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 tolose electrons, the more active the metal is.
2. (a) Iodine is oxidized. Its oxidation number increases from 0 to
(b) Chlorine is reduced. Its oxidation number decreases from 0 to
3. The higher metal on the list is more reactive.
(a) Al (b) Ba (c) Ni
4. If the free element is higher on the list than the ion with which it is paired, the reactionoccurs.
(a) Yes.
(b) No reaction
(c) Yes.
(d) No reaction
(e) Yes.
(f) No reaction
(g) Yes.
(h) Yes.
5. Copper is more active than silver. Therefore, copper undergoes oxidation more easilythan silver. Accordingly, it is more difficult for copper ion to undergo reduction than it isfor silver ion. When a silver wire is placed in a solution of copper (II) nitrate one mightpredict that copper crystals would form on the silver wire. However for copper to go froman oxidation state of to an oxidation state of 0 it would have to gain electrons(reduction) and silver would have to lose electrons (oxidation). This will not happenbecause copper is more active than silver.
6. (a)
(b) Al is above Fe in the activity series, which indicates Al is more active than Fe.
(c) No. Iron is less active than aluminum and will not displace aluminum from itscompounds.
(d) Yes. Aluminum is above chromium in the activity series and will displace from its compounds.
Cr3+
2 Al + Fe2O3 ¡ Al2O3 + 2 Fe + Heat
+2
2 Al(s) + 3 CuSO4(aq) ¡ Al2(SO4)3(aq) + 3 Cu(s)
Ni(s) + Hg(NO3)2(aq) ¡ Ni(NO3)2(aq) + Hg(l)
Ba(s) + FeCl2(aq) ¡ BaCl2(aq) + Fe(s)
Sn(s) + 2 Ag+(aq) ¡ Sn2+(aq) + 2 Ag(s)
Zn(s) + Cu2+(aq) ¡ Zn2+(aq) + Cu(s)
-1.
+5.
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7. (a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
8. The oxidation number for an atom in an ionic compound is the same as the charge of theion that resulted when that atom lost or gained electrons to form an ionic bond. In acovalently bonded compound electrons are shared between the two atoms making up thebond. Those shared electrons are assigned to the atom in the bond with a higherelectronegativity giving it a negative oxidation number.
9. In an electrolytic cell the anode is the positively charged electrode and attracts negativelycharged ions (anions). The cathode is the negatively charged electrode and attracts positivelycharge ions (cations). In a voltaic cell the anode is the negatively changed electrode whereoxidation occurs. The cathode is the positively charged electrode where reduction occurs.
10. (a) Oxidation occurs at the anode. The reaction is
(b) Reduction occurs at the cathode. The reaction is
(c) The net chemical reaction is
Ni2+(aq) + 2 Cl-(aq) electrical
energy " Ni(s) + Cl2(g)
Ni2+(aq) + 2 e- ¡ Ni(s)
2 Cl-(aq) ¡ Cl2(g) + 2 e-
Zn(s) + H2SO4(aq) ¡ ZnSO4(aq) + H2(g)
Zn(s) + 2 HCl(aq) ¡ ZnCl2(aq) + H2(g)
Hg(l) + H2SO4(aq) ¡ no reaction
Hg(l) + HCl(aq) ¡ no reaction
Mg(s) + H2SO4(aq) ¡ MgSO4(aq) + H2(g)
Mg(s) + 2 HCl(aq) ¡ MgCl2(aq) + H2(g)
Cu(s) + H2SO4(aq) ¡ no reaction
Cu(s) + HCl(aq) ¡ no reaction
Fe(s) + H2SO4(aq) ¡ FeSO4(aq) + H2(g)
Fe(s) + 2 HCl(aq) ¡ FeCl2(aq) + H2(g)
Au(s) + H2SO4(aq) ¡ no reaction
Au(s) + HCl(aq) ¡ no reaction
2 Cr(s) + 3 H2SO4(aq) ¡ Cr2(SO4)3(aq) + 3 H2(g)
2 Cr(s) + 6 HCl(aq) ¡ 2 CrCl3(aq) + 3 H2(g)
2 Al(s) + 3 H2SO4(aq) ¡ Al2(SO4)3(aq) + 3 H2(g)
2 Al(s) + 6 HCl(aq) ¡ 2 AlCl3(aq) + 3 H2(g)
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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) It would not be possible to monitor the voltage produced, but the reactions in thecell would still occur.
(b) If the salt bridge were removed, the reaction would stop. Ions must be mobile tomaintain an electrical neutrality of ions in solution. The two solutions would beisolated with no complete electrical circuit.
13. Oxidation and reduction are complementary processes because one does not occurwithout the other. The loss of in oxidation is accompanied by a gain of in reduction.
14. cathode reaction, reduction
anode reaction, oxidation
15. During electroplating of metals, the metal is plated by reducing the positive ions of themetal in the solution. The plating will occur at the cathode, the source of the electrons. Withan alternating current, the polarity of the electrode would be constantly changing, so at oneinstant 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 bridgesin the cells of a lead storage battery.
17. The electrolyte in a lead storage battery is dilute sulfuric acid. In the discharge cycle,is removed from solution as it reacts with and to form and
Therefore, the electrolyte solution contains less and becomes less dense.
18. If ions are reduced to metallic mercury, this would occur at the cathode, becausereduction takes place at the cathode.
19. In both electrolytic and voltaic cells, oxidation and reduction reactions occur. In anelectrolytic cell an electric current is forced through the cell causing a chemical change tooccur. 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.
Hg2+
H2SO4H2O.PbSO4(s)H+PbO2SO4
2-,
2 Br - ¡ Br2 + 2 e-Ca2+ + 2 e- ¡ Ca
e-e-
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CHAPTER 17
SOLUTIONS TO EXERCISES
1. The oxidation number of the underlined element is indicated by the number following theformula.
(a) (c) (e)
(b) (d) (f)
2. The oxidation number of the underlined element is indicated by the number following theformula.
(a) (c) (e)
(b) 0 (d) (f)
3. The oxidation number of the underlined element is indicated by the number following theformula.
(a) (c)
(b) (d)
4. The oxidation number of the underlined element is indicated by the number following theformula.
(a) 0 (c)
(b) (d)
Changing Type of5. Balanced half-reaction Element reaction
(a) Zn reduction
(b) Br oxidation
(c) Mn reduction
(d) Ni oxidation
Changing Type of6. Balanced half-reactions Element reaction
(a) S oxidation
(b) N reduction
(c) S oxidation
(d) Fe oxidationFe2+ ¡ Fe3+ + 1 e-S2O4
2- + 2 H2O ¡ 2 SO3
2- + 4 H+ + 2 e-NO3
- + 4 H+ + 3 e- ¡ NO + 2 H2O
SO3
2- + H2O ¡ SO4
2- + 2 H+ + 2 e-
Ni ¡ Ni2+ + 2 e-MnO4
- + 8 H+ + 5 e- ¡ Mn2+ + 4 H2O
2 Br - ¡ Br2 + 2 e-Zn2+ + 2 e- ¡ Zn
+5IO3
-+5AsO4
3--2Fe(OH)3O2
+3Bi3++3NO2
--1Na2O2-2S2-
+6K2Cr2O7+5KClO3I2
+6K2CrO4-3NH3+7KMnO4
-3NH4Cl+5NaNO3-1FeCl3
+4H2SO3+4PbO2+1NaCl
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7. (1)
(a) Cr is oxidized, H is reduced
(b) HCl is the oxidizing agent, Cr the reducing agent
(2)
(a) I is oxidized, S is reduced
(b) is the oxidizing agent, the reducing agent
8. (1)
(a) As is oxidized, Ag is reduced
(b) is the oxidizing agent, the reducing agent
(2)
(a) Br is oxidized, Cl is reduced
(b) is the oxidizing agent, NaBr the reducing agent
9. (a) correctly balanced
(b) correctly balanced
(c) incorrectly balanced
(d) incorrectly balanced
10. (a) incorrectly balanced
(b) correctly balanced
(c) correctly balanced
(d) incorrectly balanced
11. Balancing oxidation-reduction equations
(a)
Add half-reactions
the 2 e- cancel
Zn + S ¡ ZnS
Zn0 ¡ Zn2+ + 2 e-
S0 + 2 e- ¡ S2-
Zn + S ¡ ZnS
ox
red
8 H2O(l) + 2 MnO4
- (aq) + 7 S2-
(aq) ¡ 2 MnS(s) + 16 OH- (aq) + 5 S(s)
3 MnO2(s) + 4 Al(s) ¡ 3 Mn(s) + 2 Al2O3(s)
3 CH3OH(aq) + Cr2O7
2- (aq) + 8 H+
(aq) ¡ 2 Cr3+ (aq) + 3 CH2O(aq) + 7 H2O(l)
Mg(s) + 2 HCl(aq) ¡ Mg2+ (aq) + 2Cl-
(aq) + H2(g)
Cl2
Cl2 + NaBr ¡ NaCl + Br2
AsH3Ag+
AsH3 + Ag+ + H2O ¡ H3AsO4 + Ag + H+
I-SO4
2-
SO4
2- + I- + H+ ¡ H2S + I2 + H2O
Cr + HCl ¡ CrCl3 + H2
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(b)
Transfer the coefficients to the original equation and complete the balancingby inspection.
(c)
Transfer the coefficients to the original equation (the coefficient 2 in front ofthe becomes the subscript 2 in ). Complete the balancing byinspection.
(d)
Transfer the coefficients to the original equations and complete the balancing byinspection.
(e)
Transfer the coefficients to the original equation. The coefficient 2 in front of thebecomes the subscript 2 in the Also, 2 more ions are required to
account for the ions that do not change oxidation numbers. These 2 are partof the compound
MnO2 + 4 HBr ¡ MnBr2 + Br2 + 2 H2O
MnBr2.2 Br -
Br -Br2.Br -
Multiply by 2
Add equations and the 2 e- cancel
MnO2 + HBr ¡ MnBr2 + Br2 + H2O
Br - ¡ Br0 + 1 e-
Mn4+ + 2 e- ¡ Mn2+ Mn4+ + 2 Br - ¡ Mn2+ + 2 Br0
3 H2S + 2 HNO3 ¡ 3 S + 2 NO + 4 H2O
Multiply by 3
Multiply by 2, add, the 6 e-
H2S + HNO3 ¡ S + NO + H2O
S2- ¡ S0 + 2 e-
N5+ + 3 e- ¡ N2+ 3 S2- + 2 N5+ ¡ 3 S + 2 N2+
Fe2O3 + 3 CO ¡ 2 Fe + 3 CO2
Fe2O3Fe3+
Multiply by 3
Multiply by 2, add, the 6 e- cancel
Fe2O3 + CO ¡ Fe + CO2
C2+ ¡ C4+ + 2 e-
Fe3+ + 3 e- ¡ Fe0 3 C2+ + 2 Fe3+ ¡ 3 C4+ + 2 Fe
ox
red
2 AgNO3 + Pb ¡ Pb(NO3)2 + 2 Ag
Multiply by 2, add the half-reactions
the 2 e- cancel
AgNO3 + Pb ¡ Pb(NO3)2 + Ag
Pb0 ¡ Pb2+ + 2 e-
Ag+ + 1 e- ¡ Ag0 Pb + 2 Ag+ ¡ Pb2+ + 2 Ag
ox
red
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12. (a) Balancing oxidation-reduction equations
Transfer the coefficients to the original equations and complete the balancing byinspection.
(b)
Transfer the coefficients to the original equations and complete the balancing byinspection.
(c)
Transfer the coefficients to the original equations and complete the balancing byinspection.
(d)
Transfer the coefficients to the original equations and complete the balancing byinspection.
(e) H2O2 + KMnO4 + H2SO4 ¡ O2 + MnSO4 + K2SO4 + H2O
O2
2- ¡ O2
0 + 2 e- Multiply by 5
Mn7+ + 5 e- ¡ Mn2+ Multiply by 2, add, the 10 e - cancel
5 O2
2- + 2 Mn7+ ¡ 5 O2 + 2 Mn2+
3 PbO2 + 2 Sb + 2 NaOH ¡ 3 PbO + 2 NaSbO2 + H2O
PbO2 + Sb + NaOH ¡ PbO + NaSbO2 + H2O
Sb0 ¡ Sb3+ + 3 e- Multiply by 2
Pb4+ + 2 e- ¡ Pb2+ Multiply by 3, add, the 6 e- cancel
2 Sb + 3 Pb4+ ¡ 2 Sb3+ + 3 Pb2+
3 CuO + 2 NH3 ¡ N2 + 3 Cu + 3 H2O
Multiply by 2
Multiply by 3, add, the 6 e- cancel
CuO + NH3 ¡ N2 + Cu + H2O
N3- ¡ N0 + 3 e-
Cu2+ + 2 e- ¡ Cu0 2 N3- + 3 Cu2+ ¡ N2 + 3 Cu
3 Ag + 4 HNO3 ¡ 3 AgNO3 + NO + 2 H2O
Multiply by 3, add,
the 3 e- cancel
Ag + HNO3 ¡ AgNO3 + NO + H2O
Ag0 ¡ Ag+ + e-
N5+ + 3 e- ¡ N2+ 3 Ag + N5+ ¡ 3 Ag+ + N2+
3 Cl2 + 6 KOH ¡ KClO3 + 5 KCl + 3 H2O
Multiply by 5, add, the 5 e- cancel
6 Cl0 becomes 3 Cl2
Cl2 + KOH ¡ KCl + KClO3 + H2O
Cl0 ¡ Cl5+ + 5 e-
Cl0 + e- ¡ Cl- 3 Cl2 ¡ Cl5+ + 5 Cl-
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Transfer the coefficients to the original equations and complete the balancing byinspection.
13. (a) (acidic solution)
Step 1 Write half-reaction equations. Balance except H and O.
Step 2 Balance H and O using and
Step 3 Balance electrically with electrons
Step 4 Equalize the loss and gain of electrons
Step 5 Add the half-reactions–electrons cancel
(b) (acidic solution)
Step 1 Write half-reaction equations. Balance except H and O.
Step 2 Balance H and O using and
Step 3 Balance electrically with electrons
2 H+ + NO3
- + e- ¡ NO2 + H2O
4 H2O + S ¡ SO4
2- + 8 H+ + 6 e-
2 H+ + NO3
- ¡ NO2 + H2O
4 H2O + S ¡ SO4
2- + 8 H+
H+H2O
NO3
- ¡ NO2
S ¡ SO4
2-
NO3
- + S ¡ NO2 + SO4
2-
10 H+ + 4 Zn + NO3
- ¡ 4 Zn2+ + NH4
+ + 3 H2O
10 H+ + NO3
- + 8 e- ¡ NH4
+ + 3 H2O
4 (Zn ¡ Zn2+ + 2 e-)
10 H+ + NO3
- + 8 e- ¡ NH4
+ + 3 H2O
Zn ¡ Zn2+ + 2 e-
10 H+ + NO3
- ¡ NH4
+ + 3 H2O
Zn ¡ Zn2+
H+H2O
NO3
- ¡ NH4
+
Zn ¡ Zn2+
Zn + NO3
- ¡ Zn2+ + NH4
+
5 H2O2 + 2 KMnO4 + 3 H2SO4 ¡ 5 O2 + 2 MnSO4 + K2SO4 + 8 H2O
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Step 4 and 5 Equalize the loss and gain of electrons; add the half-reactions
(c) (acidic solution)
Step 1 Write half-reaction equations. Balance except H and O.
Step 2 Balance H and O using and
Step 3 Balance electrically with electrons
Step 4 and 5 Equalize the loss and gain of electrons; add the half-reactions
(d) (acidic solution)
Step 1 Write half-reaction equations. Balance except H and O.
Step 2 Balance H and O using and
Step 3 Balance electrically with electrons
4H+ + NO3
- + 3 e- ¡ NO + 2 H2O
Cu ¡ Cu2+ + 2 e-
4H+ + NO3
- ¡ NO + 2 H2O
Cu ¡ Cu2+
H+H2O
NO3
- ¡ NO
Cu ¡ Cu2+
Cu + NO3
- ¡ Cu2+ + NO
2 H2O + PH3 ¡ H3PO2 + 4 H+ + 4 e-
2 (I2 + 2 e- ¡ 2 I-)
PH3 + 2 H2O + 2 I2 ¡ H3PO2 + 4 I- + 4 H+
I2 + 2 e- ¡ 2 I-
2 H2O + PH3 ¡ H3PO2 + 4 H+ + 4 e-
I2 ¡ 2 I-
2 H2O + PH3 ¡ H3PO2 + 4 H+
H+H2O
I2 ¡ 2I-
PH3 ¡ H3PO2
PH3 + I2 ¡ H3PO2 + I-
and 6 e- canceled from each side4 H2O, 8 H +
4 H2O + S ¡ SO4
2- + 8 H+ + 6 e-
6 (2 H+ + NO3
- + e- ¡ NO2 + H2O) 4 H+ + S + 6 NO3
- ¡ 6 NO2 + SO4
2- + 2 H2O
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Step 4 and 5 Equalize the loss and gain of electrons; add the half-reactions
(e) (acidic solution)
Step 1 Write half-reaction equations. Balance except H and O.
Step 2 Balance H and O using and
Step 3 Balance electrically with electrons
Step 4 and 5 Equalize the loss and gain of electrons; add the half-reactions
14. (a) (acidic solution)
Step 1 Write half-reaction equations. Balance except H and O.
Step 2 Balance H and O using and
Step 3 Balance electrically with electrons
6 H+ + ClO3
- + 6 e- ¡ Cl- + 3 H2O
2 I- ¡ I2 + 2 e-
6 H+ + ClO3
- ¡ Cl- + 3 H2O
2 I- ¡ I2
H+H2O
ClO3
- ¡ Cl-
2 I- ¡ I2
ClO3
- + I- ¡ I2 + Cl-
5 (Cl- ¡ Cl0 + e-)
6 H+ + ClO3
- + 5 e- ¡ Cl0 + 3 H2O
6 H+ + ClO3
- + 5 Cl- ¡ 3 Cl2 + 3 H2O
6 H+ + ClO3
- + 5 e- ¡ Cl0 + 3 H2O
Cl- ¡ Cl0 + e-
6 H+ + ClO3
- ¡ Cl0 + 3 H2O
Cl- ¡ Cl0H+H2O
ClO3
- ¡ Cl0Cl- ¡ Cl0
ClO3
- + Cl- ¡ Cl2
3 (Cu ¡ Cu2+ + 2 e-)
2 (4 H+ + NO3
- + 3 e- ¡ NO + 2 H2O) 3 Cu + 8 H+ + 2 NO3
- ¡ 3 Cu2+ + 2 NO + 4 H2O
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Step 4 and 5 Equalize the loss and gain of electrons; add the half-reactions
(b) (acidic solution)
Step 1 Write half-reaction equations. Balance except H and O.
Step 2 Balance H and O using and
Step 3 Balance electrically with electrons
Step 4 and 5 Equalize the loss and gain of electrons; add the half-reactions
(c) (acidic solution)
Step 1 Write half-reaction equations. Balance except H and O.
Step 2 Balance H and O using and
Step 3 Balance electrically with electrons
8 H+ + MnO4
- + 5 e- ¡ Mn2+ + 4 H2O
2 H2O + SO2 ¡ SO4
2- + 4 H+ + 2 e-
8 H+ + MnO4
- ¡ Mn2+ + 4 H2O
2 H2O + SO2 ¡ SO4
2- + 4 H+
H+H2O
MnO4
- ¡ Mn2+
SO2 ¡ SO4
2-
MnO4
- + SO2 ¡ Mn2+ + SO4
2-
6 (Fe2+ ¡ Fe3+ + e-)
14 H+ + Cr2O7
2- + 6 e- ¡ 2 Cr3+ + 7 H2O
14 H+ + Cr2O7
2- + 6 Fe2+ ¡ 2 Cr3+ + 6 Fe3+ + 7 H2O
14 H+ + Cr2O7
2- + 6 e- ¡ 2 Cr3+ + 7 H2O
Fe2+ ¡ Fe3+ + e-
14 H+ + Cr2O7
2- ¡ 2 Cr3+ + 7 H2O
Fe2+ ¡ Fe3+
H+H2O
Cr2O7
2- ¡ 2 Cr3+
Fe2+ ¡ Fe3+
Cr2O7
2- + Fe2+ ¡ Cr3+ + Fe3+
3 (2 I- ¡ I2 + 2 e-)
6 H+ + ClO3
- + 6 e- ¡ Cl- + 3 H2O 6 H+ + ClO3
- + 6 I- ¡ 3 I2 + Cl- + 3 H2O
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Step 4 and 5 Equalize the loss and gain of electrons; add the half-reactions
(d) (acidic solution)
Step 1 Write half-reaction equations. Balance except H and O.
Step 2 Balance H and O using and
Step 3 Balance electrically with electrons
Step 4 and 5 Equalize the loss and gain of electrons; add the half-reactions
(e) (acidic solution)
Step 1 Write half-reaction equations. Balance except H and O.
Step 2 Balance H and O using and
Step 3 Balance electrically with electrons
14 H+ + Cr2O7
2- + 6 e- ¡ 2 Cr3+ + 7 H2O
H2O + H3AsO3 ¡ 2 H+ + H3AsO4 + 2 e-
14 H+ + Cr2O7
2- ¡ 2 Cr3+ + 7 H2O
H2O + H3AsO3 ¡ 2 H+ + H3AsO4
H+H2O
Cr2O7
2- ¡ 2 Cr3+
H3AsO3 ¡ H3AsO4
Cr2O7
2- + H3AsO3 ¡ Cr3+ + H3AsO4
5 (H2O + H3AsO3 ¡ 2 H+ + H3AsO4 + 2 e-)
2 (8 H+ + MnO4
- + 5 e- ¡ Mn2+ + 4 H2O) 6 H+ + 5 H3AsO3 + 2 MnO4
- ¡ 5 H3AsO4 + 2 Mn2+ + 3 H2O
5 H2O, 10 H+, and 10 e- canceled from each side
8 H+ + MnO4
- + 5 e- ¡ Mn2+ + 4 H2O
H2O + H3AsO3 ¡ 2 H+ + H3AsO4 + 2 e-
8 H+ + MnO4
- ¡ Mn2+ + 4 H2O
H2O + H3AsO3 ¡ 2 H+ + H3AsO4
H+H2O
MnO4
- ¡ Mn2+
H3AsO3 ¡ H3AsO4
H3AsO3 + MnO4
- ¡ H3AsO4 + Mn2+
5 (2 H2O + SO2 ¡ SO4
2- + 4 H+ + 2 e-)
2 (8 H+ + MnO4
- + 5 e- ¡ Mn2+ + 4 H2O) 2 H2O + 2 MnO4
- + 5 SO2 ¡ 4 H+ + 2 Mn2+ + 5 SO4
2-
8 H2O, 16 H+, and 10 e- canceled from each side
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Step 4 and 5 Equalize the loss and gain of electrons; add the half-reactions
15. (a) (basic solution)
Step 1 Write half-reaction equations. Balance except H and O.
Step 2 Balance H and O using and
Step 3 Add ions to both sides (same number as ions)
Step 4 Combine and to form cancel where possible
Step 5 Balance electrically with electrons
Step 6 Electron loss and gain is balanced
Step 7 Add half-reactions
(b) (basic solution)
Step 1 Write half-reaction equations. Balance except H and O.
MnO4
- ¡ MnO2
ClO2
- ¡ ClO4
-
MnO4
- + ClO2
- ¡ MnO2 + ClO4
-
2 OH- + IO3
- + Cl2 ¡ IO4
- + 2 Cl- + H2O
Cl2 + 2 e- ¡ 2 Cl-
2 OH- + IO3
- ¡ IO4
- + H2O + 2 e-
Cl2 ¡ 2 Cl-
(1 H2O cancelled)2 OH- + IO3
- ¡ IO4
- + H2O
Cl2 ¡ 2 Cl-
2 OH- + H2O + IO3
- ¡ IO4
- + 2 H2O
H2OH2O;OH-H+
Cl2 ¡ 2 Cl-
2 OH- + H2O + IO3
- ¡ IO4
- + 2 H+ + 2 OH-
H+OH-
Cl2 ¡ 2 Cl-
H2O + IO3
- ¡ IO4
- + 2 H+
H+H2O
Cl2 ¡ 2 Cl-
IO3
- ¡ IO4
-
Cl2 + IO3
- ¡ Cl- + IO4
-
3 (H2O + H3AsO3 ¡ 2 H+ + H3AsO4 + 2 e-)
14 H+ + Cr2O7
2- + 6 e- ¡ 2 Cr3+ + 7 H2O 8 H+ + Cr2O7
2- + 3 H3AsO3 ¡ 2 Cr3+ + 3 H3AsO4 + 4 H2O
3 H2O, 6 H+, and 6 e- canceled from each side
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Step 2 Balance H and O using and
Step 3 Add ions to both sides (same number as ions)
Step 4 Combine and to form cancel where possible
Step 5 Balance electrically with electrons
Step 6 and 7 Equalize gain and loss of electrons; add half-reactions
(c) (basic solution)
Step 1 Write half-reaction equations. Balance except H and O.
Step 2 Balance H and O using and
Step 3 Add ions to both sides (same number as ions)
Se ¡ Se2-
6 OH- + 3 H2O + Se ¡ SeO3
2- + 6 H+ + 6 OH-
H+OH-
Se ¡ Se2-
3 H2O + Se ¡ SeO3
2- + 6 H+
H+H2O
Se ¡ Se2-
Se ¡ SeO3
2-
Se ¡ SeO3
2- + Se2-
3 (4 OH- + ClO2
- ¡ ClO4
- + 2 H2O + 4 e-)
4 (2 H2O + MnO4
- + 3 e- ¡ MnO2 + 4 OH-) 2 H2O + 4 MnO4
- + 3 ClO2
- ¡ 4 MnO2 + 3 ClO4
- + 4 OH-
6 H2O, 12 OH-, and 12 e- canceled from each side
2 H2O + MnO4
- + 3 e- ¡ MnO2 + 4 OH-
4 OH- + ClO2
- ¡ ClO4
- + 2 H2O + 4 e-
(2 H2O cancelled)2 H2O + MnO4
- ¡ MnO2 + 4 OH-
(2 H2O cancelled)4 OH- + ClO2
- ¡ ClO4
- + 2 H2O
4 H2O + MnO4
- ¡ MnO2 + 2 H2O + 4 OH-
4 OH- + 2 H2O + ClO2
- ¡ ClO4
- + 4 H2O
H2OH2O;OH-H+
4 OH- + MnO4
- + 4 H+ ¡ MnO2 + 2 H2O + 4 OH-
4 OH- + 2 H2O + ClO2
- ¡ ClO4
- + 4 H+ + 4 OH-
H+OH-
MnO4
- + 4 H+ ¡ MnO2 + 2 H2O
2 H2O + ClO2
- ¡ ClO4
- + 4 H+
H+H2O
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Step 4 Combine and to form cancel where possible
Step 5 Balance electrically with electrons
Step 6 and 7 Equalize gain and loss of electrons; add half-reactions
(d) (basic solution)
Step 1 Write half-reaction equations. Balance except H and O.
Step 2 Balance H and O using and
Step 3 Add ions to both sides (same number as ions)
Step 4 Combine and to form cancel where possible
Step 5 Balance electrically with electrons
2 H2O + MnO4
- + 3 e- ¡ MnO2 + 4 OH-
2 OH- + 2 Fe3O4 ¡ 3 Fe2O3 + H2O + 2 e-
(2 H2O cancelled)2 H2O + MnO4
- ¡ MnO2 + 4 OH-
(1 H2O cancelled)2 OH- + 2 Fe3O4 ¡ 3 Fe2O3 + H2O
4 H2O + MnO4
- ¡ MnO2 + 2 H2O + 4 OH-
2 OH- + H2O + 2 Fe3O4 ¡ 3 Fe2O3 + 2 H2O
H2OH2O;OH-H+
4 OH- + 4 H+ + MnO4
- ¡ MnO2 + 2 H2O + 4 OH-
2 OH- + H2O + 2 Fe3O4 ¡ 3 Fe2O3 + 2 H+ + 2 OH-
H+OH-
4 H+ + MnO4
- ¡ MnO2 + 2 H2O
H2O + 2 Fe3O4 ¡ 3 Fe2O3 + 2 H+
H+H2O
MnO4
- ¡ MnO2
2 Fe3O4 ¡ 3 Fe2O3
Fe3O4 + MnO4
- ¡ Fe2O3 + MnO2
6 OH- + Se ¡ SeO3
2- + 3 H2O + 4 e-
2 (Se + 2 e- ¡ Se2-) 6 OH- + 3 Se ¡ SeO3
2- + 2 Se2- + 3 H2O
Se + 2 e- ¡ Se2-
6 OH- + Se ¡ SeO3
2- + 3 H2O + 4 e-
(3 H2O cancelled)6 OH- + Se ¡ SeO3
2- + 3 H2O
Se ¡ Se2-
6 OH- + 3 H2O + Se ¡ SeO3
2- + 6 H2O
H2OH2O;OH-H+
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Step 6 and 7 Equalize gain and loss of electrons; add half-reactions
(e) (basic solution)
Step 1 Write half-reaction equations. Balance except H and O.
Step 2 Balance H and O using and
Step 3 Add ions to both sides (same number as ions)
Step 4 Combine and to form cancel where possible
Step 5 Balance electrically with electrons
Step 6 and 7 Equalize gain and loss of electrons; add half-reactions
16. (a) (basic solution)
Step 1 Write half-reaction equations. Balance except H and O.
MnO4
- ¡ MnO2
SO3
2- ¡ SO4
2-
MnO4
- + SO3
2- ¡ MnO2 + SO4
2-
2 (4 OH- + Cr(OH)4
- ¡ CrO4
2- + 4 H2O + 3 e-)
3 (H2O + BrO- + 2 e- ¡ Br - + 2 OH-) 2 OH- + 3 BrO- + 2 Cr(OH)4
- ¡ 3 Br - + 2 CrO4
2- + 5 H2O
3 H2O, 6 OH- and 6 e- canceled from each side
H2O + BrO- + 2 e- ¡ Br - + 2 OH-4 OH- + Cr(OH)4
- ¡ CrO4
2- + 4 H2O + 3 e-
(1 H2O cancelled)H2O + BrO- ¡ Br - + 2 OH-2 H2O + BrO- ¡ Br - + H2O + 2 OH-4 OH- + Cr(OH)4
- ¡ CrO4
2- + 4 H2O
H2OH2O;OH-H+2 OH- + 2 H+ + BrO- ¡ Br - + H2O + 2 OH-4 OH- + Cr(OH)4
- ¡ CrO4
2- + 4 H+ + 4 OH-H+OH-
2 H+ + BrO- ¡ Br - + H2O
Cr(OH)4
- ¡ CrO4
2- + 4 H+H+H2O
BrO- ¡ Br -Cr(OH)4
- ¡ CrO4
2-
BrO- + Cr(OH)4
- ¡ Br - + CrO4
2-
3 (2 OH- + 2 Fe3O4 ¡ 3 Fe2O3 + H2O + 2 e-)
2 (2 H2O + MnO4
- + 3 e- ¡ MnO2 + 4 OH-) H2O + 6 Fe3O4 + 2 MnO4
- ¡ 9 Fe2O3 + 2 MnO2 + 2 OH-
3 H2O, 6 OH-, and 6 e- canceled from each side
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Step 2 Balance H and O using and
Step 3 Add ions to both sides (same number as ions)
Step 4 Combine and to form cancel where possible
Step 5 Balance electrically with electrons
Step 6 and 7 Equalize gain and loss of electrons; add half-reactions
(b) (basic solution)
Step 1 Write half-reaction equations. Balance except H and O.
Step 2 Balance H and O using and
Step 3 Add ions to both sides (same number as ions)
ClO2 ¡ ClO2
-
2 OH- + 4 H2O + SbO2
- ¡ Sb(OH)6
- + 2 H+ + 2 OH-
H+OH-
ClO2 ¡ ClO2
-
4 H2O + SbO2
- ¡ Sb(OH)6
- + 2 H+
H+H2O
ClO2 ¡ ClO2
-
SbO2
- ¡ Sb(OH)6
-
ClO2 + SbO2
- ¡ ClO2
- + Sb(OH)6
-
3 (2 OH- + SO3
2- ¡ SO4
2- + H2O + 2 e-)
2 (MnO4
- + 2 H2O + 3 e- ¡ MnO2 + 4 OH-) H2O + 2 MnO4
- + 3 SO3
2- ¡ 2 MnO2 + 3 SO4
2- + 2 OH-
3 H2O, 4 OH-, and 6 e- canceled from each side
3 e- + MnO4
- + 2 H2O ¡ MnO2 + 4 OH-
2 OH- + SO3
2- ¡ SO4
2- + H2O + 2 e-
(2 H2O cancelled)MnO4
- + 2 H2O ¡ MnO2 + 4 OH-
(1 H2O cancelled)2 OH- + SO3
2- ¡ SO4
2- + H2O
MnO4
- + 4 H2O ¡ MnO2 + 2 H2O + 4 OH-
2 OH- + H2O + SO3
2- ¡ SO4
2- + 2 H2O
H2OH2O;OH-H+
4 OH- + MnO4
- + 4 H+ ¡ MnO2 + 2 H2O + 4 OH-
2 OH- + H2O + SO3
2- ¡ SO4
2- + 2 H+ + 2 OH-
H+OH-
MnO4
- + 4 H+ ¡ MnO2 + 2 H2O
H2O + SO3
2- ¡ SO4
2- + 2 H+
H+H2O
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Step 4 Combine and to form cancel where possible
Step 5 Balance electrically with electrons
Step 6 and 7 Equalize gain and loss of electrons; add half-reactions
(c) (basic solution)
Step 1 Write half-reaction equations. Balance except H and O.
Step 2 Balance H and O using and
Step 3 Add ions to both sides (same number as ions)
Step 4 Combine and to form cancel where possible
Step 5 Balance electrically with electrons
6 H2O + NO3
- + 8 e- ¡ NH3 + 9 OH-
4 OH- + Al ¡ Al(OH)4
- + 3 e-
(3 H2O cancelled)6 H2O + NO3
- ¡ NH3 + 9 OH-
(4 H2O cancelled)4 OH- + Al ¡ Al(OH)4
-
9 H2O + NO3
- ¡ NH3 + 3 H2O + 9 OH-
4 OH- + 4 H2O + Al ¡ Al(OH)4
- + 4 H2O
H2OH2O;OH-H+
9 OH- + 9 H+ + NO3
- ¡ NH3 + 3 H2O + 9 OH-
4 OH- + 4 H2O + Al ¡ Al(OH)4
- + 4 H+ + 4 OH-
H+OH-
9 H+ + NO3
- ¡ NH3 + 3 H2O
4 H2O + Al ¡ Al(OH)4
- + 4 H+
H+H2O
NO3
- ¡ NH3
Al ¡ Al(OH)4
-
Al + NO3
- ¡ NH3 + Al(OH)4
-
2 H2O + 2 OH- + SbO2
- ¡ Sb(OH)6
- + 2 e-
2 (ClO2 + e- ¡ ClO2
-) 2 H2O + 2 ClO2 + 2 OH- + SbO2
- ¡ 2 ClO2
- + Sb(OH)6
-
ClO2 + e- ¡ ClO2
-
2 OH- + 2 H2O + SbO2
- ¡ Sb(OH)6
- + 2 e-
(2 H2O cancelled)2 OH- + 2 H2O + SbO2
- ¡ Sb(OH)6
-
ClO2 ¡ ClO2
-
2 OH- + 4 H2O + SbO2
- ¡ Sb(OH)6
- + 2 H2O
H2OH2O;OH-H+
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Step 6 and 7 Equalize gain and loss of electrons; add half-reactions
(d) (basic solution)
Step 1 Write half-reaction equations. Balance except H and O.
Step 2 Balance H and O using and
Step 3 Add ions to both sides (same number as ions)
Step 4 Combine and to form cancel where possible
Step 5 Balance electrically with electrons
Step 6 and 7 Loss and gain of electrons are equal; add half-reactions
Divide equation by 2
(e) (basic solution)
Step 1 Write half-reaction equations. Balance except H and O.
OH- ¡ H2
Al ¡ Al(OH)4
-
Al + OH- ¡ Al(OH)4
- + H2
4 OH- + 2 H2O + P4 ¡ 2 HPO3
2- + 2 PH3
8 OH- + 4 H2O + 2 P4 ¡ 4 HPO3
2- + 4 PH3
12 H2O + P4 + 12 e- ¡ 4 PH3 + 12 OH-
20 OH- + P4 ¡ 4 HPO3
2- + 8 H2O + 12 e-
20 OH- + P4 ¡ 4 HPO3
2- + 8 H2O (12 H2O cancelled)
12 H2O + P4 ¡ 4 PH3 + 12 OH-
20 OH- + 12 H2O + P4 ¡ 4 HPO3
2- + 20 H2O
H2OH2O;OH-H+
12 OH- + 12 H+ + P4 ¡ 4 PH3 + 12 OH-
20 OH- + 12 H2O + P4 ¡ 4 HPO3
2- + 20 H+ + 20 OH-
H+OH-
12 H+ + P4 ¡ 4 PH3
12 H2O + P4 ¡ 4 HPO3
2- + 20 H+
H+H2O
P4 ¡ 4 PH3
P4 ¡ 4 HPO3
2-
P4 ¡ HPO3
2- + PH3
8 (4 OH- + Al ¡ Al(OH)4
- + 3 e-)
3 (6 H2O + NO3
- + 8 e- ¡ NH3 + 9 OH-) 8 Al + 3 NO3
- + 18 H2O + 5 OH- ¡ 3 NH3 + 8 Al(OH)4
-
27 OH- and 24 e- canceled from each side
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Step 2 Balance H and O using and
Step 3 Add ions to both sides (same number as ions)
Step 4 Combine and to form cancel where possible
Step 5 Balance electrically with electrons
Step 6 and 7 Equalize gain and loss of electrons; add half-reactions
17. (a) (acidic solution)
Step 1 Write half-reaction equations. Balance except H and O.
Step 2 Balance H and O using and
Step 3 Balance electrically with electrons
2 I- ¡ I2 + 2 e-
12 H+ + 2 IO3
- + 10 e- ¡ I2 + 6 H2O
2 I- ¡ I2
12 H+ + 2 IO3
- ¡ I2 + 6 H2O
H+H2O
2 I- ¡ I2
2 IO3
- ¡ I2
IO3
- + I- ¡ I2
2 (4 OH- + Al ¡ Al(OH)4
- + 3 e-)
3 (2 H2O + OH- + 2 e- ¡ H2 + 3 OH-) 2 Al + 6 H2O + 2 OH- ¡ 2 Al(OH)4
- + 3 H2
9 OH- and 6 e- canceled on each side
2 H2O + OH- + 2 e- ¡ H2 + 3 OH-
4 OH- + Al ¡ Al(OH)4
- + 3 e-
(1 H2O cancelled)2 H2O + OH- ¡ H2 + 3 OH-
(4 H2O cancelled)4 OH- + Al ¡ Al(OH)4
-
3 H2O + OH- ¡ H2 + H2O + 3 OH-
4 OH- + 4 H2O + Al ¡ Al(OH)4
- + 4 H2O
H2OH2O;OH-H+
3 OH- + 3 H+ + OH- ¡ H2 + H2O + 3 OH-
4 OH- + 4 H2O + Al ¡ Al(OH)4
- + 4 H+ + 4 OH-
H+OH-
3 H+ + OH- ¡ H2 + H2O
4 H2O + Al ¡ Al(OH)4
- + 4 H+
H+H2O
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Step 4 and 5 Equalize the loss, and gain of electrons; add the half-reaction.
(b) (acid solution)
Step 1 Write half-reaction equations. Balance except H and O
Step 2 Balance H and O using and
Step 3 Balance electrically with electrons
Step 4 and 5 Equalize the loss and gain of electrons; add the half-reactions
Each side has 2 Mn, 10 S, 16 H, and 48 O and a charge.
(c) (acidic solution)
Step 1 Write half-reaction equations. Balance except H and O.
Step 2 Balance H and O using and
Step 3 Balance electrically with
5 e- + 8 H+ + MnO4
- ¡ Mn2+ + 4 H2O
6 H2O + Co(NO2)6
3- ¡ Co2+ + 6 NO3
- + 12 H+ + 11 e-
e-
8 H+ + MnO4
- ¡ Mn2+ + 4 H2O
6 H2O + Co(NO2)6
3- ¡ Co2+ + 6 NO3
- + 12 H+
H+H2O
MnO4
- ¡ Mn2+
Co(NO2)6
3- ¡ Co2+ + 6 NO3
-
Co(NO2)6
3- + MnO4
- ¡ Co2+ + Mn2+ + NO3
-
-6
2 (4 H2O + Mn2+ ¡ MnO4
- + 8 H+ + 5 e-)
5 (2 e- + S2O8
2- ¡ 2 SO4
2-) 2 Mn2+ + 5 S2O8
2- + 8 H2O ¡ 2 MnO4
- + 10 SO4
2- + 16 H+
2 e- + S2O8
2- ¡ 2 SO4
2-
4 H2O + Mn2+ ¡ MnO4
- + 8 H+ + 5 e-
S2O8
2- ¡ 2 SO4
2-
4 H2O + Mn2+ ¡ MnO4
- + 8 H+
H+H2O
S2O8
2- ¡ 2 SO4
2-
Mn2+ ¡ MnO4
-
Mn2+ + S2O8
2- ¡ MnO4
- + SO4
2-
12 H+ + 2 IO3
- + 10 e- ¡ I2 + 6 H2O
5 (2 I- ¡ I2 + 2 e-) 12 H+ + 2 IO3
- + 10 I- ¡ 6 I2 + 6 H2O
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Step 4 Equalize the loss and gain of electrons.
Step 5 Add the half-reactions
Each side has 5 Co, 30 N, 11 Mn, 28 H, 104 O and charge.
18. (a) (acid solution)
Step 1 Write half-reactions equations. Balance except H and O
Step 2 Balance H and O using and
Step 3 Balance electrically with electrons
Steps 4 and 5 Equalize the loss and gain of electrons; add the half-reactions.
(b) (basic solution)
Step 1 Write half-reaction equation. Balance except H and O
Step 2 Balance H and O using and
Step 3 Add ions to both sides (same number as )
4 OH- + Cr(OH)4
- ¡ CrO4
2- + 4 H+ + 4 OH-
2 OH- + 2 H+ + BrO- ¡ Br - + H2O + 2 OH-
H+OH-
Cr(OH)4
- ¡ CrO4
2- + 4 H+
2 H+ + BrO- ¡ Br - + H2O
H+H2O
Cr(OH)4
- ¡ CrO4
2-
BrO- ¡ Br -
BrO- + Cr(OH)4
- ¡ Br - + CrO4
2-
5 (3 H2O + Mo2O3 ¡ 2 MoO3 + 6 H+ + 6 e-)
6 (5 e- + 8 H+ + MnO4
- ¡ Mn2+ + 4 H2O) 5 Mo2O3 + 6 MnO4
- + 18 H+ ¡ 10 MoO3 + 6 Mn2+ + 9 H2O
5 e- + 8 H+ + MnO4
- ¡ Mn2+ + 4 H2O
3 H2O + Mo2O3 ¡ 2 MoO3 + 6 H+ + 6 e-
8 H+ + MnO4
- ¡ Mn2+ + 4 H2O
3 H2O + Mo2O3 ¡ 2 MoO3 + 6 H+
H+H2O
MnO4
- ¡ Mn2+
Mo2O3 ¡ 2 MoO3
Mo2O3 + MnO4
- ¡ MoO3 + Mn2+
a + 2
5 Co(NO2)6
3- + 11 MnO4
- + 28 H+ ¡ 5 Co2+ + 30 NO3
- + 11 Mn2+ + 14 H2O
11 (5 e- + 8 H+ + MnO4
- ¡ Mn2+ + 4 H2O)
5 (6 H2O + Co(NO2)6
3- ¡ Co2+ + 6 NO3
- + 12 H+ + 11 e-)
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Step 4 Combine and to form cancel where possible
Step 5 Balance electrically with electrons
Steps 6 and 7 Equalize loss and gain of electrons; add the half-reactions
(c) (basic solution)
Step 1 Write half-reaction equations. Balance except H and O.
Step 2 Balance H and O using and
Step 3 Add ions to both sides (same number as )
Step 4 Combine and to form cancel where possible
Step 5 Balance electrically with electrons
3 e- + 2 H2O + MnO4
- ¡ MnO2 + 4 OH-
10 OH- + S2O3
2- ¡ 2 SO4
2- + 5 H2O + 8 e-
(2 H2O cancelled)2 H2O + MnO4
- ¡ MnO2 + 4 OH-
(5 H2O cancelled)10 OH- + S2O3
2- ¡ 2 SO4
2- + 5 H2O
4 H2O + MnO4
- ¡ MnO2 + 2 H2O + 4 OH-
10 OH- + 5 H2O + S2O3
2- ¡ 2 SO4
2- + 10 H2O
H2OH2O;OH-H+
4 OH- + 4 H+ + MnO4
- ¡ MnO2 + 2 H2O + 4 OH-
10 OH- + 5 H2O + S2O3
2- ¡ 2 SO4
2- + 10 H+ + 10 OH-
H+OH-
4 H+ + MnO4
- ¡ MnO2 + 2 H2O
5 H2O + S2O3
2- ¡ 2 SO4
2- + 10 H+
H+H2O
MnO4
- ¡ MnO2
S2O3
2- ¡ 2 SO4
2-
S2O3
2- + MnO4
- ¡ SO4
2- + Mn2+
3 (2 e- + H2O + BrO- ¡ Br - + 2 OH-)
2 (4 OH- + Cr(OH)4
- ¡ CrO4
2- + 4 H2O + 3 e-) 3 BrO- + 2 Cr(OH)4
- + 2 OH- ¡ 3 Br - + 2 CrO4
2- + 5 H2O
4 OH- + Cr(OH)4
- ¡ CrO4
2- + 4 H2O + 3 e-
2 e- + H2O + BrO- ¡ Br - + 2 OH-
(1 H2O cancelled)H2O + BrO- ¡ Br - + 2 OH-
4 OH- + Cr(OH)4
- ¡ CrO4
2- + 4 H2O
2 H2O + BrO- ¡ Br - + H2O + 2 OH-
H2OH2O;OH-H+
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Step 6 and 7 Equalize loss and gain of electrons; add half-reactions.
Each side has 6 S, 8 Mn, 2 H, 42 O and a
19.
20. (a)
(b) The first reaction is oxidation ( is oxidized to ).
The second reaction is reduction ( is reduced to ).
(c) The first reaction (oxidation) occurs at the anode of the battery.
21. (a) The oxidizing agent is
(b) The reducing agent is HCl.
(c) 5 moles of electrons
22. Zinc is a more reactive metal than copper so when corrosion occurs the zinc preferentiallyreacts. Zinc is above hydrogen in the Activity series of metals; copper is below hydrogen.
23. (balanced)
(25.0 g Ag)a 1 mol Ag
107.9 g Agb a1 mol NO
3 mol Agb = 0.0772 mol NO
g Ag ¡ mol Ag ¡ mol NO
3 Ag + 4 HNO3 ¡ 3 AgNO3 + NO + 2 H2O
¢ 5 mol e-
mol KMnO4≤ ¢ 6.022 * 1023 e-
mol e- ≤ = 3.011 * 1024 electrons
mol KMnO4
5 e- + Mn7+ ¡ Mn2+
KMnO4.
Pb2+Pb4+Pb2+Pb0
PbO2 + SO4
2- + 4 H+ + 2 e- ¡ PbSO4 + 2 H2O
Pb + SO4
2- ¡ PbSO4 + 2 e-
Voltagesource
Anode (+)
– +
Cathode (–)
Solution of HBr
Br–
H3O+
-14 charge.
3 (10 OH- + S2O3
2- ¡ 2 SO4
2- + 5 H2O + 8 e-)
8 (3 e- + 2 H2O + MnO4
- ¡ MnO2 + 4 OH-) 3 S2O3
2- + 8 MnO4
- + H2O ¡ 6 SO4
2- + 8 MnO2 + 2 OH
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24.
25.
26.
27.
28.
29. (a) is an oxidation, but when electrons are gained reduction should occur.
(b) When is reduced, it requires two individual electrons. Anelectron has only a single negative charge (e-).
Pb2+ + 2 e- ¡ Pb0.Pb2+Cu+ + e- ¡ Cu0 or Cu+ ¡ Cu2+ + e-Cu+ ¡ Cu2+
(100.0 g Al)a1 mol Al
26.98 gb ¢3 mol H2
2 mol Al≤ = 5.560 mol H2
g Al ¡ mol Al ¡ mol H2
2 Al + 2 OH- + 6 H2O ¡ 2 Al(OH)4
- + 3 H2
= 10.0 mL of 0.200 M K2Cr2O7
(60.0 mL FeSO4)a0.200 mol
1000 mLb ¢1 mol Cr2O7
2-
6 mol FeSO4≤ a 1000 mL
0.200 molb
mL FeSO4 ¡ mol FeSO4 ¡ mol Cr2O7
2- ¡ mL Cr2O7
2-
Cr2O7
2- + 6 Fe2+ + 14 H+ ¡ 2 Cr3+ + 6 Fe3+ + 7 H2O
(5.00 g H3AsO3)a 1 mol
125.9 gb ¢1 mol Cr2O7
2-
3 mol H3AsO3≤ a 1000 mL
0.200 molb = 66.2 mL of 0.200 M K2Cr2O7
g H3AsO3 ¡ mol H3AsO3 ¡ mol Cr2O7
2- ¡ mL Cr2O7
2-
Cr2O7
2- + 3 H3AsO3 + 8 H+ ¡ 2 Cr3+ + 3 H3AsO4 + 4 H2O
= 17 g KMnO4a2 mol KMnO4
5 mol H2O2b a158.0 g
molb
(100. mL H2O2 solution)a1.031 g
mLb ¢ 9.0 g H2O2
100. g H2O2 solution≤ a 1 mol
34.02 gb
mL H2O2 ¡ g H2O2 ¡ mol H2O2 ¡ mol KMnO4 ¡ g KMnO4
5 H2O2 + 2 KMnO4 + 3 H2SO4 ¡ 5 O2 + 2 MnSO4 + K2SO4 + 8 H2O
(0.300 mol KClO3)¢ 3 mol Cl21 mol KClO3
≤ a 22.4 L
1 molb = 20.2 L Cl2
mol KClO3 ¡ mol Cl2 ¡ L Cl2
3 Cl2 + 6 KOH ¡ KClO3 + 5 KCl + 3 H2O
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30. The electrons lost by the species undergoing oxidation must be gained (or attracted) byanother species which then undergoes reduction.
31. cannot take from A
cannot take from A
takes from D
takes 2 from B
Therefore, is least able to attract then then then
32. can only be an oxidizing agent.
can only be a reducing agent.
can be both oxidizing and reducing.
33. is the best oxidizing agent of the group, since its greater
oxidation number makes it very attractive to electrons.
34. Equations (a) and (b) represent oxidation
(a)
(b)
35. (a)
(b)
(c)
V = a0.005 mol
1.4 atmb a 0.0821 L atm
mol Kb (323 K) = 0.09 L Br2 vapor
V =nRT
PPV = nRT
(100.0 mL Mn2+)a 0.05 mol
1000 mLb a 1 mol Br2
1 mol Mn2+ b = 0.005 mol Br2
(100.0 mL Mn2+)a 0.05 mol
1000 mLb ¢1 mol MnO2
1 mol Mn2+ ≤ a86.94 g
molb = 0.4 g MnO2
mL Mn2+ ¡ mol Mn2+ ¡ mol MnO2 ¡ g MnO2
MnO2 + 2 Br - + 4 H+ ¡ Mn2+ + Br2 + 2 H2O
SO2 ¡ SO3; (S4+ ¡ S6+ + 2 e-)
Mg ¡ Mg2+ + 2 e-
+7KMnO4
+6K2MnO4
+4MnO2
(+7)+3MnF3
KMnO4+2Mn(OH)2
Sn2+ + 2 e- ¡ Sn0 (oxidizing)
Sn2+ ¡ Sn4+ + 2 e- (reducing)Sn2+
Sn0 ¡ Sn2+ + 2 e-
Sn0 ¡ Sn4+ + 4 e-Sn0
Sn4+ + 2 e- ¡ Sn2+
Sn4+ + 4 e- ¡ Sn0Sn4+
A+C+,D2+,e-,B2+e-D2+B(s) + D2+(aq) ¡ D(s) + B2+(aq)
e-C+D(s) + 2 C+(aq) ¡ 2C(s) + D2+(aq)
e-C+A(s) + C+(aq) ¡ NR
e-B2+A(s) + B2+(aq) ¡ NR
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36. (a)
(b)
(c)
(d)
37.
38. See Exercise 13(a).
39. Equation 1 2 3 4 5
a C oxidized S oxidized N oxidized S oxidized oxidized
b reduced N reduced Cu reduced reduced reduced
c O.A. O.A. CuO, O.A. O.A. O.A.
d R.A. R.A. R.A. R.A. R.A.
e
f
40.
(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.
Ag
Saltbridge
Pb
Pb2+
NO–3
Ag+
NO–3
Pb + 2 Ag+ ¡ 2 Ag + Pb2+
R.A. = Reducing agent
O.A. = Oxidizing agent
O2
2- ¡ O2-O2
2- ¡ O2-Cu2+ ¡ Cu0N5+ ¡ N2+O0 ¡ O2-
O2
2- ¡ O2
0S4+ ¡ S6+N3- ¡ N2
0S2- ¡ S0C223+ ¡ C4+
H2O2,Na2SO3,NH3,H2S,C3H8,
H2O2,H2O2,HNO3,O2,
O2
2-O2
2-O2
O2
2-
4 Zn + NO3
- + 10 H+ ¡ 4 Zn2+ + NH4
+ + 3 H2O
Mn(s) + 2 HCl(aq) ¡ Mn2+(aq) + H2(g) + 2 Cl-(aq)
Br2 + 2 I- ¡ 2 Br - + I2
I2 + Cl- ¡ NR
Br2 + Cl- ¡ NR
F2 + 2 Cl- ¡ 2 F- + Cl2
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41.
start with grams and work towards g KI
42.
V =(0.167 mol NO)(0.0821 L atm>mol K)(301 K)
(0.979 atm)= 4.22 L NO
T = 301 K
P = (744 torr)a 1 atm
760. torrb = 0.979 atm
V =nRT
PPV = nRT
(0.500 mol Ag)a1 mol NO
3 mol Agb = 0.167 mol NO
mol Ag ¡ mol NO
3 Ag + 4 HNO3 ¡ 3 AgNO3 + NO + 2 H2O
a 3.65 g KI
4.00 g sampleb11002 = 91.3% KI
(2.79 g I2)a 1 mol
253.8 gb ¢8 mol KI
4 mol I2≤ a 166.0 g
molb = 3.65 g KI in sample
g I2 ¡ mol I2 ¡ mol KI ¡ g KI
I2
8 KI + 5 H2SO4 ¡ 4 I2 + H2S + 4 K2SO4 + 4 H2O
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