Chapter 5
Practice Exercises
5.1
2Na(s) + O2(g) J Na2O2(s)
Oxygen is reduced since it gains electrons.
Sodium is oxidized since it loses electrons.
5.2
2Al(s) + 3Cl2(g) J 2AlCl3(aq)
Aluminum is oxidized and is, therefore, the reducing agent.
Chlorine is reduced and is, therefore, the oxidizing agent.
5.3
ClO2–: O –2 Cl +3
5.4
(a) Ni +2; Cl –1
(b) Mg +2; Ti +4; O –2
(c) K +1; Cr +6; O –2
(d) H +1; P +5, O –2
(e) V +3; C 0; H +1; O –2
5.5
There is a total charge of +8, divided over three atoms, so the average charge is +8/3.
5.6
First the oxidation numbers of all atoms must be found.
N2O5 + 3NaHCO3 J 2NaNO3 + 2CO2 + H2O
Reactants:
N = +5
O = –2
Products:
N = +5
O = –2
Na = +1
Na = +1
H = +1
C = +4
O = –2
C = +4
O = –2
H = +1
O = –2
None of the oxidation numbers change, therefore it is not an oxidation reaction.
KClO3 + 3HNO2 J KCl + 3HNO3
Reactants:
K = +1
Cl = +5
O = –2
Products:
K = +1
Cl = –1
H = +1
N = +3
O = –2
H = +1
N = +5
O = –2
The oxidation numbers for K and Na do not change. However, the oxidation numbers for all chlorines
atoms decrease. The oxidation numbers for nitrogen increase.
Therefore, KClO3 is reduced and HNO2 is oxidized.
This means KClO3 is the oxidizing agent and HNO2 is the reducing agent.
This reaction is the redox reaction. In the other reaction, the oxidation numbers of the atoms do not
change.
97
Chapter 5
5.7
First the oxidation numbers of all atoms must be found.
Cl2 + 2NaClO2 J 2ClO2 + 2NaCl
Reactants:
Cl = 0
Products:
Cl = +4
O = –2
Na = +1
Cl = +3
O = –2
Na = +1
Cl = –1
The oxidation numbers for O and Na do not change. However, the oxidation numbers for all chlorine
atoms change. There is no simple way to tell which chlorines are reduced and which are oxidized in this
reaction.
One analysis would have the Cl in Cl2 end up as the Cl in NaCl, while the Cl in NaClO2 ends up as the Cl
in ClO2. In this case Cl2 is reduced and is the oxidizing agent, while NaClO2 is oxidized and is the
reducing agent.
5.8
If H2O2 acts as an oxidizing agent, it gets reduced itself in the process. Examining the oxidation numbers:
H2O2 H = +1, O = –1
H2O
H = +1, O = –2
O2
O=0
If H2O2 is reduced it must form water, since the oxidation number of oxygen drops from –1 to –2 in the
formation of water (a reduction).
The product is therefore water.
5.9
Al(s) + Cu2+(aq) J Al3+(aq) + Cu(s)
First, we break the reaction above into half-reactions:
Al(s) J Al3+(aq)
Cu2+(aq) J Cu(s)
Each half-reaction is already balanced with respect to atoms, so next we add electrons to balance the
charges on both sides of the equations:
Al(s) J Al3+(aq) + 3e–
2e– + Cu2+(aq) J Cu(s)
Next, we multiply both equations so that the electrons gained equals the electrons lost,
2(Al(s) J Al3+(aq) + 3e–)
3(2e– + Cu2+(aq) J Cu(s))
which gives us:
2Al(s) J 2Al3+(aq) + 6e–
6e– + 3Cu2+(aq) J 3Cu(s)
Now, by adding the half-reactions back together, we have our balanced equation:
2Al(s) + 3Cu2+(aq) J 2Al3+(aq) + 3Cu(s)
5.10
TcO4− + Sn2+ J Tc4+ + Sn4+
First, we break the reaction above into half-reactions:
TcO4– J Tc4+
Sn2+ J Sn4+
Each half-reaction is already balanced with respect to atoms other than O and H, so next we balance the O
atoms by using water:
TcO4– J Tc4+ + 4H2O
Sn2+ J Sn4+
98
Chapter 5
Now we balance H by using H+:
8H+ + TcO4– J Tc4+ + 4H2O
Sn2+ J Sn4+
Next, we add electrons to balance the charges on both sides of the equations:
3e– + 8H+ + TcO4– J Tc4+ + 4H2O
Sn2+ J Sn4+ + 2e–
We multiply the equations so that the electrons gained equals the electrons lost,
2(3e– + 8H+ + TcO4– J Tc4+ + 4H2O)
3(Sn2+ J Sn4+ + 2e–)
which gives us:
6e– + 16H+ + 2TcO4– J 2Tc4+ + 8H2O
3Sn2+ J 3Sn4+ + 6e–
Now, by adding the half-reactions back together, we have our balanced equation:
3Sn2+ + 16H+ + 2TcO4– J 2Tc4+ + 8H2O + 3Sn4+
5.11
(Cu J Cu2+ + 2e–) × 4
2NO3– + 10H+ + 8e– J N2O + 5H2O
4Cu + 2NO3– + 10H+ J 4Cu2+ + N2O +5H2O
5.12
2H2O + SO2 J SO42– + 4H+ + 2e–
4OH– + 2H2O + SO2 J SO42– + 4H+ + 2e– + 4OH–
4OH– + 2H2O + SO2 J SO42– + 2e– + 4H2O
4OH– + SO2 J SO42– + 2e– + 2H2O
5.13
(MnO4– + 4H+ + 3e– J MnO2 + 2H2O) × 2
(C2O42– + 2H2O J 2CO32– + 4H+ + 2e–) × 3
2MnO4– + 3C2O42– + 2H2O J 2MnO2 + 6CO32– + 4H+
Adding 4OH– to both sides of the above equation we get:
2MnO4– + 3C2O42– + 2H2O + 4OH– J 2MnO2 + 6CO32– + 4H2O
which simplifies to give:
2MnO4– + 3C2O42– + 4OH– J 2MnO2 + 6CO32– + 2H2O
5.14
Zn + H+ J Zn2+ + H2
Divide the reaction into two half reactions and balance the number of atoms
Zn J Zn2+
2H+ J H2
Balance the charges with electrons
Zn J Zn2+ + 2e–
2H+ + 2e– J H2
5.15
HNO3 + Mg J NH4+ + Mg2+
Divide the reaction into two half reactions and balance the number of atoms
HNO3 J NH4+
Mg J Mg2+
HNO3 J NH4+ + 3H2O
Mg J Mg2+
HNO3 + 9H+ J NH4+ + 3H2O
Mg J Mg2+
Balance the charges with electrons
8e– + HNO3 + 9H+ J NH4+ + 3H2O
Mg J Mg2+ + 2e–
Multiply by a factor to make the number of electrons the same
8e– + HNO3 + 9H+ J NH4+ + 3H2O
(Mg J Mg2+ + 2e–) × 4
Add together
HNO3 + 9H+ + 4Mg J NH4+ + 4Mg2+ + 3H2O
99
Chapter 5
5.16
(a)
molecular: Mg(s) + 2HCl(aq) J MgCl2(aq) + H2(g)
ionic: Mg(s) + 2H+(aq) + 2Cl–(aq) J Mg2+(aq) + 2Cl–(aq) + H2(g)
net ionic: Mg(s) + 2H+(aq) J Mg2+(aq) + H2(g)
(b)
molecular: 2Al(s) + 6HCl(aq) J 2AlCl3(aq) + 3H2(g)
ionic: 2Al(s) + 6H+(aq) + 6Cl–(aq) J 2Al3+(aq) + 6Cl–(aq) + 3H2(g)
net ionic: 2Al(s) + 6H+(aq) J 2Al3+(aq) + 3H2(g)
5.17
Cu2+(aq) + Mg(s) J Cu(s) + Mg2+(aq)
5.18
(a) 2Al(s) + 3Cu2+(aq) J 2Al3+(aq) + 3Cu(s)
(b) Ag(s) + Mg2+(aq) J No reaction
5.19
2C20H42(s) + 21O2(g) J 40C(s) + 42H2O(g)
5.20
2C4H10(g) + 13O2(g) J 8CO2(g) + 10H2O(g)
5.21
C2H5OH(l) + 3O2(g) J 2CO2(g) + 3H2O(g)
5.22
2Sr(s) + O2(g) J 2SrO(s)
5.23
4Fe(s) + 3O2(g) J 2Fe2O3(s)
5.24
First we need a balanced equation:
C2O42– J CO2
MnO4– J Mn2+
Balance the atoms
C2O42– J 2CO2
8H+ + MnO4– J Mn2+ + 4H2O
Balance the charges
C2O42– J 2CO2 + 2e–
5e– + 8H+ + MnO4– J Mn2+ + 4H2O
Add the reactions together
5C2O42– + 16H+ + 2MnO4– J 10CO2 + 2Mn2+ + 8H2O
⎛ 1 L KMnO4 ⎞ ⎛ 0.02000 mol KMnO4 ⎞ ⎛ 5 mol C2 O4 −2 ⎞
⎟ =
mol C2O42– = (18.30 mL KMnO4) ⎜
⎟⎜
⎟ ⎜⎜
1 L KMnO 4
⎝ 1000 mL KMnO4 ⎠ ⎝
⎠ ⎝ 2 mol KMnO4 ⎟⎠
0.0009150 mol C2O42–
⎛ 0.0009150 mol C O 2− ⎞ ⎛ 1000 mL C O 2− ⎞ ⎛ 1 mol H C O ⎞
2 4
2 4
2 2 4 ⎟ = 0.06100 M H C O
⎟⎜
⎟⎜
M H2C2O4 = ⎜
2 2 4
⎜ 15.00 mL C O 2− ⎟ ⎜ 1 L C O 2−
⎟ ⎜ 1 mol C O 2− ⎟
2 4
2 4
2 4
⎝
⎠⎝
⎠⎝
⎠
5.25
First we need a balanced equation:
Cl2 + 2e– J 2Cl–
S2O32– + 5H2O J 2SO42– + 10H+ + 8e–
4Cl2 + S2O32– +5H2O J 8Cl– + 2SO42– + 10H+
⎛ 1 mol Cl2 ⎞ ⎛ 1 mol Na 2S2 O3 ⎞ ⎛ 158.132 g Na 2S2 O3 ⎞
g Na2S2O3 = (4.25 g Cl2) ⎜
⎟ = 2.37 g Na2S2O3
⎟⎜
⎟⎜
⎝ 70.906 g Cl2 ⎠ ⎝ 4 mol Cl2 ⎠ ⎝ 1 mol Na 2S2 O3 ⎠
5.26
(a)
(Sn2+ J Sn4+ + 2e–) × 5
(MnO4– + 8H+ + 5e– J Mn2+ + 4H2O) × 2
5Sn2+ + 2MnO4– + 16H+ J 5Sn4+ + 2Mn2+ + 8H2O
100
Chapter 5
(b)
(c)
(d)
⎛ 0.0500 mol KMnO 4
g Sn = (8.08 mL KMnO4 soln) ⎜
⎝ 1000 mL KMnO 4
⎞ ⎛ 1 mol MnO4 −
⎟ ⎜⎜
⎠ ⎝ 1 mol KMnO 4
⎞ ⎛ 5 mol Sn 2+
⎟⎜
⎟ ⎜ 2 mol MnO −
4
⎠⎝
⎞
⎟
⎟
⎠
⎛ 1 mol Sn ⎞ ⎛ 118.71 g Sn ⎞
× ⎜
⎟⎜
⎟ = 0.120 g Sn
⎝ 1 mol Sn 2+ ⎠ ⎝ 1 mol Sn ⎠
0.120 g Sn
× 100% = 40.0% Sn
% Sn =
0.300 g sample
⎛ 0.0500 mol KMnO 4
g SnO2 = (8.08 mL KMnO4 soln) ⎜
⎝ 1000 mL KMnO 4
⎞ ⎛ 1 mol MnO4 −
⎟ ⎜⎜
⎠ ⎝ 1 mol KMnO 4
⎞ ⎛ 5 mol Sn 2+
⎟⎜
⎟ ⎜ 2 mol MnO −
4
⎠⎝
⎞
⎟
⎟
⎠
⎛ 1 mol SnO 2 ⎞ ⎛ 150.71 g SnO 2 ⎞
× ⎜
⎟ = 0.152 g SnO2
⎟⎜
⎝ 1 mol Sn 2+ ⎠ ⎝ 1 mol SnO2 ⎠
0.152 g SnO 2
× 100% = 50.7% SnO2
% SnO2 =
0.300 g sample
Review Questions
5.1
(a)
(b)
Oxidation is the loss of one or more electrons.
Reduction is the gain of one or more electrons.
The oxidation number decreases in a reduction and increases in an oxidation.
5.2
The number of electrons involved in both the reduction and the oxidation must be the same; only those
electrons that come from the reductant and go to the oxidant are involved. No electrons from external or
uninvolved sources are allowed to enter the process, and there cannot be any electrons left unreacted at the
end of the process. An oxidizing agent is the species that is reduced or gains electrons in an oxidation
reduction reaction. A reducing agent is the species that is oxidized or loses electrons in an oxidation
reduction reaction.
5.3
The oxidation number of the Mg changes from 0 to +2. The oxidation number of the oxygen changes from
0 in O2 to –2 in MgO. The magnesium is oxidized and the oxygen is reduced. Consequently, Mg is the
reducing agent and the O2 is the oxidizing agent.
5.4
Oxidation state and oxidation number are synonyms. Thus, the As is in the +3 oxidation state.
5.5
The oxidation state of Cr is +6 in both reactants and products. No other element changes oxidation state.
Therefore this is not a redox reaction.
5.6
This change in oxidation number represents a reduction of nitrogen by 5 units, and it requires that nitrogen
gain 5 electrons.
5.7
This is not a redox reaction since there is no change in oxidation number.
5.8
By inspection,
2Cr3+ + 3Zn J 2Cr + 3Zn2+
5.9
The equation is not balanced since the charge is different on either side of the arrow. It is easily balanced
by inspection to give:
2Ag+ + Fe J 2Ag + Fe2+
5.10
(a)
(b)
+9 charge on the left, +1 charge on the right; add 8 electrons to the left side.
0 charge on the left, +6 charge on the right; add 6 electrons to the right side.
101
Chapter 5
5.11
(a)
(b)
5.12
It is a reaction in which one element replaces another in a compound.
5.13
NO3–(aq)
5.14
A "nonoxidizing acid" is one in which the H+ ion is the strongest oxidizing agent. That is, the anion of the
acid is not itself a better oxidizing agent than H+. Examples are HCl and H2SO4. The oxidizing agent in a
nonoxidizing acid is the H+.
5.15
5.16
is a reduction
is an oxidation
The best reducing agents are those that are most easily oxidized and are found at the bottom of the activity
series. The best oxidizing agents are those that are most easily reduced and are found at the top of the
activity series. (See Table 5.2 for clarification.)
The most active types of metals will react with water for example:
2Cs(s) + 2H2O J 2CsOH(aq) + H2(g)
2Rb(s) + 2H2O J 2RbOH(aq) + H2(g)
Ba(s) + 2H2O J Ba(OH)2(aq) + H2(g)
Up to
Pb(s) + 2H2O J Pb(OH)2(aq) + H2(g)
5.17
The metal must be below hydrogen in the activity series in order for it to react with HCl.
5.18
Based on the activity series, the manganese is oxidized and therefore is the reducing agent.
5.19
This would be any metal higher (less reactive) than hydrogen, i.e. gold, mercury, silver, and copper.
5.20
Combustion is the rapid reaction of a substance with oxygen, which is accompanied by the evolution of
light and heat.
5.21
Historically, the reaction of a substance with oxygen was termed oxidation. Now we realize that reaction
with oxygen most typically means that oxygen acquires electrons from the substance with which it reacts.
The oxidation of a substance is, therefore, taken to represent the loss of electrons by a substance, whether
the substance has reacted with oxygen or with another oxidizing agent.
5.22
The products will be CO2, H2O, and SO2:
2C2H6S(l)+ 9O2(g) J 4CO2(g) + 2SO2(g) + 6H2O(g)
5.23
a)
b)
c)
5.24
The other product is water:
4NH3(g) + 3O2(g) J 2N2(g) + 6H2O(g)
CO2(g) and H2O(g)
CO(g), CO2(g) and H2O(g)
C(s) and H2O(g)
Review Problems
5.25
The sum of the oxidation numbers should be equal to the total charge:
(a)
O: –2
(c)
O: –2
Na: +1
Na: +1
Cl: +1
Cl: +5
(b)
O: –2
(d)
O: –2
Na: +1
Na: +1
Cl: +3
Cl: +7
102
Chapter 5
5.26
The sum of the oxidation numbers should be equal to the total charge.
(a)
(b)
O: –2
Ca: +2
V: +5
Cl: –1
Sn: +4
(c)
O: –2
Mn: +6
(d)
O: –2
Mn: +4
5.27
The sum of the oxidation numbers should be equal to the total charge:
(a)
S2–: –2
(b)
SO2: S +4, O –2
(c)
P4: P 0
(d)
PH3: P –3, H +1
5.28
The sum of the oxidation numbers should be equal to the total charge:
(a)
ClO4–: Cl +7, O –2
(b)
CrCl3: Cr +3, Cl –1
(c)
SnS2: Sn +4, S –2
(d)
Au(NO3)3: Au +3, N +5,O –2
5.29
(a)
(b)
(c)
(d)
5.30
(a)
(b)
(c)
(d)
5.31
5.32
substance reduced (and oxidizing agent): H2SO4
substance oxidized (and reducing agent): Cu
substance reduced (and oxidizing agent): HNO3
substance oxidized (and reducing agent): SO2
substance reduced (and oxidizing agent): H2SO4
substance oxidized (and reducing agent): Zn
substance reduced (and oxidizing agent): HNO3
substance oxidized (and reducing agent): I2
The sum of the oxidation numbers should be zero:
(a)
S: –2
(c)
Cs +1
Pb: +2
O –1/2 (The Cs can only have an oxidation number of +1 or 0.)
(b)
Cl: –1
Ti: +4
(d)
F –1
O +1
(a)
Sr: +2
O: –2
I: +5
Cr: +3
S: –2
(c)
O +2
F –1
(d)
H +1
F –1
O 0
(b)
5.33
substance reduced (and oxidizing agent): HNO3
substance oxidized (and reducing agent): H3AsO3
substance reduced (and oxidizing agent): HOCl
substance oxidized (and reducing agent): NaI
substance reduced (and oxidizing agent): KMnO4
substance oxidized (and reducing agent): H2C2O4
substance reduced (and oxidizing agent): H2SO4
substance oxidized (and reducing agent): Al
H+(aq) + Cl–(aq) + HOCl(aq)
Cl2(aq) + H2O
In the forward direction: The oxidation number of the chlorine atoms decreases from 0 to –1. Therefore
Cl2 is reduced. However, in HOCl, chlorine has an oxidation number of +1, so Cl2 also oxidized! (One
atom is reduced, the other is oxidized.)
103
Chapter 5
In the reverse direction: The Cl– ion begins with an oxidation number of –1 and ends with an oxidation
number of 0. Therefore the Cl– ion is oxidized: This means Cl– is the reducing agent. Since the
oxidation number of H+ does not change, HOCl must be the oxidizing agent.
5.34
N is reduced from +4 to +2 and N is oxidized from +4 to +5.
5.35
(a)
2S2O32– J S4O62– + 2e–
OCl– + 2H+ + 2e– J Cl– + H2O
OCl– + 2S2O32– + 2H+ J S4O62– + Cl– + H2O
(b)
(NO3– + 2H+ + e– J NO2 + H2O) × 2
Cu J Cu2+ + 2e–
2NO3– + Cu + 4H+ J 2NO2 + Cu2+ + 2H2O
(c)
IO3– + 6H+ + 6e– J I– + 3H2O
(H2O + AsO33– J AsO43– + 2H+ + 2e–) × 3
IO3– + 3AsO33– + 6H+ + 3H2O J I– + 3AsO43– + 3H2O + 6H+
which simplifies to give:
3AsO33– + IO3– J I– + 3AsO43–
(d)
SO42– + 4H+ + 2e– J SO2 + 2H2O
Zn J Zn2+ + 2e–
Zn + SO42– + 4H+ J Zn2+ + SO2 + 2H2O
(e)
NO3– + 10H+ + 8e– J NH4+ + 3H2O
(Zn J Zn2+ + 2e–) × 4
NO3– + 4Zn + 10H+ J 4Zn2+ + NH4+ + 3H2O
(f)
2Cr3+ + 7H2O J Cr2O72– + 14H+ + 6e–
(BiO3– + 6H+ + 2e– J Bi3+ + 3H2O) × 3
2Cr3+ + 3BiO3– + 18H+ + 7H2O J Cr2O72– + 14H+ + 3Bi3+ + 9H2O
which simplifies to give:
2Cr3+ + 3BiO3– + 4H+ J Cr2O72– + 3Bi3+ + 2H2O
(g)
I2 + 6H2O J 2IO3– + 12H+ + 10e–
(OCl– + 2H+ + 2e– JCl– + H2O) × 5
I2 + 5OCl– + H2O J 2IO3– + 5Cl– + 2H+
(h)
(Mn2+ + 4H2O J MnO4– + 8H+ + 5e–) × 2
(BiO3– + 6H+ + 2e– J Bi3+ + 3H2O) × 5
2Mn2+ + 5BiO3– + 30H+ + 8H2O J 2MnO4– + 5Bi3+ + 16H+ + 15H2O
which simplifies to:
2Mn2+ + 5BiO3– + 14H+ J 2MnO4– + 5Bi3+ + 7H2O
(i)
(H3AsO3 + H2O J H3AsO4 + 2H+ + 2e–) × 3
Cr2O72– + 14H+ + 6e– J 2Cr3+ + 7H2O
3H3AsO3 + Cr2O72– + 3H2O + 14H+ J 3H3AsO4 + 2Cr3+ + 6H+ + 7H2O
which simplifies to give:
3H3AsO3 + Cr2O72– + 8H+ J 3H3AsO4 + 2Cr3+ + 4H2O
(j)
2I– J I2 + 2e–
HSO4– + 3H+ + 2e– J SO2 + 2H2O
2I– + HSO4– + 3H+ J I2 + SO2 + 2H2O
104
Chapter 5
5.36
5.37
(a)
(Sn + 2H2O J SnO2 + 4H+ + 4e–) × 3
(NO3– + 4H+ + 3e– J NO + 2H2O) × 4
3Sn + 4NO3– + 16H+ + 6H2O J 3SnO2 + 12H+ + 4NO + 8H2O
which simplifies to:
3Sn + 4NO3– + 4H+ J 3SnO2 + 4NO + 2H2O
(b)
PbO2 + 2Cl– + 4H+ + 2e– J PbCl2 + 2H2O
2Cl– J Cl2 + 2e–
PbO2 + 4Cl– + 4H+ J PbCl2 + Cl2 + 2H2O
(c)
Ag J Ag+ + e–
NO3– + 2H+ + e– J NO2 + H2O
Ag + 2H+ + NO3– J Ag+ + NO2 + H2O
(d)
(Fe3+ + e– J Fe2+) × 4
2NH3OH+ J N2O + H2O + 6H+ + 4e–
4Fe3+ + 2NH3OH+ J 4Fe2+ + N2O + 6H+ + H2O
(e)
2I– J I2 + 2e–
(HNO2 + H+ + e– J NO + H2O) × 2
2I– + 2HNO2 + 2H+ J I2 + 2NO + 2H2O
(f)
C2O42– J 2CO2 + 2e–
(HNO2 + H+ + e– J NO + H2O) × 2
C2O42– + 2HNO2 + 2H+ J 2CO2 + 2NO + 2H2O
(g)
(HNO2 + H2O J NO3– + 3H+ + 2e–) × 5
(MnO4– + 8H+ + 5e– J Mn2+ + 4H2O) × 2
5HNO2 + 2MnO4– + 16H+ + 5H2O J 5NO3– + 2Mn2+ + 15H+ + 8H2O
which simplifies to give:
5HNO2 + 2MnO4– + H+ J 5NO3– + 2Mn2+ + 3H2O
(h)
(H3PO2 + 2H2O J H3PO4 + 4H+ + 4e–) × 3
(Cr2O72– + 14H+ + 6e– J 2Cr3+ + 7H2O) × 2
3H3PO2 + 2Cr2O72– + 28H+ + 6H2O J 3H3PO4 + 4Cr3+ + 14H2O + 12H+
which simplifies to:
3H3PO2 + 2Cr2O72– + 16H+ J 3H3PO4 + 4Cr3+ + 8H2O
(i)
(VO2+ + 2H+ + e– J VO2+ + H2O) × 2
Sn2+ J Sn4+ + 2e–
2VO2+ + Sn2+ + 4H+ J 2VO2+ + Sn4+ + 2H2O
(j)
XeF2 + 2e– J Xe + 2F–
2Cl– J Cl2 + 2e–
XeF2 + 2Cl– J Xe + Cl2 + 2F–
For redox reactions in basic solution, we proceed to balance the half reactions as if they were in acid
solution, and then add enough OH– to each side of the resulting equation in order to neutralize (titrate) all
of the H+. This gives a corresponding amount of water (H+ + OH– → H2O) on one side of the equation, and
an excess of OH– on the other side of the equation, as befits a reaction in basic solution.
(a)
(CrO42– + 4H+ + 3e– J CrO2– + 2H2O) × 2
(S2– J S + 2e–) × 3
2CrO42– + 3S2– + 8H+ J 2CrO2– + 3S + 4H2O
Adding 8OH– to both sides of the above equation we obtain:
2CrO42– + 3S2– + 8H2O J 2CrO2– + 8OH– + 3S + 4H2O
which simplifies to:
2CrO42– + 3S2– + 4H2O J 2CrO2– + 3S + 8OH–
105
Chapter 5
5.38
(b)
(C2O42– J 2CO2 + 2e–) × 3
(MnO4– + 4H+ + 3e– J MnO2 + 2H2O) × 2
3C2O42– + 2MnO4– + 8H+ J 6CO2 + 2MnO2 + 4H2O
Adding 8OH– to both sides of the above equation we get:
3C2O42– + 2MnO4– + 8H2O J 6CO2 + 2MnO2 + 4H2O + 8OH–
which simplifies to give:
3C2O42– + 2MnO4– + 4H2O J 6CO2 + 2MnO2 + 8OH–
(c)
(ClO3– + 6H+ + 6e– J Cl– + 3H2O) × 4
(N2H4 + 2H2O J 2NO + 8H+ + 8e–) × 3
4ClO3– + 3N2H4 + 24H+ + 6H2O J 4Cl– + 6NO + 12H2O + 24H+
which needs no OH–, because it simplifies directly to:
4ClO3– + 3N2H4 J 4Cl– + 6NO + 6H2O
(d)
NiO2 + 2H+ + 2e– J Ni(OH)2
2Mn(OH)2 J Mn2O3 + H2O + 2H+ + 2e–
NiO2 + 2Mn(OH)2 J Ni(OH)2 + Mn2O3 + H2O
(e)
(SO32– + H2O J SO42– + 2H+ + 2e–) × 3
(MnO4– + 4H+ + 3e–J MnO2 + 2H2O) × 2
3SO32– + 3H2O + 8H+ + 2MnO4– J 3SO42– + 6H+ + 2MnO2 + 4H2O
Adding 8OH– to both sides of the equation we obtain:
3SO32– + 11H2O + 2MnO4– J 3SO42– + 10H2O + 2MnO2 + 2OH–
which simplifies to:
3SO32– + 2MnO4– + H2O J 3SO42– + 2MnO2 + 2OH–
(a)
(CrO2– + 2H2O J CrO42– + 4H+ + 3e–) × 2
(S2O82– + 2e– J 2SO42–) × 3
3S2O82– + 2CrO2– + 4H2O J 2CrO42– + 6SO42– + 8H+
Adding 8OH– to both sides of this equation:
3S2O82– + 2CrO2– + 4H2O + 8OH– J 2CrO42– + 6SO42– + 8H2O
which simplifies to give:
3S2O82– + 2CrO2– + 8OH– J 2CrO42– + 6SO42– + 4H2O
(b)
(SO32– + H2O J SO42– + 2H+ + 2e–) × 3
(CrO42– + 4H+ + 3e– J CrO2– + 2H2O) × 2
3SO32– + 2CrO42– + 8H+ + 3H2O J 3SO42– + 2CrO2– + 6H+ + 4H2O
Adding 8OH– to both sides of the equation we get:
3SO32– + 2CrO42– + 11H2O J 3SO42– + 2CrO2– + 2OH– + 10H2O
which simplifies to:
3SO32– + 2CrO42– + H2O J 3SO42– + 2CrO2– + 2OH–
(c)
(O2 + 2H+ + 2e– J H2O2) × 2
N2H4 J N2 + 4H+ + 4e–
2O2 + 4H+ + N2H4 J 2H2O2 + N2 + 4H+
which simplifies to:
2O2 + N2H4 J 2H2O2 + N2
(d)
(Fe(OH)2 + OH– J Fe(OH)3 + e–) × 4
O2 + 2H2O + 4e– J 4OH–
4Fe(OH)2 + O2 + 4OH– + 2H2O J 4Fe(OH)3 + 4OH–
which simplifies to:
4Fe(OH)2 + O2 + 2H2O J 4Fe(OH)3
(e)
(Au + 4CN– J Au(CN)4– + 3e–) × 4
(O2 + 2H2O + 4e– J 4OH–) × 3
4Au + 16CN– + 3O2 + 6H2O J 4Au(CN)4– + 12OH–
106
Chapter 5
5.39
O3 + 6H+ + 6e– J 3H2O
Br– + 3H2O J BrO3– + 6H+ + 6e–
O3 + Br– + 3H2O + 6H+ J BrO3– + 3H2O + 6H+
which simplifies to:
O3 + Br– J BrO3–
5.40
(Cl2 + 2e– J 2Cl–) × 4
S2O32– + 5H2O → 2SO42– + 10H+ + 8e–
4Cl2 + S2O32– + 5H2O J 8Cl– + 2SO42– + 10H+
5.41
(OCl– + 2H+ + 2e– J Cl– + H2O) × 4
S2O32– + 5H2O J 2SO42– + 10H+ + 8e–
4OCl– + S2O32– + 5H2O + 8H+ J 4Cl– + 2SO42– + 10H+ + 4H2O
which simplifies to:
4OCl– + S2O32– + H2O J 4Cl– + 2SO42– + 2H+
5.42
(H2C2O4 J 2CO2 + 2H+ + 2e–) × 3
K2Cr2O7 + 14H+ + 6e– J 2K+ + 2Cr3+ + 7H2O
3H2C2O4 + K2Cr2O7 + 14H+ J 6CO2 + 2K+ + 2Cr3+ + 6H+ + 7H2O
which simplifies to:
3H2C2O4 + K2Cr2O7 + 8H+ J 6CO2 + 2K+ + 2Cr3+ + 7H2O
5.43
(a)
(b)
5.44
Cu(s) → Cu2+(aq) + 2e–
H2SO4(aq) + 2H+(aq) + 2e– J SO2(g) + 2H2O
Cu(s) + H2SO4(aq)+ 2H+(aq) J Cu2+(aq) + SO2(g) + 2H2O
5.45
(a)
m: Mn(s) + 2HCl(aq) J MnCl2(aq) + H2(g)
I: Mn(s) + 2H+(aq)+ 2Cl–(aq) J Mn2+(aq) + 2Cl–(aq) + H2(g)
NI: Mn(s) + 2H+(aq) J Mn2+(aq) + H2(g)
(b)
m: Cd(s) + 2HCl(aq) J CdCl2(aq) + H2(g)
I: Cd(s) + 2H+(aq) + 2Cl–(aq) J Cd2+(aq) +2Cl–(aq) + H2(g)
NI: Cd(s) + 2H+(aq) J Cd2+(aq) + H2(g)
(c)
m: Sn(s) + 2HCl(aq) J SnCl2(aq) + H2(g)
I: Sn(s) + 2H+(aq) + 2Cl–(aq) J Sn2+(aq) + 2Cl–(aq) + H2(g)
NI: Sn(s) + 2H+(aq) J Sn2+(aq) + H2(g)
(a)
m: Ni(s) + H2SO4(aq) J NiSO4(aq) + H2(g)
I: Ni(s) + 2H+(aq)+ SO42–(aq) J Ni2+(aq) + SO42–(aq) + H2(g)
NI: Ni(s) + 2H+(aq) J Ni2+(aq)+ H2(g)
(b)
m: 2Cr(s) + 3H2SO4(aq) J Cr2(SO4)3(aq) + 3H2(g)
I: 2Cr(s) + 6H+(aq) + 3SO42–(aq) J 2Cr3+(aq) + 3SO42–(aq) + 3H2(g)
NI: 2Cr(s) + 6H+(aq) J 2Cr3+(aq) + 3H2(g)
(c)
m: 2Al(s) + 3H2SO4(aq) J Al2(SO4)3(aq) + 3H2(g)
I: 2Al(s) + 6H+(aq) + 3SO42–(aq) J 2Al3+(aq) + 3SO42–(aq) + 3H2(g)
NI: 2Al(s) + 6H+(aq) J 2Al3+(aq) + 3H2(g)
5.46
3Ag(s) + 4HNO3(aq) J 3AgNO3(aq) + 2H2O + NO(g)
Ag(s) + 2HNO3(aq) J AgNO3(aq) + H2O + NO2(aq)
5.47
Increasing ease of oxidation: Pt, Ru, Tl, Pu
5.48
Increasing ease of oxidation: Ni, Y, Mo
107
Chapter 5
5.49
In each case, the reaction should proceed to give the less reactive of the two metals, together with the ion of
the more reactive of the two metals. The reactivity is taken from the reactivity series table 5.2.
(a)
N.R.
(b)
2Cr(s) + 3Pb2+(aq) J 2Cr3+(aq) + 3Pb(s)
(c)
2Ag+(aq) + Fe(s) J 2Ag(s) + Fe2+(aq)
(d)
3Ag(s) + Au3+(aq) J Au(s) + 3Ag+(aq)
5.50
In each case, the reaction should proceed to give the less reactive of the two metals, together with the ion of
the more reactive of the two metals. The reactivity is taken from the reactivity series table 5.2.
(a)
Mn(s) + Fe2+ J Mn2+ + Fe(s)
(b)
N.R.
(c)
Mg(s) + Co2+ J Mg2+ + Co(s)
(d)
2Cr(s) + 3Sn2+ J 2Cr3+ + 3Sn(s)
5.51
The equation given shows that Cd is more active than Ru. Coupled with the information in Review
Problem 5.47, we also see that Cd is more active than Tl. This means that in a mixture of Cd and Tl+, Cd
will be oxidized and Tl+ will be reduced:
Cd(s) + 2TlCl(aq) J CdCl2(aq) + 2Tl(s)
(The Tl(s) and the Cd(NO3)2(aq) will not react.)
5.52
The observation shows that Mg is more active than Ni, however, the information is not sufficient to
determine which is easier to oxidize, Mg or Mo. Therefore it cannot be determined which reaction will
occur spontaneously.
5.53
(a)
(b)
(c)
2C6H6(l) + 15O2(g) J 12CO2(g) + 6H2O(g)
C3H8(g) + 5O2(g) J 3CO2(g) + 4H2O(g)
C21H44(s) + 32O2(g) J 21CO2(g) + 22H2O(g)
5.54
(a)
(b)
(c)
2C12H26(l) + 37O2(g) J 24CO2(g) + 26H2O(g)
C18H36(l) + 27O2(g) J 18CO2(g) + 18H2O(g)
C7H8(l) + 9O2(g) J 7CO2(g) + 4H2O(g)
5.55
2CH3OH(l) + 3O2(g) J 2CO2(g) + 4H2O(g)
5.56
C6H12O6(s) + 6O2(g) J 6CO2(g) + 6H2O(g)
5.57
(a)
2C6H6(l) + 9O2(g) J 12CO(g) + 6H2O(g)
2C3H8(g) + 7O2(g) J 6CO(g) + 8H2O(g)
2C21H44(s) + 43O2(g) J 42CO(g) + 44H2O(g)
(b)
2C6H6(l) + 3O2(g) J 12C(s) + 6H2O(g)
C3H8(g) + 2O2(g) J 3C(s) + 4H2O(g)
C21H44(s) + 11O2(g) J 21C(s) + 22H2O(g)
(a)
2C12H26(l) + 25O2(g) J 24CO(g) + 26H2O(g)
C18H36(l) + 18O2(g) J 18CO(g) + 18H2O(g)
2C7H8(l) + 11O2(g) J 14CO(g) + 8H2O(g)
(b)
2C12H26(l) + 13O2(g) J 24C(s) + 26H2O(g)
C18H36(l) + 9O2(g) J 18C(s) + 18H2O(g)
C7H8(l) + 2O2(g) J 7C(s) + 4H2O(g)
(a)
(b)
2Zn(s) + O2(g) J 2ZnO(s)
4Al(s) + 3O2(g) J 2Al2O3(s)
5.58
5.59
108
Chapter 5
(c)
(d)
2Mg(s) + O2(g) J 2MgO(s)
4Fe(s) + 3O2(g) J 2Fe2O3(s)
5.60
(a)
(b)
(c)
(d)
2Be(s) + O2(g) J 2BeO(s)
4Li(s) + O2(g) J 2Li2O(s)
2Ba(s) + O2(g) J 2BaO(s)
4Bi(s) + 3O2(g) J 2Bi2O3(s)
5.61
2(CH3)2S(g) + 9O2(g) J 4CO2(g) + 6H2O(g) + 2SO2(g)
5.62
C4H4S(l) + 6O2(g) J 4CO2(g) + 2H2O(g) + SO2(g)
5.63
Cu + 2Ag+ J Cu2+ + 2Ag
⎛ 1 mol Ag ⎞ ⎛ 1 mol Cu ⎞ ⎛ 63.546 g Cu ⎞
g Cu = (12.0 g Ag) ⎜
⎟⎜
⎟⎜
⎟ = 3.53 g Cu
⎝ 107.868 g Ag ⎠ ⎝ 2 mol Ag ⎠ ⎝ 1 mol Cu ⎠
5.64
Al(s) + 3AgNO3(aq) J 3Ag(s) + Al(NO3)3(aq)
⎛ 1 mol AgNO3 ⎞ ⎛ 1 mol Al ⎞ ⎛ 26.98 g Al ⎞
g Al = (25.0 g AgNO3) ⎜
⎟⎜
⎟⎜
⎟ = 1.32 g Al
⎝ 169.9 g AgNO3 ⎠ ⎝ 3 mol AgNO3 ⎠ ⎝ 1 mol Al ⎠
5.65
(a)
(b)
IO3– + 6H+ + 6e– J I– + 3H2O
[SO32– + H2O J SO42– + 2H+ + 2e–] × 3
IO3– + 3SO32– + 6H+ + 3H2O J I– + 3SO42– + 3H2O + 6H+
Which simplifies to:
IO3– + 3SO32– J I– + 3SO42–
⎛ 1 mol NaIO3 ⎞ ⎛ 3 mol Na 2SO3 ⎞ ⎛ 126.0 g Na 2SO3 ⎞
g Na2SO3 = (5.00 g NaIO3) ⎜
⎟⎜
⎟⎜
⎟
⎝ 197.9 g NaIO3 ⎠ ⎝ 1 mol NaIO3 ⎠ ⎝ 1 mol Na 2SO3 ⎠
= 9.55 g Na2SO3
5.66
(a)
(b)
[Mn2+(aq) + 4H2O J MnO4–(aq) + 8H+(aq) + 5e–] × 2
[BiO3–(aq) + 6H+(aq) + 2e– J Bi3+(aq) + 3H2O(l)] × 5
2Mn2+(aq) + 5BiO3–(aq) + 8H2O + 30H+(aq) J
2MnO4–(aq) + 5Bi3+(aq) + 15H2O(l) + 16H+(aq)
which simplifies to give:
2Mn2+(aq) + 5BiO3–(aq) + 14H+(aq) J 2MnO4–(aq) + 5Bi3+(aq) + 7H2O(l)
⎛ 1 g ⎞ ⎛ 1 mol MnSO 4
g NaBiO3 = (18.5 mg MnSO4) ⎜
⎟⎜
⎝ 1000 mg ⎠ ⎝ 151.0 g MnSO4
⎞ ⎛ 1 mol Mn 2+
⎟ ⎜⎜
⎠ ⎝ 1 mol MnSO 4
⎞
⎟
⎟
⎠
⎛ 5 mol BiO3− ⎞ ⎛ 1 mol NaBiO3 ⎞ ⎛ 280.0 g NaBiO3 ⎞ ⎛ 1000 mg NaBiO3 ⎞
⎟⎜
⎟
× ⎜
⎜ 2 mol Mn 2+ ⎟ ⎜ 1 mol BiO − ⎟ ⎜⎝ 1 mol NaBiO3 ⎟⎠ ⎜⎝ 1 g NaBiO3 ⎟⎠
3 ⎠
⎝
⎠⎝
= 85.9 g NaBiO3
5.67
(a)
⎛ 0.02000 mol KMnO4
g H2O2 = (17.60 mL KMnO4) ⎜
⎝ 1000 mL KMnO4
⎛ 5 mol H O
2 2
× ⎜
⎜ 2 mol MnO −
4
⎝
(b)
⎞ ⎛ 1 mol MnO4 −
⎟ ⎜⎜
⎠ ⎝ 1 mol KMnO4
⎞ ⎛ 34.02 g H O ⎞
2 2 = 0.02994 g H O
⎟
2 2
⎟ ⎜⎝ 1 mol H 2 O2 ⎟⎠
⎠
(0.02994 g H2O2/1.000 g sample) × 100% = 2.994% H2O2
109
⎞
⎟
⎟
⎠
Chapter 5
5.68
⎛ 0.01000 mol KMnO 4
g NaNO2 = (12.15 mL KMnO4) ⎜
⎝ 1000 mL KMnO 4
⎞ ⎛ 1 mol MnO 4 2−
⎟ ⎜⎜
⎠ ⎝ 1 mol KMnO 4
⎞
⎟
⎟
⎠
⎛ 5 mol HNO ⎞ ⎛ 1 mol NaNO ⎞ ⎛ 68.995 g NaNO ⎞
2 ⎟
2
2
× ⎜
⎜ 2 mol MnO 2− ⎟ ⎜⎝ 1 mol HNO2 ⎟⎠ ⎜⎝ 1 mol NaNO 2 ⎟⎠
4
⎝
⎠
=2.096 × 10–2 g NaNO2
% NaNO2 = (2.096 × 10–2 g NaNO2 / 1.000 g sample) × 100% = 2.096%
5.69
5.70
(a)
The stoichiometry for calcium is as follows:
1 mol C2O42– = 1 mol Ca2+ = 1 mol CaCl2
Thus the number of grams of CaCl2 is given simply by:
5.405 × 10–3 mol CaCl2 × 110.98 g/mol = 0.5999 g CaCl2
(c)
(0.5999 g/2.463 g) × 100 = 24.35% CaCl2
(a)
(c)
(a)
(b)
5.72
⎞ ⎛ 5 mol C2 O4 2−
⎟ ⎜⎜
⎠ ⎝ 2 mol KMnO 4
(b)
(b)
5.71
⎛ 0.1000 mol KMnO 4
mol C2O42– = (21.62 mL KMnO4) ⎜
⎝ 1000 mL KMnO 4
–3
2–
= 5.405 × 10 mol C2O4
(a)
⎞
⎟
⎟
⎠
⎞
⎛ 0.3000 mol Na 2S2 O3 ⎞ ⎛ 1 mol I3−
⎟
mol I3– = (29.25 mL Na2S2O3) ⎜
⎟ ⎜⎜
⎝ 1000 mL Na 2S2 O3 ⎠ ⎝ 2 mol Na 2S2 O3 ⎟⎠
= 4.388 × 10–3 mol I3–
⎛ 2 mol NO − ⎞
2 ⎟ = 8.776 × 10–3 mol NO –
mol NO2– = (4.388 × 10–3 mol I3–) ⎜
2
⎜ 1 mol I − ⎟
3
⎝
⎠
⎛ 1 mol NaNO ⎞ ⎛ 68.995 g NaNO ⎞
2⎟
2 = 0.6055 g NaNO
g NaNO2 = (8.776 × 10–3 mol NO2–) ⎜
2
⎜ 1 mol NO − ⎟ ⎜⎝ 1 mol NaNO2 ⎟⎠
2 ⎠
⎝
% NaNO2 = (0.6055 g NaNO2 / 1.104 g sample) × 100% = 54.85% NaNO2
[MnO4– + 8H+ + 5e– J Mn2+ + 4H2O] × 2
[Sn2+ J Sn4+ + 2e–] × 5
2MnO4– +5Sn2+ + 16H+ J 2Mn2+ + 5Sn4+ + 8H2O
⎛ 0.250 mol SnCl2
mL KMnO4 = (40.0 mL SnCl2) ⎜
⎝ 1000 mL SnCl2
⎞ ⎛ 1 mol Sn 2+
⎟⎜
⎠ ⎝⎜ 1 mol SnCl2
⎞ ⎛ 2 mol MnO 4−
⎟⎜
2+
⎟⎜
⎠ ⎝ 5 mol Sn
⎛ 1 mol KMnO ⎞ ⎛ 1000 mL KMnO ⎞
4⎟
4 = 17.4 mL KMnO
× ⎜
4
⎜ 1 mol MnO − ⎟ ⎜⎝ 0.230 mol KMnO4 ⎟⎠
4 ⎠
⎝
[HSO3– + H2O J SO42– + 3H+ + 2e–] × 3
ClO3– + 6H+ + 6e– J Cl– + 3H2O
3HSO3– + ClO3– + 3H2O + 6H+ J 3SO42– + 9H+ + Cl– + 3H2O
Which simplifies to:
3HSO3– + ClO3– J 3SO42– + 3H+ + Cl–
110
⎞
⎟
⎟
⎠
Chapter 5
(b)
⎛ 0.450 mol NaHSO3 ⎞ ⎛ 1 mol HSO3− ⎞
⎟
mL NaClO3 = (30.0 mL NaHSO3) ⎜
⎟⎜
⎝ 1000 mL NaHSO3 ⎠ ⎜⎝ 1 mol NaHSO3 ⎟⎠
⎛ 1 mol ClO −
3
× ⎜
⎜ 3 mol HSO −
3
⎝
5.73
(a)
(b)
⎞⎛ 1 mol NaClO ⎞ ⎛ 1000 mL NaClO ⎞
3⎟
3
⎟⎜
= 30.0 mL NaClO3
⎟⎜ 1 mol ClO − ⎟ ⎜⎝ 0.150 mol NaClO3 ⎟⎠
3 ⎠
⎠⎝
2CrO42– + 3SO32– + H2O J 2CrO2– + 3SO42– + 2OH–
⎛ 1 mol Na 2SO3 ⎞
mol CrO42– = (3.18 g Na2SO3) ⎜
⎟
⎝ 126.04 g Na 2SO3 ⎠
⎛ 1 mol SO32− ⎞ ⎛ 2 mol CrO 2−
4
⎟⎜
× ⎜
⎜ 1 mol Na 2SO3 ⎟ ⎜ 3 mol SO 2−
3
⎝
⎠⎝
⎞
⎟ = 1.68 × 10–2 mol CrO42–
⎟
⎠
Since there is one mole of Cr in each mole of CrO42–, then the above number of moles of CrO42– is
also equal to the number of moles of Cr that were present:
0.0168 mol Cr × 52.00 g/mol = 0.875 g Cr in the original alloy.
5.74
(c)
(0.875 g Cr/3.450 g sample) × 100% = 25.4% Cr
(a)
(Sn2+ J Sn4+ + 2e–) × 3
Cr2O72– + 14H+ + 6e– J 2Cr3+ + 7H2O
3Sn2+ + Cr2O72– + 14H+ J 3Sn4+ + 2Cr3+ + 7H2O
⎛ 1 mol Na 2 Cr2 O7 ⎞ ⎛ 1 mol Cr2 O7 2−
g Sn = (0.368 g Na2Cr2O7) ⎜
⎟⎜
⎝ 262.0 g Na 2 Cr2 O7 ⎠ ⎜⎝ 1 mol Na 2 Cr2 O7
(b)
⎛ 3 mol Sn 2+
× ⎜
⎜ 1 mol Cr O 2−
2 7
⎝
5.75
⎞ ⎛ 1 mol Sn
⎟⎜
⎟ ⎝ 1 mol Sn 2+
⎠
⎞
⎟
⎟
⎠
⎞ ⎛ 118.7 g Sn ⎞
⎟⎜
⎟ = 0.500 g Sn
⎠ ⎝ 1 mol Sn ⎠
(c)
(0.500 g Sn / 1.50 g solder) × 100% = 33.3%
(a)
⎛ 0.02100 mol S O 2−
2 3
mol Cu2+ = (29.96 mL S2O32–) ⎜
⎜ 1000 mL S O 2−
2 3
⎝
⎞ ⎛ 1 mol I −
3
⎟⎜
⎟ ⎜ 2 mol S O 2−
2 3
⎠⎝
⎞⎛ 2 mol Cu 2+
⎟⎜
⎟⎜ 1 mol I −
3
⎠⎝
⎞
⎟
⎟
⎠
= 6.292 × 10−4 mol Cu 2+ = 6.929 × 10–2 mol Cu2+
g Cu = (6.292 × 10–4 mol Cu) × (63.546 g Cu/mol Cu)
= 3.998 × 10–2 g Cu
% Cu = (3.998 × 10–2 g Cu/0.4225 g sample) × 100 = 9.463%
(b)
⎛ 1 mol CuCO3 ⎞ ⎛ 123.56 g CuCO3 ⎞
g CuCO3 = (6.292 × 10–4 mol Cu) ⎜
⎟ = 0.07774 g CuCO3
⎟⎜
⎝ 1 mol Cu ⎠ ⎝ 1 mol CuCO3 ⎠
⎛ 0.07774 g CuCO3 ⎞
% CuCO3 = ⎜
⎟ × 100% = 18.40%
⎝ 0.4225 g sample ⎠
5.76
(a)
2+
2−
4
2−
⎛ 0.0281 mol MnO
2– ⎜
g Fe = 39.42 mL MnO4 ⎜
⎝ 1000 mL MnO4
= 0.309 g Fe2+
0.309 g
% Fe =
× 100% = 22.7% Fe
1.362 g
111
⎞ ⎛ 5 mol Fe2+
⎟⎜
⎟ ⎜ 1 mol MnO 2−
4
⎠⎝
⎞ ⎛ 55.845 g Fe2+
⎟⎜
⎟ ⎜ 1 mol Fe2+
⎠⎝
⎞
⎟
⎟
⎠
Chapter 5
(b)
5.77
(a)
(b)
(c)
⎛ 1 mol Fe ⎞ ⎛ 1 mol Fe3O 4
g Fe3O4 = (0.309 g Fe) ⎜
⎟⎜
⎝ 55.845 g Fe ⎠ ⎝ 3 mol Fe
= 0.427 g Fe3O4
0.427 g
× 100% = 31.4%
% Fe =
1.362 g
⎞ ⎛ 231.55 g Fe3O 4 ⎞
⎟
⎟⎜
⎠ ⎝ 1 mol Fe3O 4 ⎠
⎛ 1 mol NaIO3 ⎞ ⎛ 1 mol IO3– ⎞ ⎛ 3 mol I3–
⎟⎜
mol of I3– = 0.0421 g NaIO3 ⎜
⎟⎜
–
⎝ 197.89 g NaIO3 ⎠ ⎜⎝ 1 mol NaIO3 ⎟⎠ ⎜⎝ 1 mol IO3
–4
–
= 6.38 × 10 mol I3
⎛ 6.38 × 10 –4 mol I – ⎞ ⎛ 1000 mL ⎞
3 ⎟
Molarity of I3– = ⎜
= 6.38 × 10–3 M I3–
⎜
⎟ ⎜⎝ 1 L ⎟⎠
100
mL
⎝
⎠
⎛ 1 L I3 −
⎞⎛ 6.38 × 10−3 mol I3− ⎞
⎟⎜
⎟
g SO2 = 2.47 mL I3– ⎜⎜
− ⎟⎜
−
⎟
1000
mL
I
1
L
I
3 ⎠⎝
3
⎝
⎠
⎞
⎟
⎟
⎠
⎛ 1 mol SO ⎞ ⎛ 64.07 g SO ⎞
2⎟
2 = 1.01 × 10–3 g SO
⎜
2
⎜ 1 mol I − ⎟ ⎜⎝ 1 mol SO2 ⎟⎠
3 ⎠
⎝
The density of the wine was 0.96 g/mL and the SO2 concentration was 1.01 × 10–3 g SO2/in 50 mL
1.01× 10−3 g SO 2
= 2.02 × 10–5 g SO2/mL
50 mL
In 1 mL of solution there are 0.96 g of wine and 2.02 × 10–5 g SO2
Therefore the percentage of SO2 in the wine is
concentration SO2 =
2.02 × 10−5 g SO2
× 100% = 2.10 × 10–3 %
0.96 g wine
5.78
2.02 × 10−5 g SO2
× 106 ppm = 21 ppm
0.96 g wine
(d)
ppm SO2 =
(a)
⎛ 1 mol KIO3
Molarity of I3– solution = 0.462 g KIO3 ⎜
⎝ 214.00 g KIO3
⎞ ⎛ 1 mol IO3 – ⎞ ⎛ 3 mol I3 –
⎟⎜
⎟ ⎜⎜
–
⎠ ⎝ 1 mol KIO3 ⎟⎠ ⎜⎝ 1 mol IO3
1
⎛
⎞
–
×⎜
⎟ = 0.0259 M I3
0.250
L
⎝
⎠
(b)
−
⎛ 1 L ⎞ ⎛ 0.0259 mol I3
⎜
g (NH4)2S2O3 = 27.99 mL I3– solution ⎜
⎟
1L
⎝ 1000 mL ⎠ ⎜⎝
⎞ ⎛ 2 mol S2 O32−
⎟⎜
⎟ ⎜ 1 mol I −
3
⎠⎝
⎛ 1 mol (NH 4 )2S2 O3 ⎞ ⎛ 148.24 g (NH 4 )2S2 O3 ⎞
⎟
×⎜
= 0.2149 g (NH4)2S2O3
⎜ 1 mol S O 2– ⎟ ⎜⎝ 1 mol (NH 4 ) 2S2 O3 ⎟⎠
2 3
⎝
⎠
(c)
⎛ 0.2149 g (NH 4 )2 S2 O3 ⎞
% by mass = ⎜
⎟ × 100% = 98.6% (NH4)2S2O3 in sample
0.218 g sample
⎝
⎠
112
⎞
⎟
⎟
⎠
⎞
⎟
⎟
⎠
Chapter 5
Additional Exercises
5.79
Cu + 2Ag+ J 2Ag + Cu2+
The number of moles of Ag+ available for the reaction is
0.125 M × 0.255 L = 0.0319 mol Ag+
Since the stoichiometry is 2/1, the number of moles of Cu2+ ion that are consumed is 0.0319 ÷ 2 = 0.0159
mol. The mass of copper consumed is 0.0159 mol × 63.546 g/mol = 1.01 g. The amount of unreacted
copper is thus: 12.340 g – 1.01 g = 11.33 g Cu. The mass of Ag that is formed is: 0.0319 mol × 108 g/mol
= 3.45 g Ag. The final mass of the bar is: 11.33 g + 3.45 g = 14.78 g.
5.80
⎛ 0.0500 mol S2 O32– ⎞ ⎛ 1 mol I3–
⎟⎜
mg KIO3 = 22.61 mL S2O3 ⎜
⎟ ⎜ 2 mol S O 2–
1000
mL
2 3
⎝
⎠⎝
–⎜
⎞⎛ 1 mol IO3 –
⎟⎜
⎟⎜ 3 mol I –
3
⎠⎝
⎞
⎟
⎟
⎠
⎛ 1 mol KIO ⎞ ⎛ 214.00 g KIO ⎞ ⎛ 1000 mg KIO ⎞
3⎟
3
3
× ⎜
= 40.32 mg KIO3
⎜ 1 mol IO – ⎟ ⎜⎝ 1 mol KIO3 ⎟⎠ ⎜⎝ 1 g KIO3 ⎟⎠
3 ⎠
⎝
5.81
No, the first, fourth, and fifth reactions were not necessary.
5.82
The balanced equation for the reaction is:
5H2C2O4 + 2MnO4– + 6H+ J 10CO2 + 2Mn2+ + 8H2O
M KMO4 =
5.83
⎛ 1 mol K 2 C2 O4 ⎞ ⎛ 2 mol KMnO 4 ⎞
(0.1244 g K 2 C2 O 4 ) ⎜
⎟⎜
⎟
⎝ 166.2 g K 2 C2 O 4 ⎠ ⎝ 5 mol K 2 C2 O 4 ⎠ = 0.02149 M KMnO
4
⎛ 1L ⎞
(13.93 mL KMnO4 ) ⎜
⎟
⎝ 1000 mL ⎠
The balanced equation for the oxidation-reduction reaction is:
3H2C2O4 + Cr2O72– + 8H+ J 6CO2 + 2Cr3+ + 7H2O
⎛ 0.200 moles K 2 Cr2 O7 ⎞ ⎛ 3 moles H 2 C2 O4 ⎞
mol H2C2O4 = (6.25 mL K2Cr2O7) ⎜
⎟⎜
⎟
⎝ 1000 mL K 2 Cr2 O7 ⎠ ⎝ 1 mole K 2 Cr2 O7 ⎠
= 3.75 × 10–3 mol H2C2O4
So, if we titrate the same oxalic acid solution using NaOH we will need:
⎛ 2 moles NaOH
ml NaOH = (3.75 × 10–3 mol H2C2O4) ⎜
⎝ 1 mole H 2 C2 O4
5.84
⎞ ⎛ 1000 mL NaOH ⎞
⎟⎜
⎟ = 16.7 mL NaOH
⎠ ⎝ 0.450 moles NaOH ⎠
It is first necessary to write a balanced equation for the reaction of MnO4– with Sn2+.
Sn2+ J Sn4+ + 2e–
MnO4– + 8H+ + 5e– J Mn2+ + 4H2O
5Sn2+ + 2MnO4– + 16H+ J 5Sn4+ + 2Mn2+ + 8H2O
Then, we calculate the original moles of Sn2+ in the 50.0 mL of 0.0300 M SnCl2 solution.
⎛ 0.0300 mol Sn 2+ ⎞
mol Sn2+ = (50.0 mL Sn2+) ⎜
⎟ = 0.0015 mol Sn2+
⎜
⎟
1000 mL
⎝
⎠
113
Chapter 5
The moles of Sn2+ that were titrated are calculated by multiplying (remembering to include stoichiometry)
molarity (0.0100 M) by volume of titrant (0.02728 L).
⎛ 0.0100 mol MnO − ⎞ ⎛ 5 mol Sn 2+ ⎞
4 ⎟⎜
⎟ = 6.82 × 10–4 mol Sn2+
mol Sn2+ = (0.02728 L MnO4–) ⎜
⎜
⎟ ⎜ 2 mol MnO − ⎟
1
L
4 ⎠
⎝
⎠⎝
This number of moles of tin ion remaining is subtracted from the total that was available in the 50.0 mL
portion that was titrated,
mol Sn2+ remaining = 0.0015 mol Sn2+ – 6.82 × 10–4 mol Sn2+ = 8.18 × 10–4 mol Sn2+
and the answer is converted to the number of moles of MnO4– that had reacted with this number of moles of Sn2+.
⎛ 2 mol MnO − ⎞
4 ⎟ = 3.27 × 10–4 mol MnO –
mol MnO4– = (8.18 × 10–4 mol Sn2+) ⎜
4
⎜ 5 mol Sn 2+ ⎟
⎝
⎠
Multiply this number by 10 to get the moles of MnO4– that had not reacted with the SO2 in the original 500
mL of 0.0200 M KMnO4.
3.27 × 10–4 mol MnO4– × 10 = 3.27 × 10–3 mol MnO4–
By difference, calculate the moles of SO2 that had reacted, which is equal to the number of moles of S in
the original sample.
⎛ 0.0200 mol MnO − ⎞
4 ⎟
mol MnO4– added to SO2 = (500 mL MnO4) ⎜
⎜
⎟
1000
mL
⎝
⎠
= 0.0100 mol MnO4–
Balanced reaction of SO2 and MnO4
SO2 + 2H2O J SO42– + 4H+ + 2e–
MnO4– + 8H+ + 5e– J Mn2+ + 4H2O
5SO2 + 2MnO4– + 2H2O J 2Mn2+ + 5SO42– + 4H+
⎛ 5 mol SO
2
mol SO2 = (0.0100 mol MnO4– – 3.27 × 10–3 mol MnO4–) ⎜
⎜ 2 mol MnO −
4
⎝
The mass of S is calculated by dividing moles by atomic mass,
⎛ 1 mol S ⎞ ⎛ 32.067 g S ⎞
g SO2 = (0.0168 mol SO2) ⎜
⎟⎜
⎟ = 0.540 g S
⎝ 1 mol SO 2 ⎠ ⎝ 1 mol S ⎠
⎞
⎟ = 0.0168 mol SO2
⎟
⎠
and the percentage of S in the original sample is the mass of S divided by the total sample mass, times 100.
0.540 g S
× 100% = 51.7% S
%S=
1.045 g sample
5.85
(a)
–2
(b)
0
(c)
5.86
(a)
2NBr3 + 6e– J N2 + 6Br–
2NBr3 + 6H2O J N2 + 6HOBr + 6H+ + 6e–
4NBr3 + 6H2O J 2N2 + 6HOBr + 6Br– + 6H+
Add 6OH– to both sides of the above equation:
4NBr3 + 6H2O + 6OH– J 2N2 + 6HOBr + 6Br– + 6H2O
which simplifies to give:
4NBr3 + 6OH– J 2N2 + 6HOBr + 6Br–
or
2NBr3 + 3OH– J N2 + 3HOBr + 3Br–
(b)
(Cl2 + 2e– J 2Cl–) × 5
Cl2 + 6H2O J 2ClO3– + 12H+ + 10e–
6Cl2 + 6H2O J 2ClO3– + 10Cl– + 12H+
Adding 12OH– to both sides gives:
6Cl2 + 6H2O + 12OH– J 2ClO3– + 10Cl– + 12H2O
which simplifies to:
6Cl2 + 12OH– J 2ClO3– + 10Cl– + 6H2O
or
3Cl2 + 6OH– J ClO3– + 5Cl– + 3H2O
114
+4
(d)
+4
Chapter 5
(c)
H2SeO3 + 4H+ + 4e– J Se + 3H2O
(H2S J S + 2H+ + 2e–) × 2
2H2S + H2SeO3 + 4H+ J 2S + 4H+ + Se + 3H2O
which simplifies to give:
2H2S + H2SeO3 J 2S + Se + 3H2O
(d)
MnO2 + 4H+ + 2e– J Mn2+ + 2H2O
2SO32– J S2O62– + 2e–
2SO32– + MnO2 + 4H+ J Mn2+ + S2O62– + 2H2O
(e)
XeO3 + 6H+ + 6e– J Xe + 3H2O
(2I– J I2 + 2e–) × 3
XeO3 + 6I– + 6H+ J 3I2 + Xe + 3H2O
(f)
(CN)2 + 2e– J 2CN–
(CN)2 + 2H2O J 2OCN– + 4H+ + 2e–
2(CN)2 + 2H2O J 2CN– + 2OCN– + 4H+
Adding 4OH– to both sides gives:
2(CN)2 + 4OH– + 2H2O J 2CN– + 2OCN– + 4H2O
which simplifies to:
(CN)2 + 2OH–J CN–+ OCN– + H2O
5.87
Any metal that is lower than hydrogen in the activity series shown in Table 5.2 of the text will react with
H+: (c) zinc and (d) magnesium.
5.88
Total charge = –2 = (charge of sulfur atoms) + (charge of oxygen atoms)
Charge of oxygens = 6(–2) = –12
Charge of sulfur atoms = –2 –(–12) = +10
+10 spread out over 4 sulfur atoms gives a charge of: +10/4 per S atom, or:
Oxidation number of S = +2.5
5.89
The balanced equation is the place to start.
6Fe2+(aq) + Cr2O72–(aq) + 14H+(aq) → 6Fe3+(aq) + 2Cr3+(aq) + 7H2O(A)
Because amounts of both reagents are specified, we must work a limiting reactant problem to find out
which of the two reactants is completely consumed. From the number of moles of this reactant that
disappear, we can calculate the number of moles of H+ that react. This amount is subtracted from the initial
number of moles of hydrogen ion, and the amount of titrant is calculated by dividing moles by molarity of
NaOH solution.
If Fe2+ is the limiting reactant:
⎛
⎛ 0.060 mol Fe 2+
mL NaOH = ⎜ 0.400 mol H + − 400 mL Fe2+ ⎜
⎜ 1000 mL Fe2+
⎜
⎝
⎝
⎛ 1000 mL NaOH ⎞
× ⎜
⎟ = 34,400 mL NaOH
⎝ 0.0100 mol NaOH ⎠
(
)
115
⎞⎛ 14 mol H +
⎟⎜
⎟⎜ 6 mol Fe2+
⎠⎝
⎞ ⎞ ⎛ 1 mol NaOH ⎞
⎟⎟⎜
⎟ ⎟ ⎝ 1 mol H + ⎠⎟
⎠⎠
Chapter 5
If Cr2O72– is the limiting reactant:
mL NaOH =
⎛
⎜ 0.400 mol H + − 300 mL Cr2 O7 2−
⎜
⎝
(
2−
2 7
2−
2 7
⎛
mol Cr O
) ⎜⎜ 0.0200
1000 mL Cr O
⎝
⎞ ⎛ 14 mol H +
⎟⎜
⎟ ⎜ 1 mol Cr O 2−
2 7
⎠⎝
⎞ ⎞ ⎛ 1 mol NaOH ⎞
⎟⎟⎜
⎟ ⎟ ⎝ 1 mol H + ⎟⎠
⎠⎠
⎛ 1000 mL NaOH ⎞
× ⎜
⎟ = 31,600 mL NaOH
⎝ 0.0100 mol NaOH ⎠
Na2Cr2O7 is the limiting reagent, therefore 31,600 mL of NaOH is needed.
5.90
To begin, determine balanced equations for the reaction of SO32– with CrO42– and for S2O32– with CrO42–.
3SO32– + 2CrO42– + H2O J 3SO42– + 2CrO2– + 2OH–
3S2O32– + 8CrO42– + H2O J 6SO42– + 8CrO2– + 2OH–
We know that the amount of CrO42– that reacted is
⎛ 0.0500 moles CrO 2− ⎞
4
mol CrO42– = (80 mL CrO42–) ⎜
⎟ = 4.00 × 10–3 mol CrO42–
⎜ 1000 mL CrO 2− ⎟
4
⎝
⎠
We also know the amount of SO42– produced
⎛ 1 mol BaSO 4 ⎞ ⎛ 1 mol SO 42− ⎞
mol SO42– = (0.9336 g BaO4–) ⎜
⎟ = 4.00 × 10–3 mol SO42
⎟ ⎜⎜
⎟
233.39
g
BaSO
1
mol
BaSO
4
4
⎝
⎠⎝
⎠
Let x = moles SO32– and y = moles S2O32– in the 100 mL sample, from the balanced equations and the
known quantities we can write:
⎛ 1 mol SO 2− ⎞
⎛ 2 mol SO 2− ⎞
4
4
⎟ + y⎜
⎟
4.00 × 10–3 mol SO42– = x ⎜
⎜ 1 mol SO 2− ⎟
⎜ 1 mol S O 2− ⎟
3
2 3 ⎠
⎝
⎠
⎝
⎛ 2 mol CrO 2−
4
4.00 × 10–3 mol CrO42– = x ⎜
⎜ 3 mol SO 2−
3
⎝
⎞
⎛ 8 mol CrO 2−
4
⎟ + y⎜
⎟
⎜ 3 mol S O 2−
2 3
⎠
⎝
⎞
⎟
⎟
⎠
We can solve for x and y from these and calculate the initial concentration of SO32– and S2O32–.
⎛ 1 mol SO32− ⎞ ⎡ ⎛
2 mol SO 42− ⎞ ⎛ 1 mol SO32− ⎞ ⎤
⎟
–3
2– ⎜
⎟⎜
⎟⎥
x = 4.00 × 10 mol SO4 ⎜
2− ⎟ – ⎢ y ⎜
⎝ 1 mol SO4 ⎠ ⎢ ⎜⎝ 1 mol S2 O32− ⎟⎠ ⎜⎝ 1 mol SO 42− ⎟⎠ ⎥
⎣
⎦
⎛ 1 mol SO32−
x = 4.00 × 10–3 ⎜
⎜
1
⎝
⎞ ⎡ ⎛ 2 mol SO32−
⎟ – ⎢ y⎜
⎟ ⎢ ⎜ 1 mol S O 2−
2 3
⎠ ⎣ ⎝
⎞⎤
⎟⎥
⎟⎥
⎠⎦
⎡ ⎛ 2 mol SO 2− ⎞ ⎤
3
⎟⎥
x = 4.00 × 10–3 mol SO32– – ⎢ y ⎜
⎢⎣ ⎜⎝ 1 mol S2 O32− ⎟⎠ ⎥⎦
Substitute x
⎛ 2 mol CrO 2− ⎞
⎛ 8 mol CrO 2−
4
4
⎟ + y⎜
4.00 × 10–3 mol CrO42– = x ⎜
⎜ 3 mol SO 2− ⎟
⎜ 3 mol S O 2−
3
2 3
⎝
⎠
⎝
116
⎞
⎟
⎟
⎠
Chapter 5
4.00 × 10–3 mol CrO42– =
⎧
⎡ ⎛ 2 mol SO 2−
⎪
−3
2−
3
⎢ y⎜
4.00
10
mol
SO
×
−
⎨
3
2−
⎜
⎢
⎣ ⎝ 1 mol S2 O3
⎩⎪
⎞ ⎤ ⎫⎪ ⎛ 2 mol CrO 2−
4
⎟⎥ ⎜
⎟ ⎥ ⎪⎬ ⎜ 3 mol SO 2−
3
⎠⎦ ⎭ ⎝
⎞
⎛ 8 mol CrO 2−
4
⎟ + y⎜
⎟
⎜ 3 mol S O 2−
2 3
⎠
⎝
4.00 × 10–3 mol CrO42– =
⎛ 2 mol CrO 2−
4
4.00 × 10−3 mol SO32− ⎜
⎜ 3 mol SO 2−
3
⎝
⎛ 8 mol CrO 2−
4
+ y⎜
⎜ 3 mol S O 2−
2 3
⎝
4.00 × 10–3 mol CrO42– =
⎞ ⎡
⎟−⎢
⎟ ⎢
⎠ ⎣
⎛ 2 mol SO 2−
3
y⎜
⎜ 1 mol S O 2−
2 3
⎝
⎞ ⎤ ⎛ 2 mol CrO 2−
4
⎟⎥ ⎜
⎟ ⎥ ⎜ 3 mol SO 2−
3
⎠⎦ ⎝
⎞
⎟
⎟
⎠
⎞
⎟
⎟
⎠
⎛ 1 mol CrO 2−
4
2.667 × 10−3 mol CrO 42− − 1.333 y ⎜
⎜ 1 mol S O 2−
2 3
⎝
⎛ 1 mol CrO 2−
4
1.333 × 10–3 mol CrO42– = 1.333 y ⎜
⎜ 1 mol S O 2−
2 3
⎝
y = 1.000 × 10–3 mol S2O32–
Substitute y back into the x equation:
⎡ ⎛ 2 mol SO 2
3
x = 4.00 × 10–3 mol SO32– – ⎢ y ⎜
⎢⎣ ⎜⎝ 1 mol S2 O32
⎞
⎛ 8 mol CrO 2−
4
⎟ + y⎜
⎟
⎜ 3 mol S O 2−
2 3
⎠
⎝
⎞
⎟
⎟
⎠
⎞
⎟
⎟
⎠
⎞⎤
⎟⎥
⎟⎥
⎠⎦
⎡
x = 4.00 × 10–3 mol SO32– – ⎢1.000 × 10-3 mol S2 O32
⎢⎣
–3
2–
x = 4.00 × 10 mol SO3 – 2.00 × 10–3 mol SO32–
x = 2.00 × 10–3 mol SO32–
Therefore the concentration of SO32– is:
⎛ 2 mol SO 2
3
⎜
⎜ 1 mol S O 2
2 3
⎝
⎞⎤
⎟⎥
⎟⎥
⎠⎦
2.00 × 10−3 mol SO32−
= 0.0200 M SO32–
0.100 L solution
And the concentration of S2O32– is
1.00 × 10−3 mol S2 O32−
= 0.0100 M S2O32–
0.100 L solution
5.91
We choose the metal that is lower (more reactive) in the activity series shown in Table 5.2: (a) aluminum
(b) zinc (c) magnesium
5.92
The first reaction demonstrates that Al is more readily oxidized than Cu. The second reaction demonstrates
that Al is more readily oxidized than Fe. Reaction 3 demonstrates that Fe is more readily oxidized than Pb.
Reaction 4 demonstrates that Fe is more readily oxidized than Cu. The fifth reaction demonstrates that Al
is more readily oxidized than Pb. The last reaction demonstrates that Pb is more readily oxidized than Cu.
Altogether, the above facts constitute the following trend of increasing ease of oxidation:
Cu < Pb < Fe < Al
5.93
C12H22O11(s) + 12O2(g) J 12CO2(g) + 11H2O(l)
117
⎞
⎟
⎟
⎠
Chapter 5
5.94
The oxidation state of cerium in the reactant ion is +4. The number of moles of this ion in the reactant
solution is:
0.0150 M × 0.02500 L = 3.75 × 10–4 mol Ce4+
The number of moles of electrons that come from the Fe2+ reducing agent is:
0.0320 M × 0.02344 L = 7.50 × 10–4 mol e–
The ratio of moles of electrons to moles of Ce4+ reactant is therefore 2:1, and we conclude that the product
is Ce2+.
5.95
(a)
(b)
(c)
(d)
(e)
5.96
The reaction that occurs is 2Ag+(aq) + Cu(s) J 2Ag(s) + Cu2+(aq). If we assume that there is excess copper
available, we need to determine the number of moles of Ag that will be produced.
Zn + Sn2+ J Zn2+ + Sn
2Cr + 6H+ J 2Cr3+ + 3H2
Pb + Cd2+ J N. R.
Zn + Co2+ J Zn2+ + Co
Mn + Pb2+ J Mn2+ + Pb
The number of moles of Ag+ available for the reaction is
0.250 M × 0.0500 L = 0.0125 mol Ag+
We can determine the amount of copper consumed from the balanced equation. Since the stoichiometry is
2/1, the number of moles of Cu2+ ion that are consumed is 0.0125 ÷ 2 = 0.00625 mol. Convert this nmber
of moles to a number of grams: 0.00625 mol × 63.546 g/mol = 0.397 g. The amount of unreacted copper is
thus: 32.00 g – 0.397 g = 31.60 g Cu.
The mass of Ag that is formed is: 0.0125 mol × 107.9 g/mol = 1.35 g Ag.
The final mass of the bar will include the unreacted copper and the silver that is formed:
31.60 g + 1.35 g = 32.95 g.
5.97
First, balance the equation:
4H+ + 2e– + PbO2 J Pb2+ + 2H2O
2Cl– J Cl2 +2e–
4H+ + PbO2 +2Cl– J Pb2+ + 2H2O + Cl2
Then calculate the number of grams of PbO2.
⎛ 1 mol Cl2 ⎞⎛ 1 mol PbO 2 ⎞⎛ 239.2 g PbO2 ⎞
g PbO2 = 15.0 g Cl2 ⎜
⎟⎜
⎟⎜
⎟ = 50.6 g PbO2
⎝ 70.91 g Cl2 ⎠⎝ 1 mol Cl2 ⎠⎝ 1 mol PbO 2 ⎠
5.98
Element oxidized:
Element reduced:
Oxidizing agent:
Reducing agent:
Cl
Cl
NaOCl
NaClO2
118