Sample Problems for the Final CH142, Spring 2016
1. Calculate the standard state cell voltage, standard state reaction Gibbs energy, and
equilibrium constant at 25.0°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°r ed(Zn2+,Zn)
= 0.77 V – (-0.76 V) = 1.53 V
∆G° = – n F E°c ell = – 2(96485 C mol-1)(1.53 V)(1 kJ/1000 J) = -295. 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:
∆G°= -2.952x105 J mol-1 = – RT ln Ka = – 8.314 J K-1 mol-1(298.2 K) ln Ka
RT
0.02569 V
or
E°c ell = 1.53 V =
ln Ka =
ln Ka
nF
2
Ka = 5.37x1051
(your answer may differ slightly because of method or round-off error, e.g. 5.14x1051)
2. Calculate the electrochemical cell voltage at 25.0°C for the reaction:
H2O2 + 2 H+ + Cu (s) → 2 Η2O + Cu2+
given that [H2O2] is 0.500 M, the [Cu2+] is 1.00x10-3 M, and the pH is 7.00.
Answer: E°c ell = E°R – E°L = E°r ed(cathode) – E°r ed(anode) = E°r ed(H2O2,H2O) – E°r ed(Cu2+,Cu)
= 1.763 V – 0.34 V = 1.423 V
Nernst Equation relates the non-standard state and standard state cell potentials with n = 2:
2+
0.05916 V
0.05916 V
 [Cu ] 
E = E° –
log
Q
=
1.423
V
–
log


n
2
[H2O2][H+]2
0.05916 V
1.00x10–3
0.05916 V


= 1.423 V –
log (0.500)(1.00x10-7)2 = 1.423 V –
(11.301)
2
2


= 1.09 V
3. Calculate the solubility of LaF3 in pure water, Ksp = 2.0x10-19.
Answer:
Ksp = [La3+][F–]3 = (s)(3s)3 = 27 s4
s = [2.0x10-19/27]1/4 = 9.3x10-6 M
4. 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
+
–
with Ka = 2.5x10-9
reaction that determines the pH: HBrO →
← 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 M
0.0500 L
0.0500 L
0.0500 L
[H+][BrO–]
x2
x2
Ka =
=
≅
[HBrO]
[HBrO]o – x [HBrO]o
x2
+
-5
giving Ka = 2.5x10-9 =
or pH = 4.95
0.0500 with x = [H ] = 1.12x10 M
5. 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–
-3
-3
initial: 3.75x10
2.50x10 mol
final:
1.25x10-3 ~ 0
[H+] = Ka
OAc– + H2O
OH– is limiting
2.50x10-3 mol
-3
[HOAc]
-5 1.25x10 mol
-6
- = 1.8x10
2.50x10-3mol = 9.0x10
[OAc ]
or pH = 5.05
6. For the first-order reaction 2N2O5 → 2N2O4 + O2, the activation energy is 106. kJ/mol.
How many times faster will the reaction go at 100.0°C than at 25.0°C?
Ea  1 1 
kT2
Answer: lnk  = – R T – T 
 T1
 2
1
Ea = 106. kJ mol-1
R = 8.314 J K-1 mol-1
T2 = 373.2 K
T1 = 298.2 K
106.x103 J mol-1  1
1 
kT2
lnk  = – 8.314 J K-1 mol-1  373.2 K – 298.2 K = 8.59
 T1


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
ln 2 0.693
t½ = k = k
k=Ae
-Ea/RT
Ea  1 1 
k T2
ln k = – R T – T  or
 2
T1
1
-b ± b2- 4ac
2a
1
t½ = [A] k
o
E
 a 1
ln k= –  R  T + ln A
 
∆S = nR ln(V2/V1)
k T 1 Ea  1 1 
ln k = R T – T 
 2 1
T2
Kp = Kc (RT)∆n
x=