EXAM FIVE
CHM 205 (Dr. Mattson)
18 APRIL 2008
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Instructions: Show all work whenever a calculation is required! You will receive credit for how you worked each problem as
well as for the correct answer. If you need more space, you may use the back of your periodic table — Write: “See PT” in box
and then attach the periodic table. BOX YOUR ANSWERS! Write legibly.
Chapter 17. Electrochemistry
1 (4 pts) Balance this reaction in acidic solution.
ClO2- +
Mg 
Cl- +
Mg+2
2. (5 pts) Check these reactions (unbalanced) if they are
spontaneous under standard conditions (> 1 possible):
Cu+2 + Ag  Cu + Ag+
Al+3 + Ca  Al + Ca+2
4(a). (5 pts) Consider the galvanic cell:
Pb(s)|Pb+2(1 M)||Au+(1 M)|Au(s) Eo = +1.82 V
Determine the standard reduction potential for
Au+(aq) + e-  Au(s) half cell.
4(b) (5 pts) Determine E for:
Pb(s)|Pb+2(1 M)||Au+(0.0080 M)|Au(s) Eo = +1.82 V
Cd + Zn+2  Cd+2 + Zn
Cd + Sn+2  Cd+2 + Sn
Ni + Cu+2  Ni+2 + Cu
3. Consider the galvanic reaction between the two
half cells, Zn|Zn+2(aq) and Sn|Sn+2(aq).
3(a) (5 pts) Determine the spontaneous reaction. Write
a balanced overall equation and determine Eo.
5. Consider the galvanic cell:
Co(s)|Co+2(aq)||Ni+2(aq)|Ni(s) Eo = +0.020 V
5(a) (5 pts) Determine the equilibrium constant, Kc?
3(b) (4 pts) Write the Galvanic cell using cell notation.
5(b) (5 pts) What is the value of ΔGo?
3(c) (2 pts) Which electrode is gaining mass? Zn or Sn
3(d) (2 pts) The cathode cell is: Zn|Zn+2 or Sn|Sn+2
3(e) (2 pts) What is the reducing agent?
(a) Zn
(b) Zn+2
(c) Sn
(d) Sn+2
6. (5 pts) What mass of rhodium (MM = 102.9 g/mol)
can be obtained from an aqueous solution of
rhodium(III) nitrate if the solution is electrolyzed for
15.0 min with a current of 22.0 A?
12. (6 pts) What is the coordination number and
geometry name for each of these complexes? (“en” is
the bidentate chelate NH2CH2CH2NH2)
[Co(en)(H2O)2Br2]+2
Cr(NH3)3Cl3
K[FeCl4]
13. (4 pts) How many isomers are possible for each of
these complexes?
(a) [FeCl (SCN) ]-3
3
3
(b) [V(NH3)2Cl4](c) [Cr(NH3)5Br]+2
Chapter 20. Transition Metals
(d) [PtCl2Br2]-2 (square planar)
7. (8 pts) Sketch Lewis dot diagrams in order to
determine which of these are Lewis bases. Check
those that are.
CH4
SH2
CO2
Chapter 22. Nuclear Chemistry
14. (6 pts) Write balanced nuclear equations for the
following processes:
(a) α-emission of
NH3
163
66
8. (2 pt) Transition metal cations are:
(a) Lewis bases
Dy
(b) β-emission of
(b) Lewis acids (c) neither
€
9(a) (2 pts) Write the electron configuration for V+3.
[Ar]
9(b) (2 pts) How many unpaired electrons are in V+3 ?
€
186
73
Ta
15. (3 pts) The half life of indium-111, a radioactive
isotope used in studying the distribution of white
blood cells, is t 1/2 = 2.805 days. Approximately what
percent of the isotope remains after three days?
(a) <10% (b) 10-25% (c) 25-50% (d) 50-75% (e) >75%
9(c) (2 pts) Is V+3 paramagnetic?
Yes or No
10. (5 pts) Consider the following periodic trends as
they apply to the transition metals in order to answer
these questions.
(a) Which element is smallest? Cr
Fe
Cu
(b) Which is largest?
Fe
Fe+2
Fe+3
(c) Which has smallest 1st ionization energy?
Sc
Mn
Cu
(d) Which has largest density?
Cu
V
Mn
16. (5 pts) How old is a clam shell (mostly calcium
carbonate) whose C-14 content is found to be 89%
that of a live clam? The half-life of 14C is 5730 years.
(e) Which has largest effective nuclear charge?
Ti
Fe
Cu
11. (6 pts) Determine the oxidation state of the metal
ion in each of these complexes. Box your answers!
K2[Ru(CN)6]
Sign the Academic Integrity pledge (on the front) and print your
name here:
Cr(NH3)3Cl3
K[Ti(NH3)3Br3]
Your exam score (100 possible):
Determine your grade:
A+ > 95; A > 90; B+ > 85; B > 80; C+ > 75; C > 70; D > 60
Standard Table of Reduction
Useful equations:
Potentials
Eo (V)
Cl2 + 2 e-  2Cl-
1.36
O 2 + 4 H + + 4 e -  2 H 2O
1.23
Br2 + 2 e-  2 Br-
1.09
Ag+ + e-  Ag
0.80
Fe+3 + e-  Fe+2
0.77
I2 + 2 e -  2 I-
0.54
O2 + 2 H2O + 4 e-  4 OH-
0.40
Cu+2 + 2 e-  Cu
0.34
2 H+ + 2 e-  H2
0.00
Fe+3 + 3 e-  Fe
-0.036
Pb+2 + 2 e-  Pb
-0.13
Sn+2 + 2 e-  Sn
-0.14
Ni+2 + 2 e-  Ni
-0.26
Co+2 + 2 e-  Co
-0.28
Cd+2 + 2 e-  Cd
-0.40
Fe+2 + 2 e-  Fe
-0.44
Cr+3 + e-  Cr+2
-0.50
Cr+3 + 3 e-  Cr
-0.73
Zn+2 + 2 e-  Zn
-0.76
2 H2O + 2 e-  H2 + 2OH-
-0.83
Mn+2 + 2 e-  Mn
-1.18
Al+3 + 3 e-  Al
-1.66
Mg+2 + 2 e- Mg
-1.66
Na+ + e-  Na
-2.71
Ca+2 + 2 e-  Ca
-2.76
K+ + e-  K
-2.92
Li+ + e-  Li
-3.05
E = Eo −
Eo =
€
0.0592
logQ
n
0.0592
logK
n
ΔG = −nFE
€
ΔG o = −nFE o
F = 96500 C /mol
€
Charge(coul) = Current(amps) × time(s)
€ 1 faraday = 1 mol e− = 96500 coul
€
ln
€
N0
= kt
Nt
t1/ 2 =
€
0.693
k
1
2
H
He
3
4
5
6
7
8
9
10
Li
Be
B
C
N
11
12
13
14
15
O
F
Ne
16
17
Na
Mg
Al
Si
P
18
S
Cl
Ar
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
K
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Ga
Ge
As
Se
Br
Kr
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
Rb
Sr
Y
Zr
Tc
Ru
Rh
Pd
Ag
Cd
In
Sn
Sb
Te
I
Xe
55
56
57
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
Cs
Ba
La
Hf
Ta
W
Re
Os
Ir
Pt
Au
Hg
Tl
Pb
Bi
Po
At
Rn
110
111
112
Nb Mo
87
88
89
104
105
106
107
108
109
Fr
Ra
Ac
Rf
Db
Sg
Bh
Hs
Mt
58
59
Ce
Pr
60
61
62
90
91
92
93
Th
Pa
U
Np
63
Nd Pm Sm Eu
94
114
116
118
64
65
66
67
68
69
70
71
Gd
Tb
Dy
Ho
Er
Tm
Yb
Lu
96
97
98
99
100
101
102
103
Cf
Es
No
Lr
95
Pu Am Cm Bk
Fm Md
Answers:
1 ClO2- + 2 Mg + 4 H+ Cl- + 2 Mg+2+ 2 H2O
2. Check these reactions (unbalanced) if they are spontaneous under standard conditions (> 1 possible):
Cu+2 + Ag  Cu + Ag+
√ Al+3 + Ca  Al + Ca+2
Cd + Zn+2  Cd+2 + Zn
√ Cd + Sn+2  Cd+2 + Sn
√ Ni + Cu+2  Ni+2 + Cu
3. (a) Eo = 0.62 v
(b) Zn|Zn+2||Sn+2|Sn
4(a) Eo = +1.69 V
(d) Sn|Sn+2
(c) Sn
(e) Zn
(b) Eo = +1.70 V
(b) ΔGo = -3.86 kJ
5(a) Determine the equilibrium constant, Kc = 4.74
6. 7.04 g Rh
7. CH4: AB4 (not a ligand)
SH2: AB2E2 (a ligand)
CO2: AB2 on central atom (not a ligand)
NH3: AB3E (a ligand)
8. (b)
9(a) V+3 4s0 3d2
10. (a) Cu
(b) two unpaired electrons
(b) Fe
(c) Sc
(c) Yes
(d) Cu
(e) Cu
11. +4, +3, +2
12. Coord. Number = 6, octahedral; Coord. Number = 6, octahedral; Coord. Number = 4, tetrahedral
13. 2, 2, 1, 2
14. (a)
163
66
Dy→42 α +
159
64
Gd
15. (c)
16. 960 years
€
€
(b)
186
73
Ta → −10β +
186
74
W