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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
|
Answer: NO is unstable with respect to disproportionation under standard conditions at 25°C.
The right-hand standard reduction potential is greater than the left-hand standard reduction
potential to either HNO2 or NO3-.
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
3+
2+
2MnO4  MnO4  MnO2 (s)  Mn  Mn  Mn (s)
|
1.69 V
|
1.23 V
|
-
Answer: The following two half cells add to give the desired reaction. The Gibbs free energies
add to give the Gibbs free energy of the overall reaction:
E°red
MnO4 + 4 H + 3 e → MnO2 (s) + 2 H2O
MnO2 (s) + 4 H+ + 2 e– → Mn2+ + 2 H2O
+
–
MnO4- + 8H+ + 5e-
–
→ Mn2+ + 4H2O
1.69 V
1.23 V
∆G°
∆G° = – 3 FE°red = -489.2 kJ mol-1
∆G° = – 2 FE°red = -237.4 kJ mol-1
∆G° = – 5 FE°red = -726.6 kJ mol-1
After adding the Gibbs free energies, the overall voltage is given by solving ∆G° = – 5 FE°red:
E°red =
∆G°
-726.6x103 J mol-1
=
= 1.51 V
– 5 F – 5 (96485 C mol-1)
with 1 J = 1 C 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
23+
2+
MnO4  MnO4  MnO2 (s)  Mn  Mn  Mn (s)
|
1.69 V
|
1.23 V
|
-
Answers: (a). Give the best oxidizing agent under standard conditions: the most positive
reduction potential is for MnO42- . (b). Give the best reducing agent: the most negative reduction
potential is for Mn2+ +2 e– → Mn, so Mn (s) is the best reducing agent. (c). Is Mn(s) a good
oxidizing agent? No, metals don’t have negative oxidation states in aqueous solution. For
example to act as a reducing agent, a half reaction might be Mn (s) + e– → Mn– . (d). What are
the products of the disproportionation of Mn3+ ? The products are the species on either side of
Mn3+ giving the answers as MnO2 (s) and Mn2+.
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
Answer:
acidic solution:
1.745 V
1.49 V
1.584 V
1.078 V
BrO4-  BrO3-  HOBr  Br2  Br–
|
1.513 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 ____potassium perchlorate __________
(b). sulfurous acid _____H2SO3_____________
(c). [Al(OH)4]– _____tetrahydroxoaluminate (III) ion _________
(d). diaquadiamminecobalt(II) chloride ___ [Co(H2O)2(NH3)2]Cl2 ______
2. (a). Name a good oxidizing agent ____many possibilities including F2, ClO4–, NO3–, O2
(b). Name a good reducing agent ____many possibilities including S2O32– (thiosulfate),
SO32–, Na, Li, Ca, Cu+ _________
(c). Name or draw the structural formula for a good multi-dentate ligand:
some possibilities are: porphyrin, CO32-, C2O42- (oxalate), EDTA,
bipyridine, H2NCH2CH2NH2 (ethylenediamine)
(d). An element that has several allotropes and name two of the allotropes:
some possibilities are: O2 and O3 ; C(graphite) and C(diamond) and C60;
S8 and linear polymeric S-S-S-S-S, white and red P: P4 and -P4-P4-P4- polymers
3. (a). Give the balanced chemical reaction for Zn(OH)2 (s) dissolving in a strong acid:
Zn(OH)2 (s) + 2 H+ → Zn2+ + H2O
(b). Give the balanced chemical reaction for Zn(OH)2 (s) dissolving in a strong base:
Zn(OH)2 (s) + 2 OH– → [Zn(OH)4]2–
4. (a). Give the balanced chemical reaction for Fe3+ acting as an acid:
[Fe(H2O)6]3+ → [Fe(H2O)5(OH)]2+ + H+
(b). Write a balanced chemical reaction for the disproportionation of hydrogen peroxide:
Combine the standard reduction reactions listed at the end of the test:
H2O2 + 2 H+ + 2 e- → 2 Η2O
–{ O2 (g) + 2 H+ + 2 e- → Η2O2}
to give: 2 H2O2 → O2 (g) + 2 Η2O
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:
SO42- + H2O → HSO4- + OHor
HSO4- + OH- → SO42- + 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? ___acidic (1.19 V)___
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:
0.35 V
0.65 V
0.40 V
1.36 V
0.36 V
ClO4  ClO3  ClO2  OCl  Cl2  Cl|
0.88 V
|
7. For solid [Cu(H2O)4(NH3)2]Cl2:
(a). The coordination number is: ___6_____
(b). The oxidation state of the metal is: ___ (II)____
(c). The coordination geometry is : ___octahedral ___
(d). The ions produced in aqueous solution of the compound:
[Cu(H2O)4(NH3)2]Cl2 → [Cu(H2O)4(NH3)2]2+ + 2 Cl–
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+.
Answer: The Zn half cell is Zn → Zn2+ + 2e–. As written n = 2. The cathode is the reduction of
Fe3+ to Fe2+, and the anode the oxidation of Zn to Zn2+. The cell potential is:
E°c ell = E°R – E°L = E°r ed(cathode) – E°r ed(anode) = E°r ed(Fe3+,Fe2+) – E°red(Zn2+,Zn)
= 0.771 V – (-0.762 V) = 1.533 V
∆G° = – n F E°c ell = – 2(96485 C mol-1)(1.533 V)(1 kJ/1000 J) = -295.8 kJ mol-1
Remember that cell potentials are intensive, so they don’t depend on the stoichiometric
coefficients.There are two equivalent ways to find the equilibrium constant:
or
∆G°= -2.958x105 J mol-1 = – RT ln Ka = – 8.3145 J K-1 mol-1(298.15 K) ln Ka
RT
0.02569 V
E°c ell = 1.533 V =
ln Ka =
ln Ka
nF
2
51
Ka = 6.78x10
2. Calculate the solubility of PbBr2 in pure water, Ksp = 2.1x10-6.
Answer:
Ksp = [Pb2+][Br-]2 = (s)(2s)2 = 4 s3
s = [2.1x10-6/4]1/3 = 0.0081 M
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.
Answer:
titration reaction: BrO– + H+ → HBrO predominate species at the equiv. pt. is HBrO
+
–
reaction that determines the pH: HBrO →
with Ka = 2.5x10-9
← H + BrO
volume of titrant at equivalence point : 25.0 mL and total volume: 50.0 mL
nominal concentration at equivalence point:
moles HBrO moles BrO– 0.0250 L(0.100 M)
[HBrO]o = 0.0500 L = 0.0500 L =
= 0.0500 M
0.0500 L
+
–
2
2
[H ][BrO ]
x
x
=
≅
Ka =
[HBrO]
[HBrO]o – x [HBrO]o
x2
-9
giving Ka = 2.5x10 =
with x = [H+] = 1.1x10-5 M
or pH = 4.95
0.0500
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.
Answer:
initial moles of HOAc: 0.0250 L(0.150 M) = 3.75x10-3 mol
initial moles of OH– : 0.0250 L(0.100 M) = 2.50x10-3 mol
reaction: HOAc + OH–
→
initial: 3.75x10-3 2.50x10-3 mol
final:
1.25x10-3 ~ 0
OAc– + H2O
OH– is limiting
-3
2.50x10 mol
[HOAc]
1.25x10-3 mol
[H+] = Ka [OAc-] = 1.8x10-5 2.50x10-3mol = 9.0x10-6
or pH = 5.05
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?
Ea  1 1 
kT2
Answer: lnk  = – R T – T 
 2
 T1
1
Ea = 106. kJ mol-1
R = 8.314 J K-1 mol-1
T2 = 373.1 K
T1 = 298.2 K
106.x103 J mol-1  1
1 
kT2
ln  = –
–
= 8.59
-1
-1 
8.314 J K mol  373.1 K 298.2 K
kT1
k373.2 K
k
 = e8.59 = 5.38x103 times as fast
 298.2 K
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