oxidation-reductionfaculty.chemeketa.edu/lemme/solutions/ch17.pdf · chapter 17 oxidation-reduction...

C H A P T E R 1 7

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) Iodine is oxidized. Its oxidation number increases from 0 to þ5.(b) 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) 2Al(s)þ 3ZnCl2(aq)! 3Zn(s)þ 2AlCl3(aq)

(b) Sn(s)þ 2HCl(aq)! H2(g)þ SnCl2(aq)

(c) Ag(s)þH2SO4(aq)! no reaction H2 is more reactive than Ag

(d) Fe(s)þ 2AgNO3(aq)! 2Ag(s)þ Fe(NO3)2(aq)

(e) 2 Cr(s)þ 3Ni2þ(aq)! 3Ni(s)þ 2Cr3þ(aq)(f) Mg(s)þCa2þ(aq)! no reaction Ca is more reactive than Mg

(g) Cu(s)þHþ(aq)! no reaction H2 is more reactive than Cu

(h) Ag(s)þAl3þ(aq)! no reaction Al is more reactive than Ag

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) 2Alþ Fe2O3! A12O3þ 2 FeþHeat

(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 its compounds.

(d) Yes. Aluminum is above chromium in the activity series and will displace Cr3þ from its

compounds.

7. (a) 2Al(s)þ 6HCl(aq)! 2AlCl3(aq)þ 3H2(g)

2Al(s)þ 3H2SO4(aq)! Al2(SO4)3(aq)þ 3H2(g)

(b) 2 Cr(s)þ 6HCl(aq)! 2CrCl3(aq)þ 3H2(g)

2 Cr(s)þ 3H2SO4(aq)! Cr2(SO4)3(aq)þ 3H2(g)

(c) Au(s)þHCl(aq)! no reaction

Au(s)þH2SO4(aq)! no reaction

(d) Fe(s)þ 2HCl(aq)! FeCl2(aq)þH2(g)

Fe(s)þH2SO4(aq)! FeSO4(aq)þH2(g)

(e) Cu(s)þHCl(aq)! no reaction

Cu(s)þH2SO4(aq)! no reaction

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(f) Mg(s)þ 2HCl(aq)! MgCl2(aq)þH2(g)

Mg(s)þH2SO4(aq)! MgSO4(aq)þH2(g)

(g) Hg(l)þHCl(aq)! no reaction

Hg(l)þH2SO4(aq)! no reaction

(h) Zn(s)þ ! ZnCl2(aq)þH2(g)

Zn(s)þH2SO4(aq)! ZnSO4(aq)þH2(g)

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

2Cl�ðaqÞ!Cl2ðgÞ þ 2 e�

(b) Reduction occurs at the cathode. The reaction is

Ni2þðaqÞ þ 2 e�!NiðsÞ(c) The net chemical reaction is

Ni2þðaqÞ þ 2Cl�ðaqÞ!electricalenergy

NiðsÞ þ Cl2ðgÞ

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 the cell would still

occur.

(b) 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. Ca2þþ 2 e� ! Ca cathode reaction, reduction

2Br� ! Br2þ 2 e� 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.

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2HCl(aq)

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.

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SOLUTIONS TO EXERCISES

1. Oxidation numbers of each element in the compound

(a) CuCO3 Cu¼þ2 C¼þ4 O¼�2(b) CH4 C¼�4 H¼þl(c) IF I¼þ1 F¼�l(d) CH2Cl2 C¼0 H¼þ1 Cl¼�1(e) SO2 S¼þ4 O¼�2(f) Rb2CrO4 Rb¼þ1 Cr¼þ6 O¼�2

2. Oxidation numbers of each element in the compound.

(a) CHF3 C¼þ2 H¼þl F¼�1(b) P2O5 P¼þ5 O¼�2(c) SF6 S¼þ6 F¼�1(d) SnSO4 Sn¼þ2 S¼þ6 O¼�2(e) CH3OH C¼�2 H¼þl O¼�2(f) H3PO4 H¼þl P¼þ5 O¼�2

3. The oxidation number of the underlined element is indicated by the number following the formula.

(a) _PO 3�3 þ 3 (c) NaHCO3 þ 4

(b) CaSO4 þ 6 (d) _BrO �4 þ 7

4. The oxidation number of the underlined element is indicated by the number following the formula.

(a) _CO 2�3 þ 4 (c) NaH2PO4 þ 5

(b) H2SO4 þ 6 (d) _Cr2O2�7 þ 6

5. Balanced half-reaction Element

Changing

Type of

Reaction

(a) Na! Naþþ l e� Na oxidation

(b) C2O2�4 ! 2CO2 þ 2 e� C oxidation

(c) 2 I� ! I2þ 2 e� I oxidation

(d) Cr2O2�7 þ 14Hþ þ 6 e� ! 2Cr3þ þ 7H2O Cr reduction

6. Balanced half-reaction Element

Changing

Type of

Reaction

(a) Cu2þ þ 1e� ! Cu1þ Cu reduction

(b) F2 þ 2 e� ! 2 F� F reduction

(c) 2 IO �4 þ 16Hþ þ 14 e� ! I2 þ 8H2O I reduction

(d) Mn! Mn2þ þ 2e� Mn oxidation

7. (a) Cu is oxidized, Ag is reduced;

Cu is the reducing agent, AgNO3 is the oxidizing agent

(b) Zn is oxidized, H is reduced

Zn is the reducing agent, HCl is the oxidizing agent

8. (a) C is oxidized, O is reduced

CH4 is the reducing agent, O2 is the oxidizing agent

(b) Mg is oxidized, Fe is reduced

Mg is the reducing agent, FeCl3 is the oxidizing agent

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9. (a) correctly balanced

(b) correctly balanced

(c) incorrectly balanced

MgðsÞ þ 2HClðaqÞ!Mg2þðaqÞ þ 2Cl�ðaqÞ þ H2ðgÞ(d) incorrectly balanced

3CH3OHðaqÞ þ Cr2O2�7 ðaqÞ þ 8HþðaqÞ! 2Cr3þðaqÞ þ 3CH2OðaqÞ þ 7H2OðlÞ

10. (a) incorrectly balanced

3MnO2ðsÞ þ 4AlðsÞ! 3MnðsÞ þ 2Al2O3ðsÞ(b) correctly balanced

(c) correctly balanced

(d) incorrectly balanced

8H2OðlÞ þ 2MnO �4 ðaqÞ þ 7 S2�ðaqÞ! 2MnSðsÞ þ 16OH�ðaqÞ þ 5 SðsÞ

11. Balancing oxidation-reduction equations using the change-in-oxidation number method:

(a) Cuþ O2 ! CuO

ox Cu0 ! Cu2þ þ 2 e� Multiply by 2;

red 2O0 þ 4 e� ! 2O2� Add the equations; the 4 e� cancel

2 Cu0 þ 2O0 ! 2Cu2þ þ 2O2�

Transfer the coefficients to the original equation appropriately.

2 Cuþ O2 ! 2CuO

(b) KClO3 ! KClþ O2

ox 3O2� ! O0 þ 6 e� Multiply by 2;

red Cl5þ þ 6 e� ! Cl� Multiply by 2, add; the 12 e� cancel

2 Cl5þ þ 6O2� ! 2Cl� þ 3O2�

Transfer the coefficients to the original equation appropriately.

2KClO3 ! 2KClþ 3O2

(c) Caþ H2O! CaðOHÞ2 þ H2

ox Ca! Ca2þ þ 2 e�

red 2Hþ þ 2 e� ! H2 Add equations together; the 2 e� cancel:

Caþ 2Hþ ! Ca2þ þ H2

Balance the equation by inspection.

Caþ 2H2O! CaðOHÞ2 þ H2

(d) PbSþ H2O2 ! PbSO4 þ H2O

ox S2� ! Sþ6 þ 8 e�

red 2O� þ 2 e� ! 2O2� Multiply by 4, add; the 8 e� cancel:

S2� þ 8O� ! Sþ6 þ 4O2�

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- Chapter 17 -

Transfer the coefficients to the original equation and complete the balancing

by inspection.

PbSþ 4H2O2 ! PbSO4 þ 4H2O

(e) CH4 þ NO2 ! N2 þ CO2 þ H2O

ox C4� ! C4þ þ 8 e�

red N4þ þ 4 e� ! N0 Multiply by 2; add the 8 e� cancel

C4� þ 2N4þ ! C4þ þ 2N0


by inspection.

CH4 þ 2NO2 ! N2 þ CO2 þ 2H2O

12. Balancing oxidation-reduction equations using the change-in-oxidation number method:

(a) Cuþ AgNO3 ! Agþ CuðNO3Þ2ox Cu0 ! Cu2þ þ 2 e�

red Agþ þ 1 e� ! Ag0 Multiply by 2, add; the 2 e� cancel

Cu0 þ 2Agþ ! Cu2þ þ 2Ag0

Transfer the coefficients to the original equation.

Cuþ 2AgNO3 ! 2Agþ CuðNO3Þ2(b) MnO2 þ HCl! MnCl2 þ Cl2 þ H2O

ox Cl� ! Cl0 þ 1 e� Multiply by 2, add; the 2 e� cancel

red Mn4þ þ 2 e� ! Mn2þ

2Cl� þMn4þ ! 2Cl0 þMn2þ


by inspection.

MnO2 þ 4HCl! MnCl2 þ Cl2 þ 2H2O

(c) HClþ O2 ! Cl2 þ H2O

red 2O0 þ 4 e� ! 2O2�

ox 2Cl� ! 2Cl0 þ 2 e� Multiply by 2, add; the 4 e� cancel

4 Cl� þ 2O0 ! 4Cl0 þ 2O2�


by inspection.

4HClþ O2 ! 2Cl2 þ 2H2O

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- Chapter 17 -

(d) Agþ H2Sþ O2 ! Ag2Sþ H2O

red 2O0 þ 4 e� ! 2O2�

ox Ag! Agþ þ 1 e� Multiply by 4, add; the 4 e� cancel

2Agþ 2O0 ! 2Agþ þ 2O2�


by inspection.

4Agþ 2H2Sþ O2 ! 2Ag2Sþ 2H2O

(e) KMnO4 þ CaC2O4 þ H2SO4 ! K2SO4 þMnSO4 þ CaSO4 þ CO2 þ H2O

red Mn7þ þ 5 e� ! Mn2þ Multiply by 2;

ox 2C3þ ! 2C4þ þ 2 e� Multiply by 5, add; the 10 e� cancel

10C3þ þ 2Mn7þ ! 10C4þ þ 2Mn2þ


by inspection.

2KMnO4 þ 5CaC2O4 þ 8H2SO4 ! K2SO4 þ 2MnSO4 þ 5CaSO4

þ 10CO2 þ 8H2O

13. Balancing ionic redox equations(a) Znþ NO �

3 !Zn2þ þ NH þ4 ðacidic solutionÞ

Step 1 Write half-reaction equations. Balance except H and O.

Zn!Zn2þ

NO �3 !NH þ

4

Step 2 Balance H and O using H2O and Hþ

Zn!Zn2þ

10Hþ þ NO �3 !NH þ

4 þ 3H2O

Step 3 Balance electrically with electrons

Zn!Zn2þ þ 2 e�

10Hþ þ NO �3 þ 8 e�!NH þ

4 þ 3H2O

Step 4 Equalize the loss and gain of electrons

4ðZn!Zn2þ þ 2 e�Þ10Hþ þ NO �

3 þ 8 e�!NH þ4 þ 3H2O

Step 5 Add the half-reactions-electrons cancel

10Hþ þ 4 Znþ NO �3 ! 4Zn2þ þ NH þ

4 þ 3H2O

(b) NO �3 þ S!NO2 þ SO 2�

4 ðacidic solutionÞStep 1 Write half-reaction equations. Balance except H and O.

S! SO 2�4

NO �3 !NO2

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- Chapter 17 -


4H2Oþ S! SO 2�4 þ 8Hþ

2Hþ þ NO �3 !NO2 þ H2O


4H2Oþ S! SO 2�4 þ 8Hþ þ 6 e�

2Hþ þ NO �3 þ e�!NO2 þ H2O

Step 4 and 5 Equalize the loss and gain of electrons; add the half-reactions

4H2Oþ S! SO 2�4 þ 8Hþ þ 6 e�

6 ð2Hþ þ NO �3 þ e�!NO2 þ H2OÞ

4Hþ þ Sþ 6NO �3 ! 6NO2 þ SO 2�

4 þ 2H2O

4H2O, 8Hþ and 6e� canceled from each side

(c) PH3 þ I2!H3PO2 þ I� ðacidic solutionÞStep 1 Write half-reaction equations. Balance except H and O.

PH3!H3PO2

I2! 2 I�


2H2Oþ PH3!H3PO2 þ 4Hþ

I2! 2 I�


2H2Oþ PH3!H3PO2 þ 4Hþ þ 4 e�

I2 þ 2 e�! 2 I�


2H2Oþ PH3!H3PO2 þ 4Hþ þ 4 e�

2 ðI2 þ 2 e�! 2 I�ÞPH3 þ 2H2Oþ 2 I2!H3PO2 þ 4 I� þ 4Hþ

(d) Cuþ NO �3 !Cu2þ þ NO ðacidic solutionÞ


Cu!Cu2þ

NO �3 !NO


Cu!Cu2þ

4Hþ þ NO �3 !NOþ 2H2O


Cu!Cu2þ þ 2 e�

4Hþ þ NO �3 þ 3 e�!NOþ 2H2O

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- Chapter 17 -


3 ðCu!Cu2þ þ 2 e�Þ2 ð4Hþ þ NO �

3 þ 3 e�!NOþ 2H2OÞ3Cuþ 8Hþ þ 2NO �

3 ! 3Cu2þ þ 2NOþ 2H2O

(e) ClO �3 þ Cl�!Cl2 ðacidic solutionÞ


Cl�!Cl0

ClO �3 !Cl0


Cl�!Cl0

6Hþ þ ClO �3 !Cl0 þ 3H2O


Cl�!Cl0 þ e�

6Hþ þ ClO �3 þ 5 e�!Cl0 þ 3H2O


5ðCl�!Cl0 þ e�Þ6Hþ þ ClO �

3 þ 5 e�!Cl0 þ 3H2O

6Hþ þ ClO �3 þ 5Cl�! 3Cl2 þ 3H2O

14. (a) ClO �3 þ I�! I2 þ Cl� ðacidic solutionÞ


2 I�! I2

ClO �3 !Cl�


2 I�! I2

6Hþ þ ClO �3 !Cl� þ 3H2O


2 I�! I2 þ 2 e�

6Hþ þ ClO �3 þ 6 e�!Cl� þ 3H2O


3 ð2 I�! I2 þ 2 e�Þ6Hþ þ ClO �

3 þ 6 e�!Cl� þ 3H2O

6Hþ þ ClO �3 þ 6 I�! 3 I2 þ Cl� þ 3H2O

(b) Cr2O2�7 þ Fe2þ!Cr3þ þ Fe3þ ðacidic solutionÞ


Fe2þ! Fe3þ

Cr2O2�7 ! 2Cr3þ

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- Chapter 17 -


Fe2þ!Fe3þ

14Hþ þ Cr2O2�7 ! 2Cr3þ þ 7H2O


Fe2þ!Fe3þ þ e�

14Hþ þ Cr2O2�7 þ 6 e�! 2Cr3þ þ 7H2O


6ðFe2þ!Fe3þ þ e�Þ14Hþ þ Cr2O

2�7 þ 6 e�! 2Cr3þ þ 7H2O

14Hþ þ Cr2O2�7 þ 6 Fe2þ! 2Cr3þ þ 6 Fe3þ þ 7H2O

(c) MnO �4 þ SO2!Mn2þ þ SO 2�

4 ðacidic solutionÞStep 1 Write half-reaction equations. Balance except H and O.

SO2!SO 2�4

MnO �4 !Mn2þ


2H2Oþ SO2! SO 2�4 þ 4Hþ

8Hþ þMnO �4 !Mn2þ þ 4H2O


2H2Oþ SO2! SO 2�4 þ 4Hþ þ 2 e�

8Hþ þMnO �4 þ 5 e�!Mn2þ þ 4H2O


5ð2H2Oþ SO2! SO 2�4 þ 4Hþ þ 2 e�Þ

2ð8Hþ þMnO �4 þ 5 e�!Mn2þ þ 4H2OÞ

2H2Oþ 2MnO �4 þ 5 SO2! 4Hþ þ 2Mn2þ þ 5 SO 2�

4

8 H2O, 16 Hþ, and 10 e� canceled from each side

(d) H3AsO3 þMnO �4 !H3AsO4 þMn2þ ðacidic solutionÞ


H3AsO3!H3AsO4

MnO �4 !Mn2þ


H2Oþ H3AsO3! 2Hþ þ H3AsO4

8Hþ þMnO �4 !Mn2þ þ 4H2O


H2Oþ H3AsO3! 2Hþ þ H3AsO4 þ 2 e�

8Hþ þMnO �4 þ 5 e�!Mn2þ þ 4H2O

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- Chapter 17 -


5ðH2Oþ H3AsO3! 2Hþ þ H3AsO4 þ 2 e�Þ2ð8Hþ þMnO �

4 þ 5 e�!Mn2þ þ 4H2OÞ6Hþ þ 5H3AsO3 þ 2MnO �

4 ! 5H3AsO4 þ 2Mn2þ þ 3H2O

5 H2O, 10 Hþ, and 10 e� canceled from each side

(e) Cr2O2�7 þ H3AsO3!Cr3þ þ H3AsO4ðacidic solutionÞ


H3AsO3!H3AsO4

Cr2O2�7 ! 2Cr3þ


H2Oþ H3AsO3! 2Hþ þ H3AsO4

14Hþ þ Cr2O2�7 ! 2Cr3þ þ 7H2O


H2Oþ H3AsO3! 2Hþ þ H3AsO4 þ 2 e�

14Hþ þ Cr2O2�7 þ 6 e�! 2Cr3þ þ 7H2O


3 ðH2Oþ H3AsO3! 2Hþ þ H3AsO4 þ 2 e�Þ14Hþ þ Cr2O

2�7 þ 6 e�! 2Cr3þ þ 7H2O

8Hþ þ Cr2O2�7 þ 3H3AsO3! 2Cr3þ þ 3H3AsO4 þ 4H2O

3H2O, 6Hþ, and 6 e� canceled from each side

15. (a) Cl2 þ IO �3 !Cl� þ IO �

4 ðbasic solutionÞ


IO �3 ! IO �

4

Cl2! 2Cl�


H2Oþ IO �3 ! IO �

4 þ 2Hþ

Cl2! 2Cl�

Step 3 Add OH� ions to both sides (same number as Hþ ions)

2OH� þ H2Oþ IO �3 ! IO �

4 þ 2Hþ þ 2OH�

Cl2! 2Cl�

Step 4 Combine Hþ and OH� to form H2O; cancel H2O where possible

2OH� þ H2Oþ IO �3 ! IO �

4 þ 2H2O

Cl2! 2Cl�

2OH� þ IO �3 ! IO �

4 þ H2O ð1 H2O cancelledÞCl2! 2Cl�

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- Chapter 17 -


2OH� þ IO �3 ! IO �

4 þ H2Oþ 2 e�

Cl2 þ 2 e�! 2Cl�

Step 6 Electron loss and gain is balanced

Step 7 Add half-reactions

2OH� þ IO �3 þ Cl2! IO �

4 þ 2Cl� þ H2O

(b) MnO �4 þ ClO �

2 !MnO2 þ ClO �4 ðbasic solutionÞ


ClO �2 !ClO �

4

MnO �4 !MnO2


2H2Oþ ClO �2 !ClO �

4 þ 4Hþ

MnO �4 þ 4Hþ!MnO2 þ 2H2O


4OH� þ 2H2Oþ ClO �2 !ClO �

4 þ 4Hþ þ 4OH�

4OH� þMnO �4 þ 4Hþ!MnO2 þ 2H2Oþ 4OH�


4OH� þ 2H2Oþ ClO �2 !ClO �

4 þ 4H2O

4H2OþMnO �4 !MnO2 þ 2H2Oþ 4OH�

4OH� þ ClO �2 !ClO �

4 þ 2H2O ð2H2O cancelledÞ2H2OþMnO �

4 !MnO2 þ 4OH� ð2H2O cancelledÞStep 5 Balance electrically with electrons

4OH� þ ClO �2 !ClO �

4 þ 2H2Oþ 4 e�

2H2OþMnO �4 þ 3 e�!MnO2 þ 4OH�

Step 6 and 7 Equalize gain and loss of electrons; add half-reactions

3 ð4OH� þ ClO �2 !ClO �

4 þ 2H2Oþ 4 e�Þ4 ð2H2OþMnO �

4 þ 3 e�!MnO2 þ 4OH�Þ2H2Oþ 4MnO �

4 þ 3ClO �2 ! 4MnO2 þ 3ClO �

4 þ 4OH�

6 H2O, 12 OH�, and 12 e� canceled from each side

(c) Se!SeO 2�3 þ Se2� ðbasic solutionÞ


Se! SeO 2�3

Se! Se2�


3H2Oþ Se!SeO 2�3 þ 6Hþ

Se! Se2�

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- Chapter 17 -


6OH� þ 3H2Oþ Se!SeO 2�3 þ 6Hþ þ 6OH�

Se!Se2�


6OH� þ 3H2Oþ Se!SeO 2�3 þ 6H2O

Se!Se2�

6OH� þ Se!SeO 2�3 þ 3H2O ð3H2O cancelledÞ


6OH� þ Se!SeO 2�3 þ 3H2Oþ 4 e�

Seþ 2 e�!Se2�


6OH� þ Se!SeO 2�3 þ 3H2Oþ 4 e�

2ðSeþ 2 e�!Se2�Þ6OH� þ 3 Se! SeO 2�

3 þ 2 Se2� þ 3H2O

(d) Fe3O4 þMnO �4 !Fe2O3 þMnO2 ðbasic solutionÞ


2 Fe3O4! 3 Fe2O3

MnO �4 !MnO2


H2Oþ 2 Fe3O4! 3 Fe2O3 þ 2Hþ

4Hþ þMnO �4 !MnO2 þ 2H2O


2OH� þ H2Oþ 2 Fe3O4! 3 Fe2O3 þ 2Hþ þ 2OH�

4OH� þ 4Hþ þMnO �4 !MnO2 þ 2H2Oþ 4OH�


2OH� þ H2Oþ 2 Fe3O4! 3 Fe2O3 þ 2H2O

4H2OþMnO �4 !MnO2 þ 2H2Oþ 4OH�

2OH� þ 2 Fe3O4! 3 Fe2O3 þ H2O ð1 H2O cancelledÞ2H2OþMnO �

4 !MnO2 þ 4OH� ð2 H2O cancelledÞStep 5 Balance electrically with electrons

2OH� þ 2 Fe3O4! 3 Fe2O3 þ H2Oþ 2 e�

2H2OþMnO �4 þ 3 e�!MnO2 þ 4OH�


3ð2OH� þ 2 Fe3O4! 3 Fe2O3 þ H2Oþ 2 e�Þ2ð2H2OþMnO �

4 þ 3 e�!MnO2 þ 4OH�ÞH2Oþ 6 Fe3O4 þ 2MnO �

4 ! 9 Fe2O3 þ 2MnO2 þ 2OH�Þ3 H2O, 6 OH

�, and 6 e� canceled from each side

- 238 -

- Chapter 17 -

(e) BrO� þ CrðOHÞ �4 !Br� þ CrO 2�4 ðbasic solutionÞ


CrðOHÞ �4 !CrO 2�4

BrO�!Br�


CrðOHÞ �4 !CrO 2�4 þ 4Hþ

2Hþ þ BrO�!Br� þ H2O


4OH� þ CrðOHÞ �4 !CrO 2�4 þ 4Hþ þ 4OH�

2OH� þ 2Hþ þ BrO�!Br� þ H2Oþ 2OH�


4OH� þ CrðOHÞ �4 !CrO 2�4 þ 4H2O

2H2Oþ BrO�!Br� þ H2Oþ 2OH�

H2Oþ BrO�!Br� þ 2OH� ð1 H2O cancelledÞStep 5 Balance electrically with electrons

4OH� þ CrðOHÞ �4 !CrO 2�4 þ 4H2Oþ 3 e�

H2Oþ BrO� þ 2 e�!Br� þ 2OH�


2 ð4OH� þ CrðOHÞ �4 !CrO 2�4 þ 4H2Oþ 3 e�Þ

3 ðH2Oþ BrO� þ 2 e�!Br� þ 2OH�Þ2OH� þ 3BrO� þ 2CrðOHÞ �4 ! 3Br� þ 2CrO 2�

4 þ 5H2O

3 H2O, 6 OH� and 6 e� canceled from each side

16. (a) MnO �4 þ SO 2�

3 !MnO2 þ SO 2�4 ðbasic solutionÞ


SO 2�3 ! SO 2�

4

MnO �4 !MnO2


H2Oþ SO 2�3 !SO 2�

4 þ 2Hþ

MnO �4 þ 4Hþ!MnO2 þ 2H2O


2OH� þ H2Oþ SO 2�3 !SO 2�

4 þ 2Hþ þ 2OH�

4OH� þMnO �4 þ 4Hþ!MnO2 þ 2H2Oþ 4OH�

- 239 -

- Chapter 17 -


2OH� þ H2Oþ SO 2�3 ! SO 2�

4 þ 2H2O

MnO �4 þ 4H2O!MnO2 þ 2H2Oþ 4OH�

2OH� þ SO 2�3 !SO 2�

4 þ H2O ð1 H2O cancelledÞMnO �

4 þ 2H2O!MnO2 þ 4OH� ð2 H2O cancelledÞStep 5 Balance electrically with electrons

2OH� þ SO 2�3 !SO 2�

4 þ H2Oþ 2 e�

3 e� þMnO �4 þ 2H2O!MnO2 þ 4OH�


3 ð2OH� þ SO 2�3 !SO 2�

4 þ H2Oþ 2 e�Þ2 ðMnO �

4 þ 2H2Oþ 3 e�!MnO2 þ 4OH�ÞH2Oþ 2MnO �

4 þ 3 SO 2�3 ! 2MnO2 þ 3 SO 2�

4 þ 2OH�

3H2O, 4OH�, and 6 e� canceled from each side

(b) ClO2 þ SbO �2 !ClO �

2 þ Sb OHð Þ �6 þ 2 H2O ðbasic solutionÞ


SbO �2 !SbðOHÞ �6

ClO2!ClO �2


4H2Oþ SbO �2 !SbðOHÞ �6 þ 2Hþ

ClO2!ClO �2


2OH� þ 4H2Oþ SbO �2 !SbðOHÞ �6 þ 2Hþ þ 2OH�

ClO2!ClO �2


2 OH� þ 4 H2Oþ SbO �2 !Sb OHð Þ �6 þ 2 H2O

ClO2!ClO �2

2 OH� þ 2 H2Oþ SbO �2 !Sb OHð Þ �6 2 H2O cancelledð Þ


2 OH� þ 2 H2Oþ SbO �2 !Sb OHð Þ �6 þ 2 e�

ClO2 þ e�!ClO �2


2 H2Oþ 2 OH� þ SbO �2 !Sb OHð Þ �6 þ 2 e�

2 ClO2 þ e�!ClO �2

� �2 H2Oþ 2 ClO2 þ 2 OH� þ SbO �

2 ! 2 ClO �2 þ Sb OHð Þ �6

- 240 -

- Chapter 17 -

(c) Alþ NO �3 !NH3 þ Al OHð Þ �4 basic solutionð Þ


Al!Al OHð Þ �4NO �

3 !NH3


4 H2Oþ Al!Al OHð Þ �4 þ 4 Hþ

9 Hþ þ NO �3 !NH3 þ 3 H2O


4 OH� þ 4 H2Oþ Al!Al OHð Þ �4 þ 4 Hþ þ 4 OH�

9 OH� þ 9 Hþ þ NO �3 !NH3 þ 3 H2Oþ 9 OH�


4 OH� þ 4 H2Oþ Al!Al OHð Þ �4 þ 4 H2O

9 H2Oþ NO �3 !NH3 þ 3 H2Oþ 9 OH�

4 OH� þ Al!Al OHð Þ �4 4 H2O cancelledð Þ6 H2Oþ NO �

3 !NH3 þ 9 OH� 3 H2O cancelledð ÞStep 5 Balance electrically with electrons

4 OH� þ Al!Al OHð Þ �4 þ 3 e�

6 H2Oþ NO �3 þ 8 e�!NH3 þ 9 OH�


8 4 OH� þ Al!Al OHð Þ �4 þ 3 e��

3 6 H2Oþ NO �3 þ 8 e�!NH3 þ 9 OH�

� �8 Alþ 3 NO �

3 þ 18 H2Oþ 5 OH�! 3 NH3 þ 8 Al OHð Þ �427 OH� and 24 e� canceled from each side

(d) P4!HPO 2�3 þ PH3 basic solutionð Þ


P4! 4 HPO 2�3

P4! 4 PH3


12 H2Oþ P4! 4 HPO 2�3 þ 20 Hþ

12 Hþ þ P4! 4 PH3


20 OH� þ 12 H2Oþ P4! 4 HPO 2�3 þ 20 Hþ þ 20 OH�

12 OH� þ 12 Hþ þ P4! 4 PH3 þ 12 OH�

- 241 -

- Chapter 17 -


20 OH� þ 12 H2Oþ P4! 4 HPO 2�3 þ 20 H2O

12 H2Oþ P4! 4 PH3 þ 12 OH�

20 OH� þ P4! 4 HPO 2�3 þ 8 H2O 12 H2O cancelledð Þ


20 OH� þ P4! 4 HPO 2�3 þ 8 H2Oþ 12 e�

12 H2Oþ P4 þ 12 e�! 4 PH3 þ 12 OH�

Step 6 and 7 Loss and gain of electrons are equal; add half-reactions

8 OH� þ 4 H2Oþ 2 P4! 4 HPO 2�3 þ 4 PH3

Divide equation by 2

4 OH� þ 2 H2Oþ P4! 2 HPO 2�3 þ 2 PH3

(e) Alþ OH�!Al OHð Þ �4 þ H2 basic solutionð Þ


Al!Al OHð Þ �4OH�!H2


4 H2Oþ Al!Al OHð Þ �4 þ 4 Hþ

4 Hþ þ OH�!H2 þ H2O


4 OH� þ 4 H2Oþ Al!Al OHð Þ �4 þ 4 Hþ þ 4 OH�

3 OH� þ 3 Hþ þ OH�!H2 þ H2Oþ 3 OH�


4 OH� þ 4 H2Oþ Al!Al OHð Þ �4 þ 4 H2O

3 H2Oþ OH�!H2 þ H2Oþ 3 OH�

4 OH� þ Al!Al OHð Þ �4 4 H2O cancelledð Þ2 H2Oþ OH�!H2 þ 3 OH� 1 H2O cancelledð Þ


4 OH� þ Al!Al OHð Þ �4 þ 3 e�

2 H2Oþ OH� þ 2 e�!H2 þ 3 OH�

Step 6 and 7 Equalize gain and loss of electrons; and half-reactions

2 4 OH� þ Al!Al OHð Þ �4 þ 3 e��

3 2 H2Oþ OH� þ 2 e�!H2 þ 3 OH�ð Þ2 Alþ 6 H2Oþ 2 OH�! 2 Al OHð Þ �4 þ 3 H2

9 OH� and 6 e� canceled on each side

- 242 -

- Chapter 17 -

17. (a) IO �3 þ I�! I2 acidic solutionð Þ


2 IO �3 ! I2

2 I�! I2


12 Hþ þ 2 IO �3 ! I2 þ 6 H2O

2 I�! I2


12 Hþ þ 2 IO �3 þ 10 e�! I2 þ 6 H2O

2 I�! I2 þ 2 e�

Step 4 and 5 Equalize the loss and gain of electrons; add the half-reaction.

12 Hþ þ 2 IO �3 þ 10 e�! I2 þ 6 H2O

5 ð2 I�! I2 þ 2 e�Þ12 Hþ þ 2 IO �

3 þ 10 I�! 6 I2 þ 6 H2O�

(b) Mn2þ þ S2O2�8 !MnO �

4 þ SO 2�4 (acid solution)

Step 1 Write half-reaction equations. Balance except H and O

Mn2þ!MnO �4

S2O2�8 ! 2 SO 2�

4


4 H2OþMn2þ!MnO �4 þ 8 Hþ

S2O2�8 ! 2 SO 2�

4


4 H2OþMn2þ!MnO �4 þ 8 Hþ þ 5 e�

2 e� þ S2O2�8 ! 2 SO 2�

4


2 4 H2OþMn2þ!MnO �4 þ 8 Hþ þ 5 e�

� �5 2 e�! S2O

2�8 ! 2 SO 2�

4

� �2Mn2þ þ 5 S2O

2�8 þ 8 H2O! 2MnO �

4 þ 10 SO 2�4 þ 16 Hþ�

Each side has 2 Mn, 10 S, 16 H, and 48 O and a �6 charge.(c) CoðNO2Þ 3�

6 þMnO�4!Co2þ þMn2þ þ NO �3 (acidic solution)


CoðNO2Þ 3�6 !Co2þ þ 6 NO �

3

MnO �4 !Mn2þ

- 243 -

- Chapter 17 -


6 H2Oþ CoðNO2Þ 3�6 !Co2þ þ 6 NO �

3 þ 12 Hþ

8 Hþ þMnO �4 !Mn2þ þ 4 H2O

Step 3 Balance electrically with e�

6 H2Oþ CoðNO2Þ 3�6 !Co2þ þ 6 NO �

3 þ 12 Hþ þ 11 e�

5 e� þ 8 Hþ þMnO �4 !Mn2þ þ 4 H2O

Step 4 Equalize the loss and gain of electrons.

5 ð6 H2Oþ CoðNO2Þ 3�6 !Co2þ þ 6 NO �

3 þ 12 Hþ þ 11 e�Þ11 ð5 e� þ 8 Hþ þMnO �

4 !Mn2þ þ 4 H2OÞStep 5 Add the half-reactions

5 CoðNO2Þ 3�6 þ 11MnO �

4 þ 28 Hþ! 5 Co2þþ 30 NO �3 þ 11Mn2þþ 14 H2O

Each side has 5 Co, 30 N, 11 Mn, 28 H, 104 O and aþ 2 charge.

18. (a) Mo2O3 þMnO �4 !MoO3 þMn2þ (acid solution)

Step 1 Write half-reactions equations. Balance except H and O

Mo2O3! 2MoO3

MnO �4 !Mn2þ


3 H2OþMo2O3! 2MoO3 þ 6 Hþ

8 Hþ þMnO �4 !Mn2þ þ 4 H2O


3 H2OþMo2O3! 2MoO3 þ 6 Hþ þ 6 e�

5 e� þ 8 Hþ þMnO �4 !Mn2þ þ 4 H2O

Steps 4 and 5 Equalize the loss and gain of electrons; add the half-reactions.

5 ð3 H2OþMo2O3! 2MoO3 þ 6 Hþ þ 6 e�Þ6 ð5 e� þ 8 Hþ þMnO �

4 !Mn2þ þ 4 H2OÞ5Mo2O3 þ 6MnO �

4 þ 18 Hþ! 10MoO3 þ 6Mn2þ þ 9 H2O

(b) BrO� þ CrðOHÞ �4 !Br� þ CrO 2�4 (basic solution)

Step 1 Write half-reaction equation. Balance except H and O

BrO�!Br�

CrðOHÞ �4 !CrO 2�4


2 Hþ þ BrO�!Br� þ H2O

CrðOHÞ�4!CrO 2�4 þ 4 Hþ

- 244 -

- Chapter 17 -

Step 3 Add OH� ions to both sides (same number as Hþ)

2 OH� þ 2 Hþ þ BrO�!Br� þ H2Oþ 2 OH�

4 OH� þ CrðOHÞ �4 !CrO 2�4 þ 4 Hþ þ 4 OH�


2 H2Oþ BrO�!Br� þ H2Oþ 2 OH�

4 OH� þ CrðOHÞ �4 !CrO 2�4 þ 4 H2O

H2Oþ BrO�!Br� þ 2 OH� ð1 H2O cancelledÞStep 5 Balance electrically with electrons

2 e� þ H2Oþ BrO�!Br� þ 2 OH�

4 OH� þ CrðOHÞ �4 !CrO 2�4 þ 4 H2Oþ 3 e�

Steps 6 and 7 Equalize loss and gain of electrons; add the half-reactions

3 ð2 e� þ H2Oþ BrO�!Br� þ 2 OH�Þ2 ð4 OH� þ CrðOHÞ �4 !CrO 2�

4 þ 4 H2Oþ 3 e�Þ3 BrO� þ 2 CrðOHÞ �4 þ 2 OH�! 3 Br� þ 2 CrO 2�

4 þ 5 H2O�

(c) S2O2�3 þMnO �

4 ! SO 2�4 þMnO2 (basic solution)


S2O2�3 ! 2 SO 2�

4

MnO �4 !MnO2


5 H2Oþ S2O2�3 ! 2 SO 2�

4 þ 10 Hþ

4 Hþ þMnO �4 !MnO2 þ 2 H2O

Step 3 Add OH� ions to both sides (same number as Hþ)

10 OH� þ 5 H2Oþ S2O2�3 ! 2 SO 2�

4 þ 10 Hþ þ 10 OH�

4 OH� þ 4 Hþ þMnO �4 !MnO2 þ 2 H2Oþ 4 OH�


10 OH� þ 5 H2Oþ S2O2�3 ! 2 SO 2�

4 þ 10 H2O

4 H2OþMnO �4 !MnO2 þ 2 H2Oþ 4 OH�

10 OH� þ S2O2�3 ! 2 SO 2�

4 þ 5 H2O ð5 H2O cancelledÞ2 H2OþMnO �

4 !MnO2 þ 4 OH� ð2 H2O cancelledÞStep 5 Balance electrically with electrons

10 OH� þ S2O2�3 ! 2 SO 2�

4 þ 5 H2Oþ 8 e�

3 e� þ 2 H2OþMnO �4 !MnO2 þ 4 OH�

Step 6 and 7 Equalize loss and gain of electrons; add half-reactions.

3 ð10 OH� þ S2O2�3 ! 2 SO 2�

4 þ 5 H2Oþ 8 e�Þ8 ð3 e� þ 2 H2OþMnO �

4 !MnO2 þ 4 OH�Þ�3 S2O

2�3 þ 8MnO �

4 þ H2O! 6 SO 2�4 þ 8MnO2 þ 2 OH�

Each side has 6 S, 8 Mn, 2 H, 42 O and a �14 charge.

- 245 -

- Chapter 17 -

19.

Voltagesource

Solution of HBr

Cathode (–) Anode (+)

Br–

H3O+

+–

20. (a) Pbþ SO 2�4 ! PbSO4 þ 2 e�

PbO2 þ SO 2�4 þ 4 Hþ þ 2 e�! PbSO4 þ 2 H2O

(b) The first reaction is oxidation (Pb0 is oxidized to Pb2þ).The second reaction is reduction (Pb4þ is reduced to Pb2þ).

(c) The first reaction (oxidation) occurs at the anode of the battery.

21. (a) The oxidizing agent is KMnO4.

(b) The reducing agent is HCl.

(c) 5 moles of electrons 5 e�þMn7þ ! Mn2þ

5 mol e�

mol KMnO4

� �6:022� 1023e�

mol e�

� �¼ 3:011� 1024

electrons

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 Cuþ 8 HNO3 ! 3 Cu(NO3)2þ 2 NOþ 4 H2O

ð75:5 g CuÞ 1 mol Cu

63:55 g

� �2 mol NO

3mol Cu

� �22:4 L

mol

� �¼ 17:7 L NO

ð55:0 g HNO3Þ 1 mol HNO3

63:02 g

� �2 mol NO

8mol HNO3

� �22:4 L

mol

� �¼ 4:89 L NO

4.89 L of NO gas at STP will be produced; HNO3 is limiting.

Cu is oxidized; N is reduced.

24. K2S2O8þH2C2O4 ! K2SO4þH2SO4þ 2 CO2

ð25:5 g K2S2O8Þ 1 mol K2S2O8

270:3 g

� �2 mol CO2

1 mol HNO3

� �22:4 L

mol

� �¼ 4:23 L CO2

ð35:5 g H2C2O4Þ 1 mol H2C2O4

90:04 g

� �2 mol CO2

1 mol H2C2O4

� �22:4 L

mol

� �¼ 17:7 L CO2

4.23 L CO2 will be produced; K2S2O8 is limiting.

C is oxidized; S is reduced.

- 246 -

- Chapter 17 -

25. 5 H2O2þ 2 KMnO4þ 3 H2SO4 ! 5 O2þ 2 MnSO4þK2SO4þ 8H2O

mL H2O2 ! g H2O2 ! mol H2O2 ! mol KMnO4 ! g KMnO4

ð100:mLH2O2 solutionÞ 1:031 g

mL

� �9:0 g H2O2

100: g H2O2 solution

� �1 mol

34:02 g

� �2 mol KMnO4

5 mol H2O2

� �158:0 g

mol

� �¼ 17 g KMnO4

26. 3 Znþ 2 Fe3þ ! 3 Zn2þþ 2 Fe

ð0:0250 L FeCl3Þ 1:2 mol FeCl3

L

� �3 mol Zn

2 mol FeCl3

� �65:39 g

mol

� �¼ 2:94 g Zn

27. Cr2O2�7 þ 6 Fe2þ þ 14 Hþ! 2 Cr3þ þ 6 Fe3þ þ 7 H2O

mL FeSO4!mol FeSO4!mol Cr2O2�7 !mLCr2O

2�7

ð60:0 mL FeSO4Þ 0:200 mol

1000 mL

� �1 mol Cr2O

2�7

6 mol FeSO4

� �1000 mol

0:200 mol

� �¼ 10:0 mL of 0:200MK2Cr2O7

28. 2 Alþ 2 OH� þ 6 H2O! 2 AlðOHÞ �4 þ 3 H2

g Al!mol Al!mol H2

ð100:0 g AlÞ 1 mol Al

26:98 g

� �3 mol H2

2 mol Al

� �¼ 5:560 mol H2

29. (a) Cuþ ! Cu2þ is an oxidation, but when electrons are gained reduction should occur.

Cuþ þ e�!Cu0 or Cuþ!Cu2þ þ e�

(b) When Pb2þ is reduced, it requires two individual electrons. Pb2þþ 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

D(s) þ 2 Cþ(aq)! 2C(s)þD2þ(aq) Cþ takes e� from D

B(s) þ D2þ(aq)! D(s)þB2þ(aq) D2þ takes 2 e� from B

Therefore, B2þ is least able to attract e�, then D2þ, then Cþ, then Aþ

32.Sn4þ can only be an oxidizing agent.

Sn4þþ 2 e� ! Sn2þ

Sn4þþ 4 e� ! Sn0

Sn0 can only be a reducing agent.Sn0! Sn2þþ 2 e�

Sn0! Sn4þþ 4 e�

Sn2þ can be both oxidizing and reducing.Sn2þþ 2 e� ! Sn0

Sn2þ ! Sn4þþ 2 e�(oxidizing)

(reducing)

- 247 -

- Chapter 17 -

33. Mn(OH)2 þ2 KMnO4 is the best oxidizing agent of the group, since its greater

MnF3 þ3 oxidation number (þ7) makes it very attractive to electrons.

MnO2 þ4K2MnO4 þ6KMnO4 þ7

34. Equations (a) and (b) represent oxidation

(a) Mg! Mg2þþ 2 e�

(b) SO2! SO3; (S4þ ! S6þþ 2 e�)

35. (a) MnO2þ 2 Br� þ 4 Hþ ! Mn2þþBr2þ 2 H2O

(b) mL Mn2þ ! mol Mn2þ ! mol MnO2! g MnO2

ð100:0 mLMn2þÞ 0:05 mol

1000 mL

� �1 mol MnO2

1 mol Mn2þ

� �86:94 g

mol

� �¼ 0:4 gMnO2

(c) ð100:0 mLMn2þÞ 0:05 mol

1000 mL

� �1 mol Br2

1 mol Mn2þ

� �¼ 0:005 mol Br2

PV ¼ nRT V ¼ nRT

P

V ¼ 0:005 mol

1:4 atm

� �0:0821 L atm

mol K

� �ð323 KÞ ¼ 0:09 L Br2 vapor

36. (a) F2þ 2 Cl� ! 2 F�þCl2(b) Br2þCl� ! NR

(c) I2þCl� ! NR

(d) Br2þ 2 I� ! 2 Br�þ I2

37. Mn(s)þ 2 HCl(aq)! Mn2þ(aq)þH2(g)þ 2 Cl�(aq)

38. 4 Znþ NO �3 þ 10 Hþ! 4 Zn2þ þ NH þ

4 þ 3 H2O See Exercise 13(a).

39.

Equation 1 2 3 4 5

a C oxidized S oxidized N oxidized S oxidized O 2�2 oxidized

b O2 reduced N reduced Cu reduced O 2�2 reduced O 2�

2 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 C22

3þ!C4þ S2! S0 N3�!N 0

2 S4þ ! S6þ O 2�2 !O 0

2

f O0! O2� N5þ ! N2þ Cu2þ ! Cu0 O 2�2 !O2� O 2�

2 !O2�

O.A.¼Oxidizing agentR.A.¼Reducing agent

- 248 -

- Chapter 17 -

40. Pbþ 2 Agþ ! 2 Agþ 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.

PbAg

Pb2+

NO–3Ag+

NO–3

Saltbridge

41. 8KIþ 5H2SO4! 4 I2 þ H2Sþ 4K2SO4 þ 4H2Ostart with grams I2 and work towards g KI

g I2!mol I2!mol KI! g KI

ð2:79 g I2Þ 1 mol

253:8 g

� �8 mol KI

4 mol I2

� �166:0 g

mol

� �¼ 3:65 g KI in sample

3:65 g KI

4:00 g sample

� �ð100Þ ¼ 91:3%KI

42. 3Agþ 4HNO3! 3AgNO3 þ NOþ 2H2O

mol Ag!mol NO

ð0:500 mol AgÞ 1 mol NO

3mol Ag

� �¼ 0:167 mol NO

PV ¼ nRT V ¼ nRT

P

P ¼ ð744 torrÞ 1 atm

760: torr

� �¼ 0:979 atm

T ¼ 301 K

V ¼ ð0:167 mol NOÞð0:0821 L atm=mol KÞð301 KÞð0:979 atmÞ ¼ 4:22 L NO

- 249 -

- Chapter 17 -

oxidation-reductionfaculty.chemeketa.edu/lemme/solutions/ch17.pdf · chapter 17 oxidation-reduction...

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