Practice Final CH142, Spring 2012
First here are a group of practice problems on Latimer Diagrams:
1. The Latimer diagram for nitrogen oxides in given below. Is NO stable with respect to
disproportionation under standard conditions at 25°C?
0.96 V
|
| 1.59 V
0.79 V
1.12 V
1.00 V
1.77 V
0.27 V
NO3  NO2(g)  HNO2  NO  N2O  N2  NH4+
|
1.25 V
|
2. The Latimer diagram for manganese in acidic solution is given below at 25°C. Find the
standard reduction potential for the reduction of permanganate ion, MnO4-, to Mn2+ from the
potentials listed.
0.56 V
2.26 V
0.95 V
1.51 V
-1.18 V
MnO4-  MnO42-  MnO2 (s)  Mn3+  Mn2+  Mn (s)
|
1.69 V
|
1.23 V
|
3. The Latimer diagram for manganese in acidic solution is given below at 25°C. (a). Give the
best oxidizing agent under standard conditions. (b). Give the best reducing agent. (c). Is Mn(s) a
good oxidizing agent? (d). What are the products of the disproportionation of Mn3+ ?
0.56 V
2.26 V
0.95 V
1.51 V
-1.18 V
3+
2+
2MnO4  MnO4  MnO2 (s)  Mn  Mn  Mn (s)
|
1.69 V
|
1.23 V
|
-
4. Given the following standard reduction potentials, construct the Latimer diagram:
BrO4 + 2H + 2e → BrO3 + H2O
BrO3– + 5H+ + 4e– → HBrO + 2H2O
BrO3– + 6H+ + 5e– → ½ Br2 (l) + 3H2O
HOBr + H++ e– → ½ Br2 (l) + H2O
Br2 (l) + 2e- → 2Br–
–
+
–
–
E°r ed (V)
1.745 V
1.49 V
1.513 V
1.584 V
1.078 V
Chemistry 142
Final
Name ______________________
Part 1: Answer 6 of the following 7 questions. If you answer more than 7 cross out the problem
that you don’t wish to have graded. (10 points each)
1. Name the following compounds or supply the formula:
(a). KClO4 ______________
(b). sulfurous acid __________________
(c). [Al(OH)4]– ______________
(d). diaquadiamminecobalt(II) chloride _________
2. (a). Name a good oxidizing agent ______________
(b). Name a good reducing agent _____________
(c). Name or draw the structural formula for a good multi-dentate ligand:
(d). An element that has several allotropes and name two of the allotropes:
3. (a). Give the balanced chemical reaction for Zn(OH)2 (s) dissolving in a strong acid:
(b). Give the balanced chemical reaction for Zn(OH)2 (s) dissolving in a strong base:
4. (a). Give the balanced chemical reaction for Fe3+ acting as an acid:
(b). Write a balanced chemical reaction for the disproportionation of hydrogen peroxide:
5. Choose the spontaneous direction for the following aqueous reactions at 298 K:
(a). at standard state:
Zn (s) + Cr3+ → Zn2+ + Cr2+ (s)
or
Zn2+ + Cr2+ (s) → Zn (s) + Cr3+
(b). at 0.10 M concentrations for each species:
[Cu(NH3)4]2+ → Cu2+ + 4 NH3
or
Cu2+ + 4 NH3 → [Cu(NH3)4]2+
(c). at 0.10 M concentrations for each species:
PO43- + H2O → HPO42- + OHor
HPO42- + OH- → PO43- + H2O
(d). BaSO4 (s) with 0.10 M Na2SO4:
BaSO4 (s) → Ba2+ + SO42–
Ba2+ + SO42–→ BaSO4 (s)
or
6. Is perchlorate ion a better oxidizing agent in acidic or basic solution? ____________
In acidic solution:
1.19 V
1.21 V
1.65 V
1.63 V
1.36 V
ClO4  ClO3  HClO2  HOCl  Cl2  Cl|
|
1.47 V
In basic solution:
0.36 V
0.35 V
0.65 V
0.40 V
1.36 V
ClO4-  ClO3-  ClO2-  OCl-  Cl2  Cl|
0.88 V
|
7. For solid [Cu(H2O)4(NH3)2]Cl2:
(a). The coordination number is: ________
(b). The oxidation state of the metal is: _______
(c). The coordination geometry is : ______
(d). The ions produced in aqueous solution of the compound:
Part 2: Answer 4 of the following 5 problems. If you answer more than 4 cross out the problem
that you don’t wish to have graded. (10 points each)
1. Calculate the standard state cell voltage, standard state reaction Gibbs energy, and
equilibrium constant at 25°C for the reaction: Zn (s) + 2 Fe3+ → Zn2+ + 2 Fe2+.
2. Calculate the solubility of PbBr2 in pure water, Ksp = 2.1x10-6.
3. Calculate the pH at the equivalence point of a titration of 25.0 mL of 0.100 NaBrO with 0.100
M HCl. Ka for HBrO is 2.5x10-9. Show the reaction that determines the pH.
4. Calculate the pH of a solution prepared from 25.0 mL of 0.150 M acetic acid and 25.0 mL of
0.100 M NaOH. Ka for acetic acid is 1.8x10-5.
5. For the first-order reaction 2N2O5 → 2N2O4 + O2, the activation energy is 106. kJ/mol.
How many times faster will the reaction go at 100.°C than at 25.0°C?
Standard Reduction Potentials at 25°C
F2 (g) + 2e- → 2FH2O2 + 2 H+ + 2 e- → 2 Η2O
PbO2 (s) + 4H+ + SO42- + 2e- → PbSO4(s) + 2H2O
MnO4- + 8H+ + 5e- → Mn2+ + 4H2O
PbO2 (s) + 4H+ + 2e- → Pb2+ + 2H2O
Cl2 (g) + 2e- → 2ClCr2O72- + 14H+ + 6e- → 2Cr3+ + 7H2O
O2 (g) + 4H+ + 4e- → 2H2O
Br2 (g) + 2e- → 2BrNO3- + 4H+ + 3e- → NO (g) + 2H2O
Hg2+ + 2e- → Hg (l)
Ag+ + e- → Ag (s)
Fe3+ + e- → Fe2+
O2 (g) + 2 H+ + 2 e- → Η2O2
I2 (s) + 2e- → 2ICu+ + e- → Cu (s)
Cu2+ + 2e- → Cu (s)
AgCl (s) + 1 e- → Ag (s) + ClCu2+ + e- → Cu+
Sn4+ + 2e- → Sn2+
2H+ + 2e- → H2 (g)
Fe3+ + 3e- → Fe (s)
Pb2+ + 2e- → Pb (s)
Sn2+ + 2e- → Sn (s)
Ni2+ + 2e- → Ni (s)
PbSO4 (s) + 2e- → Pb + SO42Cr3+ + e- → Cr2+
Cd2+ + 2e- → Cd (s)
Fe2+ + 2e- → Fe (s)
Cr3+ + 3e- → Cr (s)
Zn2+ + 2e- → Zn (s)
V2+ + 2e- → V (s)
Mn2+ + 2e- → Mn (s)
Al3+ + 3e- → Al (s)
Mg2+ + 2e- → Mg (s)
Na+ + e- → Na (s)
Ca2+ + 2e- → Ca (s)
K+ + e- → K (s)
Li+ + e- → Li (s)
OCl– + H2O (l) + 2 e- → O2 (g) + 2 OH–
O2 (g) + 2 H2O (l) + 4 e- → 4 ΟΗ−
2H2O + 2e- → H2 (g) + 2OH–
E°r ed (V)
2.87
1.763
1.69
1.49
1.46
1.36
1.33
1.23
1.078
0.96
0.85
0.80
0.77
0.695
0.54
0.52
0.34
0.222
0.15
0.15
0.00
-0.04
-0.13
-0.14
-0.25
-0.359
-0.40
-0.40
-0.41
-0.74
-0.76
-1.18
-1.18
-1.66
-2.37
-2.714
-2.866
-2.925
-3.045
+0.890
+0.401
-0.828
Formulas and Constants Given on the ACS Test
R = 8.314 J mol-1⋅K-1
R = 0.0821 L atm mol-1
1 F = 96,485. C mol-1
1 F = 96,485. J V-1 mol
Arrhenius Equation:
-E /RT
k=Ae a
NA = 6.022x1023 mol-1
h = 6.626x10-34 J s
c = 2.998x1010 m s-1
0°C = 273.15 K
Nernst Equation:
RT
E = E° –
ln Q
nF
Nernst Equation at 25°C:
0.05916 V
E = E° –
log Q
n
Integrated Rate Laws:
zero: [A] = [A]o – kt
first: ln [A] = ln [A]o – k t
1
1
second:
= kt+
[A]t
[A]o
Additional Formulas and Constants Given on the Colby Test
t½ =
ln 2 0.693
=
k
k
k = A e-Ea/RT
ln
Ea 1 1 
kT2
= –  –  or
kT1
R T2 T1
1
[A]o k
Ea 1
ln k= –   + ln A
R T
t½ =
ln
kT1 Ea 1 1 
=
–
kT2 R T2 T1
x=
-b ± b2- 4ac
2a
∆S = nR ln(V2/V1)
Kp = Kc (RT)∆n