CHAPTER 17
OXIDATION-REDUCTION
SOLUTIONS TO REVIEW QUESTIONS
1. Oxidation of a metal occurs when the metal loses electrons. The easier it is for a metal to lose electrons,
the more active the metal is.
2. (a)
(b)
Iodine is oxidized. Its oxidation number increases from 0 to þ5.
Chlorine is reduced. Its oxidation number decreases from 0 to 1.
3. The higher metal on the activity series list is more reactive.
(a) Ca
(b) Fe
(c) Zn
4. If the free element is higher on the activity series than the ion with which it is paired, the reaction occurs.
The higher metal on the activity series is more reactive.
(a) 2 Al(s) þ 3 ZnCl2(aq) ! 3 Zn(s) þ 2 AlCl3(aq)
(b) Sn(s) þ 2 HCl(aq) ! H2(g) þ SnCl2(aq)
(c) Ag(s) þ H2SO4(aq) ! no reaction H2 is more reactive than Ag
(d) Fe(s) þ 2 AgNO3(aq) ! 2 Ag(s) þ Fe(NO3)2(aq)
(e) 2 Cr(s) þ 3 Ni2þ(aq) ! 3 Ni(s) þ 2 Cr3þ(aq)
(f) Mg(s) þ Ca2þ(aq) ! no reaction
Ca is more reactive than Mg
(g) Cu(s) þ Hþ(aq) ! no reaction
H2 is more reactive than Cu
Al is more reactive than Ag
(h) Ag(s) þ Al3þ(aq) ! no reaction
5. If a copper wire is placed into a solution of lead (II) nitrate, no reaction will occur. Lead is more active
than copper and undergoes oxidation more readily than copper. Lead is already oxidized, therefore will
stay oxidized in the presence of copper.
6. (a)
(b)
(c)
(d)
2 Al þ Fe2O3 ! A12O3 þ 2 Fe þ Heat
Al is above Fe in the activity series, which indicates Al is more active than Fe.
No. Iron is less active than aluminum and will not displace aluminum from its compounds.
Yes. Aluminum is above chromium in the activity series and will displace Cr3þ from its
compounds.
7. (a)
2 Al(s) þ 6 HCl(aq) ! 2 AlCl3(aq) þ 3 H2(g)
2 Al(s) þ 3 H2SO4(aq) ! Al2(SO4)3(aq) þ 3 H2(g)
2 Cr(s) þ 6 HCl(aq) ! 2 CrCl3(aq) þ 3 H2(g)
2 Cr(s) þ 3 H2SO4(aq) ! Cr2(SO4)3(aq) þ 3 H2(g)
Au(s) þ HCl(aq) ! no reaction
Au(s) þ H2SO4(aq) ! no reaction
Fe(s) þ 2 HCl(aq) ! FeCl2(aq) þ H2(g)
Fe(s) þ H2SO4(aq) ! FeSO4(aq) þ H2(g)
Cu(s) þ HCl(aq) ! no reaction
Cu(s) þ H2SO4(aq) ! no reaction
(b)
(c)
(d)
(e)
- 226 -
- Chapter 17 Mg(s) þ 2 HCl(aq) ! MgCl2(aq) þ H2(g)
Mg(s) þ H2SO4(aq) ! MgSO4(aq) þ H2(g)
(g) Hg(l) þ HCl(aq) ! no reaction
Hg(l) þ H2SO4(aq) ! no reaction
(h) Zn(s) þ 2 HCl(aq) ! ZnCl2(aq) þ H2(g)
Zn(s) þ H2SO4(aq) ! ZnSO4(aq) þ H2(g)
(f)
8. The oxidation number for an atom in an ionic compound is the same as the charge of the ion that resulted
when that atom lost or gained electrons to form an ionic bond. In a covalently bonded compound
electrons are shared between the two atoms making up the bond. Those shared electrons are assigned to
the atom in the bond with a higher electronegativity giving it a negative oxidation number.
9. In an electrolytic cell the anode is the positively charged electrode and attracts negatively charged ions
(anions). The cathode is the negatively charged electrode and attracts positively charge ions (cations). In
a voltaic cell the anode is the negatively changed electrode where oxidation occurs. The cathode is the
positively charged electrode where reduction occurs.
10. (a)
Oxidation occurs at the anode. The reaction is
2 Cl ðaqÞ ! Cl2 ðgÞ þ 2 e
(b)
Reduction occurs at the cathode. The reaction is
Ni2þ ðaqÞ þ 2 e ! NiðsÞ
(c)
The net chemical reaction is
electrical
Ni2þ ðaqÞ þ 2 Cl ðaqÞ! NiðsÞ þ Cl2 ðgÞ
energy
11. In Figure 17.3, electrical energy is causing chemical reactions to occur. In Figure 17.4, chemical
reactions are used to produce electrical energy.
12. (a)
(b)
It would not be possible to monitor the voltage produced, but the reactions in the cell would still
occur.
If the salt bridge were removed, the reaction would stop. Ions must be mobile to maintain an
electrical neutrality of ions in solution. The two solutions would be isolated with no complete
electrical circuit.
13. Oxidation and reduction are complementary processes because one does not occur without the other. The
loss of e in oxidation is accompanied by a gain of e in reduction.
14.
Ca2þ þ 2 e ! Ca
2 Br ! Br2 þ 2 e
cathode reaction, reduction
anode reaction, oxidation
15. During electroplating of metals, the metal is plated by reducing the positive ions of the metal in the
solution. The plating will occur at the cathode, the source of the electrons. With an alternating current, the
polarity of the electrode would be constantly changing, so at one instant the metal would be plating and
the next instant the metal would be dissolving.
16. Since lead dioxide and lead(II) sulfate are insoluble, it is unnecessary to have salt bridges in the cells of a
lead storage battery.
- 227 -
- Chapter 17 17. The electrolyte in a lead storage battery is dilute sulfuric acid. In the discharge cycle, SO42, is removed
from solution as it reacts with PbO2 and Hþ to form PbSO4(s) and H2O. Therefore, the electrolyte
solution contains less H2SO4 and becomes less dense.
18. If Hg2þ ions are reduced to metallic mercury, this would occur at the cathode, because reduction takes
place at the cathode.
19. In both electrolytic and voltaic cells, oxidation and reduction reactions occur. In an electrolytic cell an
electric current is forced through the cell causing a chemical change to occur. In voltaic cells,
spontaneous chemical changes occur, generating an electric current.
20. In some voltaic cells, the reactants at the electrodes are in solution. For the cell to function, these
reactants must be kept separated. A salt bridge permits movement of ions in the cell. This keeps the
solution neutral with respect to the charged particles (ions) in the solution.
- 228 -
- Chapter 17 -
SOLUTIONS TO EXERCISES
1. Oxidation numbers of each element in the compound
Cu ¼ þ2
C ¼ þ4
O ¼ 2
(a) CuCO3
(b) CH4
C ¼ 4
H ¼ þl
(c) IF
I ¼ þ1
F ¼ l
(d) CH2Cl2
C¼0
H ¼ þ1
Cl ¼ 1
(e) SO2
S ¼ þ4
O ¼ 2
Rb ¼ þ1
Cr ¼ þ6
O¼2
(f) Rb2CrO4
2. Oxidation numbers of each element in the compound.
C ¼ þ2
H ¼ þl
F ¼ 1
(a) CHF3
(b) P2O5
P ¼ þ5
O ¼ 2
S ¼ þ6
F ¼ 1
(c) SF6
(d) SnSO4
Sn ¼ þ2
S ¼ þ6
O ¼ 2
C ¼ 2
H ¼ þl
O ¼ 2
(e) CH3OH
(f) H3PO4
H ¼ þl
P ¼ þ5
O ¼ 2
3. The oxidation number of the underlined element is indicated by the number following the formula.
(a) P
_ O33 þ 3
(c) NaHCO3 þ 4
(d) B
_ rO4 þ 7
(b) CaSO4 þ 6
4. The oxidation number of the underlined element is indicated by the number following the formula.
(c) NaH2PO4 þ 5
(a) C
_ O32 þ 4
(b) H2SO4 þ 6
(d) C
_ r 2 O72 þ 6
5. Balanced half-reaction
(a)
(b)
(c)
(d)
Na ! Naþ þ l e
C2 O42 ! 2 CO2 þ 2 e
2 I ! I2 þ 2 e
Cr2 O72 þ 14 Hþ þ 6 e ! 2 Cr3þ þ 7 H2 O
Element
Changing
Na
C
I
Cr
6. Balanced half-reaction
(a)
(b)
(c)
(d)
7. (a)
(b)
8. (a)
(b)
Type of
Reaction
oxidation
oxidation
oxidation
reduction
Element
Changing
Cu
F
I
Mn
Cu2þ þ 1e ! Cu1þ
F 2 þ 2 e ! 2 F
2 IO4 þ 16 Hþ þ 14 e ! I2 þ 8 H2 O
Mn ! Mn2þ þ 2e
Cu is oxidized, Ag is reduced;
Cu is the reducing agent, AgNO3 is the oxidizing agent
Zn is oxidized, H is reduced
Zn is the reducing agent, HCl is the oxidizing agent
C is oxidized, O is reduced
CH4 is the reducing agent, O2 is the oxidizing agent
Mg is oxidized, Fe is reduced
Mg is the reducing agent, FeCl3 is the oxidizing agent
- 229 -
Type of
Reaction
reduction
reduction
reduction
oxidation
- Chapter 17 9. (a)
(b)
(c)
correctly balanced
correctly balanced
incorrectly balanced
MgðsÞ þ 2 HClðaqÞ ! Mg2þ ðaqÞ þ 2 Cl ðaqÞ þ H2 ðgÞ
(d)
incorrectly balanced
3 CH3 OHðaqÞ þ Cr2 O72 ðaqÞ þ 8 Hþ ðaqÞ ! 2 Cr3þ ðaqÞ þ 3 CH2 OðaqÞ þ 7 H2 OðlÞ
10. (a)
incorrectly balanced
3 MnO2 ðsÞ þ 4 AlðsÞ ! 3 MnðsÞ þ 2 Al2 O3 ðsÞ
(b)
(c)
(d)
correctly balanced
correctly balanced
incorrectly balanced
8 H2 OðlÞ þ 2 MnO4 ðaqÞ þ 7 S2 ðaqÞ ! 2 MnSðsÞ þ 16 OH ðaqÞ þ 5 SðsÞ
11. Balancing oxidation-reduction equations using the change-in-oxidation number method:
(a) Cu þ O2 ! CuO
ox
Cu0 ! Cu2þ þ 2 e
red
2O þ 4e ! 2O
0
2
Multiply by 2;
Add the equations; the 4 e cancel
2 Cu0 þ 2 O0 ! 2 Cu2þ þ 2 O2
Transfer the coefficients to the original equation appropriately.
2 Cu þ O2 ! 2 CuO
KClO3 ! KCl þ O2
(b)
ox
red
3 O2 ! O0 þ 6 e
5þ
Cl
þ 6 e ! Cl
Multiply by 2;
Multiply by 2, add; the 12 e cancel
2 Cl5þ þ 6 O2 ! 2 Cl þ 3 O2
Transfer the coefficients to the original equation appropriately.
2 KClO3 ! 2 KCl þ 3 O2
Ca þ H2 O ! CaðOHÞ2 þ H2
(c)
ox
Ca ! Ca2þ þ 2 e
red
2 Hþ þ 2 e ! H2
Add equations together; the 2 e cancel:
Ca þ 2 Hþ ! Ca2þ þ H2
Balance the equation by inspection.
Ca þ 2 H2 O ! CaðOHÞ2 þ H2
PbS þ H2 O2 ! PbSO4 þ H2 O
(d)
ox
S2 ! Sþ6 þ 8 e
red
2O þ 2 e ! 2 O2 Multiply by 4, add; the 8 e cancel:
S2 þ 8 O ! Sþ6 þ 4 O2
- 230 -
- Chapter 17 Transfer the coefficients to the original equation and complete the balancing
by inspection.
PbS þ 4 H2 O2 ! PbSO4 þ 4 H2 O
CH4 þ NO2 ! N2 þ CO2 þ H2 O
(e)
ox
C4 ! C4þ þ 8 e
red
N4þ þ 4 e ! N0
Multiply by 2; add the 8 e cancel
C4 þ 2 N4þ ! C4þ þ 2 N0
Transfer the coefficients to the original equation and complete the balancing
by inspection.
CH4 þ 2 NO2 ! N2 þ CO2 þ 2 H2 O
12. Balancing oxidation-reduction equations using the change-in-oxidation number method:
(a)
Cu þ AgNO3 ! Ag þ CuðNO3 Þ2
ox
Cu0 ! Cu2þ þ 2 e
red
Agþ þ 1 e ! Ag0
Multiply by 2, add; the 2 e cancel
Cu0 þ 2 Agþ ! Cu2þ þ 2 Ag0
Transfer the coefficients to the original equation.
Cu þ 2 AgNO3 ! 2 Ag þ CuðNO3 Þ2
MnO2 þ HCl ! MnCl2 þ Cl2 þ H2 O
(b)
ox
Cl ! Cl0 þ 1 e
red
Mn4þ þ 2 e ! Mn2þ
Multiply by 2, add; the 2 e cancel
2 Cl þ Mn4þ ! 2 Cl0 þ Mn2þ
Transfer the coefficients to the original equation and complete the balancing
by inspection.
MnO2 þ 4 HCl ! MnCl2 þ Cl2 þ 2 H2 O
HCl þ O2 ! Cl2 þ H2 O
(c)
red
2 O0 þ 4 e ! 2 O2
ox
2 Cl ! 2 Cl0 þ 2 e
Multiply by 2, add; the 4 e cancel
4 Cl þ 2 O0 ! 4 Cl0 þ 2 O2
Transfer the coefficients to the original equation and complete the balancing
by inspection.
4 HCl þ O2 ! 2 Cl2 þ 2 H2 O
- 231 -
- Chapter 17 Ag þ H2 S þ O2 ! Ag2 S þ H2 O
(d)
red
2 O0 þ 4 e ! 2 O2
ox
Ag ! Agþ þ 1 e
Multiply by 4, add; the 4 e cancel
2 Ag þ 2 O0 ! 2 Agþ þ 2 O2
Transfer the coefficients to the original equation and complete the balancing
by inspection.
4 Ag þ 2 H2 S þ O2 ! 2 Ag2 S þ 2 H2 O
KMnO4 þ CaC2 O4 þ H2 SO4 ! K2 SO4 þ MnSO4 þ CaSO4 þ CO2 þ H2 O
(e)
red
ox
Mn7þ þ 5 e ! Mn2þ
2C
3þ
! 2C
4þ
þ 2e
Multiply by 2;
Multiply by 5, add; the 10 e cancel
10 C3þ þ 2 Mn7þ ! 10 C4þ þ 2 Mn2þ
Transfer the coefficients to the original equation and complete the balancing
by inspection.
2 KMnO4 þ 5 CaC2 O4 þ 8 H2 SO4 ! K2 SO4 þ 2 MnSO4 þ 5 CaSO4
þ 10 CO2 þ 8 H2 O
13. Balancing ionic redox equations
(a) Zn þ NO3 ! Zn2þ þ NH4þ ðacidic solutionÞ
Step 1
Write half-reaction equations. Balance except H and O.
Zn ! Zn2þ
NO3 ! NH4þ
Step 2
Balance H and O using H2O and Hþ
Zn ! Zn2þ
Step 3
10 Hþ þ NO3 ! NH4þ þ 3 H2 O
Balance electrically with electrons
Zn ! Zn2þ þ 2 e
Step 4
10 Hþ þ NO3 þ 8 e ! NH4þ þ 3 H2 O
Equalize the loss and gain of electrons
4ðZn ! Zn2þ þ 2 e Þ
Step 5
10 Hþ þ NO3 þ 8 e ! NH4þ þ 3 H2 O
Add the half-reactions-electrons cancel
10 Hþ þ 4 Zn þ NO3 ! 4 Zn2þ þ NH4þ þ 3 H2 O
(b)
NO3 þ S ! NO2 þ SO42
Step 1
ðacidic solutionÞ
Write half-reaction equations. Balance except H and O.
S ! SO42
NO3 ! NO2
- 232 -
- Chapter 17 Step 2
Balance H and O using H2O and Hþ
4 H2 O þ S ! SO42 þ 8 Hþ
Step 3
2 Hþ þ NO3 ! NO2 þ H2 O
Balance electrically with electrons
4 H2 O þ S ! SO42 þ 8 Hþ þ 6 e
Step 4 and 5
2 Hþ þ NO3 þ e ! NO2 þ H2 O
Equalize the loss and gain of electrons; add the half-reactions
4 H2 O þ S ! SO42 þ 8 Hþ þ 6 e
6 ð2 Hþ þ NO3 þ e ! NO2 þ H2 OÞ
4 Hþ þ S þ 6 NO3 ! 6 NO2 þ SO42 þ 2 H2 O
4 H2O, 8 Hþ and 6e canceled from each side
(c)
PH3 þ I2 ! H3 PO2 þ I
Step 1
ðacidic solutionÞ
Write half-reaction equations. Balance except H and O.
PH3 ! H3 PO2
Step 2
I2 ! 2 I
Balance H and O using H2O and Hþ
2 H2 O þ PH3 ! H3 PO2 þ 4 Hþ
Step 3
I2 ! 2 I
Balance electrically with electrons
2 H2 O þ PH3 ! H3 PO2 þ 4 Hþ þ 4 e
Step 4 and 5
I2 þ 2 e ! 2 I
Equalize the loss and gain of electrons; add the half-reactions
2 H2 O þ PH3 ! H3 PO2 þ 4 Hþ þ 4 e
2 ðI2 þ 2 e ! 2 I Þ
PH3 þ 2 H2 O þ 2 I2 ! H3 PO2 þ 4 I þ 4 Hþ
(d)
Cu þ NO3 ! Cu2þ þ NO
ðacidic solutionÞ
Step 1
Write half-reaction equations. Balance except H and O.
Step 2
Cu ! Cu2þ
NO3 ! NO
Balance H and O using H2O and Hþ
Cu ! Cu2þ
4 Hþ þ NO3 ! NO þ 2 H2 O
Step 3
Balance electrically with electrons
Cu ! Cu2þ þ 2 e
4 Hþ þ NO3 þ 3 e ! NO þ 2 H2 O
- 233 -
- Chapter 17 Step 4 and 5
Equalize the loss and gain of electrons; add the half-reactions
3 ðCu ! Cu2þ þ 2 e Þ
2 ð4 Hþ þ NO3 þ 3 e ! NO þ 2 H2 OÞ
3 Cu þ 8 Hþ þ 2 NO3 ! 3 Cu2þ þ 2 NO þ 2 H2 O
(e)
ClO3 þ Cl ! Cl2
Step 1
ðacidic solutionÞ
Write half-reaction equations. Balance except H and O.
Cl ! Cl0
Step 2
ClO3 ! Cl0
Balance H and O using H2O and Hþ
Cl ! Cl0
Step 3
6 Hþ þ ClO3 ! Cl0 þ 3 H2 O
Balance electrically with electrons
Cl ! Cl0 þ e
Step 4 and 5
6 Hþ þ ClO3 þ 5 e ! Cl0 þ 3 H2 O
Equalize the loss and gain of electrons; add the half-reactions
5ðCl ! Cl0 þ e Þ
6 Hþ þ ClO3 þ 5 e ! Cl0 þ 3 H2 O
6 Hþ þ ClO3 þ 5 Cl ! 3 Cl2 þ 3 H2 O
14. (a)
ClO3 þ I ! I2 þ Cl
Step 1
Step 2
ðacidic solutionÞ
Write half-reaction equations. Balance except H and O.
2 I ! I2
ClO3 ! Cl
Balance H and O using H2O and Hþ
2 I ! I2
Step 3
Step 4 and 5
6 Hþ þ ClO3 ! Cl þ 3 H2 O
Balance electrically with electrons
2 I ! I2 þ 2 e
6 Hþ þ ClO3 þ 6 e ! Cl þ 3 H2 O
Equalize the loss and gain of electrons; add the half-reactions
3 ð2 I ! I2 þ 2 e Þ
(b)
6 Hþ þ ClO3 þ 6 e ! Cl þ 3 H2 O
6 Hþ þ ClO3 þ 6 I ! 3 I2 þ Cl þ 3 H2 O
Cr2 O72 þ Fe2þ ! Cr3þ þ Fe3þ ðacidic solutionÞ
Step 1
Write half-reaction equations. Balance except H and O.
Fe2þ ! Fe3þ
Cr2 O72 ! 2 Cr3þ
- 234 -
- Chapter 17 Step 2
Balance H and O using H2O and Hþ
Fe2þ ! Fe3þ
Step 3
14 Hþ þ Cr2 O72 ! 2 Cr3þ þ 7 H2 O
Balance electrically with electrons
Fe2þ ! Fe3þ þ e
Step 4 and 5
14 Hþ þ Cr2 O72 þ 6 e ! 2 Cr3þ þ 7 H2 O
Equalize the loss and gain of electrons; add the half-reactions
6ðFe2þ ! Fe3þ þ e Þ
14 Hþ þ Cr2 O72 þ 6 e ! 2 Cr3þ þ 7 H2 O
14 Hþ þ Cr2 O72 þ 6 Fe2þ ! 2 Cr3þ þ 6 Fe3þ þ 7 H2 O
(c)
MnO4 þ SO2 ! Mn2þ þ SO42
Step 1
ðacidic solutionÞ
Write half-reaction equations. Balance except H and O.
SO2 ! SO42
Step 2
MnO4 ! Mn2þ
Balance H and O using H2O and Hþ
2 H2 O þ SO2 ! SO42 þ 4 Hþ
Step 3
8 Hþ þ MnO4 ! Mn2þ þ 4 H2 O
Balance electrically with electrons
2 H2 O þ SO2 ! SO42 þ 4 Hþ þ 2 e
Step 4 and 5
8 Hþ þ MnO4 þ 5 e ! Mn2þ þ 4 H2 O
Equalize the loss and gain of electrons; add the half-reactions
5ð2 H2 O þ SO2 ! SO42 þ 4 Hþ þ 2 e Þ
2ð8 Hþ þ MnO4 þ 5 e ! Mn2þ þ 4 H2 OÞ
2 H2 O þ 2 MnO4 þ 5 SO2 ! 4 Hþ þ 2 Mn2þ þ 5 SO42
8 H2O, 16 Hþ, and 10 e canceled from each side
(d)
H3 AsO3 þ MnO4 ! H3 AsO4 þ Mn2þ
Step 1
Step 2
ðacidic solutionÞ
Write half-reaction equations. Balance except H and O.
H3 AsO3 ! H3 AsO4
MnO4 ! Mn2þ
Balance H and O using H2O and Hþ
H2 O þ H3 AsO3 ! 2 Hþ þ H3 AsO4
Step 3
8 Hþ þ MnO4 ! Mn2þ þ 4 H2 O
Balance electrically with electrons
H2 O þ H3 AsO3 ! 2 Hþ þ H3 AsO4 þ 2 e
8 Hþ þ MnO4 þ 5 e ! Mn2þ þ 4 H2 O
- 235 -
- Chapter 17 Step 4 and 5
Equalize the loss and gain of electrons; add the half-reactions
5ðH2 O þ H3 AsO3 ! 2 Hþ þ H3 AsO4 þ 2 e Þ
2ð8 Hþ þ MnO4 þ 5 e ! Mn2þ þ 4 H2 OÞ
6 Hþ þ 5 H3 AsO3 þ 2 MnO4 ! 5 H3 AsO4 þ 2 Mn2þ þ 3 H2 O
5 H2O, 10 Hþ, and 10 e canceled from each side
(e)
Cr2 O72 þ H3 AsO3 ! Cr3þ þ H3 AsO4 ðacidic solutionÞ
Step 1
Step 2
Write half-reaction equations. Balance except H and O.
H3 AsO3 ! H3 AsO4
Cr2 O72 ! 2 Cr3þ
Balance H and O using H2O and Hþ
H2 O þ H3 AsO3 ! 2 Hþ þ H3 AsO4
Step 3
14 Hþ þ Cr2 O72 ! 2 Cr3þ þ 7 H2 O
Balance electrically with electrons
H2 O þ H3 AsO3 ! 2 Hþ þ H3 AsO4 þ 2 e
Step 4 and 5
14 Hþ þ Cr2 O72 þ 6 e ! 2 Cr3þ þ 7 H2 O
Equalize the loss and gain of electrons; add the half-reactions
3 ðH2 O þ H3 AsO3 ! 2 Hþ þ H3 AsO4 þ 2 e Þ
14 Hþ þ Cr2 O72 þ 6 e ! 2 Cr3þ þ 7 H2 O
8 Hþ þ Cr2 O72 þ 3 H3 AsO3 ! 2 Cr3þ þ 3 H3 AsO4 þ 4 H2 O
3 H2O, 6 Hþ, and 6 e canceled from each side
15. (a)
Cl2 þ IO3 ! Cl þ IO4
ðbasic solutionÞ
Step 1
Write half-reaction equations. Balance except H and O.
Step 2
IO3 ! IO4
Cl2 ! 2 Cl
Balance H and O using H2O and Hþ
H2 O þ IO3 ! IO4 þ 2 Hþ
Step 3
Cl2 ! 2 Cl
Add OH ions to both sides (same number as Hþ ions)
2 OH þ H2 O þ IO3 ! IO4 þ 2 Hþ þ 2 OH
Step 4
Cl2 ! 2 Cl
Combine Hþ and OH to form H2O; cancel H2O where possible
2 OH þ H2 O þ IO3 ! IO4 þ 2 H2 O
Cl2 ! 2 Cl
2 OH þ IO3 ! IO4 þ H2 O
Cl2 ! 2 Cl
- 236 -
ð1 H2 O cancelledÞ
- Chapter 17 Step 5
Step 6
Step 7
(b)
Balance electrically with electrons
2 OH þ IO3 ! IO4 þ H2 O þ 2 e
Cl2 þ 2 e ! 2 Cl
Electron loss and gain is balanced
Add half-reactions
2 OH þ IO3 þ Cl2 ! IO4 þ 2 Cl þ H2 O
MnO4 þ ClO2 ! MnO2 þ ClO4
Step 1
Step 2
ðbasic solutionÞ
Write half-reaction equations. Balance except H and O.
ClO2 ! ClO4
MnO4 ! MnO2
Balance H and O using H2O and Hþ
2 H2 O þ ClO2 ! ClO4 þ 4 Hþ
Step 3
MnO4 þ 4 Hþ ! MnO2 þ 2 H2 O
Add OH ions to both sides (same number as Hþ ions)
4 OH þ 2 H2 O þ ClO2 ! ClO4 þ 4 Hþ þ 4 OH
Step 4
4 OH þ MnO4 þ 4 Hþ ! MnO2 þ 2 H2 O þ 4 OH
Combine Hþ and OH to form H2O; cancel H2O where possible
4 OH þ 2 H2 O þ ClO2 ! ClO4 þ 4 H2 O
4 H2 O þ MnO4 ! MnO2 þ 2 H2 O þ 4 OH
4 OH þ ClO2 ! ClO4 þ 2 H2 O
MnO4
Step 5
Step 6 and 7
(c)
2 H2 O þ
! MnO2 þ 4 OH
ð2 H2 O cancelledÞ
Balance electrically with electrons
4 OH þ ClO2 ! ClO4 þ 2 H2 O þ 4 e
2 H2 O þ MnO4 þ 3 e ! MnO2 þ 4 OH
Equalize gain and loss of electrons; add half-reactions
3 ð4 OH þ ClO2 ! ClO4 þ 2 H2 O þ 4 e Þ
4 ð2 H2 O þ MnO4 þ 3 e ! MnO2 þ 4 OH Þ
2 H2 O þ 4 MnO4 þ 3 ClO2 ! 4 MnO2 þ 3 ClO4 þ 4 OH
6 H2O, 12 OH, and 12 e canceled from each side
Se ! SeO32 þ Se2
Step 1
ðbasic solutionÞ
Write half-reaction equations. Balance except H and O.
Se ! SeO32
Step 2
ð2 H2 O cancelledÞ
Se ! Se2
Balance H and O using H2O and Hþ
3 H2 O þ Se ! SeO32 þ 6 Hþ
Se ! Se2
- 237 -
- Chapter 17 Step 3
Add OH ions to both sides (same number as Hþ ions)
6 OH þ 3 H2 O þ Se ! SeO32 þ 6 Hþ þ 6 OH
Step 4
Se ! Se2
Combine Hþ and OH to form H2O; cancel H2O where possible
6 OH þ 3 H2 O þ Se ! SeO32 þ 6 H2 O
Se ! Se2
Step 5
6 OH þ Se ! SeO32 þ 3 H2 O
Balance electrically with electrons
ð3 H2 O cancelledÞ
6 OH þ Se ! SeO32 þ 3 H2 O þ 4 e
Step 6 and 7
Se þ 2 e ! Se2
Equalize gain and loss of electrons; add half-reactions
6 OH þ Se ! SeO32 þ 3 H2 O þ 4 e
2ðSe þ 2 e ! Se2 Þ
6 OH þ 3 Se ! SeO32 þ 2 Se2 þ 3 H2 O
(d)
Fe3 O4 þ MnO4 ! Fe2 O3 þ MnO2
Step 1
ðbasic solutionÞ
Write half-reaction equations. Balance except H and O.
2 Fe3 O4 ! 3 Fe2 O3
Step 2
MnO4 ! MnO2
Balance H and O using H2O and Hþ
Step 3
H2 O þ 2 Fe3 O4 ! 3 Fe2 O3 þ 2 Hþ
4 Hþ þ MnO4 ! MnO2 þ 2 H2 O
Add OH ions to both sides (same number as Hþ ions)
Step 4
2 OH þ H2 O þ 2 Fe3 O4 ! 3 Fe2 O3 þ 2 Hþ þ 2 OH
4 OH þ 4 Hþ þ MnO4 ! MnO2 þ 2 H2 O þ 4 OH
Combine Hþ and OH to form H2O; cancel H2O where possible
2 OH þ H2 O þ 2 Fe3 O4 ! 3 Fe2 O3 þ 2 H2 O
4 H2 O þ MnO4 ! MnO2 þ 2 H2 O þ 4 OH
2 OH þ 2 Fe3 O4 ! 3 Fe2 O3 þ H2 O
MnO4
Step 5
Step 6 and 7
ð1 H2 O cancelledÞ
2 H2 O þ
! MnO2 þ 4 OH
ð2 H2 O cancelledÞ
Balance electrically with electrons
2 OH þ 2 Fe3 O4 ! 3 Fe2 O3 þ H2 O þ 2 e
2 H2 O þ MnO4 þ 3 e ! MnO2 þ 4 OH
Equalize gain and loss of electrons; add half-reactions
3ð2 OH þ 2 Fe3 O4 ! 3 Fe2 O3 þ H2 O þ 2 e Þ
2ð2 H2 O þ MnO4 þ 3 e ! MnO2 þ 4 OH Þ
H2 O þ 6 Fe3 O4 þ 2 MnO4 ! 9 Fe2 O3 þ 2 MnO2 þ 2 OH Þ
3 H2O, 6 OH, and 6 e canceled from each side
- 238 -
- Chapter 17 (e)
BrO þ CrðOHÞ4 ! Br þ CrO42
Step 1
ðbasic solutionÞ
Write half-reaction equations. Balance except H and O.
CrðOHÞ4 ! CrO42
Step 2
BrO ! Br
Balance H and O using H2O and Hþ
CrðOHÞ4 ! CrO42 þ 4 Hþ
Step 3
2 Hþ þ BrO ! Br þ H2 O
Add OH ions to both sides (same number as Hþ ions)
4 OH þ CrðOHÞ4 ! CrO42 þ 4 Hþ þ 4 OH
Step 4
2 OH þ 2 Hþ þ BrO ! Br þ H2 O þ 2 OH
Combine Hþ and OH to form H2O; cancel H2O where possible
4 OH þ CrðOHÞ4 ! CrO42 þ 4 H2 O
2 H2 O þ BrO ! Br þ H2 O þ 2 OH
Step 5
H2 O þ BrO ! Br þ 2 OH
Balance electrically with electrons
ð1 H2 O cancelledÞ
4 OH þ CrðOHÞ4 ! CrO42 þ 4 H2 O þ 3 e
Step 6 and 7
H2 O þ BrO þ 2 e ! Br þ 2 OH
Equalize gain and loss of electrons; add half-reactions
2 ð4 OH þ CrðOHÞ4 ! CrO42 þ 4 H2 O þ 3 e Þ
3 ðH2 O þ BrO þ 2 e ! Br þ 2 OH Þ
2 OH þ 3BrO þ 2 CrðOHÞ4 ! 3Br þ 2 CrO42 þ 5 H2 O
3 H2O, 6 OH and 6 e canceled from each side
16. (a)
MnO4 þ SO32 ! MnO2 þ SO42
Step 1
ðbasic solutionÞ
Write half-reaction equations. Balance except H and O.
SO32 ! SO42
Step 2
MnO4 ! MnO2
Balance H and O using H2O and Hþ
H2 O þ SO32 ! SO42 þ 2 Hþ
Step 3
MnO4 þ 4 Hþ ! MnO2 þ 2 H2 O
Add OH ions to both sides (same number as Hþ ions)
2 OH þ H2 O þ SO32 ! SO42 þ 2 Hþ þ 2 OH
4 OH þ MnO4 þ 4 Hþ ! MnO2 þ 2 H2 O þ 4 OH
- 239 -
- Chapter 17 Step 4
Combine Hþ and OH to form H2O; cancel H2O where possible
2 OH þ H2 O þ SO32 ! SO42 þ 2 H2 O
MnO4 þ 4 H2 O ! MnO2 þ 2 H2 O þ 4 OH
2 OH þ SO32 ! SO42 þ H2 O ð1 H2 O cancelledÞ
Step 5
MnO4 þ 2 H2 O ! MnO2 þ 4 OH
Balance electrically with electrons
ð2 H2 O cancelledÞ
2 OH þ SO32 ! SO42 þ H2 O þ 2 e
Step 6 and 7
3 e þ MnO4 þ 2 H2 O ! MnO2 þ 4 OH
Equalize gain and loss of electrons; add half-reactions
3 ð2 OH þ SO32 ! SO42 þ H2 O þ 2 e Þ
2 ðMnO4 þ 2 H2 O þ 3 e ! MnO2 þ 4 OH Þ
H2 O þ 2 MnO4 þ 3 SO32 ! 2 MnO2 þ 3 SO42 þ 2 OH
3 H2O, 4 OH, and 6 e canceled from each side
(b)
ClO2 þ SbO2 ! ClO2 þ SbðOHÞ6 þ 2 H2 O ðbasic solutionÞ
Step 1
Write half-reaction equations. Balance except H and O.
SbO2 ! SbðOHÞ6
Step 2
ClO2 ! ClO2
Balance H and O using H2O and Hþ
4 H2 O þ SbO2 ! SbðOHÞ6 þ 2 Hþ
Step 3
ClO2 ! ClO2
Add OH ions to both sides (same number as Hþ ions)
2 OH þ 4 H2 O þ SbO2 ! SbðOHÞ6 þ 2 Hþ þ 2 OH
Step 4
ClO2 ! ClO2
Combine Hþ and OH to form H2 O; cancel H2 O where possible
2 OH þ 4 H2 O þ SbO2 ! SbðOHÞ6 þ 2 H2 O
ClO2 ! ClO2
Step 5
Step 6 and 7
2 OH þ 2 H2 O þ SbO2 ! SbðOHÞ6
ð2 H2 O cancelledÞ
Balance electrically with electrons
2 OH þ 2 H2 O þ SbO2 ! SbðOHÞ6 þ 2 e
ClO2 þ e ! ClO2
Equalize gain and loss of electrons; add half-reactions
2 H2 O þ 2 OH þ SbO2 ! SbðOHÞ6 þ 2 e
2 ClO2 þ e ! ClO2
2 H2 O þ 2 ClO2 þ 2 OH þ SbO2 ! 2 ClO2 þ SbðOHÞ6
- 240 -
- Chapter 17 (c)
Al þ NO3 ! NH3 þ AlðOHÞ4
Step 1
Step 2
ðbasic solutionÞ
Write half-reaction equations. Balance except H and O.
Al ! AlðOHÞ4
NO3 ! NH3
Balance H and O using H2 O and Hþ
4 H2 O þ Al ! AlðOHÞ4 þ 4 Hþ
Step 3
9 Hþ þ NO3 ! NH3 þ 3 H2 O
Add OH ions to both sides (same number as Hþ ions)
4 OH þ 4 H2 O þ Al ! AlðOHÞ4 þ 4 Hþ þ 4 OH
Step 4
9 OH þ 9 Hþ þ NO3 ! NH3 þ 3 H2 O þ 9 OH
Combine Hþ and OH to form H2 O; cancel H2 O where possible
4 OH þ 4 H2 O þ Al ! AlðOHÞ4 þ 4 H2 O
9 H2 O þ NO3 ! NH3 þ 3 H2 O þ 9 OH
4 OH þ Al ! AlðOHÞ4
NO3
Step 5
Step 6 and 7
ð4 H2 O cancelledÞ
6 H2 O þ
! NH3 þ 9 OH ð3 H2 O cancelledÞ
Balance electrically with electrons
4 OH þ Al ! AlðOHÞ4 þ 3 e
6 H2 O þ NO3 þ 8 e ! NH3 þ 9 OH
Equalize gain and loss of electrons; add half-reactions
8 4 OH þ Al ! AlðOHÞ4 þ 3 e
3 6 H2 O þ NO3 þ 8 e ! NH3 þ 9 OH
8 Al þ 3 NO3 þ 18 H2 O þ 5 OH ! 3 NH3 þ 8 AlðOHÞ4
27 OH and 24 e canceled from each side
(d)
P4 ! HPO32 þ PH3
Step 1
ðbasic solutionÞ
Write half-reaction equations. Balance except H and O.
P4 ! 4 HPO32
Step 2
P4 ! 4 PH3
Balance H and O using H2 O and Hþ
12 H2 O þ P4 ! 4 HPO32 þ 20 Hþ
Step 3
12 Hþ þ P4 ! 4 PH3
Add OH ions to both sides (same number as Hþ ions)
20 OH þ 12 H2 O þ P4 ! 4 HPO32 þ 20 Hþ þ 20 OH
12 OH þ 12 Hþ þ P4 ! 4 PH3 þ 12 OH
- 241 -
- Chapter 17 Step 4
Combine Hþ and OH to form H2 O; cancel H2 O where possible
20 OH þ 12 H2 O þ P4 ! 4 HPO32 þ 20 H2 O
12 H2 O þ P4 ! 4 PH3 þ 12 OH
Step 5
20 OH þ P4 ! 4 HPO32 þ 8 H2 O ð12 H2 O cancelledÞ
Balance electrically with electrons
20 OH þ P4 ! 4 HPO32 þ 8 H2 O þ 12 e
Step 6 and 7
12 H2 O þ P4 þ 12 e ! 4 PH3 þ 12 OH
Loss and gain of electrons are equal; add half-reactions
8 OH þ 4 H2 O þ 2 P4 ! 4 HPO32 þ 4 PH3
Divide equation by 2
4 OH þ 2 H2 O þ P4 ! 2 HPO32 þ 2 PH3
(e)
Al þ OH ! AlðOHÞ4 þ H2 ðbasic solutionÞ
Step 1
Step 2
Write half-reaction equations. Balance except H and O.
Al ! AlðOHÞ4
OH ! H2
Balance H and O using H2O and Hþ
4 H2 O þ Al ! AlðOHÞ4 þ 4 Hþ
Step 3
4 Hþ þ OH ! H2 þ H2 O
Add OH ions to both sides (same number as Hþ ions)
4 OH þ 4 H2 O þ Al ! AlðOHÞ4 þ 4 Hþ þ 4 OH
Step 4
3 OH þ 3 Hþ þ OH ! H2 þ H2 O þ 3 OH
Combine Hþ and OH to form H2O; cancel H2O where possible
4 OH þ 4 H2 O þ Al ! AlðOHÞ4 þ 4 H2 O
3 H2 O þ OH ! H2 þ H2 O þ 3 OH
4 OH þ Al ! AlðOHÞ4
ð4 H2 O cancelledÞ
Step 5
2 H2 O þ OH ! H2 þ 3 OH ð1 H2 O cancelledÞ
Balance electrically with electrons
4 OH þ Al ! AlðOHÞ4 þ 3 e
Step 6 and 7
2 H2 O þ OH þ 2 e ! H2 þ 3 OH
Equalize gain and loss of electrons; and half-reactions
2 4 OH þ Al ! AlðOHÞ4 þ 3 e
3 ð2 H2 O þ OH þ 2 e ! H2 þ 3 OH Þ
2 Al þ 6 H2 O þ 2 OH ! 2 AlðOHÞ4 þ 3 H2
9 OH and 6 e canceled on each side
- 242 -
- Chapter 17 17. (a)
IO3 þ I ! I2
Step 1
Step 2
ðacidic solutionÞ
Write half-reaction equations. Balance except H and O.
2 IO3 ! I2
2 I ! I2
Balance H and O using H2O and Hþ
12 Hþ þ 2 IO3 ! I2 þ 6 H2 O
Step 3
2 I ! I2
Balance electrically with electrons
12 Hþ þ 2 IO3 þ 10 e ! I2 þ 6 H2 O
Step 4 and 5
2 I ! I2 þ 2 e
Equalize the loss and gain of electrons; add the half-reaction.
12 Hþ þ 2 IO3 þ 10 e ! I2 þ 6 H2 O
5 ð2 I ! I2 þ 2 e Þ
12 Hþ þ 2 IO3 þ 10 I ! 6 I2 þ 6 H2 O
(b)
Mn2þ þ S2 O82 ! MnO4 þ SO42
Step 1
(acid solution)
Write half-reaction equations. Balance except H and O
Mn2þ ! MnO4
Step 2
S2 O82 ! 2 SO42
Balance H and O using H2O and Hþ
4 H2 O þ Mn2þ ! MnO4 þ 8 Hþ
Step 3
S2 O82 ! 2 SO42
Balance electrically with electrons
4 H2 O þ Mn2þ ! MnO4 þ 8 Hþ þ 5 e
Step 4 and 5
(c)
2 e þ S2 O82 ! 2 SO42
Equalize the loss and gain of electrons; add the half-reactions
2 4 H2 O þ Mn2þ ! MnO4 þ 8 Hþ þ 5 e
5 2 e ! S2 O82 ! 2 SO42
2 Mn2þ þ 5 S2 O82 þ 8 H2 O ! 2 MnO4 þ 10 SO42 þ 16 Hþ
Each side has 2 Mn, 10 S, 16 H, and 48 O and a 6 charge.
2þ
CoðNO2 Þ63 þ MnO
þ Mn2þ þ NO3
4 ! Co
Step 1
(acidic solution)
Write half-reaction equations. Balance except H and O.
CoðNO2 Þ63 ! Co2þ þ 6 NO3
MnO4 ! Mn2þ
- 243 -
- Chapter 17 Step 2
Balance H and O using H2O and Hþ
6 H2 O þ CoðNO2 Þ63 ! Co2þ þ 6 NO3 þ 12 Hþ
Step 3
8 Hþ þ MnO4 ! Mn2þ þ 4 H2 O
Balance electrically with e
6 H2 O þ CoðNO2 Þ63 ! Co2þ þ 6 NO3 þ 12 Hþ þ 11 e
Step 4
5 e þ 8 Hþ þ MnO4 ! Mn2þ þ 4 H2 O
Equalize the loss and gain of electrons.
5 ð6 H2 O þ CoðNO2 Þ63 ! Co2þ þ 6 NO3 þ 12 Hþ þ 11 e Þ
Step 5
11 ð5 e þ 8 Hþ þ MnO4 ! Mn2þ þ 4 H2 OÞ
Add the half-reactions
5 CoðNO2 Þ63 þ 11 MnO4 þ 28 Hþ ! 5 Co2þ þ 30 NO3 þ 11 Mn2þ þ 14 H2 O
Each side has 5 Co, 30 N, 11 Mn, 28 H, 104 O and a þ 2 charge.
18. (a)
Mo2 O3 þ MnO4 ! MoO3 þ Mn2þ
Step 1
Step 2
(acid solution)
Write half-reactions equations. Balance except H and O
Mo2 O3 ! 2 MoO3
MnO4 ! Mn2þ
Balance H and O using H2O and Hþ
3 H2 O þ Mo2 O3 ! 2 MoO3 þ 6 Hþ
Step 3
8 Hþ þ MnO4 ! Mn2þ þ 4 H2 O
Balance electrically with electrons
3 H2 O þ Mo2 O3 ! 2 MoO3 þ 6 Hþ þ 6 e
Steps 4 and 5
5 e þ 8 Hþ þ MnO4 ! Mn2þ þ 4 H2 O
Equalize the loss and gain of electrons; add the half-reactions.
5 ð3 H2 O þ Mo2 O3 ! 2 MoO3 þ 6 Hþ þ 6 e Þ
6 ð5 e þ 8 Hþ þ MnO4 ! Mn2þ þ 4 H2 OÞ
5 Mo2 O3 þ 6 MnO4 þ 18 Hþ ! 10 MoO3 þ 6 Mn2þ þ 9 H2 O
(b)
BrO þ CrðOHÞ4 ! Br þ CrO42
Step 1
(basic solution)
Write half-reaction equation. Balance except H and O
BrO ! Br
Step 2
CrðOHÞ4 ! CrO42
Balance H and O using H2O and Hþ
2 Hþ þ BrO ! Br þ H2 O
2
CrðOHÞ
þ 4 Hþ
4 ! CrO4
- 244 -
- Chapter 17 Step 3
Add OH ions to both sides (same number as Hþ)
2 OH þ 2 Hþ þ BrO ! Br þ H2 O þ 2 OH
Step 4
4 OH þ CrðOHÞ4 ! CrO42 þ 4 Hþ þ 4 OH
Combine Hþ and OH to form H2O; cancel H2O where possible
2 H2 O þ BrO ! Br þ H2 O þ 2 OH
4 OH þ CrðOHÞ4 ! CrO42 þ 4 H2 O
Step 5
H2 O þ BrO ! Br þ 2 OH ð1 H2 O cancelledÞ
Balance electrically with electrons
2 e þ H2 O þ BrO ! Br þ 2 OH
Steps 6 and 7
4 OH þ CrðOHÞ4 ! CrO42 þ 4 H2 O þ 3 e
Equalize loss and gain of electrons; add the half-reactions
3 ð2 e þ H2 O þ BrO ! Br þ 2 OH Þ
2 ð4 OH þ CrðOHÞ4 ! CrO42 þ 4 H2 O þ 3 e Þ
3 BrO þ 2 CrðOHÞ4 þ 2 OH ! 3 Br þ 2 CrO42 þ 5 H2 O
(c)
S2 O32 þ MnO4 ! SO42 þ MnO2
Step 1
(basic solution)
Write half-reaction equations. Balance except H and O.
S2 O32 ! 2 SO42
Step 2
MnO4 ! MnO2
Balance H and O using H2O and Hþ
5 H2 O þ S2 O32 ! 2 SO42 þ 10 Hþ
Step 3
4 Hþ þ MnO4 ! MnO2 þ 2 H2 O
Add OH ions to both sides (same number as Hþ)
10 OH þ 5 H2 O þ S2 O32 ! 2 SO42 þ 10 Hþ þ 10 OH
Step 4
4 OH þ 4 Hþ þ MnO4 ! MnO2 þ 2 H2 O þ 4 OH
Combine Hþ and OH to form H2O; cancel H2O where possible
10 OH þ 5 H2 O þ S2 O32 ! 2 SO42 þ 10 H2 O
4 H2 O þ MnO4 ! MnO2 þ 2 H2 O þ 4 OH
10 OH þ S2 O32 ! 2 SO42 þ 5 H2 O ð5 H2 O cancelledÞ
Step 5
2 H2 O þ MnO4 ! MnO2 þ 4 OH
Balance electrically with electrons
ð2 H2 O cancelledÞ
10 OH þ S2 O32 ! 2 SO42 þ 5 H2 O þ 8 e
Step 6 and 7
3 e þ 2 H2 O þ MnO4 ! MnO2 þ 4 OH
Equalize loss and gain of electrons; add half-reactions.
3 ð10 OH þ S2 O32 ! 2 SO42 þ 5 H2 O þ 8 e Þ
8 ð3 e þ 2 H2 O þ MnO4 ! MnO2 þ 4 OH Þ
3 S2 O32 þ 8 MnO4 þ H2 O ! 6 SO42 þ 8 MnO2 þ 2 OH
Each side has 6 S, 8 Mn, 2 H, 42 O and a 14 charge.
- 245 -
- Chapter 17 19.
– Voltage
source
+
Anode (+)
Cathode (–)
Br–
H3O+
Solution of HBr
20. (a)
(b)
(c)
21. (a)
(b)
(c)
Pb þ SO42 ! PbSO4 þ 2 e
PbO2 þ SO42 þ 4 Hþ þ 2 e ! PbSO4 þ 2 H2 O
The first reaction is oxidation (Pb0 is oxidized to Pb2þ).
The second reaction is reduction (Pb4þ is reduced to Pb2þ).
The first reaction (oxidation) occurs at the anode of the battery.
The oxidizing agent is KMnO4.
The reducing agent is HCl.
5 moles of electrons 5 e þ Mn7þ ! Mn2þ
5 mol e
6:022 1023 e
electrons
¼ 3:011 1024
mol KMnO4
mol e
mol KMnO4
22. Zinc is a more reactive metal than copper so when corrosion occurs the zinc preferentially reacts.
Zinc is above hydrogen in the Activity series of metals; copper is below hydrogen.
23. 3 Cu þ 8 HNO3 ! 3 Cu(NO3)2 þ 2 NO þ 4 H2O
1 mol Cu 2 mol NO 22:4 L
¼ 17:7 L NO
ð75:5 g CuÞ
63:55 g
3 mol Cu
mol
1 mol HNO3
2 mol NO
22:4 L
ð55:0 g HNO3 Þ
¼ 4:89 L NO
63:02 g
8 mol HNO3
mol
4.89 L of NO gas at STP will be produced; HNO3 is limiting.
Cu is oxidized; N is reduced.
24. K2S2O8 þ H2C2O4 ! K2SO4 þ H2SO4 þ 2 CO2
1 mol K2 S2 O8
2 mol CO2
22:4 L
ð25:5 g K2 S2 O8 Þ
¼ 4:23 L CO2
270:3 g
1 mol HNO3
mol
1 mol H2 C2 O4
2 mol CO2
22:4 L
ð35:5 g H2 C2 O4 Þ
¼ 17:7 L CO2
90:04 g
1 mol H2 C2 O4
mol
4.23 L CO2 will be produced; K2S2O8 is limiting.
C is oxidized; S is reduced.
- 246 -
- Chapter 17 25. 5 H2O2 þ 2 KMnO4 þ 3 H2SO4 ! 5 O2 þ 2 MnSO4 þ K2SO4 þ 8 H2O
mL H2O2 ! g H2O2 ! mol H2O2 ! mol KMnO4 ! g KMnO4
1:031 g
9:0 g H2 O2
1 mol
ð100: mL H2 O2 solutionÞ
100: g H2 O2 solution 34:02 g
mL
2 mol KMnO4
158:0 g
¼ 17 g KMnO4
5 mol H2 O2
mol
26. 3 Zn þ 2 Fe3þ ! 3 Zn2þ þ 2 Fe
1:2 mol FeCl3
3 mol Zn
65:39 g
ð0:0250 L FeCl3 Þ
¼ 2:94 g Zn
L
2 mol FeCl3
mol
27. Cr2 O72 þ 6 Fe2þ þ 14 Hþ ! 2 Cr3þ þ 6 Fe3þ þ 7 H2 O
mL FeSO4 ! mol FeSO4 ! mol Cr2 O72 ! mL Cr2 O72
0:200 mol 1 mol Cr2 O72
1000 mol
ð60:0 mL FeSO4 Þ
6 mol FeSO4
1000 mL
0:200 mol
¼ 10:0 mL of 0:200 M K2 Cr2 O7
28. 2 Al þ 2 OH þ 6 H2 O ! 2 AlðOHÞ4 þ 3 H2
g Al ! mol Al ! mol H2
1 mol Al 3 mol H2
ð100:0 g AlÞ
¼ 5:560 mol H2
2 mol Al
26:98 g
29. (a)
Cuþ ! Cu2þ is an oxidation, but when electrons are gained reduction should occur.
Cuþ þ e ! Cu0
(b)
or Cuþ ! Cu2þ þ e
When Pb2þ is reduced, it requires two individual electrons. Pb2þ þ 2 e ! Pb0. An electron
has only a single negative charge (e).
30. The electrons lost by the species undergoing oxidation must be gained (or attracted) by another species
which then undergoes reduction.
31. A(s) þ B2þ(aq) ! NR
B2þ cannot take e from A
A(s) þ Cþ(aq) ! NR
Cþ cannot take e from A
þ
2þ
Cþ takes e from D
D(s) þ 2 C (aq) ! 2C(s) þ D (aq)
2þ
2þ
B(s) þ D (aq) ! D(s) þ B (aq)
D2þ takes 2 e from B
2þ
2þ
Therefore, B is least able to attract e , then D , then Cþ, then Aþ
32.
Sn4þ can only be an oxidizing agent.
Sn0 can only be a reducing agent.
Sn2þ can be both oxidizing and reducing.
Sn4þ þ 2 e ! Sn2þ
Sn4þ þ 4 e ! Sn0
Sn0 ! Sn2þ þ 2 e
Sn0 ! Sn4þ þ 4 e
Sn2þ þ 2 e ! Sn0
Sn2þ ! Sn4þ þ 2 e
- 247 -
(oxidizing)
(reducing)
- Chapter 17 33. Mn(OH)2
MnF3
MnO2
K2MnO4
KMnO4
þ2
þ3
þ4
þ6
þ7
KMnO4 is the best oxidizing agent of the group, since its greater
oxidation number (þ7) makes it very attractive to electrons.
34. Equations (a) and (b) represent oxidation
(a) Mg ! Mg2þ þ 2 e
(b) SO2 ! SO3; (S4þ ! S6þ þ 2 e)
35. (a)
(b)
(c)
36. (a)
(b)
(c)
(d)
MnO2 þ 2 Br þ 4 Hþ ! Mn2þ þ Br2 þ 2 H2O
mL Mn2þ ! mol Mn2þ ! mol MnO2 ! g MnO2
0:05 mol 1 mol MnO2
86:94 g
2þ
ð100:0 mL Mn Þ
¼ 0:4 g MnO2
1 mol Mn2þ
1000 mL
mol
0:05 mol
1 mol Br2
¼ 0:005 mol Br2
ð100:0 mL Mn2þ Þ
1000 mL 1 mol Mn2þ
nRT
PV ¼ nRT
V¼
P
0:005 mol 0:0821 L atm
ð323 KÞ ¼ 0:09 L Br2 vapor
V¼
1:4 atm
mol K
F2 þ 2 Cl ! 2 F þ Cl2
Br2 þ Cl ! NR
I2 þ Cl ! NR
Br2 þ 2 I ! 2 Br þ I2
37. Mn(s) þ 2 HCl(aq) ! Mn2þ(aq) þ H2(g) þ 2 Cl(aq)
38. 4 Zn þ NO3 þ 10 Hþ ! 4 Zn2þ þ NH4þ þ 3 H2 O
See Exercise 13(a).
39.
Equation 1
2
3
4
5
a
C oxidized
S oxidized
N oxidized
S oxidized
O22 oxidized
b
O2 reduced
N reduced
Cu reduced
O22 reduced
O22 reduced
c
O2, O.A.
HNO3, O.A.
CuO, O.A.
H2O2,O.A.
H2O2, O.A.
d
C3H8, R.A.
H2S, R.A.
NH3, R.A.
Na2SO3, R.A.
H2O2, R.A.
e
C
S2 ! S0
N3 ! N20
S4þ ! S6þ
O22 ! O20
f
O0 ! O2
N5þ ! N2þ
Cu2þ ! Cu0
O22 ! O2
O22 ! O2
2
3
2 þ
! C4þ
O.A. ¼ Oxidizing agent
R.A. ¼ Reducing agent
- 248 -
- Chapter 17 40. Pb þ 2 Agþ ! 2 Ag þ Pb2þ
(a) Pb is the anode
(b) Ag is the cathode
(c) Oxidation occurs at Pb (anode)
(d) Reduction occurs at Ag (cathode)
(e) Electrons flow from the lead electrode through the wire to the silver electrode.
(f) Positive ions flow through the salt bridge towards the negatively charged strip of silver; negative
ions flow toward the positively charged strip of lead.
Salt
bridge
Ag
Pb
Pb2+
NO–3
Ag+
NO3–
41. 8 KI þ 5 H2 SO4 ! 4 I2 þ H2 S þ 4 K2 SO4 þ 4 H2 O
start with grams I2 and work towards g KI
g I2 ! mol I2 ! mol KI ! g KI
1 mol
8 mol KI 166:0 g
¼ 3:65 g KI in sample
ð2:79 g I2 Þ
253:8 g
4 mol I2
mol
3:65 g KI
ð100Þ ¼ 91:3% KI
4:00 g sample
42. 3 Ag þ 4 HNO3 ! 3 AgNO3 þ NO þ 2 H2 O
mol Ag ! mol NO
1 mol NO
ð0:500 mol AgÞ
¼ 0:167 mol NO
3 mol Ag
nRT
PV ¼ nRT
V¼
P
1 atm
¼ 0:979 atm
P ¼ ð744 torrÞ
760: torr
T ¼ 301 K
ð0:167 mol NOÞð0:0821 L atm=mol KÞð301 KÞ
¼ 4:22 L NO
V¼
ð0:979 atmÞ
- 249 -