1
SEMESTER 2
FINAL EXAM REVIEW
Unit VIII – Solutions
I. Good things to know:
• Mixtures: homogeneous and heterogeneous
• Types of colloids: aerosol, foam, emulsion, sol
• types of solutions
• solution terms: solute, solvent, insoluble, soluble
• molecular view of solution process
• saturated, unsaturated, supersaturated; fractional crystallization
• solubility of solid solutes vs. temperature
• gas solubility vs. temperature and pressure (Henry’s Law)
• colligative properties, van’t Hoff Factor
II. Essays/Short Answer
1. Explain why ethanol (CH3CH2OH) is soluble in both water and carbon tetrachloride.
The CH3CH2 portion is nonpolar, therefore it will be more soluble in a
nonpolar solvent, such as carbon tetrachloride. The OH portion of the
molecule is more polar, thus making it also soluble in a polar solvent like water.
2. Give the estimated van’t Hoff factor for the following substances:
2
a. NaCl
1
b. CH3CH2OH (ethanol)
3
c. Ag2SO4
III. Problems
1. A solution is made by dissolving 67.8g of glucose (C6H12O6) in 50θ mL of water. Assume that the
density of the solution is 1.00 g/cm3,and that the resulting volume does not change with the addition of the
glucose. Calculate the mass percent, molarity, molality, and mole fraction of glucose.
500mL = 500 g water → 500 + 67.8 = 568g solution
67.8g gluc
%=
x100% = 11.9%
568 g sol' n
1 mol gluc 0.377 mol NaCl
=
= 0.753M
67.8g gluc ∗
0.500 L
180 g gluc
500 g H 2 O ∗
m=
1mol H 2 O
= 27.7 + 0.377 mol = 28.16mol
18 g H 2 O
0.377 mol gluc
0.377 mol NaCl
= 0.753m → X =
= 0.0134
0.500kg H 2 O
28.16mol
2
SEMESTER 2
FINAL EXAM REVIEW
2. You form a solution by dissolving 20.00g of the nonelectrolyte benzoic acid (C7H6O2) in 200.0 mL of
benzene (C6H6) (D = 0.8765 g/mL). The boiling point of benzene is 80.1oC,
with a Kb = 2.53 oC/m, and the melting point is 5.5oC, with a Kf = 5.12 oC/m.
a. calculate the molality of the solution
1 mol
= 0.164mol BA
0.164mol
122g BA
m=
= 0.936m
0.1753kg
0.8765g
200.0mL ∗
= 175.3 g = 0.1753kg
1mL
20.00 g BA ∗
b. calculate the boiling point of the solution
∆Tb = (2.53
o
C
m
)(0.936m) = 2.37 o C → 80.1 + 2.4 = 82.5 o C
c. calculate the freezing point of the solution
∆T f = (5.12
o
C
m
)(0.936m) = 4.79 o C → 5.5 − 4.8 = 0.7 o C
3. The molecular weight and formula of a hydrocarbon are to be determined through the use of the freezing
point depression method. The hydrocarbon is known to be 86% carbon and 14% hydrogen by mass. In the
experiment, 3.72g of the unknown hydrocarbon were placed into 50.0g of liquid benzene (C6H6). The
freezing point of the solution was measured to be 0.06oC. The normal freezing point of benzene is 5.50oC,
and the freezing point depression constant for benzene is 5.12 oC/m.
a. What is the molecular weight of the compound?
∆T f = 5.50 − 0.06 = 5.44 o C = (5.12
FM =
o
C
m
)m → m = 1.06 =
x
→ x = 0.053125mol
0.050kg
3.72 g
= 70.0 g mol
0.053125mol
b. What is the molecular formula of the hydrocarbon?
1C
= 7.16 C
12g C
70
÷ 7.16 → CH 2 ( FM = 14) →
= 5 → mf = C 5 H 10
1H
14
14g H ∗
= 14 H
1g H
86 g C ∗
c. What is the mole fraction of benzene in the solution?
0.0531mol C5 H10
50.0 g benzene ∗
X benzene =
0.0531 + 0.641 = 0.694mol
1 mol
= 0.641mol benzene
78 g
0.641mol
= 0.924
0.694mol
d. If the density of the solution is 875 g/L, what is the molarity of the solution?
50.0 g + 3.72 g = 53.72 g sol' n ∗
0.0531mol C 5 H 10
1L
= 0.0614 L sol' n → M =
= 0.865M
875g
0.0614 L
3
SEMESTER 2
FINAL EXAM REVIEW
Unit IX - Equilibrium
I. Good things to know
• equilibrium, physical and chemical
• mass-action statement, Keq, Kc, KP
• homogeneous vs. heterogeneous equilibria
• rules of multiple equilibria (K’ = K1K2, etc.)
Kc =
kf
•
equilibria and kinetics:
•
LeChatlier’s Principle; effect of concentration, pressure, volume, temperature, addition of an inert gas,
presence of a catalyst on equilibrium
solubility product, dissociation equation, solubility expression
solubility, molar solubility
common-ion effect
•
•
•
kr
I. Short Answer
1. Compare the solubility of iron (II) hydroxide [Fe(OH)2] in distilled water to a solution of iron(II)
chloride [FeCl2], and to a solution of barium hydroxide [Ba(OH)2].
The dissociation equation is Fe(OH)2
Fe+2 + 2 OH-, making the solubility
+2
expression Ksp = [Fe ][OH ]. In distilled water, the concentration of both ions is zero,
thus the solubility will be the greatest. Due to the common ion effect, the presence of
iron(II) ions or hydroxide ions decreases the solubility of the iron(II) hydroxide. Since
FeCl2 releases one iron(II) ion per formula unit, and Ba(OH)2 releases 2 hydroxide
ions per formula unit, there will be more common ions present in the barium hydroxide
solution, therefore the iron(II) hydroxide will be more soluble in the iron(II) chloride
solution than in the barium hydroxide solution.
II. Problems
1. Write the Kc and KP mass-action expressions for the following equilibrium:
a. 2 NH3(g)
N2(g) + 3 H2(g)
[N ][H 2 ]3
Kc = 2
[NH 3 ]2
KP =
PN 2 • PH 2
PNH 3
3
2
2. For the following equilibrium performed at standard temperature and pressure in a 2.50 L container:
A(s) + 2 B(aq)
2 C(g) + 3 D(aq)
a. If at equilibrium, you have 0.106g of A, the concentration of B is 0.225 mol/L, you have 0.153
mol of C, and [D] = 0.554 M, what are the Kc and KP? Assume conditions are at STP.
0.153mol
= 0.0611M = (C)
2.50 L
(C ) 2 ( D ) 3 (0.0611) 2 (0.554) 3
Kc =
=
= 0.0125
( B) 2
(0.225) 2
K P = K c ( RT ) ∆n = (0.0125)(0.0821 • 273) 2 = 6.29
Note that A is excluded from the K because it is a solid
Kc appears to favor the reactants, but
KP appears to favor the products Hmmmm…..
b. Does this favor the reactants or the products?
SEMESTER 2
FINAL EXAM REVIEW
3. For the following equilibrium:
A+B
C+D
If Kc = 0.32, and you begin with a 1.0 molar solutions of A & a 2.0 molar solution of B, what
will be the equilibrium concentrations of the four species?
A + B ⇔ C + D
Initial : 1
2
0
0
Change : − x − x
+x +x
Final : 1 − x 2 − x
x
x
(C)(D)
x2
Kc =
=
= 0.32
(A)(B) (1 − x)(2 − x)
0.32(2 − 3 x + x 2 ) = x 2 → 0.64 − 0.96 x + 0.32 x 2 = x 2
0 = 0.68 x 2 + 0.96 x − 0.64 → x = 0.49
(C ) = ( D) = 0.49 M → ( A) = 1 − x = 0.51M → ( B ) = 2 − x = 1.51M
N2(g) + 3 H2(g) ∆H = +90.8 kJ/mol
4. For the following equilibrium: Heat + 2 NH3(g)
What is the effect on the concentrations of the 3 species and the equilibrium constant (K) of:
a. increasing nitrogen concentration
(H2) decreases, (NH3) increases, no effect on K
b. decreasing temperature
(H2), (N2) decrease, (NH3) increases, K decreases
c. decreasing pressure
(NH3) decreases, (N2), (H2) increase, K increases
d. decreasing the volume of the container
(NH3) increases, (N2), (H2) decrease, K decreases
e. adding some argon gas to the chamber
No Effect
f. increasing the rate of the reaction by using a platinum catalyst
No Effect
5. For the following ionic compound, give the dissociation equation, solubility expression, molar
solubility, and solubility:
barium hydroxide (Ksp = 5.0 x 10-3)
Ba(OH)2(s)
Ba+2 + 2 OH-1
Ksp = (Ba+2)(OH-1)2
2
3
0.0050 = (x)(2x)
0.0050 = 4x x=0.11M
0.1077 mol Ba(OH) 2 171g Ba(OH) 2
∗
= 18 g L
1L
1mol Ba(OH) 2
4
SEMESTER 2
FINAL EXAM REVIEW
5
6. The Ksp for iron(III) hydroxide is 4.0 x 10-38. What is the solubility of the iron(III) hydroxide in a
5.05 x 10-6M NaOH solution?
Ksp = (Fe+3)(OH-1)3 4.0x10-38 = (x)(5.05x10-6)3
3.1x10 −22 mol Fe(OH) 3 107 g Fe(OH) 3
∗
= 3.3 x10 − 20 g L
1L
1mol Fe(OH) 3
7. If you mix 50.0 mL of a 0.0100M Fe(NO3)3 solution with 50.0 mL of a 5.05 x 10-6M NaOH solution,
will a precipitate form? The Ksp for iron(III) hydroxide is 4.0 x 10-38.
Vtot = 50.0mL + 50.0mL = 100.0mL = 0.100 L → K sp = 4.0 x10 −38
n Fe + 3 = (0.0500 L)(0.0100 mol L) = 5.00 x10 − 4 mol ÷ 0.100 L = 0.00500 M = (Fe +3 )
nOH −1 = (0.0500 L)(5.05 x10 −6
mol
L
) = 2.525 x10 − 7 mol ÷ 0.100 L = 2.525 x10 −6 M = (OH -1 )
Q sp = (Fe + 3 )(OH −1 ) 3 = (0.00500)(2.525 x10 − 6 ) 3 = 8.05 x10 − 20
Q > K , so YES ppt will form
Unit X – Acids & Bases
I. Good things to know
• electrolyte, ionization, dissociation, salt
• 6 strong acids (HCl, HBr, HI, HNO3, HClO4, and H2SO4 and 5 strong bases (Li, Na, K, Ba, Sr OH)
• Arrhenius, Lowry-Bronsted, and Lewis definitions of acids/bases, amphoteric, conjugate pairs
• buffer system, common ion effect, hydrolysis
• acid-base indicators
II. Multiple Choice
Questions 1-3
The diagram below shows the titration of a weak monoprotic acid by a strong base:
B
D
E
1. At this point in the titration, the pH of the solution is equal to the pKa of the acid.
2. This is the equivalence point of the titration.
3. Of the points shown on the graph, this is the point when the solution is most basic.
SEMESTER 2
FINAL EXAM REVIEW
Questions 4-5
a. HNO3
b. HCN
a
e
b
c
c. H2CO3
d. HF
6
e. H2O
4. This is a strong electrolyte
5. A 0.1-molar solution of this substance will have the highest pH of the substances listed.
6. A 0.5-molar solution of which of the following salts will have a pH = 7?
a. KC2H3O2 b. NaClO4 c. NH4Cl d. (NH4)2SO4 e. LiCN
7. Which of the following species is amphoteric?
a. H+1 b. CO3-2 c. HCO3-1 d. H2CO3
e. H2
III. Short Answer
1. Tell which one of the acids listed in pairs below is the stronger of the two and explain why.
a. HF and HCl
HCl. The anion is larger, thus causing the H to be further away from
the nucleus, making it easier to remove. The easier it is to remove an
H, the stronger the acid
b. H2SO3 and HSO3-1
H2SO3. The first hydrogen on a diprotic acid is always the easiest to
remove. After the first H is removed, the attractive force from the sulfite
ion is distributed over one less H.
c. HNO3 and HNO2
HNO3. Oxoacids with more oxygens have easier H’s to remove. This is
because O is so electronegative. The more O’s on the molecule, the
more electron density is pulled toward those atoms, thus away from the
H, thus making it easier to remove
d. HClO4 and HBrO4
HClO4. Since the number of O’s is the same, the central atom with the
greatest electronegativity will correspond to the strongest acid. Same
reason as #1c.
2. Why are buffer systems made with weak acids instead of strong acids?
HX H+ + X-. In order for a buffer system to work, high
concentrations of HX and X- (through the addition of a salt) must be
present. Strong acids dissociate completely, thus making
the concentration of HX effectively 0.
3. Will a solution of sodium fluoride (NaF) be acidic or basic? Explain using a reaction that supports
your conclusion.
basic. the fluoride ion is the conjugate of the weak acid HF, therefore it
will act as a base:
F-1 + H+1
HF
SEMESTER 2
FINAL EXAM REVIEW
4.
H+ + H2PO4-1
H+ + HPO4-2
H+ + PO4-3
H3PO4
H2PO4-1
HPO4-2
K1 = 7.5 x 10-3
K2 = 6.2 x 10-8
K3 = 2.2 x 10-13
a. Of the acids listed above, which would be the most useful in creating a buffer system with a
pH of 7.5?
H2PO4- is the best choice. pKa ~ 7.5. Create it by mixing salts such as
KH2PO4 and K2HPO4
b. Sketch the titration curve that results when H3PO4 is titrated with excess NaOH and label the
two axes.
note 3 equivalence
points; one for each
proton in the acid
pH
volume NaOH
IV. Problems
1. What is the pH of a solution in which you dissolve 5.00g of NaOH in 25θ mL of water?
1mol NaOH 0.125mol
=
= 0.500 M = (OH −1 )
0
.
250
L
40 g NaOH
pOH = − log(OH ) = 0.30 → pH = 14 − 0.30 = 13.7
5.00 g NaOH ∗
2. What is the pH of a 0.100 M HF solution (Ka = 7.2 x 10-4)?
Ka =
( H + )( A− )
x2
=
= 7.2 x10 − 4 → x = 0.0085M = ( H + ) → pH = 2.1
( HA)
0.100
3. What is the pH of a 0.100 M H2SO4 solution (K1 = large, K2 = 1.2 x 10-2)?
H2SO4 H+ + HSO4- This occurs to ~100% ionization (strong acid), therefore
(H+)1 = 0.20M = (HSO4-)
HSO 4
−
initial : 0.100
final : 0.100 − x
Ka =
⇔
H+
+ SO 4
0.100
0
0.100 + x
x
−2
(0.100 + x)( x)
= 0.012 → 0.100 x + x 2 = 0.0012 − 0.012 x
(0.100 − x)
x 2 + 0.112 x − 0.0012 = 0 → x = 0.00985M = (H + ) 2
(H + ) tot = (H + )1 + (H + ) 2 = 0.10 + 0.00985 = 0.10985 → pH = 0.96
4. What is the pH of a 0.100 M NH3 solution (Kb = 1.8 x 10-5)
NH3 will hydrolyze water to form OH- ions, so it is a weak base
x2
= 1.8 x10 −5 → x = 0.0013M = (OH − ) → pOH = 2.87 → pH = 11.1
0.100
7
SEMESTER 2
FINAL EXAM REVIEW
8
5. What is the pH of a mixture of 0.100 M NH4Cl and 0.200 M NH3? See the Kb above.
pOH = pK b + log
(HB)
 0.100 
= − log(1.8 x10 −5 ) + log
 = 4.44 → pH = 9.56
(B)
 0.200 
6. Calculate the pH at the equivalence point for a titration in which you begin with 100.0 mL of
0.500 M acetic acid (Ka = 1.8 x 10-5) and titrate it with 0.300 M sodium hydroxide.
(0.500 M )(0.1000 L) = 0.0500mol HC 2 H 3 O 2 (acid − HAc) → reacts w/ 0.0500mol NaOH(base)
V NaOH = 0.0500mol ∗
1L
= 0.167 L → Vtot = 0.1000 L + 0.167 = 0.267 L
0.300mol
HAc + OH − ⇔ Ac −
+ HOH
0.0500
~
0.0500
0
− 0.0500 − 0.0500
~
+ 0.0500
~
~0
~0
0.0500mol
Ac − is a weak base, so find concentration & K b :
0.0500mol
= 0.187 M
0.267 L
1.00 x10 −14
Kb =
= 5.6 x10 −10
−5
1.8 x10
(Ac - ) =
x2
= 5.6 x10 −10 → x = (OH − ) = 1.0 x10 −5 M → pOH = 5.0 → pH = 9.0
0.187
7. A 10θ mL sample of 0.100-molar NH4Cl solution was added to 8θ mL of a 0.200-molar soution of
NH3. The value of Kb for ammonia is 1.79 x 10-5.
a. What is the pH of this solution?
nNH + = (0.100 M )(0.100 L) = 0.0100mol → nNH 3 = (0.200 M )(0.0800 L) = 0.0160mol
4
pOH = pK b + log
(HB)
 0.0100 
= − log(1.79 x10 −5 ) + log
 = 4.54 → pH = 9.46
(B)
 0.0160 
b. If 0.200g of NaOH were added to the solution, what would be the new pH of the solution?
Assume the volume of the solution does not change.
0.200 g NaOH ∗
1mol NaOH
= 0.00500mol
40 g NaOH
+
NH 4 + OH − ⇔ NH 3 + HOH
0.016
0.00500
0.0100
− 0.00500 − 0.00500 + 0.00500
0.021
~0
0.00500
pOH = pK b + log
(HB)
 0.00500 
= − log(1.79 x10 −5 ) + log
 = 4.12 → pH = 9.88
(B)
 0.021 
c. If equal molar quantities of NH3 and NH4+1 were mixed in solution, what would be the pH of
the solution?
pOH = pKb = -log(1.79x10-5)=4.75 pH=9.25
SEMESTER 2
FINAL EXAM REVIEW
8. The Ksp for cobalt(III) hydroxide is 2.5 x 10-43 at a temperature of 25oC.
a. Write the dissociation equation
Co(OH)3(s)
Co+3 + 3 OH-1
b. Write the solubility expression
Ksp = (Co+3)(OH-1)3
c. Calculate the molar solubility of cobalt(III) hydroxide
2.5 x10
− 43
3
4
= ( x)(3 x) = 27 x → x =
4
2.5 x10 −43
= 9.8 x10 −12 M
27
d. Calculate the molar solubility of cobalt(III) hydroxide in a solution with a pH of 4.50.
Is it more or less soluble than it was in distilled water?
pH = 4.50 → pOH = 9.50 → (OH − ) = 10 -9.50 M = 3.16 x10 −10 M
2.5 x10 − 43
= 7.9 x10 −15 M
−10
3.16 x10
−12
< 9.8 x10 , so salt is less soluble in the basic solution
2.5 x10 − 43 = ( x)(3.16 x10 −10 ) 3 → x =
7.9 x10 −15
Unit XI – Thermodynamics
I. Good Things to Know
• 3 Laws of Thermodynamics
• endothermic, exothermic
• insulator, conductor
• heat of solution, lattice energy, heat of hydration
• heat of dilution
• enthalpy (H), Hess’ Law, enthalpy of formation (∆Hfo)
• entropy (S) – examples of changes that create more order (-) or more disorder (+)
• Gibbs Free Energy (G) – how ∆H and ∆S relate (∆G = ∆H – T∆S)
• thermodynamic properties at standard conditions (∆Ho, ∆So, ∆Go)
II. Problems
1.
C(s) + O2(g) CO2(g)
2 C(s) + H2(g) C2H2(g)
H2(g) + ½O2(g) H2O(l)
∆H = -394 kJ/mol multiply by 4
∆H = +226.6 kJ/mol switch and double
∆H = -286 kJ/mol double
Based on the above information, what is ∆H for the following reaction?
2 C2H2(g) + 5 O2(g) 4 CO2(g) + 2 H2O(l)
4 C + 4 O2 4 CO2
∆H = -1576
2C2H2 4 C + 2 H2
∆H = -453.2
2H2 + O2 2 H2O
∆H = -572
-------------------------------------------------------------2 C2H2 + 5 O2 2 H2O + 4 CO2
∆H = -2601kJ
9
10
SEMESTER 2
FINAL EXAM REVIEW
Substance
H2O(l)
H2O(s)
H2O(g)
Al(s)
Hg(l)
C(s)
specific heat
(J/g-oC)
4.18
2.03
2.02
0.89
0.14
0.71
For Water:
heat of fusion (qfus)
heat of vaporization(qvap)
q (kJ/mol)
5.99
40.7
2. A 44.0g sample of an unknown metal at 99.0 oC was placed in a constant-pressure calorimeter containing 80.0g of water at 24.0oC. The final temperature of the system was found to be 28.4oC. Calculate the specific heat of the metal. The heat capacity of the calorimeter is 12.4 J/oC
− ∆H metal = ∆H water → ∆Tmetal = 28.4 − 99.0 = −70.6 o C → ∆Twater = 28.4 − 24.0 = 4.4 o C
∆H water = mc∆T + C p ∆T = (80.0 g )(4.18 J g •o C )(4.4 o C ) + (12.4 J o C )(4.4 o C )
= 1530 J → −1530 J = (44.0 g )c (−70.6 o C ) → c =
− 1530 J
= 0.49 J g •oC
(44.0 g )(−70.6 o C )
3. Calculate the ∆H, ∆S, and ∆G for the following reaction: (use the table on p.3 in your notes)
6 CO2(g) + 6 H2O(l) C6H12O6(s) + 6 O2(g)
(photosynthesis)
- then label each reaction as:
* endothermic or exothermic
* favoring more disorder or less disorder
* spontaneous or nonspontaneous
∆H = (-1275) + 6(0) – 6(-394) – 6(-286) = +2805kJ, endothermic
∆S = (212) + 6(205) – 6(213.6) – 6(69.9) = -259 J/K, less disorder
∆G = (-911) + 6(0) – 6(-394) – 6(-237) = +2875kJ, nonspontaneous
4. Explain using Kinetic Theory why NaOH evolves so much heat when it dissolved in water.
The lattice energy (energy to separate the ions) is significantly less than the
hydration energy (the energy released when the ions bond with water
molecules) net result: energy given off
5. Which of the following species, Na+1 or K+1 has a greater hydration energy?
Na+1 has a greater hydration energy because it has a smaller ionic radius than
potassium, therefore, based on Coulomb’s Law, the force of attraction is greater
11
SEMESTER 2
FINAL EXAM REVIEW
6. Given that a reaction has ∆Ho = -38 kJ/mol and ∆So = -187 J/K, ∆S = -0.187 kJ/K
a. calculate the ∆Go for this reaction at 298K.
∆G = ∆H – T∆S = (-38kJ) – (298K)(-0.187kJ/K) = +17.7 kJ
b. Is the reaction spontaneous?
nonspontaneous
c. At what temperature will it become spontaneous? this occurs when ∆G<0
0 = -38kJ – T(-0.187kJ/K)
38 = 0.187T
T = 203K
d. What is the value for Keq for this reaction?
∆G o = − RT ln K → 17,700 J = −(8.31 J mol • K )(298 K ) ln K
ln K = −7.15 → K = e −7.15 = 7.87 x10 − 4
Unit XII – Electrochemistry
I. Good things to know
• oxidation number
• oxidation, reduction; oxidizing agent reducing agent
• galvanic cell; electrode (cathode, anode), salt bridge, electromotive force
• electrolytic cell, concentration cell
• batteries: dry cell, alkaline dry cell, lead battery, Ni-Cad battery, fuel cell
• corrosion how does it work? how can it be prevented? galvanized steel, stainless steel
• electrolysis of metals and water
II. Problems
1) Give the oxidation numbers for the following:
+2
+4
-6
+3
+1 + 4- 2
a) H 2 C O 3
+3
-6
+6
+ 1 + 3- 2
b) H 3 As O 3
−8
+3 −2
c) C 2 O 4 − 2
2) Balance the following in acidic solution: Name the oxidizing and reducing agents.
+
+8 −2
-1.5 +1
+4 +4
-1
4 H + Os O 4 + 2 C 4 H 8 (OH) 2 Os
(gain 4)
-1
+1 -2
+ 2 C 4 H 8 O 2 + 4 H2O
2(lose 2)
3) Balance the following in basic solution: Name the oxidizing and reducing agents.
+2 -3 −1
5 H2O + 3 C N
+6 −2
−2
+4 -3 -2 −1
+ 2 Cr O 4 3 C N O
2(gain 3)
3(lose 2)
+3
−1
+ 2 Cr (OH) 4
−1
+ 2 OH-1
12
SEMESTER 2
FINAL EXAM REVIEW
4) Answer the following questions using the table of standard potentials:
a. Is Fe+3 capable of oxidizing I-1? Explain your answer. means Fe is reduced Fe+3 Fe+2
2 I-1 I2 + 2 e- Eo = -0.54V
Fe+3 + e- Fe+2 Eo = +0.77V
--------------Eo = +0.23V spontaneous, so YES
b. Is Fe+2 capable of reducing Cr+3 to its metallic form?
Cr+3 + 3e- Cr Eo = -0.74V
Fe+2 Fe+2+ e- Eo = -0.77V
--------------Eo = -1.51V nonspontaneous, so NO
5) For the following cell:
* calculate the standard potential (Eo) for the cell
* calculate the standard free energy (∆Go) and equilibrium constant (Keq) for the reactions
* write out the complete, balanced redox reaction for the cell
* draw a basic picture of the cell, labeling the cathode, anode, and direction of electron flow
Ag/Ag+1 : Ni/Ni+2
2 Ag+ + 2e- 2 Ag
Ni Ni+2 + 2e-
Eo = +0.80V doubled the rxn, didn’t double Eo
Eo = +0.26V
Eo = +0.80 + 0.26 = +1.06V
∆G = −nFE o = −(2mol )(96,500 J mol •V )(+1.06V ) = −204,580 J = −205kJ
∆G = − RT ln K → −204,580 J = −(8.31 J mol • K )(298 K ) ln K
82.6 = ln K → K = e82.6 = 7.55 x1035
1.06V
Ag Metal
Ni Metal
Salt Bridge
Ni+2 Ni+2
Ag+1
Ag+1
+1
Ag
Ni+2
Ni+2
Ag+1
Ni+2
Ni+2
6) Examine the galvanic cell below.
Ag
[Ag+] = 1.0 M
Ag
+1
Ag+1
SEMESTER 2
FINAL EXAM REVIEW
13
Calculate the cell potential at 25oC when the conditions in the right half are as follows:
3.4 x 10-3M AgNO3 and Ag electrode
* Note that for Ag Ag+, n=1
0.0592
0.0592  3.4 x10 −3 
 = +0.15V
E=E −
log Q = 0 −
log
n
1
1
.
0


o
* Note that the setup for Q makes E positive
7) How long will it take to plate out 1.0 g Au from aqueous Au+3 with a current of 10.0 A?
1.0 g Au •
1mol Au 3mol e - 96,500C
1s
∗
∗
∗
= 147 s
197 g Au 1mol Au 1mol e
10.0C
8) What mass of zinc metal (Zn) can be produced in 1.0 hour from molten ZnO with a current of 7.5 A?
* 1.0hr = 60min = 3600s
* ZnO: the Zn has an oxidation of +2, so Zn+2 Zn, n=2
3600 s •
7.5C 1mol e - 1mol Zn 65.4 g Zn
∗
∗
∗
= 9.1g Zn
1s
96,500C 2mol e - 1mol Zn
9) What volume of F2 gas, at 25oC and 1.00 atm, is produced when molten KF is electrolyzed by a current
of 10.0 A for 2.00 hours? What mass of potassium metal is produced? At which electrode does
this reaction occur?
KF(l) contains K+1 and F-1, both in liquid form
K+1 + e- K
2F-1 F2 + 2e2hrs = 7200s * 10 C/s = 72,000 C used in the process:
1mol e - 1mol K 39.1g K
72,000C ∗
∗
∗
= 29.2 g K
96,500C 1mol e - 1mol K
1mol e - 1mol F2
72,000C ∗
∗
= 0.373mol F2
96,500C 2mol e atm • L
nRT (0.373mol )(0.0821 mol
• K )( 298 K )
PV = nRT → VF2 =
=
= 9.13L F2
P
(1.00atm)
K is gaining electrons, so it is the cathode
F2 is losing electrons, so it is the anode
14
SEMESTER 2
FINAL EXAM REVIEW
Unit III – Nomenclature & Chemical Equations
1) Give the name or formula for the following:
H2SO4
a. sulfuric acid
Ca(OH)2
XeF6
b. calcium hydroxide
c. xenon hexafluoride
FeCr2O7
d. iron(II) dichromate
ammonia (nitrogen trihydride)
k. NH3
nickel(II) sulfide
l. NiS
ammonium carbonate
m. (NH4)2CO3
aluminum chloride
n. AlCl3
HCN
e. hydrocyanic acid
chromium(III) hydroxide
o. Cr(OH)3
Rb2O
f. rubidium oxide
carbon tetrachloride
p. CCl4
acetic acid
r. HC2H3O2
sodium hydride
s. NaH
magnesium fluoride
t. MgF2
Sn3(PO4)4
CO
h. tin(IV) phosphate
i. carbon monoxide
BaSO4-8H2O
j. barium sulfate
octohydrate
2. Write the formulas to show the reactants and products for all the laboratory situations described below.
In all cases, a reaction occurs. Assume that all solutions are aqueous unless otherwise indicated. Represent
substances in solution as ions if the substances are extensively ionized. Omit formulas for any ions or
molecules that are unchanged by the reaction. You need not balance the equations.
a.
b.
c.
d.
e.
f.
g.
h.
i.
A bar of nickel metal is placed in a solution of copper(II) sulfate.
Sodium metal is added to water.
Propanone (C3H6O) is burned in air.
A solution of lead(II) nitrate is added to a solution of potassium sulfate.
A solution of ammonia is mixed with a solution of acetic acid.
Magnesium metal is burned in air.
Excess concentrated potassium cyanide solution is added to a solution of nickel chloride.
Solid sodium acetate is added to 1.0M hydrobromic acid.
An acidified solution of iron(II) sulfate is mixed with an aqueous potassium permanganate
solution.
a) Ni + Cu+2 Ni+2 + Cu
b) Na + H2O Na+1 + OH-1 + H2
c) C3H6O + O2 CO2 + H2O
d) Pb+2 + SO4-2 PbSO4
e) NH3 + HC2H3O2 NH4+1 + C2H3O2-1
f) Mg + O2 MgO
g) CN-1 + Ni+2 Ni(CN)4-2
h) NaC2H3O2 + H+ HC2H3O2 + Na+1
i) H+ + Fe+2 + MnO4-1 Fe+3 + Mn+2 + H2O