bold type) and divide coefficients by 2:
NH4+(aq) + NO3−(aq) → N2O(g) + 2H2O(l)
14.82
The oxidation number we assign to the carbon atoms in propane is not an integer. The value (−2 2/3, or
–8/3) reflects the fact that two of the carbons are bonded to three hydrogens and one of the carbons is
bonded to two hydrogens. In this reaction carbon is oxidized (from −2 2/3 to 4+) so C3H8 is the reducing
agent. Oxygen is reduced (from 0 to 2−) so O2 is the oxidizing agent.
–8/3
1+
0
C3H8(g) + O2(g)
4+
→
2–
1+
2–
CO2(g) + H2O(g)
14.83
The element which is highest on the activity series is the best reducing agent (Figure 14.22).
(a) Al; (b) Zn; (c) Mn; (d) Mg
14.84
Gold is at the bottom of the activity series and, therefore, is not likely to corrode.
14.85
Brass is an alloy of copper and zinc. Because sweat is generally acidic, it facilitates the oxidation of the
copper from Cu to Cu2+ (remember acids taste sour and corrode metals). There are several copper
compounds that are green (including CuCO3).
14.86
All solutions are electrically neutral (the total positive and negative charges are always equal). Magnesium
is a more active metal than copper, so the spontaneous reaction involves the oxidation of magnesium and
the reduction of copper. To understand what is happening in the salt bridge we will first focus on the
magnesium half-cell. As the magnesium electrode oxidizes, the concentration of magnesium ions (and
therefore, the positive charge) in the solution increases. Because a solution must be electrically neutral, one
of two processes can occur to counteract the increase of positive charge in the magnesium half-cell:
14−53
positive ions can flow out of the magnesium half-cell or negative ions can flow into the magnesium halfcell (to cancel the positive charge of the added magnesium ions). The ions needed to maintain electrical
neutrality travel through the salt bridge. Sulfate ions from Na2SO4 flow into the magnesium half-cell.
Similarly, as copper is reduced in the copper half-cell, the positive charge in the solution surrounding the
copper electrode decreases. To compensate for this, cations (Na+ in Na2SO4) flow into the copper half-cell.
Notice that the effects of the copper and magnesium half-cells compliment each other.
14.87
Cu(s) + H2SO4(aq) → Cu2+(aq) + SO2(g)
The copper half-reaction is balance simply by balancing charge:
Cu(s) → Cu2+(aq) + 2e−
H2SO4(aq) → SO2(g)
add electrons
H2SO4(aq) + 2e− → SO2(g)
add H+
H2SO4(aq) + 2H+(aq) + 2e− → SO2(g)
+
−
H2SO4(aq) + 2H (aq) + 2e (g) → 2H2O(l) + SO2
add H2O
Then we add the two half-reactions and cancel the two electrons that appear on both sides of the equation.
H2SO4(aq) + 2H+(aq) + 2e – → 2H2O(l) + SO2(g)
Cu(s) → Cu2+(aq) + 2e –
H2SO4(aq) + 2H (aq) + Cu(s) → 2H2O(l) + SO2(g) + Cu2+(aq)
+
14.88
The maximum and miniumum oxidation numbers of an element are based on the element’s position in the
periodic table. Bromine has seven valence electrons and fills its valence shell by picking up one electron.
The next electron would have to enter the next higher energy level (5s). Thus, Br− cannot be reduced
because, in the process it would have to gain one or more electrons and this is not energetically feasible.
Bromide ions can be oxidized because they can lose anywhere from one to eight valence electrons.
Manganese has seven electrons in its highest energy levels (4s and 3d electrons), so it can reach an
oxidation state of 7+ in compounds (as in MnO4−), but going to the next higher oxidation state would
require removing electrons from electrons in the more stable (2p) energy level. The manganese in MnO4−
can be reduced by adding up to seven electrons.
14.89
(a) The reducing agent is the substance that loses electrons. The oxidation state of Tl+ increases to Tl3+.
This is represented by the oxidation half-reaction:
Tl+(aq) → Tl3+(aq) + 2e−
Since Tl+ is giving up electrons, it is being oxidized and it is the reducing agent.
Ce4+(aq) + e− → Ce3+(aq)
Since Ce4+ is gaining electrons, it is being reduced and it is the oxidizing agent.
(b) The nitrogen in the reactant NO3− has an oxidation state of 5+, and in the product NO2 the oxidation
state of nitrogen is 4+. The nitrogen is being reduced from 5+ to 4+.
The reduction half-reaction is NO3−(aq) + 2H+(aq) + e− → NO2(g) + H2O(l).
Nitrate, NO3−, is the oxidizing agent.
The oxidation state of bromine in the Br− reactant is 1−, and in the Br2 product the oxidation state of Br
is zero. Bromine is being oxidized from 1− to 0.
The oxidation half-reaction is represented by 2Br−(aq) → Br2(l) + 2e−.
Bromide ion, Br−, is the reducing agent.
14.90
Iron is a more active metal than copper. Consequentially, when both are placed in contact with each other,
the iron acts as a sacrificial anode. The iron will constantly be oxidized to keep the copper in its reduced
form. As a result, the iron pipe will eventually develop holes as the metal is oxidized away.
14−54
Balancing rules for redox reactions (condensed from text):
1) Separate the half-reactions.
2) For each half-reaction:
i) Balance the atom being oxidized or reduced (redox atom).
ii) Calculate oxidation number change of redox atom.
iii) Add electrons to balance oxidation number change.
iv) Add H+ to balance charge in acid solution, or OH− to balance charge in basic solution.
v) Add water to balance atoms.
3) Equalize the number of electrons exchanged in the oxidation and reduction half-reactions by
multiplying half-reactions appropriately.
4) Add half-reactions, and eliminate duplicate substances.
5) Reduce coefficients if necessary.
14.91
The activity series shows that copper is a more active metal than silver.
(a) The higher the metal on the activity series, the more easily it is oxidized, so copper is more easily
oxidized the silver.
(b) The lower a metal on the activity series, the more easily its cation form is reduced back to elemental
form. Since silver is lower than copper on the activity series, Ag+ is more easily reduced than Cu2+.
(c) Since Cu is more easily oxidized than Ag, and Ag+ is more easily reduced than Cu2+, then we would
expect the following reaction to be spontaneous:
Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s)
14.92
balanced
In the reaction shown, carbon in CH3CH2OH increases in oxidation number from 2- to 1- while the
oxidation numbers on oxygen (2- ) and hydrogen (1+) remain unchanged. The PCC reagent causes the
oxidation of carbon in ethanol so PCC is an oxidizing agent.
O
O
+
C5H5 NHCrrO3 Cl +
(PCC)
CH3CH2OH
+ C5H5 NHCl +
+ H2 CrrO3
+
C
H3C
C
H
Etthaannool)
(E
H
H
(Etthaannaal)
14.93
2CH3CH2OH + 2Na → 2Na+CH3CH2O− + H2
In this reaction Na is converted to Na+. Its oxidation number increases from zero to 1+ as a result of losing
an electron, so it is oxidized and is therefore the reducing agent. The carbons in ethanol, CH3CH2OH, and
most of the hydrogens do not undergo a change in oxidation number but the hydrogen from the –OH group
that converts to H2 is reduced from an oxidation number of 1+ to zero. Because the hydrogen being
reduced was in the ethanol reactant, then CH3CH2OH is the oxidizing agent.
14.94
4CH3SH(g) + O2(g) → 2CH3S−SCH3(l) + 2H2O(l)
(a) Carbon’s oxidation number increases from 2− to 1−, so carbon is oxidized.
(b) Oxygen’s oxidation number decreases from zero to 2−, so oxygen is reduced.
(c) The oxidizing agent is the reactant containing the element being reduced, so O2 is the oxidizing agent.
(d) The reducing agent is the reactant containing the element being oxidized, so CH3SH is the reducing
agent.
14.95
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
(a) Carbon’s oxidation number increases from 4− to 4+ so carbon is oxidized.
(b) Oxygen’s oxidation number decreases from zero to 2− so oxygen is reduced.
(c) Carbon’s oxidation number increases by 8 so for each carbon that is oxidized, 8 electrons are
transferred to oxygen. Scaling up to mole scale and taking into account that there is one mole of
carbon in every mole of CH4, 8 moles of electrons are transferred for every mole of CH4 that reacts.
14−55
14.96
Each of these compounds contains an element in its highest possible oxidation state: The oxidation state for
chromium in K2CrO4 is 6+. The oxidation state for manganese in KMnO4 is 7+. The oxidation state for
sulfur in H2SO4 is 6+.
14.97.
The unbalanced equation is: Br2(aq) → Br−(aq) + BrO−(aq). In this reaction, one bromine atom in Br2 is
oxidized while the other one is reduced.
Bromine, Br2, is therefore the reactant in both half-reactions:
Br2(aq) → Br−(aq)
and
Br2(aq) → BrO−(aq)
Starting with the first half-reaction (reduction), we first balance the bromine atoms:
Br2(aq) → 2Br−(aq)
In this half-reaction, each bromine atom’s oxidation number changes from zero to 1−, and there are two
bromine atoms reacting so it is a two-electron reduction:
2e− + Br2(aq) → 2Br−(aq)
There are no other elements, so this half-reaction is balanced. The number of bromine atoms and the
charge is the same on both sides of the equation.
In the oxidation half-reaction, we must also start by balancing bromine atoms:
Br2(aq) → 2BrO−(aq)
In this half-reaction, each bromine atom’s oxidation number changes from zero to 1+, and there are two
bromine atoms reacting so it is a two-electron oxidation:
Br2(aq) → 2BrO−(aq) + 2e−
Now we have to balance the charge. The total charge on the left is zero, and the total charge on the right is
4−. When in basic solution, this reaction occurs in the presence of hydroxide ions, so we can balance
charge with OH- ions:
4OH−(aq) + Br2(aq) → 2BrO−(aq) + 2e−
Now that the charges are balanced, the last step is to balance the hydrogen and oxygen atoms by adding
H2O molecules. There are four H atoms and four O atoms on the left and two O atoms on the right, so
there is an excess of four H atoms and two O atoms on the left. If we add two H2O molecules to the right,
the half-reaction is balanced:
4OH−(aq) + Br2(aq) → 2BrO−(aq) + 2e− + 2H2O(l)
Now that we have both half reactions balanced, we can combine them:
2e− + Br2(aq) → 2Br−(aq)
4OH−(aq) + Br2(aq) → 2BrO−(aq) + 2e− + 2H2O(l)
The number of electrons in both half-reactions are the same (electrons lost equal electrons gained), the
electrons will cancel when we combine them:
14−56
−
−
2e + Br2(aq) + 4OH−(aq) + Br2(aq) → 2Br−(aq) + 2BrO−(aq) + 2e + 2H2O(l)
Nothing else cancels but we must combine the two Br2(aq):
2Br2(aq) + 4OH−(aq) → 2Br−(aq) + 2BrO−(aq) + 2H2O(l)
All coefficients can be divided by 2 to give to give the balanced euquation:
Br2(aq) + 2OH−(aq) → Br−(aq) + BrO−(aq) + H2O(l)
14.98
The oxidation half-reaction occurs at the anode and the reduction half-reaction occurs at the cathode.
Electrons are products in oxidation reactions. Electrons are reactants in reduction reactions.
(a) Anode: Al(s) → Al3+(aq) + 3e−; Cathode: 2e− + 2H+(aq) → H2(g)
(b) Anode: Zn(s) → Zn2+(aq) + 2e−; Cathode: 3e− + Fe3+(aq) → Fe(s)
14.99
In the activity series, the element higher in the series is more active and therefore more easily oxidized.
(a) The activity series shows that aluminum is more active than zinc, so aluminum is more easily oxidized
than zinc. When these two half-reactions are combined in a voltaic cell, the aluminum half-reaction
would occur at the anode and the zinc half-reaction would occur at the cathode:
Anode: Al(s) → Al3+(aq) + 3e−; Cathode: 2e− + Zn2+(aq) → Zn(s)
(b) Chromium is more active than silver so the chromium half-reaction will occur at the anode and the
silver half-reaction will occur at the cathode:
Anode: Cr(s) → Cr3+(aq) + 3e−; Cathode: e− + Ag+(aq) → Ag(s)
(c) Cadmium is more active than hydrogen so the cadmium half-reaction will occur at the anode and the
hydrogen half-reaction will occur at the cathode:
Anode: Cd(s) → Cd2+(aq) + 2e−; Cathode: 2e− + 2H+(aq) → H2(g)
14.100 (a)
(b)
(c)
(d)
K and Cl2
Mg and I2
K and Br2
H2 and O2
CONCEPT REVIEW
14.101 Answer: D; The overall equation for this reaction is: Zn(s) + Co(NO3)2(aq) → Zn(NO3)2(aq) + Co(s). The
net ionic equation is: Zn(s) + Co2+(aq) → Zn2+(aq) + Co(s). During this reaction the Co2+ ion becomes
elemental cobalt, which has no charge. To do this each Co2+ ion must gain two electrons, which it obtains
from a zinc atom. Losing two electrons causes a zinc atom to be oxidized so Co2+ is the oxidizing agent.
A.
B.
C.
E.
Co2+ ions are reduced to form solid cobalt.
Zn atoms are oxidized to form Zn2+ ions.
Nitrate ions are spectator ions in this reaction.
The solution will become a lighter pink as more Co2+ ions react to form solid cobalt.
14.102 Answer: C; Following the rules for assigning oxidation numbers, each oxygen atom in SO2 has an
oxidation number of 2−. The total of the positive oxidation numbers plus the total negative oxidation
numbers equals the net charge, which in this case is 0:
14−57

 

 total positive   total negative 
net charge 
=
+

 oxidation numbers   oxidation numbers 
SO 2
1× S
2×O

 



 total positive 
=
0 
 + ( –4 )
 oxidation numbers 
S


The oxidation number of sulfur in SO2 is 4+.
A. The oxidation number of Na, an alkali metal is 1+ and the net charge is zero:

 

 total positive   total negative 
net charge 
=
+

 oxidation numbers   oxidation numbers 
Na 2S
2 × Na
1× S

 



 total negative 
0= ( +2 ) + 

 oxidation numbers 
S


The oxidation number of sulfur in Na2S is 2−, consistent with what is predicted from sulfur’s position
on the periodic table.
B. The oxidation number of oxygen is 2− and the net charge is zero:

 

 total positive   total negative 
net charge 
=
+

 oxidation numbers   oxidation numbers 
SO3
1× S
3× O

 



 total positive 
=
0 
 + ( –6 )
 oxidation numbers 
S


The oxidation number of sulfur in SO3 is 6+.
D. The oxidation numbers of oxygen and hydrogen are 2− and 1+, respectively. The net charge is zero.
With a total of negative oxidation numbers of 8− and contribution of 2+ from both hydrogens, sulfur
has a positive oxidation number in this compound:

 

 total positive   total negative 
+
net charge 
=


  oxidation numbers 
oxidation
numbers
H 2SO 4
 ( 2 × H ) + (1 × S)  
4×O





 total positive 
 + ( –8 )
=
0 
 oxidation numbers 


( +2 ) + S


The oxidation number of sulfur in H2SO4 is 6+.
E. The oxidation number of oxygen is 2− and the net charge is zero, so the oxidation number of sulfur in
SO is 2+.
14.103 Answer: D; We begin by assigning oxidation numbers:
14−58
–
2–
0
+
6+
+ 1166H
+ 22Crr2O72––( aaq ) +
H++( aaq )
H( aaq ) +
H5O H
33C2H
→
→
3++
+ 1111H
+ 44Crr3++( aaq ) +
H2O (l )
H( g ) +
H3CO 2H
3 CH
During this reaction each chromium in Cr2O72− goes from an oxidation number of 6+ to 3+. To do this
each chromium must gain three electrons, which it obtains from the carbon in C2H5OH. Losing electrons
causes carbon to be oxidized so Cr2O72− is the oxidizing agent.
A.
B.
C.
E.
The oxidation number for chromium changes from 6+ to 3+ .
The oxidation number for oxygen does not change from 2−.
The reducing agent is C2H5OH.
For every ion of Cr2O72− that reacts, six electrons are transferred.
14.104 Answer: A; In this voltaic cell, the oxidation number for magnesium increases from 0 to 2+. An increase in
oxidation number is oxidation, a process that occurs at the anode.
B.
C.
D.
E.
Aqueous Cu2+ ions are reduced to form solid copper at the cathode.
Positive ions from the salt bridge flow into the cathode half-cell.
Negative ions from the salt bridge flow into the anode half-cell.
Electrons are transferred through a wire from the magnesium electrode to the copper electrode.
14.105 Answer: B and D; In this reaction, cadmium is oxidized, a process that occurs at the anode. Since this halfcell is designated to be on the left, solution B should consist of aqueous Cd2+ ions. Silver is reduced in this
reaction, so this process will occur at the cathode, the half-cell on the right. Solution D should consist of
aqueous Ag+ ions.
A. cadmium metal electrode
C. this component, the salt bridge, is necessary for any voltaic cell
E. silver metal electrode
14.106 Answer: C; From the description, it can be concluded that 5.0 g of copper reacted.
1 mol Cu
mol Cu reacted = 5.0 g Cu ×
= 0.0787 mol Cu
63.55 g Cu
For each Cu atom that reacts, two electrons are transferred. Or, for each mole of Cu that reacts, two moles
of electrons are transferred:
2 mol electrons
mol electrons transferred = 0.0787 mol Cu ×
= 0.157 mol of electrons
1 mol Cu
A. 17.0 g of solid silver was produced at the cathode.
g Ag produced = 0.0787 mol Cu ×
2 mol Ag
×
107.9 g Ag
= 17.0 g Ag
1 mol Cu
1 mol Ag
B. 0.157 mol of solid silver was produced at the cathode.
mol Ag produced = 0.0787 mol Cu ×
2 mol Ag
= 0.157 mol Ag
1 mol Cu
D. 5.0 g of solid copper was oxidized.
E. The anode solution becomes a deeper blue color as aqueous Cu2+ ions form from solid copper.
14.107 Answer: E; We begin by assigning oxidation numbers:
14−59
0
+
2+
+
4+
→ 22P
+ 22H
+P
+ 22H
H2SO 4(aaq) →
Pb O2( ss) +
H2O (l)
Pb SO 4(aaq) +
PPb( ss) +
As the battery discharges, H2SO4 reacts so its concentration decreases.
A.
B.
C.
D.
The oxidizing agent is PbO2(s).
The lead in PbO2 is reduced.
Pb(s) is the reducing agent.
The pH increases as the battery discharges.
14.108 Answer: C; We begin by assigning oxidation numbers:
0
+
5+
5++
+N
NO3––(aaq)
II2( ss) +
+
4+
→ H
+N
HIIO3(aaq) +
NO2(g )
→
Iodine is oxidized and nitrogen is reduced in this reaction. We’ll begin with the oxidation half-reaction:
Balance oxidation half-reaction:
I2(s) → 2HIO3(aq)
balance atoms
I2(s) → 2HIO3(aq) + 10e−
add e− to balance oxidation numbers
I2(s) → 2HIO3(aq) + 10H+(aq) + 10e−
add H+ to balance charge
I2(s) + 6H2O(l) → 2HIO3(aq) + 10H+(aq) + 10e−
add H2O to balance oxygen and hydrogen
Balance reduction half-reaction:
NO3−(aq) + e− → NO2(g)
add e− to balance oxidation numbers
NO3−(aq) + 2H+(aq) + e− → NO2(g)
add H+ to balance charge
NO3−(aq) + 2H+(aq) + e− → NO2(g) + H2O(l)
add H2O to balance oxygen and hydrogen
Balance the electrons between the two half-reactions:
1 × [I2(s) + 6H2O(l) → 2HIO3(aq) + 10H+(aq) + 10e−]
10 × [NO3−(aq) + 2H+(aq) + e− → NO2(g) + H2O(l)]
I2(s) + 6H2O(l) → 2HIO3(aq) + 10H+(aq) + 10e –
10NO3−(aq) + 20H+(aq) + 10e – → 10NO2(g) + 10H2O(l)
I2(s) + 10NO3−(aq) + 10H+(aq) → 2HIO3(aq) + 10NO2(g) + 4H2O(l)
H+(aq) is on the left side of the equation and has a coefficient of 10.
A.
B.
D.
E.
H2O(l) is on the right side of the equation and has a coeffi cient of 4.
The coefficient for NO2 is 10.
The coefficient for NO3− is 10.
Ten electrons are transferred for every I2 molecule that reacts.
14.109 Answer: C; Of these metals, magnesium is highest on the activity series (Figure 14.22). This means that it
is the most active and capable of reducing the other metals, itself being oxidized. The order of increasing
reducing strength is D < B < E < A < C
14.110 Answer: E; When molten NaCl is electrolyzed, the following reaction occurs:
2NaCl(l) → 2Na(s) + Cl2(g)
14−60
A.
B.
C.
D.
The cathode reaction is Na+(l) + e− → Na(l).
The anode reaction is 2Cl−(l) → Cl2(g) + 2e−.
Cl− is oxidized.
Na+ is reduced.
14−61