CHAPTER 23 THE TRANSITION ELEMENTS

THE TRANSITION ELEMENTS

PRACTICE EXAMPLES

1A (E) (a) Cu O₂ should form. 2 Cu S s + 3 O g2

 

2

 

2 Cu O s + 2 SO g2

 

2

 

(b) W(s) is the reduction product. WO s + 3 H g3

 

2

 

W s + 3 H O g

 

2

 

(c) Hg(l) forms. 2 HgO s

 

^2 Hg l + O g

 

2

 

1B (E) (a) SiO s2

 

is the oxidation product of Si.

 

2 3

 

2

   

3 Si s + 2 Cr O s ^3 SiO s + 4 Cr s

(b) Roasting is simply heating in air. 2 Co OH

   

3 s ^,AirCo O s + 3H O g2 3

 

2

 

(c) MnO s2

 

forms; (acidic solution). Mn²⁺

 

aq + 2H O(l)2 MnO s + 4H aq + 2e2

 

⁺

 

^

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

Oxidation: {V³⁺

 

aq + H O(l)2 VO²⁺

 

aq + 2H aq + e } 3⁺

 

^  E^o = 0.337V Reduction: NO3^

 

aq + 4H aq + 3e⁺

 

^NO g + 2H O

 

2 E^o = +0.956V

Net: NO3^-

 

aq + 3V³⁺

 

aq + H O(l)2 NO g + 3VO

 

²⁺

 

aq + 2H aq ⁺

 

Ecell^o = +0.619V Because the cell potential is positive, nitric acid can be used to oxidize^V3+

 

^{aq to VO2+}

 

^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{VO ( )|V ( )} 0.041V2 2

E ^ 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 withE^o{V (aq)|V(s)}= 1.13V^2  .

So, -E^o 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{Cr (aq)|Cr (aq)}3 2+ 0.42 V

E ^

   ; E^o{Fe (aq)|Fe(s)}²⁺  0.440 V

o{Zn (aq)|Zn(s)}2+ 0.763 V

E   . 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: PtCl62− + 2 e⁻ → PtCl42− + 2 Cl⁻ E° = +0.68 V

Overall: 2 V²⁺ + PtCl62− → 2 V³⁺ + PtCl42− + 2 Cl⁻ E° = 0.94 V

The equilibrium constant for the overall reaction is ³¹ 2(0.94)

0.0592

10 5.7 10

K    . For the

reverse reaction to be spontaneous, we need Q > 5.7 × 10³¹. Because Q = [V³⁺]² [PtCl42−] [Cl⁻]²/[V²⁺] [PtCl62−], 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 PtC62− 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 PtCl62− 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  ⁴⁺ Ti²⁺ Cu²⁺ Sc³⁺

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

( +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, withAl , ³⁺ Sn , ²⁺ Sn , and ⁴⁺

Pb being major exceptions. On the other hand, most transition metal ions form an extensive 2+

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 4d s^{5 2} electron configuration. Removing one electron produces an electron configuration ( 3 4d s^{5 1}) 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.

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) TiCl g + 4 Na l4

   

^Ti s + 4NaCl l

   

(b) Cr O s + 2Al s2 3

   

^2 Cr l + Al O s

 

2 3

 

(c) Ag s + HCl aq

   

no reaction

(d) K Cr O aq + 2KOH aq2 2 7

   

2 K CrO aq + H O(l)2 4

 

2

(e) MnO s + 2 C s2

   

^Mn l + 2CO g

   

12. (M) (a) Cr s

 

^{+ 2HCl aq}

 

^{ CrCl}2

 

aq ^{+ H}2

 

g Virtually any first period transition metal except Cu can be substituted for Cr.

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

 

^

 

2 2 Cr OH

   

4^ aq

The oxide must be amphoteric. Thus, Sc O , TiO_{2 3 2}, ZrO₂, and ZnO could be substituted forCr O . _{2 3}

(c) 2 La s + 6 HCl aq

   

2 LaCl aq + 3 H g3

 

2

 

Any lanthanide or actinide element can be substituted for lanthanum.

13. (M) (a) Sc OH

   

3 s + 3H aq⁺

 

Sc³⁺

 

aq + 3 H O(l)2

(b) 3 Fe²⁺

 

aq + MnO4^

 

aq + 2 H O l2

 

3 Fe³⁺

 

aq + MnO s + 4 OH2

 

^(aq) (c) 2 KOH l + TiO s

 

2

 

^K TiO s + H O g2 3

 

2

 

(d) Cu s + 2 H SO conc,aq

 

2 4

 

CuSO aq + SO g + 2 H O(l)4

 

2

 

2

14. (M) (a)

   

electrolysis

   

2 3 3 6 2

2 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) Cr s + 2 HCl aq

   

Cr²⁺

 

aq + 2 Cl aq + H g^

 

2

 

(c) 4 Cr²⁺

 

aq + O g + 4 H aq2

 

⁺

 

4 Cr³⁺

 

aq + 2 H O(l)2

(d) Ag s + 2 HNO aq

 

3

 

AgNO aq + NO g + H O(l)3

 

2

 

2

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) FeS s + 2 HCl aq

   

FeCl aq + H S g2

 

2

 

       

       

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) BaCO s + 2 HCl aq3

   

BaCl aq + H O(l) + CO g2

 

2 2

 

           

2 2 2 7 4 2

2 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) CuO s

 

 2H (aq)^ Cu²⁺

 

aq + H O l2

 

(ii) Cu²⁺

 

aq  2OH (aq)^ Cu(OH) s2

 

(b)



NH4



2Cr O s2 7

 

^N g + 4 H O g + Cr O s2

 

2

 

2 3

 

     

2 3 3 2

Cr 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.

(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 G^o 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) Hf = -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 + H}₂^{O (l)}

(b) Oxidation: Cr²⁺

 

aq ^{ Cr3+}

 

^{aq + e}

22. (E) (a) Oxidation: Fe OH

   

3 s + 5 OH aq^

 

FeO4^2

 

aq + 4 H O(l) + 3e2 ^

(b) Reduction: ^{Ag CN}

   

₂^{ aq + e}Ag s + 2 CN aq

 

^

 

23. (M) (a) First we need the reduction potential for the couple VO₂⁺

 

aq ^{/ V2+}

 

^{aq . We will}

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

       

+ + 2+ o

2 2

VO aq + 2 H aq + e^ VO aq + H O(l) G = 1 F +1.000 V

       

2+ + 3+ o

VO aq + 2 H aq + e^ V aq + H O(l)2 G = 1 F +0.337 V

     

3+ 2+ o

V aq + e^V aq G = 1 F 0.255 V

     

+ + 2+ o o

2 2

VO aq + 4 H aq + 3 e^ V aq + 2 H O G = 3 FE

o 1.000 V + 0.337 V 0.255 V

= = + 0.361 V

E 3 

We next analyze the oxidation-reduction reaction.

Oxidation:{2 Br aq^

 

Br l + 2e } 32

 

^  E^o = 1.065 V Reduction: {VO2⁺

 

aq + 4H aq + 3e⁺

 

^ V²⁺

 

aq + 2H O(l)} 2 2  E^o = +0.361V

Net:6 Br aq + 2 VO^

 

2⁺

 

aq + 8 H aq⁺

 

3 Br l + 2 V2

 

²⁺

 

aq + 4 H O2 o

cell= 0.704V E 

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

(b) Oxidation: ^Fe2+

 

^{aq Fe3+}

 

^{aq + e Eo = 0.771V}

Reduction:VO2⁺

 

aq + 2 H aq + e⁺

 

^ VO²⁺

 

aq + H O(l)2 E^o = +1.000V Net: Fe²⁺

 

aq + VO2⁺

 

aq + 2 H aq⁺

 

Fe³⁺

 

aq + VO²⁺

 

aq + H O(l)2

o

cell= +0.229V

E

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

(c) Oxidation: H O2 2 2 H aq + 2 e + O g⁺

 

^ 2

 

E^o = 0.695 V Reduction: MnO s + 4 H aq + 2 e2

 

⁺

 

^ Mn²⁺

 

aq + 2 H O(l)2 E^o = +1.23V Net:H O + MnO s + 2 H aq2 2 2

 

⁺

 

O g + Mn2

 

²⁺

 

aq + 2 H O(l)2 Ecell^o = +0.54V 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{Zn (aq) / Zn(s)}2 0.763V; 0{Sn (aq) / Sn (aq)}4 2 0.154V

E ^ E ^{ }

     

o

{I (s)/I (aq)} = 0.535 V2

E ^

 

Each of these potentials is combined with the cited reduction potential. If the resulting value of E_cell^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) E^o{Cr O_{2 7}^2(aq) / Cr (aq)} = +1.33 V^3

o o 2 3 o 2

cell 2 7

o cell

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

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

E E E

E

   



o o 2 3 o 4 2+

cell 2 7

o cell

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

= +1.33 V 0.154 V = +1.18 V Yes

E E E

E

   

 

o o 2 3 o

cell 2 7 2

o cell

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

= +1.33 V 0.535 V = +0.80 V Yes

E E E

E

   

 

(b) E^o{Cr (aq)/Cr (aq)} = 0.424 V^{3+ 2+} 

o o 3+ 2+ o 2+

cell o cell

= {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 o cell

= {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+ o

cell 2

o cell

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

= 0.424 V 0.535 V = 0.959 V No

E E E

E

 

   

(c) E^o{SO₄^2(aq)/SO (g)} = +0.17 V₂

o o 2 o 2+

cell 4 2

o cell

= {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

o cell

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

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

E E E

E

 

 

o o 2 o

cell 4 2 2

o cell

= {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

withE^o{VO ( )|V ( )} 0.337 V^2 aq ^3 aq  and a negative sum must be produced when this half-cell potential is combined withE^o{V (aq)|V (aq)}= 0.255V^{3 2+}  .

So, -E^o 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{Sn (aq)|Sn(s)}2 0.137 V

E ^

   ; E^o{H (aq)|H (g)] 0.000 V⁺ ₂ 

o{Pb (aq)|Pb(s)]2+ 0.125 V

E   ; 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⁺

E_MnO4-/ Mn2+ = 3 (1.70 V) + 2 (1.23 V)

5 = 1.512 V ~1.51 V Method 2: MnO4-  MnO42 MnO2  Mn³⁺ Mn²⁺

E_MnO4-/ 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 E_MnO4-/ Mn2+ = 1.51 V

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

2- 3+

2 7

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

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) Cr O /Cr : E = _{2 7}^{2- 2+} 3(1.33V) - 0.424 V

4 = 0.892 V

(ii) Cr /Cr : E = ³⁺ 0.424 V + 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 = 1.000 V + 0.337 V

2 = 0.669 V

(ii) VO2+/V²⁺: E = 1.000 V + 0.337 V + (-0.255 V)

3 = 0.361 V

(iii) VO2+/V: E = 1.000 V + 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 4

Cr 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. Pb²⁺

 

aq + CrO4^2

 

aq PbCrO s4

 

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⁺

 

²

   

4 2 7 2

2 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 7²

 

⁺

 

2

 

³⁺

 

2

6Cl aq + Cr O^{ } aq +14H aq 3Cl g + 2Cr aq + 7H O(l)

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 H



₂O



₄^Cl2⁺.

31. (D) Oxidation: {Zn s

 

^{ Zn2+}

 

^{aq + 2e} 3}

Reduction: Cr O2 7^2



aq,orange +14H aq + 6e



⁺

 

^ 2 Cr³⁺



aq,green + 7 H O(l)



2

Net: 3 Zn s + Cr O

 

2 7^2

 

aq +14 H aq⁺

 

3 Zn²⁺

 

aq + 2 Cr³⁺

 

aq + 7 H O(l)2

Oxidation: Zn s

 

^{ Zn2+}

 

^{aq + 2e}

Reduction: {Cr³⁺



aq,green



^{+ e Cr2+}



^aq,blue



^{} 2}

Net: Zn s + 2 Cr

 

³⁺

 

aq Zn²⁺

 

aq + 2 Cr²⁺

 

aq

The green color is most likely due to a chloro complex of Cr³⁺, such asCr H O Cl



2



4 2 . ⁺ Oxidation: {Cr²⁺



aq,blue



^{ Cr3+}



^aq,green



^{+ e} 4}

Reduction: O g + 4 H aq + 4e2

 

⁺

 

^ 2 H O(l)2

Net: 4 Cr²⁺

 

aq + O g + 4 H aq2

 

⁺

 

4 Cr³⁺

 

aq + 2 H O(l)2

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 + 2

4 2 7 2

2 CrO ^ aq + 2 H aq Cr O ^ aq + H O(l)

 

⁺

   

2 2 3

2 H O(l) + 2CO aq 2H aq + 2 HCO ^ aq

       

2 2

4 2 2 2 7 3

2 CrO ^ aq + 2 CO aq + H O(l)Cr O ^ aq + 2 HCO ^ aq

33. (M) Simple substitution into Equation (23.19) yields Cr O_{2 7}^2 in each case. In fact, the expression is readily solved for the desired concentration as follows:

2 2

2 14 + 2

2 7 4

Cr O ^ = 3.2 10 H CrO ^

      

    . In each case, we use the value of pH to determine

+ pH

H = 10^

 

  .

(a) Cr O_{2 7}^2= 3.2 10 (10 ^{14 6.62 2}) (0.20)² 0.74 M (b) ^{Cr O2 72}= 3.2 10 10 ¹⁴



^8.85



²

^

0.20 = 2.6 10 M

^

²  ^5

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

Cr₂O₇^2

 

CrO₄^2

 ² = 3.2 10¹⁴ H^{+ 2} = 3.2 10¹⁴



10^7.55



^{2 = 3.2 1014}



^{2.8 108}



^{2 = 0.254}

2

2 2 4 2 4 4

4 initial

2 4 2 4

1.505 g Na CrO 1 mol Na CrO 1 mol CrO

CrO = = 0.0269 M

0.345 L soln 161.97 g Na CrO 1mol Na CrO

 

   

 

Reaction: 2CrO4^2

 

aq + 2H aq⁺

 

   Cr₂O₇^2

 

aq ^{+ H}2O (l) Initial: 0.0269 M 2.8 10 ^8(fixed) 0 M 

Changes: 2 Mx 0 + Mx 

Equil:



0.0269 2 M x



2.8 10 ^8(fixed) x M 

   

2

2 7 2

2 2

2 4

Cr O = 0.254 = = 0.254 0.000724 0.108 + 4

0.0269 2 CrO

x x x x

x





 

  

  

 

2 2

= 0.000184 0.0274 +1.016 1.016 1.0274 + 0.000184 = 0

x  x x x  x

2 4 1.0274 1.0556 0.000736

= = = 1.0111 M, 0.00016 M

2 2.032

b b ac

x a

    

We choose the second root because the first root gives a negative CrO₄^2 .

2 2

2 7 4

Cr O ^ = = 0.00016 Mx CrO ^ = 0.0269 2 = 0.0266 Mx

    

   

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 Cr

1 hr 1 s 96485 C 6 mol e 1mol Cr



    

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

4 2

2 2

2 3

1 cm 10 cm 7.14 g Cr 1 mol Cr

mol Cr = 0.0010 mm 0.375 m = 5.1 10 mol Cr

10 mm 1 m 1 cm Cr 52.0 g Cr

  

     

 

 

Recall that the deposition of each mole of chromium requires six moles of electrons. We now compute the time required to deposit the 5.1 10 mol Cr ^2 .

2 6 mol e 96485 C 1s 1h

time = 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- 3

2 7 2

Cr O (aq) 14 H (aq) 6e ^  ^2 Cr (aq) 7H O(l)^  Note that precipitation occurs

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 CrO₄^2 is smaller than isCr O ,_{2 7}^2 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 3

Mg 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 P_O2 = 0.20 atm

E = E -

2

3+ 4 2+ 4 + 4

O

0.0592 [Fe ]

n log [Fe ] [H ] P

 

 

 

  = 0.44 V -

[Fe ]3+ 4

0.0592

4 log [Fe ]^{2+ 4}[10^{-3.25 4}] 0.20

 

 

 

 

E = 0.24 V (spontaneous under these conditions)

40. (M) {NiO(OH)(s) + e^ + H⁺(aq)  Ni(OH)2(s) }2 Ered

{Cd(s) + 2 OH^ (aq)  Cd(OH)2(s) + 2 e^ }1 Eox

{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) Ecell  1.50 V Eox may be obtained by combining the Ksp value for Cd(OH)2(s)

2

14 spCd(OH)

(K 2.5 10 ) ^ and Ered = -0.403 V for Cd²⁺(aq) + 2 e^-  Cd(s). Hence, Cd(s)  Cd²⁺(aq) + 2 e^ Eox = 0.403 V

Cd²⁺(aq) + 2 OH^-(aq)  Cd(OH)2(s) Ecell =

-14 sp

0.0592log 1 = 0.0592log 1

n K 2 2.5 10

   

   

    

 

= 0.403 V Cd(s) + 2 OH^-(aq)  Cd(OH)2(s) + 2 e^- Eox = 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 Ered

{Cd(s) + 2 OH^ (aq)  Cd(OH)2(s) + 2 e^ } 1 Eox = 0.806 V

2 NiO(OH)(s) + Cd(s) + 2 H2O(l)  Cd(OH)2(s) + 2 Ni(OH)2(s) Ecell  1.50 V Ered = 1.50 V  0.806 V = 0.69 V

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 ^{+ H}2

 

g ^{ Cu s}

 

^{+ 2H+}

 

^aq

(b) Au⁺

 

aq ^{+ Fe2+}

 

^{aq  Au s}

 

^{+ Fe3+}

 

^aq

(c) 2Cu²⁺

 

aq + SO g + 2 H O(l)2

 

2 2 Cu aq + SO⁺

 

4^2

 

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}

 

^{+ 2H}2O (l)

Net: 3Ag s + NO

 

3^

 

aq + 4H aq⁺

 

3Ag aq + NO g + 2H O(l)⁺

   

2

Oxidation: {Ag s

 

^{ Ag+}

 

^{aq + e} 2}

Reduction: SO4^2

 

aq + 4H aq + 2e⁺

 

^ SO g + 2H O(l)2

 

2

Net: 2 Ag s + SO

 

4^2

 

aq + 4 H aq⁺

 

2 Ag aq + SO g + 2 H O(l)⁺

 

2

 

2

45. (M) The Integrative Example showed

2+

6

c + 2

= 1.2 10 = Cu

K  Cu 

 

 

or

2+ 6 + 2

Cu = 1.2 10 Cu

    

   

(a) WhenCu⁺= 0.20 M, ^Cu2+= 1.2 10 0.20 = 4.8 10 M. ⁶

 

²  ⁴ This is an impossibly high concentration. Thus Cu⁺= 0.20 M can never be achieved.

(b) WhenCu⁺= 1.0 10 ^10 M, ^Cu2+= 1.2 10 1.0 10 ⁶



 ^10



² = 1.2 10 ^14 M. This is an entirely reasonable (even though small) concentration; Cu⁺= 1.0 10 ^10 Mcan 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.

Step(a): CO2(g) + OH^ (aq)  HCO3^ (aq)

Step(b): HCO3^ (aq) + OH^ (aq)  CO32 (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 Hg22+/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 =

O2

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

PV = nRT  n P

V  RT = [Hg] =

(0.001811mmHg)( 1atm ) 760 mmHg L atm

0.08206 (298K) K mol

 

 

 

= 9.744  10^8 mol L^1

Next, convert to mg Hg m^3:

[Hg] = ^{8 mol 1000}₃ 200.59 g Hg 1000 mg

9.874 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. E_mole =N E_{A photon} E_photon =hv c v=  or v c= / Then, some algebra. E_mole = N E_{A photon} = N hv N hc_A = _A / or  =N hc E_A / _mole

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 nm For ZnO, =

290 10 J mol 1m

1.196 10 J mol nm

= = 413 nm violet light

290 10 J mol

 ^{ } _ ^





     







For CdS, 1.196 10 J mol nm⁸₃ ¹₁

= = 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).

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 (E^Sn /Sn2+ = -0.137 V vs. E^Zn /Zn2+ = -0.763 V). As well, tin is more malleable and more easily prepared in pure form (free of oxide).

56.

(M)

(aq) OH 4 (aq) Cu 2 O(l)

H 2 (g) O Cu(s) 2 :

Net

(aq) OH 4 e

4 O(l) H 2 (g) O : Reduction

2 } e 2 (aq) Cu {Cu(s)

: Oxidation

2 2

2 2 2

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) OH 2 (g) SO :

Total

) ( O H (aq) SO

(aq) OH (aq) HSO :

(b) Step

(aq) HSO (aq)

OH (g) SO :

(a) Step

2 2

4 3

2 2

4 4

4 3























 

 



 

l

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

2 2

2 2 3 4

2 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, 2 Cu^2(aq)2 OH^(aq)SO₄^2(aq)Cu₂(OH)₂SO₄(s) 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

(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

Sfusion <<Svaporization.

59. (M) Oxidation: 2I^(aq)I₂(s)2 e^ E0.535V

2

2 2 cell

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 13

1 1

2 13

2 redox

ln

2 mol e 96485 C/mol e 0.376 V

ln 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

K nFE K

RT

K

 

 

 

    

 

      

  

       



     

 

 

2

2 12 2

sp 2

2 13

redox 11

2 12)

sp

2 1/ 1/(1.1 10 ) Total: Cu (aq) 4 I (aq) 2 CuI s I (s)

2 10 2 10 Clearly this reaction will occur as written.

(1.1 10

K

K K K

K



 





  

  

    



I2 and I^- form I3- in aqueous solution. So, adding an additional mole of I^- to the last reaction yields: Cu^2(aq)5I^(aq)2CuI

 

s I₃^-(s)