Hein9thCh17- elektroliza- oksidacija

7/24/2019 Hein9thCh17- elektroliza- oksidacija

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CHAPTER 17OXIDATION-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 its

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

from its compounds.Cr 3+

2 Al + Fe 2O3 Al 2O3 + 2 Fe + Heat

+ 2

2 Al( s) + 3 CuSO 4(aq ) Al 2(SO 4)3(aq ) + 3 Cu( s)

Ni( s) + Hg(NO 3)2(aq ) Ni(NO 3)2(aq ) + Hg( l)

Ba( s) + FeCl 2(aq ) BaCl 2(aq ) + Fe( s)

Sn( s) + 2 Ag + (aq ) Sn2+ (aq ) + 2 Ag( s)

Zn( s) + Cu 2+ (aq ) Zn2+ (aq ) + Cu( s)

- 1.+ 5.

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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, reductionanode 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.

Hg 2+

H 2SO 4H 2O.PbSO 4(s)H

+PbO 2SO 4 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)


(a) (c)(b) (d)


(a) 0 (c)(b) (d)

Changing Type of 5. Balanced half-reaction Element reaction

(a) Zn reduction(b) Br oxidation(c) Mn reduction(d) Ni oxidation

Changing Type of 6. Balanced half-reactions Element reaction

(a) S oxidation(b) N reduction(c) S oxidation(d) Fe oxidationFe 2+ Fe 3+ + 1 e -

S2O4 2- + 2 H 2O 2 SO 3

2- + 4 H + + 2 e -NO 3

- + 4 H + + 3 e - NO + 2 H 2OSO 3

2- + H 2O SO 4 2- + 2 H + + 2 e -

Ni Ni2+ + 2 e -MnO 4

- + 8 H + + 5 e - Mn 2+ + 4 H 2O2 Br - Br2 + 2 e

-

Zn2+ + 2 e - Zn

+ 5IO3 -+ 5AsO 4

3-- 2Fe(OH) 3O2

+ 3Bi3++ 3NO 2-

- 1Na2O

2- 2S2-

+ 6K 2Cr2O7+ 5KClO 3I2

+ 6K 2CrO 4- 3NH 3+ 7KMnO 4

- 3NH 4Cl+ 5NaNO 3- 1FeCl 3

+ 4H 2SO 3+ 4PbO 2+ 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-reactionsthe 2 e - cancel

Zn + S ZnSZn0 Zn2+ + 2 e -

S0 + 2 e - S2-

Zn + S ZnS

oxred

8 H 2O(l) + 2 MnO 4 -

(aq ) + 7 S2- (aq ) 2 MnS( s) + 16 OH - (aq ) + 5 S(s)

3 MnO 2(s) + 4 Al( s) 3 Mn( s) + 2 Al 2O3(s)

3 CH 3OH( aq ) + Cr2O7 2-

(aq ) + 8 H + (aq ) 2 Cr 3+ (aq ) + 3 CH 2O(aq ) + 7 H 2O(l)

Mg( s) + 2 HCl( aq ) Mg 2+ (aq ) + 2Cl - (aq ) + H 2(g )

Cl 2

Cl 2 + NaBr NaCl + Br2

AsH 3Ag+

AsH 3 + Ag+ + H 2O H 3AsO 4 + Ag + H

+

I -SO 4 2-

SO 4 2- + I - + H + H 2S + I2 + H 2O

Cr + HCl CrCl 3 + H 2

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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 of the 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

MnO 2 + 4 HBr MnBr 2 + Br2 + 2 H 2O

MnBr 2 .2 Br -

Br -Br2 .Br-

Multiply by 2Add equations and the 2 e - cancel

MnO 2 + HBr MnBr 2 + Br2 + H 2OBr - Br 0 + 1 e -

Mn 4+ + 2 e - Mn 2+

Mn 4+ + 2 Br - Mn 2+ + 2 Br 0

3 H 2S + 2 HNO 3 3 S + 2 NO + 4 H 2O

Multiply by 3Multiply by 2, add, the 6 e -

H 2S + HNO 3 S + NO + H 2OS2- S0 + 2 e -

N 5+ + 3 e - N 2+

3 S2- + 2 N 5+ 3 S + 2 N 2+

Fe 2O3 + 3 CO 2 Fe + 3 CO 2

Fe2O

3Fe 3+

Multiply by 3Multiply by 2, add, the 6 e - cancel

Fe 2O3 + CO Fe + CO 2C 2+ C4+ + 2 e -

Fe 3+ + 3 e - Fe 0

3 C2+ + 2 Fe 3+ 3 C 4+ + 2 Fe

oxred

2 AgNO 3 + Pb Pb(NO 3)2 + 2 Ag

Multiply by 2, add the half-reactionsthe 2 e - cancel

AgNO 3 + Pb Pb(NO 3)2 + AgPb 0 Pb 2+ + 2 e -

Ag + + 1 e - Ag 0

Pb + 2 Ag + Pb 2+ + 2 Ag

oxred

- Chapter 17 -

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12. (a) Balancing oxidation-reduction equations


(b)


(c)


(d)


(e) H 2O2 + KMnO 4 + H 2SO 4 O2 + MnSO 4 + K 2SO 4 + H 2OO2

2- O20 + 2 e - Multiply by 5

Mn 7+ + 5 e - Mn 2+ Multiply by 2, add, the 10 e - cancel5 O2

2- + 2 Mn 7+ 5 O2 + 2 Mn2+

3 PbO 2 + 2 Sb + 2 NaOH 3 PbO + 2 NaSbO 2 + H 2O

PbO 2 + Sb + NaOH PbO + NaSbO 2 + H2OSb0 Sb3+ + 3 e- Multiply by 2Pb4+ + 2 e- Pb2+ Multiply by 3, add, the 6 e - cancel2 Sb + 3 Pb4+ 2 Sb3+ + 3 Pb2 +

3 CuO + 2 NH 3 N2 + 3 Cu + 3 H 2O

Multiply by 2Multiply by 3, add, the 6 e - cancel

CuO + NH 3 N2 + Cu + H 2ON 3- N 0 + 3 e -

Cu2+ + 2 e - Cu 0

2 N 3- + 3 Cu2+ N2 + 3 Cu

3 Ag + 4 HNO 3 3 AgNO 3 + NO + 2 H 2O

Multiply by 3, add,the 3 e - cancel

Ag + HNO 3 AgNO 3 + NO + H 2OAg 0 Ag + + e -

N 5+ + 3 e - N 2+

3 Ag + N 5+ 3 Ag + + N 2+

3 Cl 2 + 6 KOH KClO 3 + 5 KCl + 3 H 2O

Multiply by 5, add, the 5 e - cancel6 Cl0 becomes 3 Cl 2

Cl 2 + KOH KCl + KClO 3 + H 2OCl0 Cl5+ + 5 e -

Cl0 + e - Cl - 3 Cl 2 Cl

5+ + 5 Cl -

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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-reactionselectrons cancel

(b) (acidic solution)




2 H + + NO 3- + e - NO 2 + H 2O

4 H 2O + S SO 4 2- + 8 H + + 6 e -

2 H + + NO 3- NO 2 + H 2O

4 H 2O + S SO 4 2- + 8 H +

H +H 2O

NO 3 - NO 2

S SO42-

NO 3- + S NO 2 + SO 4

2-

10 H + + 4 Zn + NO 3 - 4 Zn2+ + NH 4

+ + 3 H 2O

10 H + + NO 3- + 8 e - NH 4

+ + 3 H 2O

4 (Zn Zn2+ + 2 e - )

10 H + + NO 3 - + 8 e - NH 4 + + 3 H 2O

Zn Zn2+ + 2 e -

10 H + + NO 3- NH 4

+ + 3 H 2O

Zn Zn2+H +H 2O

NO 3 - NH 4

+

Zn Zn2+

Zn + NO 3- Zn2+ + NH 4

+

5 H 2O2 + 2 KMnO 4 + 3 H 2SO 4 5 O2 + 2 MnSO 4 + K 2SO 4 + 8 H 2O

- Chapter 17 -

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

(c) (acidic solution)





(d) (acidic solution)




4H + + NO 3- + 3 e - NO + 2 H 2O

Cu Cu 2+ + 2 e -

4H + + NO 3- NO + 2 H 2O

Cu Cu2+

H +H 2O

NO 3- NO

Cu Cu 2+

Cu + NO 3 - Cu 2+ + NO

2 H2O + PH 3 H3PO 2 + 4 H+ + 4 e -

2 (I2 + 2 e- 2 I- )

PH3

+ 2 H2O + 2 I

2 H

3PO

2 + 4 I- + 4 H +

I2 + 2 e- 2 I -

2 H 2O + PH 3 H 3PO 2 + 4 H+ + 4 e -

I2 2 I-

2 H 2O + PH 3 H 3PO 2 + 4 H+

H +H 2O

I2 2I-

PH 3 H 3PO 2

PH 3 + I2 H 3PO 2 + I-

and 6 e - canceled from each side4 H 2O, 8 H+

4 H 2O + S SO 4 2- + 8 H + + 6 e -

6 (2 H + + NO 3 - + e - NO 2 + H 2O)

4 H + + S + 6 NO 3 - 6 NO 2 + SO 4

2- + 2 H 2O

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(e) (acidic solution)









6 H + + ClO 3 - + 6 e - Cl - + 3 H 2O

2 I - I2 + 2 e-

6 H + + ClO 3 - Cl - + 3 H 2O

2 I-

I2

H +H 2O

ClO 3 - Cl -

2 I - I2

ClO 3 - + I - I2 + Cl

-

5 (Cl - Cl0 + e - )6 H + + ClO 3

- + 5 e - Cl0 + 3 H 2O6 H + + ClO 3

- + 5 Cl - 3 Cl 2 + 3 H 2O

6 H + + ClO 3 - + 5 e - Cl0 + 3 H 2O

Cl - Cl0 + e -

6 H + + ClO 3 - Cl0 + 3 H 2O

Cl - Cl0H +H 2O

ClO 3 - Cl0

Cl - Cl0

ClO 3 - + Cl - Cl 2

3 (Cu Cu 2+ + 2 e - )2 (4 H + + NO 3

- + 3 e - NO + 2 H 2O)3 Cu + 8 H + + 2 NO 3

- 3 Cu2+ + 2 NO + 4 H 2O

- Chapter 17 -

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(b) (acidic solution)









8 H + + MnO 4 - + 5 e - Mn 2+ + 4 H 2O

2 H 2O + SO2 SO 4 2- + 4 H + + 2 e -

8 H + + MnO 4 - Mn 2+ + 4 H 2O

2 H 2O + SO2 SO 4 2- + 4 H +

H +H 2O

MnO 4 - Mn 2+

SO 2 SO 4 2-

MnO 4 - + SO 2 Mn

2+ + SO42-

6 (Fe 2+ Fe 3+ + e - )14 H + + Cr2O7

2- + 6 e - 2 Cr 3+ + 7 H 2O14 H + + Cr2O7

2- + 6 Fe 2+ 2 Cr 3+ + 6 Fe 3+ + 7 H 2O

14 H + + Cr2O7 2- + 6 e - 2 Cr 3+ + 7 H 2O

Fe 2+ Fe 3+ + e -

14 H + + Cr2O7 2- 2 Cr 3+ + 7 H 2O

Fe 2+ Fe 3+H +H 2O

Cr2O7 2- 2 Cr 3+

Fe 2+ Fe 3+

Cr2O7 2- + Fe 2+ Cr 3+ + Fe 3+

3 (2 I - I2 + 2 e- )

6 H + + ClO 3 - + 6 e - Cl - + 3 H 2O

6 H + + ClO 3 - + 6 I - 3 I2 + Cl

- + 3 H 2O

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(d) (acidic solution)





(e) (acidic solution)




14 H + + Cr2O7 2- + 6 e - 2 Cr 3+ + 7 H 2O

H 2O + H 3AsO 3 2 H+ + H 3AsO 4 + 2 e

-

14 H + + Cr2O7 2- 2 Cr 3+ + 7 H 2O

H 2O + H 3AsO 3 2 H + + H 3AsO 4

H +H 2O

Cr2O72- 2 Cr 3+

H 3AsO 3 H 3AsO 4

Cr2O7 2- + H 3AsO 3 Cr

3+ + H 3AsO 4

5 (H 2O + H 3AsO 3 2 H+ + H 3AsO 4 + 2 e

- )2 (8 H + + MnO 4

- + 5 e - Mn 2+ + 4 H 2O)6 H + + 5 H 3AsO 3 + 2 MnO 4

- 5 H 3AsO 4 + 2 Mn2+ + 3 H 2O

5 H 2O, 10 H + , and 10 e - canceled from each side

8 H + + MnO 4 - + 5 e - Mn 2+ + 4 H 2O

H 2O + H 3AsO 3 2 H+ + H 3AsO 4 + 2 e

-

8 H + + MnO 4 - Mn 2+ + 4 H 2O

H 2O + H 3AsO 3 2 H+ + H 3AsO 4

H +H 2O

MnO 4 - Mn 2+

H 3AsO 3 H 3AsO 4

H 3AsO 3 + MnO 4- H 3AsO 4 + Mn

2+

5 (2 H 2O + SO 2 SO 4 2- + 4 H + + 2 e - )

2 (8 H + + MnO 4 - + 5 e - Mn 2+ + 4 H 2O)

2 H 2O + 2 MnO 4- + 5 SO 2 4 H

+ + 2 Mn 2+ + 5 SO4 2-

8 H 2O, 16 H+ , and 10 e - canceled from each side

- Chapter 17 -

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15. (a) (basic solution)



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

Step 4 Combine and to form cancel where possible


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.

MnO 4 - MnO 2

ClO 2 - ClO 4

-

MnO 4 - + ClO 2

- MnO 2 + ClO 4 -

2 OH - + IO3 - + Cl 2 IO4

- + 2 Cl - + H 2O

Cl 2 + 2 e- 2 Cl -

2 OH - + IO3 - IO4

- + H 2O + 2 e-

Cl 2 2 Cl-

(1 H2O cancelled)2 OH - + IO

3

- IO4

- + H2O

Cl 2 2 Cl-

2 OH - + H 2O + IO3 - IO4

- + 2 H 2O

H 2OH 2O;OH-H +

Cl 2 2 Cl-

2 OH - + H 2O + IO3 - IO4

- + 2 H + + 2 OH -H

+

OH-

Cl 2 2 Cl-

H 2O + IO3- IO4

- + 2 H +H +H 2O

Cl 2 2 Cl-

IO3 - IO4

-

Cl 2 + IO3- Cl - + IO4

-

3 (H 2O + H 3AsO 3 2 H+ + H 3AsO 4 + 2 e

- )14 H + + Cr2O7

2- + 6 e - 2 Cr 3+ + 7 H 2O8 H + + Cr2O7

2- + 3 H 3AsO 3 2 Cr3+ + 3 H 3AsO 4 + 4 H 2O

3 H 2O, 6 H+ , and 6 e - canceled from each side

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- Chapter 17 -

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

(c) (basic solution)




Se Se 2-6 OH - + 3 H 2O + Se SeO 3

2- + 6 H + + 6 OH -H +OH -

Se Se 2-3 H 2O + Se SeO 3 2- + 6 H +

H +H 2O

Se Se 2-Se SeO 3

2-

Se SeO 3 2- + Se 2-

3 (4 OH - + ClO 2 - ClO 4

- + 2 H 2O + 4 e- )

4 (2 H 2O + MnO 4 - + 3 e - MnO 2 + 4 OH

- )2 H 2O + 4 MnO 4

- + 3 ClO 2- 4 MnO 2 + 3 ClO 4

- + 4 OH -

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

2 H 2O + MnO 4 - + 3 e - MnO 2 + 4 OH

-

4 OH - + ClO 2 - ClO 4 - + 2 H 2O + 4 e -

(2 H 2O cancelled)2 H 2O + MnO 4 - MnO 2 + 4 OH

-

(2 H 2O cancelled)4 OH- + ClO 2

- ClO 4- + 2 H 2O

4 H 2O + MnO 4 - MnO 2 + 2 H 2O + 4 OH

-

4 OH - + 2 H 2O + ClO 2 - ClO 4

- + 4 H 2O

H 2OH 2O;OH-H +

4 OH - + MnO 4- + 4 H + MnO 2 + 2 H 2O + 4 OH

-

4 OH - + 2 H 2O + ClO 2 - ClO 4

- + 4 H + + 4 OH -H +OH -

MnO 4 - + 4 H + MnO 2 + 2 H 2O

2 H 2O + ClO 2 - ClO 4

- + 4 H +H +H 2O

- Chapter 17 -

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(d) (basic solution)






2 H 2O + MnO 4 - + 3 e - MnO 2 + 4 OH

-

2 OH - + 2 Fe 3O4 3 Fe 2O3 + H 2O + 2 e-

(2 H 2O cancelled)2 H 2O + MnO 4 - MnO 2 + 4 OH

-

(1 H 2O cancelled)2 OH - + 2 Fe 3O4 3 Fe 2O3 + H 2O

4 H 2O + MnO 4 - MnO 2 + 2 H 2O + 4 OH

-

2 OH - + H 2O + 2 Fe 3O4 3 Fe 2O3 + 2 H 2O

H 2OH 2O;OH-H +

4 OH - + 4 H + + MnO 4 - MnO 2 + 2 H 2O + 4 OH

-

2 OH - + H 2O + 2 Fe 3O4 3 Fe 2O3 + 2 H+ + 2 OH -

H +OH -4 H + + MnO 4

- MnO 2 + 2 H 2O

H 2O + 2 Fe 3O4 3 Fe 2O3 + 2 H+

H +H 2O

MnO 4 - MnO 2

2 Fe 3O4 3 Fe 2O3

Fe 3O4 + MnO 4 - Fe 2O3 + MnO 2

6 OH - + Se SeO 3 2- + 3 H 2O + 4 e

-

2 (Se + 2 e - Se 2- )6 OH - + 3 Se SeO 3

2- + 2 Se 2- + 3 H 2O

Se + 2 e - Se 2-6 OH - + Se SeO 3

2- + 3 H 2O + 4 e-

(3 H 2O cancelled)6 OH- + Se SeO 3

2- + 3 H 2O

Se Se 2-6 OH - + 3 H 2O + Se SeO 3

2- + 6 H 2O

H 2OH 2O;OH-H +

- 270 -

- Chapter 17 -



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(e) (basic solution)







16. (a) (basic solution)


MnO 4 - MnO 2

SO 32- SO 4

2-

MnO 4 - + SO 3

2- MnO 2 + SO4 2-

2 (4 OH - + Cr(OH) 4 - CrO 4

2- + 4 H 2O + 3 e- )

3 (H 2O + BrO- + 2 e - Br - + 2 OH - )

2 OH - + 3 BrO - + 2 Cr(OH) 4 - 3 Br - + 2 CrO 4

2- + 5 H 2O

3 H 2O, 6 OH-

and 6 e-

canceled from each side

H 2O + BrO- + 2 e - Br - + 2 OH -

4 OH - + Cr(OH) 4 - CrO 4

2- + 4 H 2O + 3 e-

(1 H 2O cancelled)H 2O + BrO- Br - + 2 OH -

2 H 2O + BrO- Br - + H 2O + 2 OH

-

4 OH - + Cr(OH) 4 - CrO 4

2- + 4 H 2O

H 2OH 2O;OH-H +

2 OH - + 2 H + + BrO - Br - + H 2O + 2 OH-

4 OH - + Cr(OH) 4 - CrO 42- + 4 H + + 4 OH -H +OH -

2 H + + BrO - Br - + H 2O

Cr(OH) 4- CrO 4

2- + 4 H +H +H 2O

BrO - Br -Cr(OH) 4

- CrO 4 2-

BrO - + Cr(OH) 4 - Br - + CrO 4

2-

3 (2 OH - + 2 Fe 3O4 3 Fe 2O3 + H 2O + 2 e- )

2 (2 H 2O + MnO 4 - + 3 e - MnO 2 + 4 OH

- )H 2O + 6 Fe 3O4 + 2 MnO 4

- 9 Fe 2O3 + 2 MnO 2 + 2 OH-

3 H 2O, 6 OH- , and 6 e - canceled from each side

- Chapter 17 -

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(b) (basic solution)




ClO 2 ClO 2-

2 OH - + 4 H 2O + SbO 2 - Sb(OH) 6

- + 2 H + + 2 OH -H +OH -

ClO 2 ClO 2-

4 H 2O + SbO 2 - Sb(OH) 6 - + 2 H +H +H 2O

ClO 2 ClO 2-

SbO 2 - Sb(OH) 6

-

ClO 2 + SbO 2 - ClO 2

- + Sb(OH) 6 -

3 (2 OH - + SO 32- SO 4

2- + H 2O + 2 e- )

2 (MnO 4 - + 2 H 2O + 3 e

- MnO 2 + 4 OH- )

H 2O + 2 MnO 4 - + 3 SO 3

2- 2 MnO 2 + 3 SO 4 2- + 2 OH -

3 H 2O, 4 OH- , and 6 e - canceled from each side

3 e - + MnO 4- + 2 H 2O MnO 2 + 4 OH

-

2 OH - + SO 3 2- SO 4 2- + H 2O + 2 e -

(2 H 2O cancelled)MnO 4 - + 2 H 2O MnO 2 + 4 OH

-

(1 H 2O cancelled)2 OH- + SO 3

2- SO 4 2- + H 2O

MnO 4 - + 4 H 2O MnO 2 + 2 H 2O + 4 OH

-

2 OH - + H 2O + SO 3 2- SO 4

2- + 2 H 2O

H 2OH 2O;OH-H +

4 OH - + MnO 4- + 4 H + MnO 2 + 2 H 2O + 4 OH

-

2 OH - + H 2O + SO 3 2- SO 4

2- + 2 H + + 2 OH -H +OH -

MnO 4 - + 4 H + MnO 2 + 2 H 2O

H 2O + SO 32- SO 4

2- + 2 H +H +H 2O

- 272 -

- Chapter 17 -



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6 H 2O + NO 3- + 8 e - NH 3 + 9 OH

-

4 OH - + Al Al(OH) 4 - + 3 e -

(3 H 2O cancelled)6 H 2O + NO 3- NH 3 + 9 OH

-

(4 H 2O cancelled)4 OH- + Al Al(OH) 4

-

9 H 2O + NO 3- NH 3 + 3 H 2O + 9 OH

-

4 OH - + 4 H 2O + Al Al(OH) 4- + 4 H 2O

H 2OH 2O;OH-H +

9 OH - + 9 H + + NO 3 - NH 3 + 3 H 2O + 9 OH

-

4 OH - + 4 H 2O + Al Al(OH) 4- + 4 H + + 4 OH -

H +OH -9 H + + NO 3

- NH 3 + 3 H 2O

4 H 2O + Al Al(OH) 4- + 4 H +

H +H 2O

NO 3 - NH 3

Al Al(OH) 4-

Al + NO 3- NH 3 + Al(OH) 4

-

2 H 2O + 2 OH- + SbO 2

- Sb(OH) 6 - + 2 e -

2 (ClO 2 + e- ClO 2

- )2 H 2O + 2 ClO 2 + 2 OH

- + SbO 2 - 2 ClO 2

- + Sb(OH) 6 -

ClO 2 + e- ClO 2

-

2 OH - + 2 H 2O + SbO 2 - Sb(OH) 6

- + 2 e -

(2 H 2O cancelled)2 OH- + 2 H 2O + SbO 2

- Sb(OH) 6 -

ClO 2 ClO 2 -

2 OH - + 4 H 2O + SbO 2 - Sb(OH) 6

- + 2 H 2O

H 2OH 2O;OH-H +

- Chapter 17 -

- 273 -



19/28


(d) (basic solution)






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

Divide equation by 2

(e) (basic solution)


OH - H 2

Al Al(OH) 4 -

Al + OH - Al(OH) 4 - + H 2

4 OH-

+ 2 H 2O + P4 2 HPO 3

2-+ 2 PH 3

8 OH - + 4 H 2O + 2 P4 4 HPO 3 2- + 4 PH 3

12 H 2O + P4 + 12 e- 4 PH 3 + 12 OH

-

20 OH - + P4 4 HPO 3 2- + 8 H 2O + 12 e

-

20 OH - + P4 4 HPO 3 2- + 8 H 2O (12 H 2O cancelled)

12 H 2O + P4 4 PH 3 + 12 OH-

20 OH - + 12 H 2O + P4 4 HPO 3 2- + 20 H 2O

H 2OH 2O;OH-H +

12 OH - + 12 H + + P4 4 PH 3 + 12 OH-

20 OH - + 12 H 2O + P4 4 HPO 3 2- + 20 H + + 20 OH -

H +OH -12 H + + P4 4 PH 3

12 H 2O + P4 4 HPO 3 2- + 20 H +

H +H 2O

P4 4 PH 3

P4 4 HPO 3 2-

P4 HPO 3 2- + PH 3

8 (4 OH - + Al Al(OH) 4 - + 3 e - )

3 (6 H 2O + NO 3 - + 8 e - NH 3 + 9 OH

- )8 Al + 3 NO 3

- + 18 H 2O + 5 OH- 3 NH 3 + 8 Al(OH) 4

-

27 OH - and 24 e - canceled from each side

- 274 -

- Chapter 17 -



20/28










2 I - I2 + 2 e-

12 H + + 2 IO3 - + 10 e - I2 + 6 H 2O

2 I - I212

H

+ +2

IO3

-

I2 +

6 H 2O

H +H 2O

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 H29 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 H 2O cancelled)2 H2O + OH- H2 + 3 OH

-

(4 H 2O cancelled)4 OH- + Al Al(OH) 4

-

3 H 2O + OH- H 2 + H 2O + 3 OH

-

4 OH - + 4 H 2O + Al Al(OH) 4- + 4 H 2O

H 2OH 2O;OH-H +

3 OH - + 3 H + + OH - H 2 + H 2O + 3 OH-

4 OH - + 4 H 2O + Al Al(OH) 4- + 4 H + + 4 OH -

H +OH -3 H + + OH - H 2 + H 2O

4 H 2O + Al Al(OH) 4- + 4 H +

H +H 2O

- Chapter 17 -

- 275 -



21/28

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




Each side has 2 Mn, 10 S, 16 H, and 48 O and a charge.




Step 3 Balance electrically with

5 e - + 8 H + + MnO 4 - Mn 2+ + 4 H 2O

6 H 2O + Co(NO 2)6 3- Co 2+ + 6 NO 3

- + 12 H + + 11 e -e -

8 H + + MnO 4 - Mn 2+ + 4 H 2O

6 H 2O + Co(NO 2)6 3- Co 2+ + 6 NO 3 - + 12 H +H +H 2O

MnO 4 - Mn 2+

Co(NO 2)6 3- Co 2+ + 6 NO 3

-

Co(NO 2)6 3- + MnO 4

- Co 2+ + Mn 2+ + NO 3 -

- 6

2 (4 H 2O + Mn2+ MnO 4

- + 8 H + + 5 e - )5 (2 e - + S2O8

2- 2 SO 4 2- )

2 Mn 2+ + 5 S2O8 2- + 8 H 2O 2 MnO 4

- + 10 SO4 2- + 16 H +

2 e - + S2O82- 2 SO 4

2-

4 H 2O + Mn 2+ MnO 4

- + 8 H + + 5 e -

S2O82- 2 SO 4

2-

4 H 2O + Mn2+ MnO 4

- + 8 H +H +H 2O

S2O82- 2 SO 4

2-

Mn 2+ MnO 4 -

Mn 2+ + S2O8 2- MnO 4

- + SO 4 2-

12 H + + 2 IO3- + 10 e - I2 + 6 H 2O

5 (2 I - I2 + 2 e- )

12 H + + 2 IO3- + 10 I - 6 I2 + 6 H 2O

- 276 -

- Chapter 17 -



22/28

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



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 3 Add ions to both sides (same number as )

4 OH - + Cr(OH) 4 - CrO 4

2- + 4 H + + 4 OH -2 OH - + 2 H + + BrO - Br - + H 2O + 2 OH

-

H +OH -Cr(OH) 4

- CrO 4 2- + 4 H +

2 H + + BrO - Br - + H 2O

H +H 2O

Cr(OH) 4- CrO 4

2-

BrO - Br -

BrO - + Cr(OH) 4 - Br - + CrO 4

2-

5 (3 H 2O + Mo 2O3 2 MoO 3 + 6 H+ + 6 e - )

6 (5 e - + 8 H + + MnO 4 - Mn 2+ + 4 H 2O)

5 Mo 2O3 + 6 MnO 4 - + 18 H + 10 MoO 3 + 6 Mn 2+ + 9 H 2O

5 e - + 8 H + + MnO 4 - Mn 2+ + 4 H 2O

3 H 2O + Mo 2O3 2 MoO 3 + 6 H+ + 6 e -

8 H+

+ MnO 4

- Mn

2++ 4 H 2O

3 H 2O + Mo 2O3 2 MoO 3 + 6 H+

H +H 2O

MnO 4 - Mn 2+

Mo 2O3 2 MoO 3

Mo 2O3 + MnO 4 - MoO 3 + Mn 2+

a + 2

5 Co(NO 2)6 3- + 11 MnO 4

- + 28 H + 5 Co 2+ + 30 NO 3- + 11 Mn 2+ + 14 H 2O

11 (5 e - + 8 H + + MnO 4 - Mn 2+ + 4 H 2O)

5 (6 H 2O + Co(NO 2)6 3- Co 2+ + 6 NO 3

- + 12 H + + 11 e - )

- Chapter 17 -

- 277 -



23/28



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




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



3 e- + 2 H2O + MnO 4 - MnO 2 + 4 OH

-

10 OH - + S2O32- 2 SO4

2- + 5 H 2O + 8 e-

(2 H 2O cancelled)2 H2O + MnO 4 - MnO 2 + 4 OH

-(5 H 2O cancelled)10 OH

-+ S2O32

- 2 SO4 2

-+ 5 H 2O

4 H2O + MnO 4 - MnO 2 + 2 H2O + 4 OH

-

10 OH - + 5 H 2O + S2O3 2- 2 SO 4

2- + 10 H 2O

H 2OH 2O;OH-H +

4 OH - + 4 H+ + MnO 4- MnO 2 + 2 H2O + 4 OH

-

10 OH - + 5 H 2O + S2O3 2- 2 SO 4

2- + 10 H + + 10 OH -H +OH -

4 H+ + MnO 4- MnO 2 + 2 H2O

5 H 2O + S2O3 2- 2 SO 4

2- + 10 H +H +H 2O

MnO 4 - MnO 2

S2O32- 2 SO 4

2-

S2O3 2- + MnO 4

- SO 42- + Mn 2+

3 (2 e - + H 2O + BrO- Br - + 2 OH - )

2 (4 OH - + Cr(OH) 4 - CrO 4

2- + 4 H 2O + 3 e- )

3 BrO - + 2 Cr(OH) 4 - + 2 OH - 3 Br - + 2 CrO 4

2- + 5 H 2O

4 OH - + Cr(OH) 4 - CrO 4

2- + 4 H 2O + 3 e-

2 e - + H 2O + BrO- Br - + 2 OH -

(1 H 2O cancelled)H 2O + BrO- Br - + 2 OH -

4 OH - + Cr(OH) 4 - CrO 4

2- + 4 H 2O

2 H 2O + BrO- Br - + H 2O + 2 OH

-

H 2OH 2O;OH-H +

- 278 -

- Chapter 17 -



24/28

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) a1 mol Ag107.9 g Ag b a1 mol NO3 mol Ag b = 0.0772 mol NOg Ag mol Ag mol NO

3 Ag + 4 HNO 3 3 AgNO 3 + NO + 2 H 2O

5 mol e -mol KMnO 4 6.022 * 1023 e -

mol e - = 3.011 * 1024 electronsmol KMnO 45 e - + Mn 7+ Mn 2+

KMnO 4 .

Pb 2+Pb 4+Pb 2+Pb 0

PbO 2 + SO 4 2- + 4 H + + 2 e - PbSO 4 + 2 H 2O

Pb + SO 4 2- PbSO 4 + 2 e

-

Voltagesource

Anode (+)

+

Cathode ()

Solution of HBr

Br

H3O+

- 14 charge.

3 (10 OH - + S2O3 2- 2 SO 4

2- + 5 H2O + 8 e- )

8 (3 e- + 2 H2O + MnO 4 - MnO 2 + 4 OH

- )3 S2O3

2- + 8 MnO 4 - + H2O 6 SO 4

2- + 8 MnO 2 + 2 OH

- Chapter 17 -

- 279 -



25/28

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 - ).

Pb 2+ + 2 e - Pb 0.Pb 2+Cu + + e - Cu0 or Cu + Cu 2+ + e -Cu + Cu2+

(100.0 g Al)

a1 mol Al

26.98 g b 3 mol H 2

2 mol Al = 5.560 mol H 2

g Al mol Al mol H 2

2 Al + 2 OH - + 6 H 2O 2 Al(OH) 4 - + 3 H 2

= 10.0 mL of 0.200 M K2Cr2O7

(60.0 mL FeSO 4)a0.200 mol1000 mL b

1 mol Cr2O7 2-

6 mol FeSO 4 a1000 mL0.200 mol b

mL FeSO 4 mol FeSO 4 mol Cr2O7 2- mL Cr2O7

2-

Cr2O7 2- + 6 Fe 2+ + 14 H + 2 Cr 3+ + 6 Fe 3+ + 7 H 2O

(5.00 g H 3AsO 3)a1 mol125.9 gb 1 mol Cr2O7 2-

3 mol H 3AsO 3 a1000 mL0.200 mol b = 66.2 mL of 0.200 M K 2Cr2O7g H 3AsO 3 mol H 3AsO 3 mol Cr2O7

2- mL Cr2O7

2-

Cr2O7 2- + 3 H 3AsO 3 + 8 H

+ 2 Cr 3+ + 3 H 3AsO 4 + 4 H 2O

= 17 g KMnO 4a2 mol KMnO 45 mol H 2O2 b a158.0 g

mol b(100. mL H 2O2 solution) a1.031 gmL b 9.0 g H 2O2100. g H 2O2 solution a1 mol34.02 gbmL H

2O

2 g H

2O

2 mol H

2O

2 mol KMnO

4 g KMnO

4

5 H 2O2 + 2 KMnO 4 + 3 H 2SO 4 5 O2 + 2 MnSO 4 + K 2SO 4 + 8 H 2O

(0.300 mol KClO 3)3 mol Cl21 mol KClO 3 a22.4 L1 mol b = 20.2 L Cl2mol KClO 3 mol Cl 2 L Cl 2

3 Cl 2 + 6 KOH KClO 3 + 5 KCl + 3 H 2O

- 280 -

- Chapter 17 -



26/28

30. The electrons lost by the species undergoing oxidation must be gained (or attracted) byanother species which then undergoes reduction.

31. cannot take from Acannot take from Atakes from Dtakes 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 greateroxidation number makes it very attractive to electrons.

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

(a)(b)

35. (a)

(b)

(c)

V = a0.005 mol1.4 atm b a0.0821 L atmmol K b(323 K) = 0.09 L Br2 vaporV =

nRTP

PV = nRT

(100.0 mL Mn 2+ )a0.05 mol1000 mL b a1 mol Br21 mol Mn 2+ b = 0.005 mol Br2(100.0 mL Mn 2+ )a0.05 mol1000 mL b 1 mol MnO 21 mol Mn 2+ a86.94 gmol b = 0.4 g MnO 2mL Mn 2+ mol Mn 2+ mol MnO 2 g MnO 2

MnO 2 + 2 Br- + 4 H + Mn 2+ + Br2 + 2 H 2O

SO 2 SO 3 ; (S4+ S6+ + 2 e - )

Mg Mg 2+ + 2 e -

+ 7KMnO 4

+ 6K 2MnO 4

+ 4MnO 2

( + 7)+ 3MnF 3

KMnO 4+ 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 + ,D 2+ ,e - ,B2+e -D 2+B( s ) + D 2+ (aq ) D(s) + B2+ (aq )

e -C +D( s ) + 2 C + (aq ) 2C( s) + D 2+ (aq )e -C +A(s) + C + (aq ) NRe -B2+A(s) + B2+ (aq ) NR

- Chapter 17 -

- 281 -



27/28

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 + Pb 2+

R.A. = Reducing agentO.A. = Oxidizing agent

O2 2- O 2-O2

2- O 2-Cu 2+ Cu0N 5+ N 2+O 0 O 2-O2

2- O2 0S4+ S6+N 3- N2

0S2- S0C 223+ C4+

H 2O2 ,Na 2SO 3 ,NH 3 ,H 2S,C3H 8 ,

H 2O2 ,H 2O2 ,HNO 3 ,O2 ,

O2 2-O2

2-O2

O2 2-

4 Zn + NO 3 - + 10 H + 4 Zn2+ + NH 4

+ + 3 H 2O

Mn( s) + 2 HCl( aq ) Mn 2+ (aq ) + H 2(g ) + 2 Cl- (aq )

Br2 + 2 I- 2 Br - + I2

I2 + Cl- NR

Br2 + Cl- NR

F2 + 2 Cl- 2 F - + Cl 2

- 282 -

- Chapter 17 -



28/28

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 atm760. torr b = 0.979 atmV =

nRT

P

PV = nRT

(0.500 mol Ag) a1 mol NO3 mol Ag b = 0.167 mol NOmol Ag mol NO

3 Ag + 4 HNO 3 3 AgNO 3 + NO + 2 H 2Oa

3.65 g KI

4.00 g sample b1100

2 = 91.3% KI

(2.79 g I2)a1 mol253.8 gb 8 mol KI4 mol I2 a166.0 gmol b = 3.65 g KI in sampleg I 2 mol I2 mol KI g KI

I2

8 KI + 5 H 2SO 4 4 I2 + H 2S + 4 K 2SO 4 + 4 H 2O

- Chapter 17 -

- 283 -

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