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CHAPTER 17

OXIDATION-REDUCTION SOLUTIONS TO REVIEW QUESTIONS 1.

Oxidation of a metal occurs when the metal loses electrons. The easier it is for a metal to lose electrons, the more active the metal is.

2.

(a) (b)

3.

The higher metal on the list is more reactive. (a) Al (b) Ba (c)

Iodine is oxidized. Its oxidation number increases from 0 to +5. Chlorine is reduced. Its oxidation number decreases from 0 to -1.

Ni

4.

If the free element is higher on the list than the ion with which it is paired, the reaction occurs. (a) Yes. Zn(s) + Cu2+(aq) ¡ Zn2+(aq) + Cu(s) (b) No reaction (c) Yes. Sn(s) + 2 Ag +(aq) ¡ Sn2+(aq) + 2 Ag(s) (d) No reaction (e) Yes. Ba(s) + FeCl 2(aq) ¡ BaCl 2(aq) + Fe(s) (f) No reaction (g) Yes. Ni(s) + Hg(NO3)2(aq) ¡ Ni(NO3)2(aq) + Hg(l) (h) Yes. 2 Al(s) + 3 CuSO4(aq) ¡ Al 2(SO4)3(aq) + 3 Cu(s)

5.

Copper is more active than silver. Therefore, copper undergoes oxidation more easily than silver. Accordingly, it is more difficult for copper ion to undergo reduction than it is for silver ion. When a silver wire is placed in a solution of copper (II) nitrate one might predict that copper crystals would form on the silver wire. However for copper to go from an oxidation state of +2 to an oxidation state of 0 it would have to gain electrons (reduction) and silver would have to lose electrons (oxidation). This will not happen because copper is more active than silver.

6.

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

2 Al + Fe 2O3 ¡ Al 2O3 + 2 Fe + Heat Al is above Fe in the activity series, which indicates Al is more active than Fe. No. Iron is less active than aluminum and will not displace aluminum from its compounds. Yes. Aluminum is above chromium in the activity series and will displace Cr 3+ from its compounds. - 256 -

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7.

(a)

2 Al(s) + 6 HCl(aq) ¡ 2 AlCl 3(aq) + 3 H 2(g) 2 Al(s) + 3 H 2SO4(aq) ¡ Al 2(SO4)3(aq) + 3 H 2(g)

(b)

2 Cr(s) + 6 HCl(aq) ¡ 2 CrCl 3(aq) + 3 H 2(g) 2 Cr(s) + 3 H 2SO4(aq) ¡ Cr2(SO4)3(aq) + 3 H 2(g)

(c)

Au(s) + HCl(aq) ¡ no reaction Au(s) + H 2SO4(aq) ¡ no reaction

(d)

Fe(s) + 2 HCl(aq) ¡ FeCl 2(aq) + H 2(g) Fe(s) + H 2SO4(aq) ¡ FeSO4(aq) + H 2(g)

(e)

Cu(s) + HCl(aq) ¡ no reaction Cu(s) + H 2SO4(aq) ¡ no reaction

(f)

Mg(s) + 2 HCl(aq) ¡ MgCl 2(aq) + H 2(g) Mg(s) + H 2SO4(aq) ¡ MgSO4(aq) + H 2(g)

(g)

Hg(l) + HCl(aq) ¡ no reaction Hg(l) + H 2SO4(aq) ¡ no reaction

(h)

Zn(s) + 2 HCl(aq) ¡ ZnCl 2(aq) + H 2(g) Zn(s) + H 2SO4(aq) ¡ ZnSO4(aq) + H 2(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) (b)

Oxidation occurs at the anode. The reaction is 2 Cl-(aq) ¡ Cl 2(g) + 2 e Reduction occurs at the cathode. The reaction is Ni2+(aq) + 2 e - ¡ Ni(s)

(c)

The net chemical reaction is Ni2+(aq) + 2 Cl-(aq)

electrical energy

" Ni(s) + Cl (g) 2 - 257 -

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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) (b)

It would not be possible to monitor the voltage produced, but the reactions in the cell would still occur. If the salt bridge were removed, the reaction would stop. Ions must be mobile to maintain an electrical neutrality of ions in solution. The two solutions would be isolated with no complete electrical circuit.

13.

Oxidation and reduction are complementary processes because one does not occur without the other. The loss of e - in oxidation is accompanied by a gain of e - in reduction.

14.

Ca2+ + 2 e - ¡ Ca 2 Br - ¡ Br2 + 2 e -

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.

17.

The electrolyte in a lead storage battery is dilute sulfuric acid. In the discharge cycle, SO4 2-, is removed from solution as it reacts with PbO2 and H + to form PbSO4(s) and H 2O. Therefore, the electrolyte solution contains less H 2SO4 and becomes less dense.

18.

If Hg 2+ 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.

cathode reaction, reduction anode reaction, oxidation

- 258 -

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CHAPTER 17 SOLUTIONS TO EXERCISES 1.

The oxidation number of the underlined element is indicated by the number following the formula. PbO2 +4 (a) NaCl +1 (c) (e) H 2SO3 +4 NaNO3 +5 NH 4Cl -3 (b) FeCl 3 -1 (d) (f)

2.

The oxidation number of the underlined element is indicated by the number following the formula. (a) (b)

3.

NH 3 KClO3

-3 +5

S2-2 NO2 +3

(c) (d)

(e) (f)

K 2CrO4 K 2Cr2O7

+6 +6

Na 2O2 -1 Bi3+ +3

O2 0 3AsO4 +5

(c) (d)

Fe(OH)3 IO3 -

-2 +5 Changing Element

Balanced half-reaction (a) (b) (c) (d)

6.

(c) (d)

The oxidation number of the underlined element is indicated by the number following the formula. (a) (b)

5.

+7 0

The oxidation number of the underlined element is indicated by the number following the formula. (a) (b)

4.

KMnO4 I2

Zn2+ + 2 e - ¡ Zn 2 Br - ¡ Br2 + 2 e MnO4 - + 8 H + + 5 e - ¡ Mn2+ + 4 H 2O Ni ¡ Ni2+ + 2 e -

Balanced half-reactions (a) (b) (c) (d)

SO3 2- + H 2O ¡ SO4 2- + 2 H + + 2 e NO3 - + 4 H + + 3 e - ¡ NO + 2 H 2O S2O4 2- + 2 H 2O ¡ 2 SO3 2- + 4 H + + 2 e Fe 2+ ¡ Fe 3+ + 1 e -

- 259 -

Zn Br Mn Ni

Type of reaction reduction oxidation reduction oxidation

Changing Element

Type of reaction

S N S Fe

oxidation reduction oxidation oxidation

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7.

(1)

Cr + HCl ¡ CrCl 3 + H 2 (a) (b)

(2)

SO4 2- + I - + H + ¡ H 2S + I 2 + H 2O (a) (b)

8.

(1)

(a) (b) (c)

As is oxidized, Ag is reduced Ag + is the oxidizing agent, AsH 3 the reducing agent

Cl 2 + NaBr ¡ NaCl + Br2 (a) (b)

9.

I is oxidized, S is reduced SO4 2- is the oxidizing agent, I - the reducing agent

AsH 3 + Ag + + H 2O ¡ H 3AsO4 + Ag + H + (a) (b)

(2)

Cr is oxidized, H is reduced HCl is the oxidizing agent, Cr the reducing agent

Br is oxidized, Cl is reduced Cl 2 is the oxidizing agent, NaBr the reducing agent

correctly balanced correctly balanced incorrectly balanced Mg(s) + 2 HCl(aq) ¡ Mg 2+ (aq) + 2Cl- (aq) + H 2(g)

(d) incorrectly balanced 3 CH 3OH(aq) + Cr2O7 2- (aq) + 8 H + (aq) ¡ 2 Cr 3+ (aq) + 3 CH 2O(aq) + 7 H 2O(l) 10.

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

incorrectly balanced 3 MnO2(s) + 4 Al(s) ¡ 3 Mn(s) + 2 Al 2O3(s) correctly balanced correctly balanced incorrectly balanced 8 H 2O(l) + 2 MnO4 - (aq) + 7 S2- (aq) ¡ 2 MnS(s) + 16 OH - (aq) + 5 S(s)

11.

Balancing oxidation-reduction equations (a) ox red

Zn + S ¡ ZnS Zn0 ¡ Zn2+ + 2 e S0 + 2 e - ¡ S2Zn + S ¡ ZnS

Add half-reactions the 2 e - cancel - 260 -

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(b) ox red

AgNO3 + Pb ¡ Pb(NO3)2 + Ag Pb 0 ¡ Pb 2+ + 2 e Ag + + 1 e - ¡ Ag 0 Pb + 2 Ag + ¡ Pb 2+ + 2 Ag

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

Transfer the coefficients to the original equation and complete the balancing by inspection. 2 AgNO3 + Pb ¡ Pb(NO3)2 + 2 Ag (c) ox red

Fe 2O3 + CO ¡ Fe + CO2 C 2+ ¡ C 4+ + 2 e Fe 3+ + 3 e - ¡ Fe 0 3 C 2+ + 2 Fe 3+ ¡ 3 C 4+ + 2 Fe

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

Transfer the coefficients to the original equation (the coefficient 2 in front of the Fe 3+ becomes the subscript 2 in Fe 2O3). Complete the balancing by inspection. Fe 2O3 + 3 CO ¡ 2 Fe + 3 CO2 (d)

H 2S + HNO3 ¡ S + NO + H 2O S2- ¡ S0 + 2 e N 5+ + 3 e - ¡ N 2+ 3 S2- + 2 N 5+ ¡ 3 S + 2 N 2+

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

Transfer the coefficients to the original equations and complete the balancing by inspection. 3 H 2S + 2 HNO3 ¡ 3 S + 2 NO + 4 H 2O (e)

MnO2 + HBr ¡ MnBr2 + Br2 + H 2O Br - ¡ Br 0 + 1 e Mn4+ + 2 e - ¡ Mn2+ Mn4+ + 2 Br - ¡ Mn2+ + 2 Br 0

Multiply by 2 Add equations and the 2 e - cancel

Transfer the coefficients to the original equation. The coefficient 2 in front of the Br - becomes the subscript 2 in the Br2 . Also, 2 more Br - ions are required to account for the 2 Br - ions that do not change oxidation numbers. These 2 are part of the compound MnBr2 . MnO2 + 4 HBr ¡ MnBr2 + Br2 + 2 H 2O - 261 -

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12.

(a)

Balancing oxidation-reduction equations Cl 2 + KOH ¡ KCl + KClO3 + H 2O Cl0 ¡ Cl5+ + 5 e Cl0 + e - ¡ ClMultiply by 5, add, the 5 e - cancel 3 Cl 2 ¡ Cl5+ + 5 Cl6 Cl0 becomes 3 Cl 2 Transfer the coefficients to the original equations and complete the balancing by inspection. 3 Cl 2 + 6 KOH ¡ KClO3 + 5 KCl + 3 H 2O

(b)

Ag + HNO3 ¡ AgNO3 + NO + H 2O Ag 0 ¡ Ag + + e Multiply by 3, add, 5+ 2+ the 3 e - cancel N + 3e ¡ N 3 Ag + N 5+ ¡ 3 Ag + + N 2+ Transfer the coefficients to the original equations and complete the balancing by inspection. 3 Ag + 4 HNO3 ¡ 3 AgNO3 + NO + 2 H 2O

(c)

CuO + NH 3 ¡ N2 + Cu + H 2O N 3- ¡ N 0 + 3 e Multiply by 2 2+ 0 Cu + 2 e ¡ Cu Multiply by 3, add, the 6 e - cancel 2 N 3- + 3 Cu2+ ¡ N2 + 3 Cu Transfer the coefficients to the original equations and complete the balancing by inspection. 3 CuO + 2 NH 3 ¡ N2 + 3 Cu + 3 H 2O

(d)

PbO2 + Sb + NaOH ¡ PbO + NaSbO2 + H2O Sb0 ¡ Sb3+ + 3 eMultiply by 2 4+ 2+ Pb + 2 e ¡ Pb Multiply by 3, add, the 6 e - cancel 2 Sb + 3 Pb4+ ¡ 2 Sb3+ + 3 Pb2 + Transfer the coefficients to the original equations and complete the balancing by inspection. 3 PbO2 + 2 Sb + 2 NaOH ¡ 3 PbO + 2 NaSbO2 + H 2O

(e)

H 2O2 + KMnO4 + H 2SO4 ¡ O2 + MnSO4 + K 2SO4 + H 2O O2 2- ¡ O2 0 + 2 e Multiply by 5 7+ 2+ Mn + 5 e ¡ Mn Multiply by 2, add, the 10 e - cancel 5 O2 2- + 2 Mn7+ ¡ 5 O2 + 2 Mn2+ - 262 -

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Transfer the coefficients to the original equations and complete the balancing by inspection. 5 H 2O2 + 2 KMnO4 + 3 H 2SO4 ¡ 5 O2 + 2 MnSO4 + K 2SO4 + 8 H 2O 13.

(a)

Zn + NO3 - ¡ Zn2+ + NH 4 + Step 1

(acidic solution)

Write half-reaction equations. Balance except H and O. Zn ¡ Zn2+ NO3 - ¡ NH 4 +

Step 2

Balance H and O using H 2O and H + Zn ¡ Zn2+ 10 H + + NO3 - ¡ NH 4 + + 3 H 2O

Step 3

Balance electrically with electrons Zn ¡ Zn2+ + 2 e 10 H + + NO3 - + 8 e - ¡ NH 4 + + 3 H 2O

Step 4

Equalize the loss and gain of electrons 4 (Zn ¡ Zn2+ + 2 e -) 10 H + + NO3 - + 8 e - ¡ NH 4 + + 3 H 2O

Step 5

Add the half-reactions–electrons cancel 10 H + + 4 Zn + NO3 - ¡ 4 Zn2+ + NH 4 + + 3 H 2O

(b)

NO3 - + S ¡ NO2 + SO4 2Step 1

(acidic solution)

Write half-reaction equations. Balance except H and O. S ¡ SO4 2NO3 - ¡ NO2

Step 2

Balance H and O using H 2O and H + 4 H 2O + S ¡ SO4 2- + 8 H + 2 H + + NO3 - ¡ NO2 + H 2O

Step 3

Balance electrically with electrons 4 H 2O + S ¡ SO4 2- + 8 H + + 6 e 2 H + + NO3 - + e - ¡ NO2 + H 2O - 263 -

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Step 4 and 5

(c)

Equalize the loss and gain of electrons; add the half-reactions 4 H 2O + S ¡ SO4 2- + 8 H + + 6 e 6 (2 H + + NO3 - + e - ¡ NO2 + H 2O) 4 H + + S + 6 NO3 - ¡ 6 NO2 + SO4 2- + 2 H 2O 4 H 2O, 8 H + and 6 e - canceled from each side

PH 3 + I 2 ¡ H 3PO2 + I Step 1

(acidic solution)

Write half-reaction equations. Balance except H and O. PH 3 ¡ H 3PO2 I 2 ¡ 2I -

Step 2

Balance H and O using H 2O and H + 2 H 2O + PH 3 ¡ H 3PO2 + 4 H + I2 ¡ 2 I -

Step 3

Balance electrically with electrons 2 H 2O + PH 3 ¡ H 3PO2 + 4 H + + 4 e I2 + 2 e - ¡ 2 I -

Step 4 and 5

Equalize the loss and gain of electrons; add the half-reactions 2 H2O + PH3 ¡ H3PO2 + 4 H+ + 4 e2 (I2 + 2 e- ¡ 2 I-) PH3 + 2 H2O + 2 I2 ¡ H3PO2 + 4 I- + 4 H +

(d)

Cu + NO3 - ¡ Cu2+ + NO Step 1

(acidic solution)

Write half-reaction equations. Balance except H and O. Cu ¡ Cu2+ NO3 - ¡ NO

Step 2

Balance H and O using H 2O and H + Cu ¡ Cu2+ 4H + + NO3 - ¡ NO + 2 H 2O

Step 3

Balance electrically with electrons Cu ¡ Cu2+ + 2 e 4H + + NO3 - + 3 e - ¡ NO + 2 H 2O - 264 -

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Step 4 and 5

Equalize the loss and gain of electrons; add the half-reactions 3 (Cu ¡ Cu2+ + 2 e -) 2 (4 H + + NO3 - + 3 e - ¡ NO + 2 H 2O) 3 Cu + 8 H + + 2 NO3 - ¡ 3 Cu2+ + 2 NO + 4 H 2O

(e)

ClO3 - + Cl- ¡ Cl 2 Step 1

(acidic solution)

Write half-reaction equations. Balance except H and O. Cl- ¡ Cl0 ClO3 - ¡ Cl0

Step 2

Balance H and O using H 2O and H + Cl- ¡ Cl0 6 H + + ClO3 - ¡ Cl0 + 3 H 2O

Step 3

Balance electrically with electrons Cl- ¡ Cl0 + e 6 H + + ClO3 - + 5 e - ¡ Cl0 + 3 H 2O

Step 4 and 5

Equalize the loss and gain of electrons; add the half-reactions 5 (Cl- ¡ Cl0 + e -) 6 H + + ClO3 - + 5 e - ¡ Cl0 + 3 H 2O 6 H + + ClO3 - + 5 Cl- ¡ 3 Cl 2 + 3 H 2O

14.

(a)

ClO3 - + I - ¡ I 2 + ClStep 1

(acidic solution)

Write half-reaction equations. Balance except H and O. 2 I - ¡ I2 ClO3 - ¡ Cl-

Step 2

Balance H and O using H 2O and H + 2 I - ¡ I2 6 H + + ClO3 - ¡ Cl- + 3 H 2O

Step 3

Balance electrically with electrons 2 I - ¡ I2 + 2 e 6 H + + ClO3 - + 6 e - ¡ Cl- + 3 H 2O - 265 -

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Step 4 and 5

(b)

Equalize the loss and gain of electrons; add the half-reactions 3 (2 I - ¡ I 2 + 2 e -) 6 H + + ClO3 - + 6 e - ¡ Cl- + 3 H 2O 6 H + + ClO3 - + 6 I - ¡ 3 I 2 + Cl- + 3 H 2O

Cr2O7 2- + Fe 2+ ¡ Cr 3+ + Fe 3+ Step 1

(acidic solution)

Write half-reaction equations. Balance except H and O. Fe 2+ ¡ Fe 3+ Cr2O7 2- ¡ 2 Cr 3+

Step 2

Balance H and O using H 2O and H + Fe 2+ ¡ Fe 3+ 14 H + + Cr2O7 2- ¡ 2 Cr 3+ + 7 H 2O

Step 3

Balance electrically with electrons Fe 2+ ¡ Fe 3+ + e 14 H + + Cr2O7 2- + 6 e - ¡ 2 Cr 3+ + 7 H 2O

Step 4 and 5

Equalize the loss and gain of electrons; add the half-reactions 6 (Fe 2+ ¡ Fe 3+ + e -) 14 H + + Cr2O7 2- + 6 e - ¡ 2 Cr 3+ + 7 H 2O 14 H + + Cr2O7 2- + 6 Fe 2+ ¡ 2 Cr 3+ + 6 Fe 3+ + 7 H 2O

(c)

MnO4 - + SO2 ¡ Mn2+ + SO4 2Step 1

(acidic solution)

Write half-reaction equations. Balance except H and O. SO2 ¡ SO4 2MnO4 - ¡ Mn2+

Step 2

Balance H and O using H 2O and H + 2 H 2O + SO2 ¡ SO4 2- + 4 H + 8 H + + MnO4 - ¡ Mn2+ + 4 H 2O

Step 3

Balance electrically with electrons 2 H 2O + SO2 ¡ SO4 2- + 4 H + + 2 e 8 H + + MnO4 - + 5 e - ¡ Mn2+ + 4 H 2O - 266 -

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Step 4 and 5

(d)

Equalize the loss and gain of electrons; add the half-reactions 5 (2 H 2O + SO2 ¡ SO4 2- + 4 H + + 2 e -) 2 (8 H + + MnO4 - + 5 e - ¡ Mn2+ + 4 H 2O) 2 H 2O + 2 MnO4 - + 5 SO2 ¡ 4 H + + 2 Mn2+ + 5 SO4 28 H 2O, 16 H +, and 10 e - canceled from each side

H 3AsO3 + MnO4 - ¡ H 3AsO4 + Mn2+ Step 1

(acidic solution)

Write half-reaction equations. Balance except H and O. H 3AsO3 ¡ H 3AsO4 MnO4 - ¡ Mn2+

Step 2

Balance H and O using H 2O and H + H 2O + H 3AsO3 ¡ 2 H + + H 3AsO4 8 H + + MnO4 - ¡ Mn2+ + 4 H 2O

Step 3

Balance electrically with electrons H 2O + H 3AsO3 ¡ 2 H + + H 3AsO4 + 2 e 8 H + + MnO4 - + 5 e - ¡ Mn2+ + 4 H 2O

Step 4 and 5

Equalize the loss and gain of electrons; add the half-reactions 5 (H 2O + H 3AsO3 ¡ 2 H + + H 3AsO4 + 2 e -) 2 (8 H + + MnO4 - + 5 e - ¡ Mn2+ + 4 H 2O) 6 H + + 5 H 3AsO3 + 2 MnO4 - ¡ 5 H 3AsO4 + 2 Mn2+ + 3 H 2O 5 H 2O, 10 H +, and 10 e - canceled from each side

(e)

Cr2O7 2- + H 3AsO3 ¡ Cr 3+ + H 3AsO4 (acidic solution) Step 1

Write half-reaction equations. Balance except H and O. H 3AsO3 ¡ H 3AsO4 Cr2O7 2- ¡ 2 Cr 3+

Step 2

Balance H and O using H 2O and H + H 2O + H 3AsO3 ¡ 2 H + + H 3AsO4 14 H + + Cr2O7 2- ¡ 2 Cr 3+ + 7 H 2O

Step 3

Balance electrically with electrons H 2O + H 3AsO3 ¡ 2 H + + H 3AsO4 + 2 e 14 H + + Cr2O7 2- + 6 e - ¡ 2 Cr 3+ + 7 H 2O - 267 -

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Step 4 and 5

Equalize the loss and gain of electrons; add the half-reactions 3 (H 2O + H 3AsO3 ¡ 2 H + + H 3AsO4 + 2 e -) 14 H + + Cr2O7 2- + 6 e - ¡ 2 Cr 3+ + 7 H 2O 8 H + + Cr2O7 2- + 3 H 3AsO3 ¡ 2 Cr 3+ + 3 H 3AsO4 + 4 H 2O 3 H 2O, 6 H +, and 6 e - canceled from each side

15.

(a)

(basic solution) Cl 2 + IO3 - ¡ Cl- + IO4 Step 1 Write half-reaction equations. Balance except H and O. IO3 - ¡ IO4 Cl 2 ¡ 2 ClStep 2

Balance H and O using H 2O and H + H 2O + IO3 - ¡ IO4 - + 2 H + Cl 2 ¡ 2 Cl-

Step 3

Add OH - ions to both sides (same number as H + ions) 2 OH - + H 2O + IO3 - ¡ IO4 - + 2 H + + 2 OH Cl 2 ¡ 2 Cl-

Step 4

Combine H + and OH - to form H 2O; cancel H 2O where possible 2 OH - + H 2O + IO3 - ¡ IO4 - + 2 H 2O Cl 2 ¡ 2 Cl2 OH - + IO3 - ¡ IO4 - + H 2O Cl 2 ¡ 2 Cl

Step 5

(1 H 2O cancelled)

-

Balance electrically with electrons 2 OH - + IO3 - ¡ IO4 - + H 2O + 2 e Cl 2 + 2 e - ¡ 2 Cl-

(b)

Step 6

Electron loss and gain is balanced

Step 7

Add half-reactions 2 OH - + IO3 - + Cl 2 ¡ IO4 - + 2 Cl- + H 2O

MnO4 - + ClO2 - ¡ MnO2 + ClO4 (basic solution) Step 1 Write half-reaction equations. Balance except H and O. ClO2 - ¡ ClO4 MnO4 - ¡ MnO2 - 268 -

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Step 2

Balance H and O using H 2O and H + 2 H 2O + ClO2 - ¡ ClO4 - + 4 H + MnO4 - + 4 H + ¡ MnO2 + 2 H 2O

Step 3

Add OH - ions to both sides (same number as H + ions) 4 OH - + 2 H 2O + ClO2 - ¡ ClO4 - + 4 H + + 4 OH 4 OH - + MnO4 - + 4 H + ¡ MnO2 + 2 H 2O + 4 OH -

Step 4

Combine H + and OH - to form H 2O; cancel H 2O where possible 4 OH - + 2 H 2O + ClO2 - ¡ ClO4 - + 4 H 2O 4 H 2O + MnO4 - ¡ MnO2 + 2 H 2O + 4 OH 4 OH - + ClO2 - ¡ ClO4 - + 2 H 2O -

2 H 2O + MnO4 ¡ MnO2 + 4 OH Step 5

-

(2 H 2O cancelled) (2 H 2O cancelled)

Balance electrically with electrons 4 OH - + ClO2 - ¡ ClO4 - + 2 H 2O + 4 e 2 H 2O + MnO4 - + 3 e - ¡ MnO2 + 4 OH -

Step 6 and 7

Equalize gain and loss of electrons; add half-reactions 3 (4 OH - + ClO2 - ¡ ClO4 - + 2 H 2O + 4 e -) 4 (2 H 2O + MnO4 - + 3 e - ¡ MnO2 + 4 OH -) 2 H 2O + 4 MnO4 - + 3 ClO2 - ¡ 4 MnO2 + 3 ClO4 - + 4 OH 6 H 2O, 12 OH -, and 12 e - canceled from each side

(c)

Se ¡ SeO3 2- + Se 2Step 1

(basic solution)

Write half-reaction equations. Balance except H and O. Se ¡ SeO3 2Se ¡ Se 2-

Step 2

Balance H and O using H 2O and H + 3 H 2O + Se ¡ SeO3 2- + 6 H + Se ¡ Se 2-

Step 3

Add OH - ions to both sides (same number as H + ions) 6 OH - + 3 H 2O + Se ¡ SeO3 2- + 6 H + + 6 OH Se ¡ Se 2- 269 -

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Step 4

Combine H + and OH - to form H 2O; cancel H 2O where possible 6 OH - + 3 H 2O + Se ¡ SeO3 2- + 6 H 2O Se ¡ Se 26 OH - + Se ¡ SeO3 2- + 3 H 2O

Step 5

(3 H 2O cancelled)

Balance electrically with electrons 6 OH - + Se ¡ SeO3 2- + 3 H 2O + 4 e Se + 2 e - ¡ Se 2-

Step 6 and 7

Equalize gain and loss of electrons; add half-reactions 6 OH - + Se ¡ SeO3 2- + 3 H 2O + 4 e 2 (Se + 2 e - ¡ Se 2-) 6 OH - + 3 Se ¡ SeO3 2- + 2 Se 2- + 3 H 2O

(d)

Fe 3O4 + MnO4 - ¡ Fe 2O3 + MnO2 Step 1

(basic solution)

Write half-reaction equations. Balance except H and O. 2 Fe 3O4 ¡ 3 Fe 2O3 MnO4 - ¡ MnO2

Step 2

Balance H and O using H 2O and H + H 2O + 2 Fe 3O4 ¡ 3 Fe 2O3 + 2 H + 4 H + + MnO4 - ¡ MnO2 + 2 H 2O

Step 3

Add OH - ions to both sides (same number as H + ions) 2 OH - + H 2O + 2 Fe 3O4 ¡ 3 Fe 2O3 + 2 H + + 2 OH 4 OH - + 4 H + + MnO4 - ¡ MnO2 + 2 H 2O + 4 OH -

Step 4

Combine H + and OH - to form H 2O; cancel H 2O where possible 2 OH - + H 2O + 2 Fe 3O4 ¡ 3 Fe 2O3 + 2 H 2O 4 H 2O + MnO4 - ¡ MnO2 + 2 H 2O + 4 OH 2 OH - + 2 Fe 3O4 ¡ 3 Fe 2O3 + H 2O -

2 H 2O + MnO4 ¡ MnO2 + 4 OH Step 5

-

(1 H 2O cancelled) (2 H 2O cancelled)

Balance electrically with electrons 2 OH - + 2 Fe 3O4 ¡ 3 Fe 2O3 + H 2O + 2 e 2 H 2O + MnO4 - + 3 e - ¡ MnO2 + 4 OH - 270 -

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Step 6 and 7

(e)

Equalize gain and loss of electrons; add half-reactions 3 (2 OH - + 2 Fe 3O4 ¡ 3 Fe 2O3 + H 2O + 2 e -) 2 (2 H 2O + MnO4 - + 3 e - ¡ MnO2 + 4 OH -) H 2O + 6 Fe 3O4 + 2 MnO4 - ¡ 9 Fe 2O3 + 2 MnO2 + 2 OH 3 H 2O, 6 OH -, and 6 e - canceled from each side

(basic solution) BrO - + Cr(OH)4 - ¡ Br - + CrO4 2Step 1 Write half-reaction equations. Balance except H and O. Cr(OH)4 - ¡ CrO4 2BrO - ¡ Br Step 2

Balance H and O using H 2O and H + Cr(OH)4 - ¡ CrO4 2- + 4 H + 2 H + + BrO - ¡ Br - + H 2O

Step 3

Add OH - ions to both sides (same number as H + ions) 4 OH - + Cr(OH)4 - ¡ CrO4 2- + 4 H + + 4 OH 2 OH - + 2 H + + BrO - ¡ Br - + H 2O + 2 OH -

Step 4

Combine H + and OH - to form H 2O; cancel H 2O where possible 4 OH - + Cr(OH)4 - ¡ CrO4 2- + 4 H 2O 2 H 2O + BrO - ¡ Br - + H 2O + 2 OH H 2O + BrO - ¡ Br - + 2 OH -

Step 5

(1 H 2O cancelled)

Balance electrically with electrons 4 OH - + Cr(OH)4 - ¡ CrO4 2- + 4 H 2O + 3 e H 2O + BrO - + 2 e - ¡ Br - + 2 OH -

Step 6 and 7

Equalize gain and loss of electrons; add half-reactions 2 (4 OH - + Cr(OH)4 - ¡ CrO4 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 CrO4 2- + 5 H 2O 3 H 2O, 6 OH - and 6 e - canceled from each side

16.

(a)

MnO4 - + SO3 2- ¡ MnO2 + SO4 2- (basic solution) Step 1

Write half-reaction equations. Balance except H and O. SO3 2- ¡ SO4 2MnO4 - ¡ MnO2 - 271 -

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Step 2

Balance H and O using H 2O and H + H 2O + SO3 2- ¡ SO4 2- + 2 H + MnO4 - + 4 H + ¡ MnO2 + 2 H 2O

Step 3

Add OH - ions to both sides (same number as H + ions) 2 OH - + H 2O + SO3 2- ¡ SO4 2- + 2 H + + 2 OH 4 OH - + MnO4 - + 4 H + ¡ MnO2 + 2 H 2O + 4 OH -

Step 4

Combine H + and OH - to form H 2O; cancel H 2O where possible 2 OH - + H 2O + SO3 2- ¡ SO4 2- + 2 H 2O MnO4 - + 4 H 2O ¡ MnO2 + 2 H 2O + 4 OH 2 OH - + SO3 2- ¡ SO4 2- + H 2O -

MnO4 + 2 H 2O ¡ MnO2 + 4 OH Step 5

(1 H 2O cancelled) -

(2 H 2O cancelled)

Balance electrically with electrons 2 OH - + SO3 2- ¡ SO4 2- + H 2O + 2 e 3 e - + MnO4 - + 2 H 2O ¡ MnO2 + 4 OH -

Step 6 and 7

(b)

Equalize gain and loss of electrons; add half-reactions 3 (2 OH - + SO3 2- ¡ SO4 2- + H 2O + 2 e -) 2 (MnO4 - + 2 H 2O + 3 e - ¡ MnO2 + 4 OH -) H 2O + 2 MnO4 - + 3 SO3 2- ¡ 2 MnO2 + 3 SO4 2- + 2 OH 3 H 2O, 4 OH -, and 6 e - canceled from each side

ClO2 + SbO2 - ¡ ClO2 - + Sb(OH)6 Step 1

(basic solution)

Write half-reaction equations. Balance except H and O. SbO2 - ¡ Sb(OH)6 ClO2 ¡ ClO2 -

Step 2

Balance H and O using H 2O and H + 4 H 2O + SbO2 - ¡ Sb(OH)6 - + 2 H + ClO2 ¡ ClO2 -

Step 3

Add OH - ions to both sides (same number as H + ions) 2 OH - + 4 H 2O + SbO2 - ¡ Sb(OH)6 - + 2 H + + 2 OH ClO2 ¡ ClO2 - 272 -

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Step 4

Combine H + and OH - to form H 2O; cancel H 2O where possible 2 OH - + 4 H 2O + SbO2 - ¡ Sb(OH)6 - + 2 H 2O ClO2 ¡ ClO2 2 OH - + 2 H 2O + SbO2 - ¡ Sb(OH)6 -

Step 5

(2 H 2O cancelled)

Balance electrically with electrons 2 OH - + 2 H 2O + SbO2 - ¡ Sb(OH)6 - + 2 e ClO2 + e - ¡ ClO2 -

Step 6 and 7

(c)

Equalize gain and loss of electrons; add half-reactions 2 H 2O + 2 OH - + SbO2 - ¡ Sb(OH)6 - + 2 e 2 (ClO2 + e - ¡ ClO2 -) 2 H 2O + 2 ClO2 + 2 OH - + SbO2 - ¡ 2 ClO2 - + Sb(OH)6 -

Al + NO3 - ¡ NH 3 + Al(OH)4 Step 1

(basic solution)

Write half-reaction equations. Balance except H and O. Al ¡ Al(OH)4 NO3 - ¡ NH 3

Step 2

Balance H and O using H 2O and H + 4 H 2O + Al ¡ Al(OH)4 - + 4 H + 9 H + + NO3 - ¡ NH 3 + 3 H 2O

Step 3

Add OH - ions to both sides (same number as H + ions) 4 OH - + 4 H 2O + Al ¡ Al(OH)4 - + 4 H + + 4 OH 9 OH - + 9 H + + NO3 - ¡ NH 3 + 3 H 2O + 9 OH -

Step 4

Combine H + and OH - to form H 2O; cancel H 2O where possible 4 OH - + 4 H 2O + Al ¡ Al(OH)4 - + 4 H 2O 9 H 2O + NO3 - ¡ NH 3 + 3 H 2O + 9 OH 4 OH - + Al ¡ Al(OH)4 -

6 H 2O + NO3 ¡ NH 3 + 9 OH Step 5

(4 H 2O cancelled) -

Balance electrically with electrons 4 OH - + Al ¡ Al(OH)4 - + 3 e 6 H 2O + NO3 - + 8 e - ¡ NH 3 + 9 OH - 273 -

(3 H 2O cancelled)

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Step 6 and 7

(d)

Equalize gain and loss of electrons; add half-reactions 8 (4 OH - + Al ¡ Al(OH)4 - + 3 e -) 3 (6 H 2O + NO3 - + 8 e - ¡ NH 3 + 9 OH -) 8 Al + 3 NO3 - + 18 H 2O + 5 OH - ¡ 3 NH 3 + 8 Al(OH)4 27 OH - and 24 e - canceled from each side

P4 ¡ HPO3 2- + PH 3 Step 1

(basic solution)

Write half-reaction equations. Balance except H and O. P4 ¡ 4 HPO3 2P4 ¡ 4 PH 3

Step 2

Balance H and O using H 2O and H + 12 H 2O + P4 ¡ 4 HPO3 2- + 20 H + 12 H + + P4 ¡ 4 PH 3

Step 3

Add OH - ions to both sides (same number as H + ions) 20 OH - + 12 H 2O + P4 ¡ 4 HPO3 2- + 20 H + + 20 OH 12 OH - + 12 H + + P4 ¡ 4 PH 3 + 12 OH -

Step 4

Combine H + and OH - to form H 2O; cancel H 2O where possible 20 OH - + 12 H 2O + P4 ¡ 4 HPO3 2- + 20 H 2O 12 H 2O + P4 ¡ 4 PH 3 + 12 OH 20 OH - + P4 ¡ 4 HPO3 2- + 8 H 2O (12 H 2O cancelled)

Step 5

Balance electrically with electrons 20 OH - + P4 ¡ 4 HPO3 2- + 8 H 2O + 12 e 12 H 2O + P4 + 12 e - ¡ 4 PH 3 + 12 OH -

Step 6 and 7

(e)

Loss and gain of electrons are equal; add half-reactions 8 OH - + 4 H 2O + 2 P4 ¡ 4 HPO3 2- + 4 PH 3 Divide equation by 2 4 OH - + 2 H 2O + P4 ¡ 2 HPO3 2- + 2 PH 3

Al + OH - ¡ Al(OH)4 - + H 2 Step 1

(basic solution)

Write half-reaction equations. Balance except H and O. Al ¡ Al(OH)4 OH - ¡ H 2

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Step 2

Balance H and O using H 2O and H + 4 H 2O + Al ¡ Al(OH)4 - + 4 H + 3 H + + OH - ¡ H 2 + H 2O

Step 3

Add OH - ions to both sides (same number as H + ions) 4 OH - + 4 H 2O + Al ¡ Al(OH)4 - + 4 H + + 4 OH 3 OH - + 3 H + + OH - ¡ H 2 + H 2O + 3 OH -

Step 4

Combine H + and OH - to form H 2O; cancel H 2O where possible 4 OH - + 4 H 2O + Al ¡ Al(OH)4 - + 4 H 2O 3 H 2O + OH - ¡ H 2 + H 2O + 3 OH 4 OH - + Al ¡ Al(OH)4 -

2 H2O + OH ¡ H2 + 3 OH Step 5

(4 H 2O cancelled) -

(1 H 2O cancelled)

Balance electrically with electrons 4 OH - + Al ¡ Al(OH)4 - + 3 e 2 H2O + OH- + 2 e- ¡ H2 + 3 OH-

Step 6 and 7

17.

(a)

Equalize gain and loss of electrons; add half-reactions 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

IO3 - + I - ¡ I 2 Step 1

(acidic solution)

Write half-reaction equations. Balance except H and O. 2 IO3 - ¡ I 2 2 I - ¡ I2

Step 2

Balance H and O using H 2O and H + 12 H + + 2 IO3 - ¡ I 2 + 6 H 2O 2 I - ¡ I2

Step 3

Balance electrically with electrons 12 H + + 2 IO3 - + 10 e - ¡ I 2 + 6 H 2O 2 I - ¡ I2 + 2 e -

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Step 4 and 5

Equalize the loss, and gain of electrons; add the half-reaction. 12 H + + 2 IO3 - + 10 e - ¡ I 2 + 6 H 2O 5 (2 I - ¡ I 2 + 2 e -) 12 H + + 2 IO3 - + 10 I - ¡ 6 I 2 + 6 H 2O

(b)

Mn2+ + S2O8 2- ¡ MnO4 - + SO4 2Step 1

(acid solution)

Write half-reaction equations. Balance except H and O Mn2+ ¡ MnO4 S2O8 2- ¡ 2 SO4 2-

Step 2

Balance H and O using H 2O and H + 4 H 2O + Mn2+ ¡ MnO4 - + 8 H + S2O8 2- ¡ 2 SO4 2-

Step 3

Balance electrically with electrons 4 H 2O + Mn2+ ¡ MnO4 - + 8 H + + 5 e 2 e - + S2O8 2- ¡ 2 SO4 2-

Step 4 and 5

Equalize the loss and gain of electrons; add the half-reactions 2 (4 H 2O + Mn2+ ¡ MnO4 - + 8 H + + 5 e -) 5 (2 e - + S2O8 2- ¡ 2 SO4 2-) 2 Mn2+ + 5 S2O8 2- + 8 H 2O ¡ 2 MnO4 - + 10 SO4 2- + 16 H + Each side has 2 Mn, 10 S, 16 H, and 48 O and a -6 charge.

(c)

Co(NO2)6 3- + MnO4 - ¡ Co 2+ + Mn2+ + NO3 Step 1

(acidic solution)

Write half-reaction equations. Balance except H and O. Co(NO2)6 3- ¡ Co 2+ + 6 NO3 MnO4 - ¡ Mn2+

Step 2

Balance H and O using H 2O and H + 6 H 2O + Co(NO2)6 3- ¡ Co 2+ + 6 NO3 - + 12 H + 8 H + + MnO4 - ¡ Mn2+ + 4 H 2O

Step 3

Balance electrically with e 6 H 2O + Co(NO2)6 3- ¡ Co 2+ + 6 NO3 - + 12 H + + 11 e 5 e - + 8 H + + MnO4 - ¡ Mn2+ + 4 H 2O - 276 -

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Step 4

Equalize the loss and gain of electrons. 5 (6 H 2O + Co(NO2)6 3- ¡ Co 2+ + 6 NO3 - + 12 H + + 11 e -)

Step 5

11 (5 e - + 8 H + + MnO4 - ¡ Mn2+ + 4 H 2O) Add the half-reactions

5 Co(NO2)6 3- + 11 MnO4 - + 28 H + ¡ 5 Co 2+ + 30 NO3 - + 11 Mn2+ + 14 H 2O Each side has 5 Co, 30 N, 11 Mn, 28 H, 104 O and a + 2 charge. 18.

(a)

Mo 2O3 + MnO4 - ¡ MoO3 + Mn2+ (acid solution) Step 1 Write half-reactions equations. Balance except H and O Mo 2O3 ¡ 2 MoO3 Step 2

MnO4 - ¡ Mn2+ Balance H and O using H 2O and H + 3 H 2O + Mo 2O3 ¡ 2 MoO3 + 6 H +

Step 3

8 H + + MnO4 - ¡ Mn2+ + 4 H 2O Balance electrically with electrons 3 H 2O + Mo 2O3 ¡ 2 MoO3 + 6 H + + 6 e -

Steps 4 and 5

5 e - + 8 H + + MnO4 - ¡ Mn2+ + 4 H 2O Equalize the loss and gain of electrons; add the half-reactions. 5 (3 H 2O + Mo 2O3 ¡ 2 MoO3 + 6 H + + 6 e -) 6 (5 e - + 8 H + + MnO4 - ¡ Mn2+ + 4 H 2O) 5 Mo 2O3 + 6 MnO4 - + 18 H + ¡ 10 MoO3 + 6 Mn2+ + 9 H 2O

(b)

BrO - + Cr(OH)4 - ¡ Br - + CrO4 2Step 1

(basic solution)

Write half-reaction equation. Balance except H and O BrO - ¡ Br Cr(OH)4 - ¡ CrO4 2-

Step 2

Balance H and O using H 2O and H + 2 H + + BrO - ¡ Br - + H 2O Cr(OH)4 - ¡ CrO4 2- + 4 H +

Step 3

Add OH - ions to both sides (same number as H +) 2 OH - + 2 H + + BrO - ¡ Br - + H 2O + 2 OH 4 OH - + Cr(OH)4 - ¡ CrO4 2- + 4 H + + 4 OH - 277 -

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Step 4

Combine H + and OH - to form H 2O; cancel H 2O where possible 2 H 2O + BrO - ¡ Br - + H 2O + 2 OH 4 OH - + Cr(OH)4 - ¡ CrO4 2- + 4 H 2O H 2O + BrO - ¡ Br - + 2 OH -

Step 5

(1 H 2O cancelled)

Balance electrically with electrons 2 e - + H 2O + BrO - ¡ Br - + 2 OH 4 OH - + Cr(OH)4 - ¡ CrO4 2- + 4 H 2O + 3 e -

Steps 6 and 7

(c)

Equalize loss and gain of electrons; add the half-reactions 3 (2 e - + H 2O + BrO - ¡ Br - + 2 OH -) 2 (4 OH - + Cr(OH)4 - ¡ CrO4 2- + 4 H 2O + 3 e -) 3 BrO - + 2 Cr(OH)4 - + 2 OH - ¡ 3 Br - + 2 CrO4 2- + 5 H 2O

S2O3 2- + MnO4 - ¡ SO4 2- + Mn2+ Step 1

(basic solution)

Write half-reaction equations. Balance except H and O. S2O3 2- ¡ 2 SO4 2MnO4 - ¡ MnO2

Step 2

Balance H and O using H 2O and H + 5 H 2O + S2O3 2- ¡ 2 SO4 2- + 10 H + 4 H+ + MnO4 - ¡ MnO2 + 2 H2O

Step 3

Add OH - ions to both sides (same number as H +) 10 OH - + 5 H 2O + S2O3 2- ¡ 2 SO4 2- + 10 H + + 10 OH 4 OH- + 4 H+ + MnO4 - ¡ MnO2 + 2 H2O + 4 OH-

Step 4

Combine H + and OH - to form H 2O; cancel H 2O where possible 10 OH - + 5 H 2O + S2O3 2- ¡ 2 SO4 2- + 10 H 2O 4 H2O + MnO4 - ¡ MnO2 + 2 H2O + 4 OH-

Step 5

10 OH - + S2O3 2- ¡ 2 SO4 2- + 5 H 2O

(5 H 2O cancelled)

2 H2O + MnO4 - ¡ MnO2 + 4 OH-

(2 H2O cancelled)

Balance electrically with electrons 10 OH - + S2O3 2- ¡ 2 SO4 2- + 5 H 2O + 8 e 3 e- + 2 H2O + MnO4 - ¡ MnO2 + 4 OH- 278 -

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Step 6 and 7

Equalize loss and gain of electrons; add half-reactions. 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 Each side has 6 S, 8 Mn, 2 H, 42 O and a -14 charge.

19. – Voltage source

+ Anode (+)

Cathode (–)

Br– H3O+ Solution of HBr

20.

(c)

Pb + SO4 2- ¡ PbSO4 + 2 e PbO2 + SO4 2- + 4 H + + 2 e - ¡ PbSO4 + 2 H 2O The first reaction is oxidation (Pb 0 is oxidized to Pb 2+). The second reaction is reduction (Pb 4+ is reduced to Pb 2+). The first reaction (oxidation) occurs at the anode of the battery.

(a) (b) (c)

The oxidizing agent is KMnO4 . The reducing agent is HCl. 5 moles of electrons 5 e - + Mn7+ ¡ Mn2+

(a) (b)

21.

¢

5 mol e 6.022 * 1023 e electrons ≤¢ ≤ = 3.011 * 1024 mol KMnO4 mol e mol KMnO4

22.

Zinc is a more reactive metal than copper so when corrosion occurs the zinc preferentially reacts. Zinc is above hydrogen in the Activity series of metals; copper is below hydrogen.

23.

3 Ag + 4 HNO3 ¡ 3 AgNO3 + NO + 2 H 2O

(balanced)

g Ag ¡ mol Ag ¡ mol NO (25.0 g Ag)a

1 mol Ag 1 mol NO ba b = 0.0772 mol NO 107.9 g Ag 3 mol Ag - 279 -

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24.

3 Cl 2 + 6 KOH ¡ KClO3 + 5 KCl + 3 H 2O mol KClO3 ¡ mol Cl 2 ¡ L Cl 2 (0.300 mol KClO3) ¢

25.

3 mol Cl2 22.4 L b = 20.2 L Cl2 ≤a 1 mol KClO3 1 mol

5 H 2O2 + 2 KMnO4 + 3 H 2SO4 ¡ 5 O2 + 2 MnSO4 + K 2SO4 + 8 H 2O mL H 2O2 ¡ g H 2O2 ¡ mol H 2O2 ¡ mol KMnO4 ¡ g KMnO4 (100. mL H 2O2 solution)a a

26.

9.0 g H 2O2 1.031 g 1 mol b¢ b ≤a mL 100. g H 2O2 solution 34.02 g

2 mol KMnO4 158.0 g ba b = 17 g KMnO4 5 mol H 2O2 mol

Cr2O7 2- + 3 H 3AsO3 + 8 H + ¡ 2 Cr 3+ + 3 H 3AsO4 + 4 H 2O g H 3AsO3 ¡ mol H 3AsO3 ¡ mol Cr2O7 2- ¡ mL Cr2O7 2(5.00 g H 3AsO3)a

27.

1 mol Cr2O7 21 mol 1000 mL b¢ b = 66.2 mL of 0.200 M K 2Cr2O7 ≤a 125.9 g 3 mol H 3AsO3 0.200 mol

Cr2O7 2- + 6 Fe 2+ + 14 H + ¡ 2 Cr 3+ + 6 Fe 3+ + 7 H 2O mL FeSO4 ¡ mol FeSO4 ¡ mol Cr2O7 2- ¡ mL Cr2O7 2(60.0 mL FeSO4)a

0.200 mol 1 mol Cr2O7 21000 mL b¢ b ≤a 1000 mL 6 mol FeSO4 0.200 mol = 10.0 mL of 0.200 M K2Cr2O7

28.

2 Al + 2 OH - + 6 H 2O ¡ 2 Al(OH)4 - + 3 H 2 g Al ¡ mol Al ¡ mol H 2 (100.0 g Al)a

29.

(a) (b)

1 mol Al 3 mol H 2 b¢ ≤ = 5.560 mol H 2 26.98 g 2 mol Al

Cu+ ¡ Cu2+ is an oxidation, but when electrons are gained reduction should occur. Cu+ + e - ¡ Cu0 or Cu+ ¡ Cu2+ + e 2+ When Pb is reduced, it requires two individual electrons. Pb 2+ + 2 e - ¡ Pb 0. An electron has only a single negative charge (e -).

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30.

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

31.

A(s) + B2+(aq) ¡ NR B2+ cannot take e - from A A(s) + C +(aq) ¡ NR C + cannot take e - from A + 2+ D(s) + 2 C (aq) ¡ 2C(s) + D (aq) C + takes e - from D B(s) + D 2+(aq) ¡ D(s) + B2+(aq) D 2+ takes 2 e - from B Therefore, B2+ is least able to attract e -, then D 2+, then C +, then A+

32.

Sn4+ can only be an oxidizing agent.

Sn4+ + 2 e - ¡ Sn2+ Sn4+ + 4 e - ¡ Sn0

Sn0 can only be a reducing agent.

Sn0 ¡ Sn2+ + 2 e Sn0 ¡ Sn4+ + 4 e -

Sn2+ can be both oxidizing and reducing.

Sn2+ + 2 e - ¡ Sn0 Sn2+ ¡ Sn4+ + 2 e -

33.

Mn(OH)2 MnF3 MnO2 K 2MnO4 KMnO4

34.

Equations (a) and (b) represent oxidation (a) Mg ¡ Mg 2+ + 2 e (b) SO2 ¡ SO3 ; (S4+ ¡ S6+ + 2 e -)

35.

(a)

MnO2 + 2 Br - + 4 H + ¡ Mn2+ + Br2 + 2 H 2O

(b)

mL Mn2+ ¡ mol Mn2+ ¡ mol MnO2 ¡ g MnO2

+2 +3 +4 +6 +7

KMnO4 is the best oxidizing agent of the group, since its greater oxidation number (+7) makes it very attractive to electrons.

(100.0 mL Mn2+)a (c)

(oxidizing) (reducing)

1 mol MnO2 86.94 g 0.05 mol b¢ b = 0.4 g MnO2 ≤a 1000 mL mol 1 mol Mn2+

1 mol Br2 0.05 mol b = 0.005 mol Br2 ba 1000 mL 1 mol Mn2+ nRT PV = nRT V = P 0.005 mol 0.0821 L atm V = a ba b (323 K) = 0.09 L Br2 vapor 1.4 atm mol K (100.0 mL Mn2+)a

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F2 + 2 Cl- ¡ 2 F - + Cl 2 Br2 + Cl- ¡ NR I 2 + Cl- ¡ NR Br2 + 2 I - ¡ 2 Br - + I 2

36.

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

37.

Mn(s) + 2 HCl(aq) ¡ Mn2+(aq) + H 2(g) + 2 Cl-(aq)

38.

4 Zn + NO3 - + 10 H + ¡ 4 Zn2+ + NH 4 + + 3 H 2O

39.

Equation 1

2

3

See Exercise 13(a). 4

5

a C oxidized

S oxidized

N oxidized

S oxidized

O2 2- oxidized

b

O2 reduced

N reduced

Cu reduced

O2 2- reduced

O2 2- reduced

c

O2 , O.A.

HNO3 , O.A.

CuO, O.A.

H 2O2 , O.A.

H 2O2 , O.A.

d

C3H 8 , R.A.

H 2S, R.A.

NH 3 , R.A.

Na 2SO3 , R.A.

H 2O2 , R.A.

N 3- ¡ N2 0

S4+ ¡ S6+

O2 2- ¡ O2 0

O2 2- ¡ O 2-

O2 2- ¡ O 2-

223 +

¡ C 4+ S2- ¡ S0

e

C

f

O 0 ¡ O 2-

N 5+ ¡ N 2+ Cu2+ ¡ Cu0

O.A. = Oxidizing agent R.A. = Reducing agent 40.

Pb + 2 Ag + ¡ 2 Ag + Pb 2+ (a) Pb is the anode (b) Ag is the cathode (c) Oxidation occurs at Pb (anode) (d) Reduction occurs at Ag (cathode) (e) Electrons flow from the lead electrode through the wire to the silver electrode. (f) Positive ions flow through the salt bridge towards the negatively charged strip of silver; negative ions flow toward the positively charged strip of lead. Salt bridge Ag

Pb

Pb2+ NO–3

Ag+ NO–3

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41.

8 KI + 5 H 2SO4 ¡ 4 I 2 + H 2S + 4 K 2SO4 + 4 H 2O start with grams I 2 and work towards g KI g I 2 ¡ mol I 2 ¡ mol KI ¡ g KI (2.79 g I 2)a a

42.

1 mol 8 mol KI 166.0 g b¢ b = 3.65 g KI in sample ≤a 253.8 g 4 mol I 2 mol

3.65 g KI b11002 = 91.3% KI 4.00 g sample

3 Ag + 4 HNO3 ¡ 3 AgNO3 + NO + 2 H 2O mol Ag ¡ mol NO (0.500 mol Ag)a PV = nRT P = (744 torr)a

1 mol NO b = 0.167 mol NO 3 mol Ag V =

nRT P

1 atm b = 0.979 atm 760. torr

T = 301 K V =

(0.167 mol NO)(0.0821 L atm>mol K)(301 K) = 4.22 L NO (0.979 atm)

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