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Chemistry 162 Exam III April 22, 2015Sshowing page 11-21 out of 28

10
Chem 162-2015 Exam III
H&P Chapter 18 = Tro Chapter 20 = Electrochemistry
Oxidation-Reduction Reaction Concepts
What is the oxidation number of As in K(NH
4
)
2
AsO
4
.
6H
2
O
A.
2
B.
3
C
.
5
D.
8
E.
10
Although one can place brackets within this compound to help determine the oxidation
number, this compound is not a coordination compound, so brackets aren’t required.
K(NH
4
)
2
AsO
4
.
6H
2
O
K
+
[(NH
4
)
2
AsO
4
.
6H
2
O]
-1
2x+1
X -8
6x0
[(NH
4
)
2
AsO
4
.
6H
2
O]
-1
+2 +X -8 +0 = -1
X = +5
C
C162s15e3v1
11
11
Chem 162-2015 Exam III
H&P Chapter 18 = Tro Chapter 20 = Electrochemistry
Galvanic Cells & Reduction-Potential Concepts
E
o
(volts)
X
2+
+ 2e
-
X
-0.13
Y
2+
+ 2 e
-
Y
-0.44
Z
2+
+ 2e
-
Z
-0.76
Which of the following is true under standard conditions?
A.
Z
2+
ions oxidize X metal
B.
Y metal reduces Z
2+
ions
C.
Y is a better reducing agent than Z
D.
Z
2+
ions are better oxidizing agents than X
2+
ions
E
.
X
2+
ions oxidize Y metal
ET note: It is assumed that the reactions are spontaneous.
For oxidation potential, reverse the equations and reverse the voltage signs.
Z → Z
2+
+ 2e
-
+0.76 volts
Y → Y
2+
+ 2e
-
+0.44 volts
X → X
2+
+ 2e
-
+0.13 volts
A.
False.
if Z
2+
ions oxidize X metal, then Z
2+
would be reduced, and X would be
oxidized:
Z
2+
+ 2e
Z
-0.76
X → X
2+
+ 2e
-
+0.13
Z
2+
+ X → Z + X
2+
-0.63
-0.63 is not a spontaneous reaction
B.
False.
This is not a spontaneous reaction.
Y → Y
2+
+ 2e
-
+0.44 volts
Z
2+
+ 2e
-
Z
-0.76
-0.32
C.
False.
A better reducing agent is more easily oxidized.
But Y has a smaller oxidation
potential than Z, so it is less easily oxidized.
D.
False.
Z
2+
has a smaller reduction potential voltage than X
2+
.
Therefore, it is a weaker
oxidizing agent than X
2+
.
E.
True. The voltage is positive; therefore, the reaction is spontaneous.
X
2+
+ 2e
X
-0.13
Y → Y
2+
+ 2e
-
+0.44
+0.31 volts
E
C162s15e3v1
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12
Chem 162-2015 Exam III
H&P Chapter 16 = Tro Chapter 18 = Other Equilibria (Solubility Product, Complex Ions)
Complex Ions
AgBr(s)
Ag
+
(aq) + Br
-
(aq)
K
sp
= 5.3
10
-13
Ag
+
(aq) + 2NH
3
(aq)
Ag(NH
3
)
2
+
(aq)
K
f
= 1.7
10
7
When NH
3
is added to AgBr(s), a combination of these two reactions occur:
AgBr(s) + 2NH
3
(aq)
Ag(NH
3
)
2
+
(aq) + Br
-
How many moles of AgBr can dissolve in 1.0 L of 2.5 M NH
3
?
A.
0.012 mol
B.
3.4
10
-7
mol
C.
2.4
10
-5
mol
D
.
7.5
10
-3
mol
E.
0.048 mol
AgBr(s)
Ag
+
(aq) + Br
-
(aq)
K
sp
= 5.3
10
-13
Ag
+
(aq) + 2NH
3
(aq)
Ag(NH
3
)
2
+
(aq)
K
f
= 1.7
10
7
AgBr(s) + 2NH
3
(aq)
Ag(NH
3
)
2
+
(aq) + Br
-
(aq)
K = 9.01 x 10
-6
AgBr(s) +
2NH
3
(aq)
Ag(NH
3
)
2+
+
Br
-
Y
2.5
0
0
-X
-2X
+X
+X
Y - X
2.5 – 2X
+X
+X
[Ag(NH
3
)
2+
][Br
-
]/[NH
3
]
2
= 9.01 x 10
-6
[X][X]/[2.5 – 2X]
2
= 9.01 x 10
-6
Quadratic equation; therefore, use small K rule.
[X][X]/[2.5]
2
= 9.01 x 10
-6
X = 7.50 x 10
-3
M
D
C162s15e3v1
13
13
Chem 162-2015 Exam III
H&P Chapter 17 = Tro Chapter 19 = Thermodynamics
Entropy Calculations
Calculate ΔS°
univ
for the following reaction at 100 °C
CO(g) + H
2
O(g)
CO
2
(g) + H
2
(g)
∆H° = -41 kJ
S° (J/mol·K)
CO(g)
200
H
2
O(g)
199
CO
2
(g)
214
H
2
(g)
131
A.
18 J/K
B
.
56 J/K
C.
-54 J/K
D.
33 J/K
E.
7.0 J/K
ΔS
o
univ
= ΔS
o
sys
+ ΔS
o
surr
ΔS
o
system
= Σn
p
S
o
products
- Σn
r
S
o
reactants
ΔS
o
system
= (214 + 131) – (200 + 199) = -54J
ΔS
o
surr
= ΔH
o
surr
/T = -ΔH
o
system
/T
ΔS
o
surr
= -(-41000)/(273 + 100) = 109.92J
ΔS
o
univ
= -54J + 109.92J = 55.9J
B
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14
Chem 162-2015 Exam III
H&P Chapter 18 = Tro Chapter 20 = Electrochemistry
Free Energy, Equilibria and Cell Potential Calculations/Nernst Equation
Consider the galvanic cell based on the following reaction:
2Al(s) + 3Fe
2+
(aq, 2.0 M)
2A1
3+
(aq, 0.030M) + 3Fe(s)
E
o
cell
= 1.22 V
Calculate E
cell
for this cell.
A.
1.44V
B.
1.55V
C.
1.44V
D.
1.32V
E
.
1.26V
2(Al → Al
3+
+ 3e
-
)
3(Fe
2+
+ 2e
-
Fe)
2Al(s) + 3Fe
2+
(aq)
2A1
3+
(aq) + 3Fe(s)
Nernst equation: E
cell
= E
o
cell
-(0.0592/n)log(Q)
E
cell
= 1.22 – ((0.0592/6) x (log((0.030
2
)/(2
3
))))
E
cell
= 1.259 volts
E
C162s15e3v1
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15
Chem 162-2015 Exam III
H&P Chapter 16 = Tro Chapter 18 = Other Equilibria (Solubility Product, Complex Ions)
Solubility Product
When solid barium fluoride, BaF
2
, is placed in enough water to make 1.5 L of solution,
it is found that 1.97 g of BaF
2
(molar mass 175.3 g/mol)
dissolves to reach equilibrium
at 25
C.
Calculate K
sp
for BaF
2
at 25
C.
A
1.4
10
-5
B.
5.6
10
-5
C.
2.2
10
-4
D.
4.2
10
-7
E
.
1.7
10
-6
BaF
2
→ Ba
2+
+ 2F
-
1.97gBaF
2
/175.3g/mol = 0.011238 mol BaF
2
in 1.5L = 0.007492 mol BaF
2
/L
BaF
2
Ba
2+
+
2F
-
Initial
Y
0
0
Change
-0.007492
+0.007492
0.014984
Equilibrium
Y – 0.007492
+0.007492
0.014984
[Ba
2+
][F
-
]
2
= K
sp
[0.007492][0.014984]
2
= K
sp
1.68 x 10
-6
= K
sp
E
C162s15e3v1
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16
Chem 162-2015 Exam III
H&P Chapter 17 = Tro Chapter 19 = Thermodynamics
Free Energy and Equilibria Calculations
N
2
(g) + 3H
2
(g)
2NH
3
(g)
G
= -16.4 kJ/mol for NH
3
Calculate
G at 298 K for this reaction when:
p
N
2
= 0.012 atm, p
H
2
= 0.050 atm, and p
NH
3
= 3.00 atm
A.
-33 kJ
B.
-21 kJ
C
.
+5.9 kJ
D.
-6.5 kJ
E.
+13 kJ
ΔG = ΔG
o
+ RT ln(Q)
Q = P
2
nh3
/(P
n2
x P
3
h2
)
Note that
G
is, by definition, the
G
o
for the formation of one mole of NH
3
, but
the equation is the formation of two moles of NH
3
.
ΔG = (2 x (-16400)) + (8.314 x 298 x
(ln((3.00
2
)/(0.012 x (0.050
3
))))) = 5.87 x
10
3
J
C
C162s15e3v1
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17
Chem 162-2015 Exam III
H&P Chapter 17 = Tro Chapter 19 = Thermodynamics
Entropy Concepts
What is
W
in Boltzmann's formula,
S
=
k
ln
W
?
A.
a fraction indicating the probability of obtaining a result
B.
a random number
C
.
the number of microstates describing the system
D.
work times Avogadro’s number
E.
number of particles in the system
S is entropy.
k is the Boltzmann constant (the gas constant divided by Avogadro’s number).
W is the number of ways (i.e., microstates) to arrange the components of a system.
C
C162s15e3v1
18
18
Chem 162-2015 Exam III
H&P Chapter 18 = Tro Chapter 20 = Electrochemistry
Free Energy, Equilibria and Cell Potential Calculations/Nernst Equation
E
o
(
volts
)
Ag
+
(aq) + e
Ag(s)
0.80
Pb
2+
(aq) + 2e
-
Pb(s)
-0.13
Based on this data, what is the equilibrium constant for the reaction:
Pb(s) + 2Ag
+
(aq)
2Ag(s) + Pb
2+
(aq)
A.
3.8
10
-32
B
.
2.6
10
31
C.
2.0
10
-16 
D.
5.1
10
15
E.
2.8
10
58
2(Ag
+
(aq) + e
-
Ag(s))
0.80
Pb(s)
Pb
2+
(aq) + 2e
-
+0.13
Pb(s) + 2Ag
+
(aq)
2Ag(s) + Pb
2+
(aq)
0.93 V
E
o
cell
= (0.0592/n)log(K)
0.93 = (0.0592/2) x log(K)
K = 2.62 x 10
31
B
C162s15e3v1
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19
Chem 162-2015 Exam III
H&P Chapter 17 = Tro Chapter 19 = Thermodynamics
Free Energy Calculations
At what temperatures will the following reaction be
nonspontaneous
under standard
conditions?
2HNO
3
(
aq
) + NO(
g
)
3 NO
2
(
g
) + H
2
O(
)
ΔH
o
= +136.5 kJ
ΔS
o
= +287.5 J/K
A.
Any temperature below 276 K
B.
Any temperature above 202 K
C
.
Any temperature below 475 K
D.
Any temperature above 475 K
E.
Any temperature below 202 K
ΔG
o
= ΔH
o
- TΔS
o
The borderline for when the reaction will be spontaneous vs non-spontaneous is when
the reaction is at equilibrium.
0 = ΔH
o
- TΔS
o
T = ΔH
o
/ΔS
o
T = 136500J/287.5J = 474.78K
ΔH
o
is a positive value.
ΔS
o
is a positive value.
Therefore, -TΔS
o
is a negative value.
ΔG
o
will be negative (i.e., spontaneous) when the magnitude of -TΔS
o
is greater
than the magnitude of ΔH
o
, which is when T is larger than 474.78K.
ΔG
o
will
be positive (i.e., non-spontaneous) when the magnitude of -TΔS
o
is smaller than
the magnitude of ΔH
o
, which is when T is smaller than 474.78K.
C
C162s15e3v1
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20
Chem 162-2015 Exam III
H&P Chapter 16 = Tro Chapter 18 = Other Equilibria (Solubility Product, Complex Ions)
Solubility Product
A saturated solution of CaCO
3
(K
sp
= 5.0
10
-9
) contains CaCO
3
(s) in equilibrium with
its ions.
Which of the following will increase the molar solubility of CaCO
3
(s)?
X.
Add HCl to the saturated solution
Y.
Add Na
2
CO
3
to the saturated solution
Z.
Add additional water to the saturated solution.
A.
X only
B.
Y only
C.
Z only
D.
X and Y only
E.
X, Y, and Z
CaCO
3
→ Ca
2+
+ CO
3
2-
CaCO
3
Ca
2+
+
CO
3
2-
Y
0
0
-X
+X
+X
Y - X
+X
+X
CO
3
2-
+ H
+
→ HCO
3
-
Acid will react with carbonate (a base) removing it, shifting the reaction to the right,
thereby dissolving more CaCO
3
.
Addition of CO
3
2-
is a common ion effect.
This will move the reaction to the left, thereby
resulting in less CaCO
3
dissolving.
A saturated solution of CaCO
3
contains excess solid in the bottom of the beaker.
Addition
of water will lower the concentration of Ca
2+
and CO
3
2-
, pushing the reaction to the right,
thusly dissolving more CaCO
3
.
That is, the solubility of CaCO
3
will increase upon the
addition of water to the saturated solution.
However, note that the problem doesn’t ask for
solubility of CaCO
3
.
It asks for
molar
solubility of CaCO
3
.
This means the moles of
CaCO
3
dissolved per liter of solution.
The solubility per liter remains constant at ((5.0 x
10
-9
)
0.5
= ) 7.07 x 10
-5
M, regardless of dilution of the solution.
A
C162s15e3v1
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