CHAPTER 23 THE TRANSITION ELEMENTS

1136

CHAPTER 23

THE TRANSITION ELEMENTS

PRACTICE EXAMPLES

1A (E) (a) 2Cu O should form. 2 2 2 22 Cu S s + 3 O g 2 Cu O s + 2 SO g

(b) W(s) is the reduction product. 3 2 2WO s + 3 H g W s + 3 H O g

(c) Hg(l) forms. 22 HgO s 2 Hg l + O g

1B (E) (a) 2SiO s is the oxidation product of Si.

2 3 23 Si s + 2 Cr O s 3 SiO s + 4 Cr s

(b) Roasting is simply heating in air. ,Air2 3 23

2 Co OH s Co O s + 3H O g

(c) 2MnO s forms; (acidic solution). 2+ +2 2Mn aq + 2H O(l) MnO s + 4H aq + 2e

2A (M) We write and combine the half-equations for oxidation and reduction. If o 0E , the reaction is spontaneous.

Oxidation: 3+ 2+ + o2{V aq + H O(l) VO aq + 2H aq + e } 3 = 0.337VE

Reduction: + o3 2NO aq + 4H aq + 3e NO g + 2H O = +0.956VE

Net: - 3+ 2+ + o3 2 cellNO aq + 3V aq + H O(l) NO g + 3VO aq + 2H aq = +0.619VE

Because the cell potential is positive, nitric acid can be used to oxidize 3+ 2+V aq to VO aq

under standard conditions. 2B (M) The reducing couple must have a half-cell potential of such a size and sign that a positive

sum results when this half-cell potential is combined with o 2 2{VO ( )|V ( )} 0.041VE aq aq (this is the weighted average of the VO2+ |V3+ and V3+|V2+

reduction potentials) and a negative sum must be produced when this half-cell potential is combined with o 2{V (aq)|V(s)}= 1.13VE .

So, -Eo for the couple must be > –0.041 V and < + 1.13 V (i.e., it cannot be more positive than 1.13V, nor more negative than –0.041 V). Some possible reducing couples from Table 20-1 are:

o 3 2+{Cr (aq)|Cr (aq)} 0.42 VE ; o 2+{Fe (aq)|Fe(s)} 0.440 VE o 2+{Zn (aq)|Zn(s)} 0.763 VE . Thus Fe(s), Cr2+(aq), and Zn(s) will do the job.

INTEGRATIVE EXAMPLE

3A (M) Because the reduction potential for PtCl62− is more positive than that of V3+, the

following half-reactions occur spontaneously in the cell: Oxidation: [ V2+ + e− → V3+ ] × 2 E° = +0.255 V Reduction: PtCl6

2− + 2 e− → PtCl42− + 2 Cl− E° = +0.68 V

Overall: 2 V2+ + PtCl62− → 2 V3+ + PtCl4

2− + 2 Cl− E° = 0.94 V

Chapter 23: The Transition Elements

1137

The equilibrium constant for the overall reaction is 31

2(0.94)

0.059210 5.7 10K . For the reverse reaction to be spontaneous, we need Q > 5.7 × 1031. Because Q = [V3+]2 [PtCl4

2−] [Cl−]2/[V2+] [PtCl6

2−], the formation of PtCl62− is favored by using a very low

concentration of V2+ and very high concentrations of V3+, PtCl42−, and Cl−. In practical

terms, though, the amount of PtCl62− that could be formed spontaneously would be small.

A quick calculation shows that starting from [V3+]o = [PtCl42−]o = [Cl−]o = 1. M and [V2+]

= [PtCl62−]o = 0, the equilibrium concentration of PtC6

2− would be about 1.6 × 10−11 M. So, in practical terms, an external voltage source would be required to make a significant amount of PtCl6

2− from V3+, PtCl42−, and Cl−.

3B (M) The disproportionation reaction is 3 Ti2+ → 2 Ti3+ + Ti and E° = −1.261 V. Because E° < 0, the reaction is not spontaneous under standard conditions.

EXERCISES

Properties of the Transition Elements

1. (E) (a) Ti [Ar] 3d 4s (b) V3+ [Ar] 3d

4s

(c) Cr2+ [Ar] 3d 4s (d) Mn4+ [Ar] 3d 4s

(e) Mn2+ [Ar] 3d 4s (f) Fe3+ [Ar] 3d 4s

2. (E) We first give the orbital diagram for each of the species, and then count the number of unpaired electrons.

Fe [Ar] 3d 4s 4 unpaired electrons

Sc3+ [Ar] 3d 4s 0 unpaired electrons

Ti2+ [Ar] 3d 4s 2 unpaired electrons

Mn4+ [Ar] 3d 4s 3 unpaired electrons

Cr [Ar] 3d 4s 6 unpaired electrons

Cu2+ [Ar] 3d 4s 1 unpaired electron

Finally, arrange in order of decreasing number of unpaired e-: Cr Fe Mn Ti Cu Sc 4+ 2+ 2+ 3+

3. (M) A given main-group metal typically displays just one oxidation state, usually equal to its family number in the periodic table. Exceptions are elements such as Tl ( +1and +3), Pb


1138

( +2 and +4 ), and Sn ( +2 and +4 ), for which the lower oxidation state represents a pair of s electrons not being ionized (a so-called “inert pair”).

Main group metals do not form a wide variety of complex ions, with 3+Al , 2+Sn , 4+Sn , and 2+Pb being major exceptions. On the other hand, most transition metal ions form an extensive

variety of complex ions. Most compounds of main group metals are colorless; exceptions occur when the anion is colored. On the other hand, many of the compounds of transition metal cations are colored, due to d-d electron transitions. Virtually every main-group metal cation has no unpaired electrons and hence is diamagnetic. On the other hand, many transition metals cations have one or more unpaired electrons and therefore are paramagnetic.

4. (M) As we proceed from Sc to Cr, the valence electron configuration has an increasing number of unpaired electrons, which are capable of forming bonds to adjacent atoms. As we continue beyond Cr, however, these electrons become paired, and the resulting atoms are less able to form bonds with their neighbors. Also, smaller atoms tend to bond more tightly, making their metallic agglomerations more difficult to melt. (i.e., the smaller the metal atom, the higher the melting point). This trend of increasing melting point with decreasing size is clearly visible for the first transition series from Sc to Zn.

5. (M) When an electron and a proton are added to a main group element to create the element of next highest atomic number, it is the electron that influences the radius. The electron is added to the outermost shell (n value), which is farthest from the nucleus. Moreover, the added electron is well shielded from the nucleus and hence it is only weakly attracted to the nucleus. Thus, this electron billows out and, as a result, has a major influence on the size of the atom. However, when an electron and a proton are added to a transition metal atom to create the atom of next highest atomic number, the electron it is not added to the outermost shell. The electron is added to the d-orbitals, which are one principal quantum number lower than the outermost shell. Also, the electron in the same subshell as the added electron offers little shielding. Thus it has small effect on the size of the atom.

6. (M) The reason why the radii of Pd (138 pm) and Pt (139 pm) are so similar, and yet so

different from the radius of Ni (125 pm) is because the lanthanide series intrudes between Pd and Pt. Because of the lanthanide contraction, elements in the second transition row are almost identical in size to their congeners (family members) in the third transition row.

7. (M) Of the first transition series, manganese exhibits the greatest number of different

oxidation states in its compounds, namely, every state from +1 to +7 . One possible explanation might be its 3 45 2d s electron configuration. Removing one electron produces an electron configuration ( 3 45 1d s ) with two half-filled subshells, removing two produces one with a half-filled and an empty subshell. Then there is no point of semistability until the remaining five d electrons are removed. These higher oxidation states all are stabilized by being present in oxides ( MnO2 ) or oxoanions (e.g., MnO4

- ). 8. (M) At the beginning of the series, there are few electrons beyond the last noble gas that

can be ionized and thus the maximum oxidation state is limited. Toward the end of the series, many of the electrons are paired up or the d subshell is filled, and thus a somewhat stable situation would be disrupted by ionization.


1139

9. (M) The greater ease of forming lanthanide cations compared to forming transition metal

cations, is due to the larger size of lanthanide atoms. The valence (outer shell) electrons of these larger atoms are further from the nucleus, less strongly attracted to the positive charge of the nucleus in diffuse f-orbitals that are do not penetrate effectively and are very effectively shielded by the core electrons. As a result, they are removed much more readily.

10. (M) This is not a simple straightforward question. There are a number of factors to consider.

Proceeding across the first transition series, electrons are added to the 3d subshell which is a shell that is beneath the 4s subshell, the outer most subshell in the valence shell occupied by one or two electrons. The metallic character of the first transition element, Sc, is rather similar to that of Ca (e.g., E° = 2.03 V compared to 2.84 V). Following this, there is a fairly regular rise in E°, (e.g., 1.63 V for Ti , 1.13 V for V…), reaching a maximum of E° = 0.340 V with Cu. These trends can be related to the regular increase in Zeff and other aspects of electron configurations (half filled shell for Mn2+ and loss of half filled shell for Cr2+). Complicating matters is the fact that these ions will have different aqueous coordination numbers and geometries.

In the lanthanide series, it is the 4f subshell that fills at the same time that the 5d subshell is mostly vacant and the 6s subshell is filled. Differences in electron configurations are essentially confined to a subshell two shells removed from the outermost valence shell, which translates into close similarities among all the elements in the series, including E° values.

Recall that the lanthanide series is an inner transition series—a transition series within another transition series. This inner transition series runs its course between La (E° = 2.38 V) and Hf (E° = 1.70 V). The difference between these two E° values is about 0.7 V, which is comparable to the 0.4 V difference between the first (Sc) and second (Ti) members of the first transition series.

Reactions of Transition Metals and Their Compounds 11. (M) (a) 4TiCl g + 4 Na l Ti s + 4NaCl l

(b) 2 3 2 3Cr O s + 2Al s 2 Cr l + Al O s

(c) Ag s + HCl aq no reaction

(d) 2 2 7 2 4 2K Cr O aq + 2KOH aq 2 K CrO aq + H O(l)

(e) 2MnO s + 2 C s Mn l + 2CO g

12. (M) (a) Cr s + 2HCl aq CrCl2 aq + H2 g Virtually any first period transition

metal except Cu can be substituted for Cr.

(b) 2 3 2 4Cr O s + 2 OH aq + 3 H O(l) 2 Cr OH aq

The oxide must be amphoteric. Thus, 2 3Sc O , TiO2 , ZrO2 , and ZnO could be

substituted for 2 3Cr O .


1140

(c) 3 22 La s + 6 HCl aq 2 LaCl aq + 3 H g

Any lanthanide or actinide element can be substituted for lanthanum.

13. (M) (a) + 3+23

Sc OH s + 3H aq Sc aq + 3 H O(l)

(b) 2+ 3+4 2 23 Fe aq + MnO aq + 2 H O l 3 Fe aq + MnO s + 4 OH (aq)

(c) 2 2 3 22 KOH l + TiO s K TiO s + H O g

(d) 2 4 4 2 2Cu s + 2 H SO conc,aq CuSO aq + SO g + 2 H O(l)

14. (M) (a) electrolysis2 3 3 6 22 Sc O l,in Na ScF + 3 C s 4 Sc l + 3 CO g

[By analogy with the equation for the electrolytic production of Al] (b) 2+

2Cr s + 2 HCl aq Cr aq + 2 Cl aq + H g

(c) 2+ + 3+2 24 Cr aq + O g + 4 H aq 4 Cr aq + 2 H O(l)

(d) 3 3 2 2Ag s + 2 HNO aq AgNO aq + NO g + H O(l)

15. (M) We write some of the following reactions as total equations rather than as net ionic equations so that the reagents used are indicated.

(a) 2 2FeS s + 2 HCl aq FeCl aq + H S g

2+ + 3+2 2

3+

3

4 Fe aq + O g + 4H aq 4Fe aq + 2 H O(l)

Fe aq + 3 OH aq Fe OH s

(b) 3 2 2 2BaCO s + 2 HCl aq BaCl aq + H O(l) + CO g

2 2 2 7 4 22 BaCl aq + K Cr O aq + 2 NaOH aq 2 BaCrO s + 2 KCl aq + 2 NaCl aq + H O(l)

16. (M) (a) (i) 2+2CuO s 2H (aq) Cu aq + H O l

(ii) 2+2Cu aq 2OH (aq) Cu(OH) s

(b) 4 2 7 2 2 2 32NH Cr O s N g + 4 H O g + Cr O s

2 3 3 2Cr O s + 6 HCl aq CrCl aq + 3 H O(l)

Extractive Metallurgy 17. (E) HgS(s) + O2(g) Hg(l) + SO2(g)

4 HgS(s) + 4 CaO(s) 4 Hg(l) + 3 CaS(s) + CaSO4(s) 18. (E) (a) C(s) + O2(g) CO2(g) S = 213.7 J K-1 - (5.74 J K-1 + 205.1 J K-1) S = 2.86 J K-1

Since S 0, G will be relatively constant with temperature.


1141

(b) 2 CO(g) + O2(g) 2 CO2(g) S = (2(213.7 J K-1) - (205.1 J K-1 + 2(197.7 J K-1)) S = -173.1 J K-1 Since S is less than zero, G will increase (become less negative or more positive) as temperature increases.

19. (M) The plot of Go versus T will consist of three lines of increasing positive slope.

The first line is joined to the second line at the melting point for Ca(s), while the second

line is joined to the third at the boiling point for Ca(l).

2 Ca(s) + O2(g) 2 CaO(s) Hf = -1270.2 kJ

S = 2(39.75 J K-1) – [2(41.42 J K-1) + 205.1 J K-1] = -208.4 J K-1

The graph should be similar to that for 2 Mg(s) + O2(g) 2 MgO(s). We expect a

positive slope with slight changes in the slope after the melting point (839 C) and boiling

point (1484 C) , mainly owing to changes in entropy. The plot will be below the

G line for 2 Mg(s) + O2(g) 2 MgO(s) at all temperatures. 20. (E) 2 Na2CrO4(s) + 3 C(s) + 4HCl(aq) Cr2O3(s) + 2H2O(l) + 3 CO(g) + 4NaCl(aq)

2 Cr2O3(s) + 3 Si(s) 4 Cr(s) + 3 SiO2(s)

Oxidation-Reduction

21. (E) (a) Reduction: VO2+ aq + 2H+ aq + e V3+ aq + H2O (l)

(b) Oxidation: Cr2+ aq Cr3+ aq + e

22. (E) (a) Oxidation: 24 23

Fe OH s + 5 OH aq FeO aq + 4 H O(l) + 3e

(b) Reduction: 2Ag CN aq + e Ag s + 2 CN aq

23. (M) (a) First we need the reduction potential for the coupleVO2+ aq / V2+ aq . We will

use the half-cell addition method learned in Chapter 20.

+ + 2+ o2 2VO aq + 2 H aq + e VO aq + H O(l) = 1 +1.000 VG F

2+ + 3+ o2VO aq + 2 H aq + e V aq + H O(l) = 1 +0.337 VG F

3+ 2+ oV aq + e V aq = 1 0.255 VG F

+ + 2+ o o2 2VO aq + 4 H aq + 3 e V aq + 2 H O = 3G FE

o 1.000 V + 0.337 V 0.255 V= = + 0.361 V

3E


1142

We next analyze the oxidation-reduction reaction.

Oxidation: o2{2 Br aq Br l + 2e } 3 = 1.065 VE

Reduction: + + 2+ o2 2{VO aq + 4H aq + 3e V aq + 2H O(l)} 2 = +0.361VE

Net: + + 2+2 2 26 Br aq + 2 VO aq + 8 H aq 3 Br l + 2 V aq + 4 H O

ocell = 0.704VE

Thus, this reaction does not occur to a significant extent as written under standard conditions.

(b) Oxidation: 2+ 3+ oFe aq Fe aq + e = 0.771VE

Reduction: + + 2+ o2 2VO aq + 2 H aq + e VO aq + H O(l) = +1.000VE

Net: +2+ + 3+ 2+2 2Fe aq + VO aq + 2 H aq Fe aq + VO aq + H O(l)

ocell = +0.229VE

This reaction does occur to a significant extent under standard conditions.

(c) Oxidation: + o

2 2 2H O 2 H aq + 2 e + O g = 0.695 VE

Reduction: + 2+ o2 2MnO s + 4 H aq + 2 e Mn aq + 2 H O(l) = +1.23VE

Net: o+ 2+2 2 2 2 2 cellH O + MnO s + 2 H aq O g + Mn aq + 2 H O(l) = +0.54VE

Thus, this reaction does occur to a significant extent under standard conditions. 24. (D) When a species acts as a reducing agent, it is oxidized. From Appendix D, we obtain

the potentials for each of the following couples.

0 2 0 4 2{Zn (aq) / Zn(s)} 0.763V; {Sn (aq) / Sn (aq)} 0.154VE E

o2{I (s)/I (aq)} = 0.535 VE

Each of these potentials is combined with the cited reduction potential. If the resulting value of Ecell

o is positive, then the reducing agent will be effective in accomplishing the desired reduction, which we indicate with “yes”; if not, we write “no”.

(a) 2o 32 7{Cr O (aq) / Cr (aq)} = +1.33 VE

2o o 3 o 2cell 2 7

ocell

= {Cr O (aq) / Cr (aq)} {Zn (aq) / Zn(s)}

= +1.33 V + 0.763 V = +2.09 V Yes

E E E

E

2o o 3 o 4 2+cell 2 7

ocell

= {Cr O (aq) / Cr (aq)} {Sn (aq) / Sn (aq)}

= +1.33 V 0.154 V = +1.18 V Yes

E E E

E

2o o 3 ocell 2 7 2

ocell

= {Cr O (aq) / Cr (aq)} {I (s) / I (aq)}

= +1.33 V 0.535 V = +0.80 V Yes

E E E

E


1143

(b) o 3+ 2+{Cr (aq)/Cr (aq)} = 0.424 VE

o o 3+ 2+ o 2+cell

ocell

= {Cr (aq)/Cr (aq)} {Zn (aq)/Zn(s)}

= 0.424 V + 0.763 V = +0.339 V Yes

E E E

E

o o 3+ 2+ o 4+ 2+cell

ocell

= {Cr (aq)/Cr (aq)} {Sn (aq)/Sn (aq)}

= 0.424 V 0.154 V = 0.578 V No

E E E

E

o o 3+ 2+ ocell 2

ocell

= {Cr (aq)/Cr (aq)} {I (s)/I (aq)}

= 0.424 V 0.535 V = 0.959 V No

E E E

E

(c) 2o4 2{SO (aq)/SO (g)} = +0.17 VE

2o o o 2+cell 4 2

ocell

= {SO (aq)/SO (g)} {Zn (aq)/Zn(s)}

== +0.17 V + 0.763 V = +0.93 V Yes

E E E

E

o o 2 o 4+ 2+cell 4 2

ocell

= {SO (aq)/SO (g)} {Sn (aq)/Sn (aq)}

= +0.17 V 0.154 V = +0.02 V Yes (barely)

E E E

E

2o o ocell 4 2 2

ocell

= {SO (aq)/SO (g)} {I (s)/I (aq)}

= +0.17 V 0.535 V = 0.37 V No

E E E

E

25. (M) The reducing couple that we seek must have a half-cell potential of such a size and

sign that a positive sum results when this half-cell potential is combined with o 2 3{VO ( )|V ( )} 0.337 VE aq aq and a negative sum must be produced when this

half-cell potential is combined with+o 3 2{V (aq)|V (aq)}= 0.255VE .

So, -Eo for the couple must be> –0.337 V and <+ 0.255 V (i.e. it cannot be more positive than 0.255V, or more negative than –0.337 V)

Some possible reducing couples from Table 20.1 are: o 2{Sn (aq)|Sn(s)} 0.137 VE ; o +

2{H (aq)|H (g)] 0.000 VE o 2+{Pb (aq)|Pb(s)] 0.125 VE ; thus Pb(s), Sn(s), H2(g) , to name but a few, will do the job.

26. (M) There are two methods that can be used to determine the MnO4

-/Mn2+ reduction potential. Method 1: MnO4

1.70 V MnO2 1.23 V Mn2+

EMnO4-/ Mn2+ =

3 (1.70 V) + 2 (1.23 V)

5= 1.512 V ~1.51 V

Method 2: MnO4- MnO4

2 MnO2 Mn3+ Mn2+

EMnO4-/ Mn2+ =

0.56 V + 2 (2.27 V) + 0.95 V +1.49 V

5= 1.508 V ~1.51 V

These answers compare favorably to Table 20.1, where EMnO4-/ Mn2+ = 1.51 V


1144

27. (M) Table D-4 contains the following data: 3+ 2+Cr /Cr reduction potential = -0.424 V, 2- 3+

2 7Cr O /Cr reduction potential = 1.33 V and 2+Cr /Cr reduction potential = -0.90 V.

By using the additive nature of free energies and the fact that G = -nFE, we can determine the two unknown potentials and complete the diagram.

(i) 2- 2+2 7Cr O /Cr : E =

3(1.33V) - 0.424V

4= 0.892 V

(ii) 3+Cr /Cr : E = 0.424V + 2(-0.90)V

3

= 0.74 V

28. (M) From Table 23.4 we are given: VO2+/VO2+ reduction potential = 1.000 V, VO2+/V3+

reduction potential = 0.337 V, V3+/V2+ reduction potential = -0.255 V and V2+/V reduction potential = -1.13 V. By taking advantage of the additive nature of free energies and the fact that G = -nFE, we can determine the three unknown potentials and complete the diagram.

(i) VO2+/V3+: E =

1.000V + 0.337 V

2= 0.669 V

(ii) VO2+/V2+: E =

1.000V + 0.337 V + (-0.255 V)

3= 0.361 V

(iii) VO2+/V: E =

1.000V + 0.337 V + (-0.255 V) + 2(-1.13 V)

5= 0.24 V

Chromium and Chromium Compounds 29. (M) Orange dichromate ion is in equilibrium with yellow chromate ion in aqueous

solution.

2 2 +2 7 2 4Cr O aq + H O(l) 2CrO aq + 2H aq

The chromate ion in solution then reacts with lead(II) ion to form a precipitate of yellow

lead(II) dichromate. 2+ 24 4Pb aq + CrO aq PbCrO s

PbCrO4 s will form until H+ from the first equilibrium increases to the appropriate

level and both equilibria are simultaneously satisfied. 30. (M) The initial dissolving reaction forms orange dichromate ion.

2+ 24 2 7 22 BaCrO s + 2 HCl aq 2 Ba aq + Cr O aq + 2 Cl aq + H O(l)

Dichromate ion is a good oxidizing agent, that is strong enough to oxidize Cl aq to

Cl2 g if the concentrations of reactants are high, the solution is acidic, and the product

Cl2 g is allowed to escape.

2 + 3+2 7 2 26Cl aq + Cr O aq +14H aq 3Cl g + 2Cr aq + 7H O(l)


1145

Chromium(III) can hydrolyze in aqueous solution to produce green Cr OH 4

aq , but

the solution needs to be alkaline (pH >7) for this to occur. A more likely source of the

green color is a complex ion such as Cr H2O 4Cl2

+.

31. (D) Oxidation: {Zn s Zn2+ aq + 2e} 3

Reduction: 2 + 3+2 7 2Cr O aq,orange +14H aq + 6e 2 Cr aq,green + 7 H O(l)

Net: 2 + 2+ 3+2 7 23 Zn s + Cr O aq +14 H aq 3 Zn aq + 2 Cr aq + 7 H O(l)

Oxidation: Zn s Zn2+ aq + 2e

Reduction: {Cr3+ aq,green + e Cr2+ aq,blue } 2

Net: 3+ 2+ 2+Zn s + 2 Cr aq Zn aq + 2 Cr aq

The green color is most likely due to a chloro complex of Cr3+, such as +

2 24Cr H O Cl .

Oxidation: {Cr2+ aq,blue Cr3+ aq,green + e} 4

Reduction: +2 2O g + 4 H aq + 4e 2 H O(l)

Net: 2+ + 3+2 24 Cr aq + O g + 4 H aq 4 Cr aq + 2 H O(l)

32. (M) CO2 g , as the oxide of a nonmetal, is an acid anhydride. Its function is to make the

solution acidic. A reasonable guess of the reactions that occur follows.

2 + 24 2 7 22 CrO aq + 2 H aq Cr O aq + H O(l)

+2 2 32 H O(l) + 2CO aq 2H aq + 2 HCO aq

2 24 2 2 2 7 32 CrO aq + 2 CO aq + H O(l) Cr O aq + 2 HCO aq

33. (M) Simple substitution into Equation (23.19) yields 22 7Cr O in each case. In fact, the

expression is readily solved for the desired concentration as follows: 222 214 +

2 7 4Cr O = 3.2 10 H CrO . In each case, we use the value of pH to determine

+ pHH = 10 .

(a) 2 14 6.62 2 22 7Cr O = 3.2 10 (10 ) (0.20) 0.74 M

(b) 2 22 14 8.85 52 7Cr O = 3.2 10 10 0.20 = 2.6 10 M


1146

34. (M) We use Equation (23.19) again:.

Cr2O72

CrO42

2 = 3.2 1014 H+ 2

= 3.2 1014 107.55 2 = 3.2 1014 2.8 108 2 = 0.254

22 2 4 2 4 4

4initial

2 4 2 4

1.505 g Na CrO 1 mol Na CrO 1 mol CrOCrO = = 0.0269 M

0.345 L soln 161.97 g Na CrO 1mol Na CrO

Reaction: 2 +

42CrO aq + 2H aq Cr2O72 aq + H2O (l)

Initial: 0.0269 M 82.8 10 (fixed) 0 M Changes: 2 Mx 0 + Mx Equil: 0.0269 2 Mx 82.8 10 (fixed) Mx

22 7 2

2 224

Cr O= 0.254 = = 0.254 0.000724 0.108 + 4

0.0269 2CrO

xx x x

x

2 2= 0.000184 0.0274 +1.016 1.016 1.0274 + 0.000184 = 0x x x x x 2 4 1.0274 1.0556 0.000736

= = = 1.0111 M, 0.00016 M2 2.032

b b acx

a

We choose the second root because the first root gives a negative CrO42 .

2 22 7 4Cr O = = 0.00016 M CrO = 0.0269 2 = 0.0266 Mx x

We carry extra significant figures to avoid a significant rounding error in this problem.

35. (E) Each mole of chromium metal plated out from a chrome plating bath (i.e., CrO3 and H2SO4) requires six moles of electrons.

mass 3600 s 3.4 C 1 mol e 1 mol Cr 52.00 g Cr

Cr =1.00 h = 1.10 g Cr1 hr 1 s 96485 C 6 mol e 1mol Cr

36. (M) First we compute the amount of Cr(s) deposited. 4 2

2 22 3

1 cm 7.14 g Cr 1 mol Cr10 cmmol Cr = 0.0010 mm 0.375 m = 5.1 10 mol Cr

10 mm 52.0 g Cr1 m 1 cm Cr

Recall that the deposition of each mole of chromium requires six moles of electrons. We now compute the time required to deposit the 25.1 10 mol Cr .

2 6 mol e 96485 C 1s 1htime = 5.1 10 mol Cr = 2.3 h

1 mol Cr 1mol e 3.5C 3600s

37. (M) Dichromate ion is the prevalent species in acidic solution. Oxoanions are better oxidizing agents in acidic solution because increasing the concentration of hydrogen ion favors formation of product. The half-equation is:

2- 32 7 2Cr O (aq) 14H (aq) 6e 2 Cr (aq) 7H O(l) Note that precipitation occurs


1147

most effectively in alkaline solution. In fact, adding an acid to a compound is often an effective way of dissolving a water-insoluble compound. Thus, we expect to see the form that predominates in alkaline solution to be the most effective precipitating agent. Notice

also that CrO42 is smaller than is 2

2 7Cr O , giving it a higher lattice energy in its

compounds, which makes these compounds harder to dissolve.

38. (M) Both metal ions precipitate as hydroxides when OH is moderate.

2+ 3+

2 3Mg aq + 2 OH aq Mg OH s Cr aq + 3 OH aq Cr OH s Because chromium(III) oxides and hydroxides are amphoteric (and those of magnesium ion are not), Cr OH 3 s will dissolve in excess base.

Cr OH 3 s + NaOH aq Na+ aq + Cr OH 4

aq The Iron Triad 39. (M) 4 Fe2+(aq) + O2(g) + 4 H+ 4 Fe3+(aq) + 2 H2O(l) E = 0.44 V

[Fe2+] = [Fe3+]; pH = 3.25 or [H+] = 5.6 10-4 and PO2 = 0.20 atm

E = E - 2

3+ 4

2+ 4 + 4O

0.0592 [Fe ]log

n [Fe ] [H ] P

= 0.44 V -3+ 4[Fe ]0.0592

log4 2+ 4[Fe ] -3.25 4[10 ] 0.20

E = 0.24 V (spontaneous under these conditions)

40. (M) {NiO(OH)(s) + e + H+(aq) Ni(OH)2(s) }2 Ered {Cd(s) + 2 OH (aq) Cd(OH)2(s) + 2 e }1 Eox

{H2O(l) H+(aq) + OH-(aq)} }2

2 NiO(OH)(s) + Cd(s) + 2 H2O(l) Cd(OH)2(s) + 2 Ni(OH)2(s) Ecell 1.50 V

Eox may be obtained by combining the Ksp value for Cd(OH)2(s)

2

14spCd(OH)(K 2.5 10 ) and Ered = -0.403 V for Cd2+(aq) + 2 e- Cd(s). Hence,

Cd(s) Cd2+(aq) + 2 e Eox = 0.403 V

Cd2+(aq) + 2 OH-(aq) Cd(OH)2(s) Ecell = -14

sp

0.0592 1 0.0592 1log = log

n K 2 2.5 10

= 0.403 V

Cd(s) + 2 OH-(aq) Cd(OH)2(s) + 2 e- Eox = 0.806 V Now plug this result back into the first set of redox reactions: {NiO(OH)(s) + e + H+(aq) Ni(OH)2(s) }2 Ered {Cd(s) + 2 OH (aq) Cd(OH)2(s) + 2 e } 1 Eox = 0.806 V

2 NiO(OH)(s) + Cd(s) + 2 H2O(l) Cd(OH)2(s) + 2 Ni(OH)2(s) Ecell 1.50 V Ered = 1.50 V 0.806 V = 0.69 V


1148

41. (E) Fe3+(aq) + K4[F[II]

e(CN)6](aq) K F[III]

e [F[II]

e(CN)6](s) + 3 K+(aq)

Alternate formulation: 4Fe3+ + 3[F[II]

e(CN)6]4-

(aq) + Fe4[F[II]

e(CN)6]3

42. (E) K+(aq) + Fe2+(aq) + K3[Fe[III]

(CN)6](aq) Fe3+(aq) + K4[Fe[II]

(CN)6](aq)

Fe3+(aq) + K4[Fe[II]

(CN)6](aq) KFe[III]

[Fe[II]

(CN)6](s) + 3 K+(aq)

Group 11 Metals

43. (E) (a) Cu2+ aq + H2 g Cu s + 2H+ aq (b) Au+ aq + Fe2+ aq Au s + Fe3+ aq

(c) 22+ + +2 2 42Cu aq +SO g + 2 H O(l) 2 Cu aq +SO aq + 4 H aq

44. (M) In either case, the acid acts as an oxidizing agent and dissolves silver; the gold is unaffected. Oxidation: {Ag s Ag+ aq + e} 3

Reduction: NO3 aq + 4H+ aq + 3e NO g + 2H2O (l)

Net: + +3 23Ag s + NO aq + 4H aq 3Ag aq + NO g + 2H O(l)

Oxidation: {Ag s Ag+ aq + e} 2

Reduction: 2 +4 2 2SO aq + 4H aq + 2e SO g + 2H O(l)

Net: 2 + +4 2 22 Ag s +SO aq + 4 H aq 2 Ag aq +SO g + 2 H O(l)

45. (M) The Integrative Example showed 2+

6c 2+

Cu= 1.2 10 =

CuK

or

22+ 6 +Cu = 1.2 10 Cu

(a) When +Cu = 0.20 M, 22+ 6 4Cu = 1.2 10 0.20 = 4.8 10 M. This is an

impossibly high concentration. Thus +Cu = 0.20 M can never be achieved.

(b) When + 10Cu = 1.0 10 M, 22+ 6 10 14Cu = 1.2 10 1.0 10 = 1.2 10 M . This is

an entirely reasonable (even though small) concentration; + 10Cu = 1.0 10 M can

be maintained in solution.

46. (M) Oxidation: {Cu(s) Cu2+ + 2 e }2 Reduction: {O2(g) + 2 H2O(l) + 4 e- 4 OH-(aq) }1

Overall: 2 Cu(s) + O2(g) + 2 H2O(l) 2 Cu2+(aq) +4 OH (aq)

Note: This produces an alkaline solution. CO2 dissolves in this alkaline solution to produce carbonate ion.


1149

Step(a): CO2(g) + OH (aq) HCO3 (aq)

Step(b): HCO3 (aq) + OH (aq) CO3

2 (aq) + H2O(l)

Total: CO2(g) + 2 OH (aq) CO32 (aq) + H2O(l)

Combination of the reactions labeled "overall" and "total" gives the following result.

2 Cu(s) + O2(g) + H2O(l) + CO2(g) 2 Cu2+(aq) + 2 OH (aq) + CO32 (aq)

The ionic product of the resultant reaction combine in a precipitation reaction to form

2 Cu2+(aq) + 2 OH (aq) + CO32 (aq) Cu2(OH)2CO3 (s)

Group 12 Metals 47. (E) Given: Hg2+/Hg reduction potential = 0.854 V and Hg2

2+/Hg reduction potential = 0.796 V Using the additive nature of free energies and the fact that G = -nFE, we can determine the

Hg2+/Hg22+ potential as

2(0.854V) - 0.796V

1= 0.912 V

48. (M) G = -25,000 J mol-1 = -RTlnKeq = -(8.3145 J K-1 mol-1)(673 K)(lnKeq)

Keq = Kp = 87 atm-1 = 2O

1

P Therefore, PO2

= P

1

K =

1

87= 0.011 atm

49. (M) (a) Estimate Kp for ZnO(s) + C(s) Zn(l) + CO(g) at 800 C (Note: Zn(l) boils at 907 C) {2 C(s) + O2(g) 2 CO(g)} 1/2 G = (415 kJ) × (1/2)

{2 ZnO(s) 2 Zn(l) + O2(g)}1/2 G = (+485 kJ) × (1/2)

ZnO(s) + C(s) Zn(l) + CO(g) G = 35 kJ Use G = -RTlnKeq where T = 800 C, Keq = Kp

35 kJ mol-1 = -(8.3145 10-3 kJ K-1 mol-1)((273.15 + 800) K)(ln Kp)

lnKp = -3.9 or Kp = 0.02 (b) Kp = PCO = 0.02 Hence, PCO = 0.02 atm

50. (M) T = 298 K. Hence, log (PmmHg) =0.05223 (61,960)

298

+ 8.118 = 2.742

PmmHg = 0.001811 mmHg


1150

PV = nRT n P

V RT = [Hg] =

1atm(0.001811mmHg)( )

760mmHg

Latm0.08206 (298K)

K mol

= 9.744 10 8 mol L 1

Next, convert to mg Hg m 3:

[Hg] = 83

mol 1000 200.59g Hg 1000 mg9.874 x 10

L m 1mol Hg 1g Hg

L

= 19.5 mg Hg m 3

(This is approximately 400 times greater than the permissible level of 0.05 mg Hg m 3) 51. (M) We must calculate the wavelength of light absorbed in order to promote an electron

across each band gap.

First, a few relationships. mole A photon=E N E photon =E hv =c v or = /v c

Then, some algebra. mole A photon A A= = = /E N E N hv N hc or A mole= /N hc E 23 1 34 8 1 9

3 1

8 1

3 1

6.022 10 mol 6.626 10 J s 2.998 10 m s 10 nmFor ZnO, =

290 10 J mol 1m

1.196 10 J mol nm= = 413 nm violet light

290 10 J mol

For CdS, 8 1

3 1

1.196 10 J mol nm= = 479 nm blue light

250 10 J mol

The blue light absorbed by CdS is subtracted from the white light incident on the surface of the solid. The remaining reflected light is yellow in this case. When the violet light is subtracted from the white light incident on the ZnO surface, the reflected light appears white.

52. (D) The color that we see is the complementary color to the color absorbed. The color

absorbed, in turn, is determined by the energy separation of the band gap. If the wavelength for the absorbed color is short, the energy separation for the band gap is large.

The yellow color of CdS means that the complementary color, violet, is absorbed. Since violet light has a quite short wavelength (about 410 nm), CdS must have a very large band gap energetically.

HgS is red, meaning that the color absorbed is green, which has a moderate wavelength (about 520 nm). HgS must have a band gap of intermediate energy.

CdSe is black, meaning that visible light of all wavelengths and thus all energies in the visible region are absorbed. This would occur if CdSe has a very small band gap, smaller than that of least energetic red light (about 650 nm).


1151

INTEGRATIVE AND ADVANCED EXERCISES

53. (M) There are two reasons why Au is soluble in aqua regia while Ag is not. First, Ag+, the oxidation product, forms a very insoluble chloride, AgCl, which probably adheres to the surface of the metal and prevents further reaction. AuCl3 is not noted as being an insoluble chloride. Second, gold(III) forms a very stable complex ion with chloride ion, [AuCl4]

–. This complex ion is much more stable than the corresponding dichloroargentate ion, [AgCl2]

–.

54. (E) Sc and Ti metals form

carbides in the presence of carbon (coke). Thus, reduction of their oxides using carbon is avoided.

55. (M) Both tin and zinc are toxic metals in high concentrations. Tin plated iron is more desirable than zinc plated (galvanized) iron. The reason for this is that zinc is a very reactive metal, especially in the presence of acidic foods. Tin, on the other hand, is much less reactive ( 2+Sn /Sn

E = -0.137 V vs. 2+Zn /ZnE = -0.763 V). As well, tin is more malleable

and more easily prepared in pure form (free of oxide).

56.

(M)

(aq)OH4(aq)Cu2O(l)H2(g)OCu(s)2:Net

(aq)OH4e4O(l)H2(g)O:Reduction

2}e2(aq)Cu{Cu(s):Oxidation

222

22

2

This oxidation-reduction reaction produces an alkaline solution. SO3(g) dissolves in this alkaline solution to produce hydrogen sulfate ion in an acid-base reaction.

O(l)H(aq)SO(aq)OH2(g)SO:Total

)(OH(aq)SO(aq)OH(aq)HSO:(b)Step

(aq)HSO(aq)OH(g)SO:(a)Step

22

43

22

44

43

l

The combination of the reactions labeled “Net:” and “Total:” gives the following result.

2 22 2 3 42 Cu(s) O (g) H O(l) SO (g) 2 Cu (aq) 2 OH (aq) SO (aq)

The ionic products of this resultant reaction combine in a precipitation reaction to form

Cu2(OH)2SO4(s). Thus, (s)SO(OH)Cu(aq)SO(aq)OH2(aq)Cu2 4222

42

57. (M) The noble gas formalism requires that the number of valence electrons possessed by the metal atom plus the number of sigma electrons be equal to the number of electrons in the succeeding noble gas atom.

(a) Mo(CO)6 ; Mo has 42 electrons, 6 CO contribute another 12 for 54, which is the number of electrons in Xe


1152

(b) Os(CO)5 ; Os has 76 electrons, 5 CO contribute another 10 for 86, which is the number of electrons in Rn

(c) Re(CO)5-; Re anion has 76 electrons, 5 CO contribute another 10 for 86, which is the

number of electrons in Rn

(d) The trigonal bipyramidal shape of nickel and iron carbonyls does not fit well in crystalline lattices, either because of its symmetry or repulsions with its neighbors and, consequently, these five-coordinate carbonyls are liquids at room temperature. Compare this to octahedral complexes (Cr(CO)6 and others), which are solids because they fit well into these lattices. Weak intermolecular attractions in the low molecular mass, symmetrical nickel and iron carbonyls result in the liquid state at room temperature. Because of the two metal atoms in the higher molecular mass, less symmetrical cobalt carbonyl molecules, stronger intermolecular attractions lead to the solid state.

(e) This compound would be an ionic, salt-like material consisting of Na+ and V(CO)6- ions.

58. (M) (a) Breaks occur at the melting point and boiling point due to changes in the states of the metals (solid liquid gas).

(b) Slopes of these lines become more positive as the temperature increases due to the sign of S for these reactions. The slope should be equal to -S, where S is a negative number in these cases that becomes more negative with a change in phase (higher temperature).

(c) The break at the boiling point is sharper than the break at the melting point because Sfusion <<Svaporization.

59. (M) V0.535e2(s)I(aq)I2:Oxidation 2 E

2

2cell2

Reduction:{Cu (aq) e Cu (aq)} 2 0.159 V

Net: Cu (aq) 2 I (aq) 2 Cu (aq) I (s) 0.376 V 2

E

E n

29.3 131 1

2 132 redox

ln

2 mol e 96485 C/mol e 0.376 Vln 29.3 e 2 10

8.3145 J mol K 298.15 K

Redox: Cu (aq) 4I (aq) 2 Cu (aq) I (s) 2 I (aq) 2 10

Precipitation: {Cu (aq) I (aq) CuI(s)}

G nFE RT K

nFEK K

RT

K

2

2 12 2sp

22

1311redox

2 12)sp

2 1/ 1/(1.1 10 )

Total: Cu (aq) 4 I (aq) 2 CuI s I (s)

2 102 10 Clearly this reaction will occur as written.

(1.1 10

K

K

KK

K

I2 and I- form I3- in aqueous solution. So, adding an additional mole of I- to the last

reaction yields: (s)IsCuI2(aq)I5(aq)Cu -3

2


1153

60. (D) A “detailed calculation” would involve determining the final concentrations of all species. But we can get approximate values of the gold-containing ions by determining values of the equilibrium constants for the reactions involved. First, we determine the value of the equilibrium constant for the disproportionation of Au+(aq). To accomplish this, find the value of Eo for the couple Au+(aq)|Au(s).

3

3cell

1 1

Oxidation Au (aq) Au (aq) 2 e E 1.36 V

Reduction: {Au (aq) e Au(s)} 2 E 1.83 V

Net: 3Au (aq) Au (aq) 2 Au(s) E 0.47 V

2 mol e 96485 C/mol e 0.47 VnFEln K 36.6

RT 8.3145 J mol K 298.15 KK

36.6 15e 8 10

Then we determine the value of the equilibrium constant for dissolving and disproportionation.

K

K

K

(aq)Cl3Au(s)2(aq)AuAuCl(s)3:Net

)100.2(3(aq)}Cl(aq)Au{AuCl(s):Dissolving

108Au(s)2(aq)Au(aq)Au3:ionationDisproport

3

3133sp

15disp

3

3 15 13 3 23 3 3 3 4disp sp

2346 3

8 10 (2.0 10 ) 6.4 10 [Au ][Cl ] ( )(3 ) 27

6.4 101 10 M [Au ]

27

K K K x x x

x

This is a significant amount of decomposition, particularly when compared with [Au+] = 4.5 × 10–7 M, which is the concentration present in saturated AuCl, assuming no disproportionation.

61. (M) (a) 22Dissolution: AgO(s) 2 H (aq) Ag (aq) H O(l)

(aq)Ag4(aq)H4(g)O(aq)Ag4O(l)H2:Net

V98.1 4(aq)}Age(aq){Ag:Reduction

V229.1e4(aq)H4(g)OO(l)H2:Oxidation

22

2

2

22

E

E

(b) V 0.75 V 1.98 V 1.229 cello E spontaneous since cell

oE > 0.

62. (M)

2 2 7 24 4 a Association: 2 H (aq) 2 CrO (aq) 2 HCrO (aq) 1/ 1/ (3.2 10 )K

24 2 7 2

2 2 144 2 7 2 c

14 14 7 22 7 2

Elimination: 2 HCrO (aq) Cr O (aq) H O(l)

Eqn. (23.18) 2 H (aq) 2 CrO (aq) Cr O (aq) H O(l) 3.2 10

3.2 10 3.2 10 (3.2 10 ) 33(3.2 10 )c

a

K

K

K KK K

K


1154

63. (D) The presence of both acetate ion and acetic acid establishes a buffer and fixes the pH of the solution. This, in turn, fixes [CrO4

2–]. We then can determine the concentration of each of the cations that can coexist with this [CrO4

2–] before precipitation occurs. First we determine [H3O

+], with the aid of the Henderson-Hasselbalch equation.

52 3 2a

2 3 2

4.74 53

2 614 2 7

c 2 2 2 2 2 5 2 2 24 3 4 4

62

4

[C H O ] 1.0 M pH p log log (1.8 10 ) log 4.74

[HC H O ] 1.0 M

[H O ] 10 1.8 10 M

[Cr O ] 0.0010 M 3.1 10 3.2 10

[CrO ] [H O ] [CrO ] (1.8 10 ) [CrO ]

3.1 10 [CrO ]

K

K

5

14

10sp2 6

max 2 54

52

max 5

42

max 5

9.8 10 M3.2 10

1.2 10 [Ba ] 1.2 10 M

[CrO ] 9.8 10

2.2 10[Sr ] 0.22 M

9.8 10

7.1 10[Ca ] 7.2 M

9.8 10

K

Each cation is initially present at 0.10 M. Neither strontium ion nor calcium ion will precipitate, while BaCrO4 will precipitate until [Ba2+] has declined to far less than 0.1% of its original value. Thus, barium ion is effectively separated from the other two cations by chromate ion precipitation under these conditions.

64. (D) The equations are balanced with the ion-electron method. Oxalic acid is oxidized in each case.

2 2 4 2

22 2

22 2 4 2 2 2

Oxidation: H C O (aq) 2 H (aq) 2 CO (g) 2 e

Reduction:MnO (s) 4 H (aq) 2 e Mn (aq) 2 H O(l)

Net: H C O (aq) MnO (s) 2 H (aq) Mn (aq) 2 CO (g) 2 H O( )l

2 2 4 2

24 2

22 2 4 4 2 2

Oxidation:{H C O (aq) 2 H (aq) 2 CO (g) 2 e } 5

Reduction: {MnO (aq) 8 H (aq) 5 e Mn (aq) 4 H O(l) } 2

Net: 5 H C O (aq) 2 MnO (aq) 6 H (aq) 2 Mn (aq) 10 CO (g) 8 H O(l)

We then determine the mass of the excess oxalic acid.

O2HOCHg0.9474O2HOCHmol1

O2HOCHg126.07

MnOmol2

OCHmol5

solnL1

MnOmol0.1000L0.03006O2HOCHmass

24222422

2422

4

42242422


1155

Thus, the mass of oxalic acid that reacted with MnO2 is 1.651 g – 0.9474 g = 0.704 g H2C2O42H2O. Now we can determine the mass of MnO2 in the ore.

2 2 4 2 22 2 2 4 2

2 2 4 2 2 2 4

2 22 2

2

1 mol H C O .2H O 1 mol MnO mass MnO 0.704 g H C O 2H O

126.07 g H C O 2H O 1 mol H C O

86.94 g MnO 0.485 g MnO 0.485 g MnO %MnO 100% 82.3%

1 mol MnO 0.589 g ore

65. (D) (a) We consider reaction with 1.00 millimoles of Fe2+. In each case, we need the balanced chemical equation of the redox reaction.

solnmL2.00MnOmmol0.1000

solutionmL1

Femmol5

MnOmmol1Femmol1.00volume

titrant

(aq)Fe5OH4(aq)Mn(aq)Fe5(aq)H8(aq)MnO

solnmL1.67OCrmmol0.1000

solutionmL1

Femmol6

OCrmmol1Femmol1.00volume

titrant

(aq)Fe6O(l)H7(aq)Cr2(aq)Fe6(aq)H14(aq)OCr

42

43

32

224

272

2

2722

32

32272

More of the 0.1000 M MnO4– solution would be required.

(b) We use the reaction of each with Fe2+ to find the equivalence between them.

4

2 22 7 4

2 2MnO2 7

44

0.1000 mmol Cr O 1 mmol MnO6 mmol FeV 24.50 mL

1 mL soln 1 mmol Cr O 5 mmol Fe

1 mL soln29.40 mL of 0.1000 mmol MnO

0.1000 mmol MnO

66. (M) A high oxidation state of a metal can only be stabilized by an anion of an element that is highly electronegative. Fluorine and oxygen are the two most electronegative elements in the periodic table and thus should be most effective at stabilizing a high oxidation state of a metal.

67. (D) The precipitation reaction of dichromate ion permits us to determine from the formula of barium chromate the amount of chromium present, while the redox reaction of permanganate ion permits us to determine the amount of manganese present in the 250.0 mL solution. The 15.95 ml volume of Fe2+ titrant assumes (contrary to what the problem implies) that the chromate has been removed by precipitation. Ba(MnO4)2 is quite water

soluble, so [ 4MnO ] should not be affected.


1156

Equation: MnO4 (aq) 8 H (aq) 5 Fe2 (aq) Mn2 (aq) 4 H

2O 5 Fe3 (aq)

4 4

4 4

2- 2 32 7 2

0.549 g BaCrO 1 mol BaCrO 1 mol Cr 52.00 g Crmass of Cr 250.0 mL soln

10.00 mL sample 253.3 g BaCrO 1 mol BaCrO 1 mol Cr

mass of Cr 2.82 g Cr

equation: Cr O 14 H 6 Fe 2Cr 7 H O

3

3 22-

2 7 2-2 7

2-22 7

3 2

6 Fe

1 mol Cr 6 mol Fe10 mlvolume of titrant to reduce Cr O 2.82 g Cr

250 ml 52.00 g Cr 1 mol Cr O

1 mol Cr O 1 L titrant 1000 mL 86.78 ml Fe titrant

1 L2 mol Cr 0.0750 mol Fe

Now the volume of

- 2-4 2 7

4

titrant to reduce both MnO and Cr O must be (15.95 86.78) mL

Volume of titrant 102.73 mL.

Not just 15.95 mL, which must be the volume of titrant required for MnO .

15.mass Mn 250.0 mL soln

alone

2

42

4

1 mmol MnO95 mL titrant 0.0750 mmol Fe

10.00 mL sample 1 mL titrant 5 mmol Fe

1 mmol Mn 54.94 mg Mn 1 g Mn 0.3286 Mn

1 mmol Mn 1000 mg Mn1 mmol MnO

0.3286 g Mn 2.82 g Cr%Mn 100% 3.286% Mn %Cr

10.000 g steel 10.000 g

100% 28.2% Cr steel

68. (M) (a) We use the technique of Chapter 3 to determine the empirical formula of nickel

dimethylglyoximate.

Nmol3.9990.3461Nmol1.384Ng14.007

Nmol1N g19.39

Omol3.9990.3461Omol1.384Og15.999

Omol1Og22.15

The empirical formula of nickel dimethylglyoximate (NiDMG) is NiC8H14O4N4, with an empirical molar mass of 288.91 g/mol

(b) Nimol1

Nig58.69

NiDMGmol1

Ni mol 1

NiDMGg288.91

NiDMGmol1

mL10.00

NiDMGg0.104mL250.0Niofmass

Hmol14.00.3461Hmol4.84 Hg1.008

Hmol1H4.88

Cmol8.0010.3461Cmol2.769Cg12.011

Cmol1Cg33.26

Nimol1.0000.3461Nimol0.3461Nig58.69

Nimol1Nig20.31

g


1157

3.52%100%

steelg15.020

Nig0.528steelinNi%

Nig0.528

69. (M) A look at Appendix D indicates that Cr(OH)3,31

sp 10 6.3 K ;

Zn(OH)2,17

sp 10 1.2 K ; Fe(OH)3,38

sp 10 4 K ; and Ni(OH)2,15

sp 10 2.0 K ; all are

expected to form. But in excess NaOH, [Cr(OH)4]–, 29

f 10 8 K , which probably is green

in color; and [Zn(OH)4]2–, 17

f 10 4.6 K , also will form. (i.e., their hydroxides will dissolve). The lack of color in the solution indicates that Cr3+ is not present. The presence of a hydroxide precipitate indicates that Ni2+ and Fe3+ may be present. Because Ni2+ forms a complex ion with NH3, [Ni(NH3)6]

2+, 8f 10 5.5 K , and Fe3+ does not, the lack of a second

precipitate indicates that Fe3+ is absent. The fact that there was a hydroxide precipitate indicates that Ni2+ is present. We are sure that Ni2+ is present and that Fe2+ and Cr3+ are absent, but we are uncertain about the presence or absence of Zn2+.

70. (M) First calculate concentration of Hg(NO3)2 solution: ? mol NaCl = 2.00 × 10−3 mL × 0.0108 mol L−1 = 2.16 × 10−5 mol NaCl

? mol HgCl2 = 1

2 × 2.16 × 10−5 mol NaCl = 1.08 × 10−5 mol

C = 1.08 × 10−5 mol/1.12 × 10−3 L = 9.64 × 10−3 mol L−1 Next, calculate amount of Cl− in the serum sample: ? mol Cl− = 1.23 × 10−3 L × 9.64 × 10−3 mol Hg2+ L−1 × 2 mol Cl−/1 mol Hg2+ = 2.37 × 10−5 mol Cl−

= 0.0237 mmol This amount of Cl− came from a 0.500 mL serum sample, so the concentration of Cl− in mmol L−1 is C = 0.0237 mmol/0.500 × 10−3 L = 47 mmol L−1.

The concentration is smaller than the lower limit of the normal range.

71. (M)

Mn

O

Mn

O

O

O

O

OO

(b)(a)

Hg Hg

(c)

Os

O

O

O

O

72. (M) 19

39 3

(7)(1.60 10 C)4500 C mm

(4/3) (39 10 mm)

This value is more than 4 times the value given in Table 21.4 for Be2+ (1108 C mm−3). We know that there is significant covalent bonding in beryllium compounds. We expect


1158

the bonding in Mn2O7 to be primarily covalent. With such a large charge density, a Mn7+ ion would be very strongly polarizing, drawing electron density away from surrounding anions and leading to bonds with primarily covalent character.

73. (M) ? Ti per unit cell = 1

8Ti per corner × 8 corners per cell = 1 Ti

? Ni per unit cell = 1

The empirical formula is NiTi. The % Ti by mass is 100 × 47.88/(47.88 + 63.55) = 45%

FEATURE PROBLEMS 74. (D) (a) If ngas = 0, then S ~ 0 and G is essentially independent of temperature

(C(s) + O2(g) CO2(g))

If ngas > 0, then S > 0 and G will become more negative with increasing temperature, hence the graph has a negative slope (2 C(s) + O2(g) 2 CO (g)).

If ngas < 0, then S < 0 and G will become more positive with increasing temperature, hence the graph has a positive slope (2 CO(g) + O2(g) 2 CO2 (g).

(b) The additional blast furnace reaction, C(s) + CO2(g) 2 CO(g)), has

H = [2 110.5 kJ] [393.5 kJ + 0 kJ] = 172.5 kJ and

S = [2 197.7 J K-1] [1 5.74 J K-1 + 213.7 J K-1] = 176.0 J K-1

It can be obtained by adding reaction (b) to the reverse of reaction (c) (both appear in the provided figure)

C(s) + O2(g) CO2(g) Reaction (b) 2CO2(s) 2CO(g) + O2(g) Reverse of Reaction (c)

Net: C(s) + CO2(g) 2 CO(g) (Additional Blast Furnace Reaction)

Consequently, the plot of G for the net reaction as a function of temperature will be a straight line with a slope of [{S (Rxn b)} {S (Rxn c)}] (in kJ/K) and a y-intercept of [{H (Rxn b)} {H (Rxn c)}] (in kJ). Since Ho (Rxn b) = 393.5 kJ, Ho (Rxn c) = 566 kJ, So (Rxn b) = 2.9 J/K and So (Rxn c) = 173.1 J/K, the plot of G vs. T for the reaction C(s) + CO2(g) 2 CO(g) will follow the equation 0.176 172.5y x

From the graph, we can see that the difference in G (line b – line c at 1000 C) is ~ 40 kJ/mol. Kp is readily calculated using this value of G.

G = -RTlnKp = -40 kJ = (8.3145 10-3 kJ K-1 mol-1)( 1273 K)(ln Kp) lnKp = 3.8 Hence, Kp = 44

The equilibrium partial pressure for CO(g) is then determined by using the Kp expression:


1159

Kp

2

2

CO

CO( )

P

P44

2

CO

(0.25 atm)

P Hence, (PCO)2

= 11 and PCO = 3.3 atm or 3 atm.

Alternatively, we can determine the partial pressure for CO2 at 1000 C via the calculated H and S values for the reaction C(s) + CO2(g) 2 CO(g) to find G at 1000 C , and ultimately Kp with the relationship G = -RTlnKp (here we are making the assumption that H and S are relatively constant over the temperature range 298 K to 1273 K). The calculated values of H and S (using Appendix D) are given below:

H = [2 110.5 kJ] [393.5 kJ + 0 kJ] = 172.5 kJ

S = [2 197.7 J K-1] [1 5.74 J K-1 + 213.7 J K-1] = 176.0 J K-1

To find G at 1000oC, we simply plug x =1273 K into the straight-line equation we developed above and solve for y (G).

So, 0.176(1273K) 172.5y ; 51.5kJy Next we need to calculate the Kp for the reaction at 1000oC.

G = -RTlnKp = -51.5 kJ = -(8.3145 10-3 kJ K-1 mol-1)( 1273 K)(ln Kp)

lnKp = 4.87 Hence, Kp = 1.3 102

The equilibrium PCO(g) is then determined by using the Kp expression:

Kp

2

2

CO

CO( )

P

P

2

CO

(0.25 atm)

PHence, (PCO)2

= 32.5 atm and PCO = 5.7 atm

75. (M) (a) The amphoteric cations are Al3+, Cr3+, and Zn2+.

Cr3+(aq) + 4OH-(aq) [Cr(OH)4]

-(aq)

Al3+(aq) + 4OH-(aq) [ Al(OH)4]

-(aq)

Zn2+(aq) + 4OH-(aq) [Zn(OH)4]

2- (aq)

Fe3+(aq) + 3OH-(aq) Fe(OH)3(s)

Ni2+(aq) + 2OH-(aq) Ni(OH)2(s)

Co2+(aq) + 2OH-(aq) Co(OH)2(s)

(b) Of the three hydroxide precipitates, only Co2+ is easily oxidized:

(refer to the Standard Reduction Potential table in Appendix D).

2 Co(OH)2(s) + H2O2(aq) 2 Co(OH)3(s) (c) We know that the chromate ion is yellow. Thus,

[Cr(OH)4]-(aq) + 3/2 H2O2(aq) + OH

-(aq) CrO4

2-(aq, YELLOW) + 4 H2O(l)


1160

(d) Co3+ is reduced to Co2+ by the H2O2 in solution. The other hydroxides simply

dissolve in strong acid:

(i) Co(OH)3(s) + 3H+(aq) Co3+(aq) + 3H2O(l)

(ii) 2 Co3+(aq) + 2 Cl-(aq) 2 Co2+(aq) + Cl2(g)

(iii) Fe(OH)3(s) + 3 H+(aq) Fe3+(aq) + 3 H2O(l)

(iv) Ni(OH)2(s) + 2 H+(aq) Ni2+(aq) + 2 H2O(l) (e) The Fe3+(aq) ions in the presence of concentrated ammonia(6 M) form an insoluble

precipitate with the hydroxide ions generated by the ammonia hydrolysis reaction:

NH3(aq) + H2O(l) NH4

+(aq) + OH-(aq)

Fe3+(aq) + 3 OH-(aq) Fe(OH)3(s); Ksp Fe(OH)3 = -384 10

Both Co3+ and Ni2+ form soluble complex ions with ammonia ligands rather than hydroxide precipitates.

SELF-ASSESSMENT EXERCISES 76. (E) (a) A key feature of ferromagnetism is that in the solid state, the metal atoms are thought to be grouped together into small regions— called domains—containing rather large numbers of atoms. (b) Flotation is a process used to determine metal concentration. (c) Leaching is the process by which metal ions are extracted (leached) from the ore by a liquid. Leaching agents include water, acids, bases, and salt solutions. Oxidation– reduction reactions may also be involved. (d) Mercury alloys, called amalgams, are commonly made with most metals, and some of these amalgams are of commercial importance. 77. (E) (a) In the series of elements in which the 4f subshell is filled, atomic radii decrease somewhat. This phenomenon occurs in the lanthanide series ( Z=58 to 71) and is called the lanthanide contraction. (b) Zone melting (or zone refining or floating zone process) is a group of similar methods of purifying crystals, in which a narrow region of a crystal is molten, and this molten zone is moved along the crystal (in practice, the crystal is pulled through the heater). The molten region melts impure solid at its forward edge and leaves a wake of purer material solidified behind it as it moves through the ingot. The impurities concentrate in the melt, and are moved to one end of the ingot. (c) Basic oxygen process is a steelmaking method in which pure oxygen is blown into a bath of molten blast-furnace iron and scrap. The oxygen initiates a series of intensively exothermic (heat-releasing) reactions, including the oxidation of such impurities as carbon, silicon, phosphorus, and manganese.


1161

(d) Slag formation is used to remove impurities from ores. 78. (E) (a) Ferromagnetism is the basic mechanism by which certain materials (such as iron) form permanent magnets, or are attracted to magnets. Paramagnetism is a form of magnetism which occurs only in the presence of an externally applied magnetic field. Paramagnetic materials are also attracted to magnetic fields. (b) An ore is roasted (heated to a high temperature) to convert a metal compound to its oxide, which can then be reduced. The oxidation number of a metal decreases as a result of reduction. (c) Hydrometallurgy is part of the field of extractive metallurgy involving the use of aqueous chemistry for the recovery of metals from ores, concentrates, and recycled or residual materials. Pyrometallurgy is a branch of extractive metallurgy. It consists of the thermal treatment of minerals and metallurgical ores and concentrates to bring about physical and chemical transformations in the materials to enable recovery of valuable metals. (d) Chromates are the salts that contain CrO4

2- anions. Dichromates, on the other hand, contain Cr2O7

2- anions. In both compounds chromium is in +6 oxidation state. 79. (E) (a) Pig iron is impure iron (95% Fe) formed in the reduction of iron ore in a blast furnace. (b) Ferromanganese is an iron-manganese alloy formed by the reduction of a mixture of iron and manganese oxides. (c) Chromite ore, Fe(CrO2)2 is the principal chromium ore. (d) Brass is an allow of Zn and Cu with small amounts of Sn, Pb, and Fe. (e) Aqua regia is a mixture of HCl(aq) and HNO3(aq) that dissolves inactive metals by a combination of oxidation and complex ion formation. (f) Blister copper, formed in the reduction of a mixture of Cu2O(s) and Cu2S(s), is impure Cu(s) containing SO2(g). (g) Stainless steel is an iron alloy with varying quantities of metals such as Cr, Mn, and Ni, and a small and carefully controlled percentage of carbon. 80. (E) (c), (f), and (g). 81. (E) (b) 82. (E) (d) 83. (E) (a) 84. (E) (d) 85. (E) (c) and (e) 86. (M) CrO3(s), (b) potassium manganate, (c) chromium carbonyl, (d) BaCr2O7, (e) lanthanum(III) sulfate nonahydrate, (f) Au(CN)33H2O.


1162

87. (M) (a) 2Fe2S3(s)+3O2 (g)+6H2O(l) 4Fe(OH)3(s)+6S(s)

(b) 2Mn2+(aq)+8H2O(l)+5S2O82- (aq) 2MnO4

- (aq)+16H+(aq)+10SO42- (aq)

(c) 4Ag(s)+8CN-(aq)+O2 (g)+2H2O(l) 4[Ag(CN2 ]-(aq)+4OH-(aq) 88. (M) Atoms of Zn, Cd and Hg have configurations of 4s2 3d10 , 5s2 4d10 , and 6s2 4 f 14 5d10 , respectively. One the ns2 electrons participate in bonding, and in this regard Zn, Cd, and Hg resemble the group 2 elements. 89. (M) HNO3 is the oxidizing agent for oxidizing the metal to Au3+ but Au3+ must be stabilized in solution. In aqua regia, Cl- ions combine with Au3+ to form [AuCl4]

-, which is stable in solution. 90. (M) The Fe3+ ion forms a complex ion, Fe(H2O)6

3+, in aqueous solution. The complex behaves as a weak monoprotic acid in solution. See Equation (23.36).

Documents

CHAPTER 23 THE TRANSITION ELEMENTS