Exam III
Thermodynamics
1 , 2nd, and 3rd Laws
Enthalpy, Entropy, and Gibbs Energy
Heats of Formation
Hess’ Law
Spontaniety
Gibbs Equation
Electrochemical Cells
Balancing Redox Reactions
Nernst Equation
Batteries
Electroplating and Hydrolysis
st
Thermodynamics
1) A 100 gram block of copper was heated with a bunsen burner and then thrown into a cup of
water (250 mL) initially at 25°C. When the system reached thermal equilibrium the final
temperature of the water and the block was 61°C. What was the initial temperature of the block
of copper? Cp(Cu)=24.5 J/mol-K, Cp(H2O)=75.2 J/mol-K, FWT(Cu) = 63.55 g/mol, FWT(H2O)
= 18 g/mol, density of water = 1 g/ml
2) Given the following information calculate the heat of formation of C2H4.
C2H4 + 3 O2 → 2 CO2 + 2 H2O ΔH° = -414 kJ/mol
C + O2 → CO2
ΔH° = -393.5 kJ/mol
H2 + ½ O2 → H2O
ΔH° = -241.8 kJ/mol
3) When ethanol (C2H5OH) burns in oxygen it forms CO2 and water. If the heat of combustion
is -1237.7 kJ/mol, the heat of formation of water is -241.8 kJ/mol, and the heat of formation of
carbon dioxide is -393.5 kJ/mol, what is the heat of formation of C2H5OH?
4) Given a block of ice at 0°C how much of the ice will melt if you put a 100 gram block of
aluminum on it that is initially at 100°C? Cp(Al)=24.1 J/mol-K, Cp(H2O)=75.2 J/mol-K, Heat of
Fusion = 6.01 kJ/mol for ice, FWT(Al) = 26.98 g/mol, FWT (water) = 18 g/mol
5) 10 grams of diethyl ether was placed into a bomb calorimeter and then combusted in excess
oxygen. The balanced combustion reaction was as follows;
C4H10O + 6 O2 -----> 4 CO2 + 5 H2O
The heat released by this reaction heated 2500 mL of water 8.6°C. What is the heat of formation
of the diethyl ether? Cp(H2O) = 75.2 J/mol-K, ΔH°f(CO2) = -393.5 kJ/mol, ΔH°f(H2O) = -241.8
kJ/mol, FWT (Diethyl ether) = 74 g/mol, FWT (water) = 18 g/mol.
6) If a 25 gram block of copper at 45°C is added to 50 grams of liquid ammonia at -60°C,
calculate the final temperature of the system. For NH3: FWT(NH3) = 17 g/mol, ΔHvap = 23.4
kJ/mol, Cp(liq)= 35.1 J/mol.K and for Cu : Cp = 24.4 J/mol.K, FWT(Cu) = 63.55 g/mol
7) You have 50 grams of ice at -2.5°C sitting in a dixie cup. If you heat 5 nickels and throw
them into the cup and 8 grams of the ice melts, how hot were the nickels? Each nickel weighs 5
grams and has a heat capacity of 24.3 J/mol-K. Assume that nickels are pure nickel with an
atomic mass of 57.81 g/mol.
8) What is the ΔG and ΔS for the dilution of 4.5 M H2SO4 to 0.03 M H2SO4 at 25°C? The
enthalpy for this reaction is - 210 kJ/mol.
9) Calculate the ΔG for the making of 6M NaOH from solid NaOH. Saturated solutions of
NaOH have a concentration of 19.25 M.
10) Mathematically we write the first law of thermodynamics as U = q + w. The work, w, is
pressure-volume work. What is the sign on this work (+ or -) and why does it have this sign?
11) Please give me a definition of a heat capacity.
12) What is the ΔG° for the following reaction at 25°C?
BaBr2(aq) + CuSO4(aq) ----> BaSO4(s) + CuBr2(s)
ΔH°
BaBr2
CuSO4
BaSO4
CuBr2
S°
-186.47 kJ/mol
-201.51 kJ/mol
-345.57 kJ/mol
-25.1 kJ/mol
42.0 J/mol K
19.5 J/mol K
7.0 J/mol K
21.9 J/mol K
Circle all that apply about this reaction,
spontaneous
exothermic
entropy increases
the reaction occurs quickly
not spontaneous
endothermic
entropy decreases
the reaction occurs slowly
What is the equilibrium constant for this reaction?
13) Given the following set of thermodynamic data calculate the ΔG̊ , ΔH̊ , ΔS̊ , and Keq for the
following reaction at 25̊C (Enthalpies are in kJ/mol and entropies are in J/mol-k)
CoCl2(s) <----> Co2+ + 2 ClCoCl2
Co2+
Cl-
ΔH°
-325.5
-67.36
-167.4
S°
106.3
-155.2
55.1
Electrochemical Cells
14) Please balance each of the following redox reactions;
MnO4- + Fe2+ —> Fe3+ + Mn2+
CrO42- + I- —> I2 + Cr3+
Co2+ + NO3- —> Co2O3 + NO
I3- + S2O32- —> I- + S4O62H2S + Cr2O72- —> Cr3+ + S
Fe2+ + NiOOH —> Fe3+ + Ni(OH)2
15) Consider the following electrochemical reaction,
PbI2 + 2 e- ---> Pb + 2 IPb ---> Pb2+ + 2 ePbI2 ----> Pb2+ + 2 IHow would the voltage change if you,
a) Add KI(s) to the oxidation cell
b) Add AgNO3 to the reduction cell
c) Add heat
d) Add water to the reduction cell?
e) Add NaNO3(s) to the oxidation cell
f) Enlarge the oxidation electrode?
Inc.
Inc.
Inc.
Inc.
Inc.
Inc.
N.C.
N.C.
N.C.
N.C.
N.C.
N.C.
Dec.
Dec.
Dec.
Dec.
Dec.
Dec.
16) Consider the following electrochemical reaction,
Au ---> Au3+ + 3eAuCl4- + 3e- ---> Au + 4 Cl-----------------------------AuCl4- ---> Au3+ + 4 ClHow would the voltage change if you,
a) Enlarge the Au electrodes?
Inc.
N.C.
Dec.
b) Add KCl(s)?
Inc.
N.C.
Dec.
c) Add AgNO3?
Inc.
N.C.
Dec.
d) Add water to the reduction cell?
Inc.
N.C.
Dec.
e) Add NaNO3(s)
Inc.
N.C.
Dec.
f) Cool the reaction vessel?
Inc.
N.C.
Dec.
17) Given the following half-reactions draw an electrochemical cell, calculate the voltage of the
cell and lable the anode and cathode. You may use a carbon electrode for the I- solution.
Fe3+ + e- ----> Fe2+
I2 ----> 2 I-
+0.771 V
+0.535 V
18) What is the value of the half reaction,
Cu2+ + e- ----> Cu+
given that,
Cu2+ + 2e- ----> Cu° ° = +0.3402
Cu+ + e- ------> Cu° ° = +0.522
19) Given the following half-reactions draw an electrochemical cell that would work. Calculate
the voltage of the cell and label the anode and cathode, tell which electrode is positive and which
is negative, and where the oxidation and reduction reactions are occurring. In addition indicate
the direction of electron flow, and the concentration of any ionic species in solution.
Co3+ + e- ---> Co2+ +1.842 V
Pb2+ + 2e- ---> Pb -0.126 V
20) Given the following half-reactions, properly set up a WORKING electrochemical cell.
Ag + I- → AgI + e° = 0.1519 V
PbI2 + 2e ---> Pb + 2 I ° = 0.3580 V
Label the anode, the cathode, the direction of electron flow, and indicate the concentrations of all
ionic species. Indicate what the electrodes are made of.
20b) Draw the cell diagram for the above cell.
20c) What is the Keq for the above cell?
20d) Look very carefully at the overall reaction, and explain whether your answer for 20c) is
reasonable or not.
21) Balance the following half-reactions in a BASE and then use these reactions to set up a
WORKING electrochemical cell.
Bi(s) → Bi2O3(s)
Hg2O(s) → Hg(l)
0.460 Volts
-0.123 Volts
Label the anode, the cathode, the direction of electron flow, and indicate the concentrations of all
ionic species. Indicate the composition of the electrodes.
21b) Draw the cell diagram for the above cell.
21c) What is the Keq for the above cell?
22) How long will it take to deposit 1 ounce of gold (31.1 g) at a constant current flow of 10
amps? MW (Au) = 197.8
23) How long will it take to deposit 1 lb. of Palladium (453.6 g) from a solution of PdCl4 at a
constant current of 25 amps? MW (Pd) = 106.4
Thermodynamics
1) A 100 gram block of copper was heated with a bunsen burner and then thrown into a cup of
water (250 mL) initially at 25̊C. When the system reached thermal equilibrium the final
temperature of the water and the block was 61̊C. What was the initial temperature of the block
of copper? Cp(Cu)=24.5 J/mol-K, Cp(H2O)=75.2 J/mol-K, FWT(Cu) = 63.55 g/mol, FWT(H2O)
= 18 g/mol, density of water = 1 g/ml
-nCpTCu = nCpTH2O
100g
J
- 63.55 ቀ24.5 molKቁ൫61°C – ܶ୧൯=
Ti = 1036.3°C
250g
18
J
ቀ75.2 molKቁ(25°C - 61°C)
2) Given the following information calculate the heat of formation of C2H4.
C2H4 + 3 O2 ------> 2 CO2 + 2 H2O ΔH̊ = -414 kJ/mol
C + O2 -----> CO2 ΔH̊ = -393.5 kJ/mol
H2 + ½ O2 -----> H2O ΔH̊ = -241.8 kJ/mol
2 CO2 + 2 H2O → C2H4 + 3 O2 ΔH° = 414 kJ/mol
2 (C + O2 →CO2)
ΔH° = -393.5 kJ/mol x2
2 (H2 + ½ O2 → H2O)
ΔH° = -241.8 kJ/mol x2
2 C + 2 H2 → C2H4 ΔH° = 414 kJ/mol + 2 (-393.5 kJ/mol )+ 2(-241.8 kJ/mol)
ΔH° = -856.6 kJ/mol
3) When ethanol (C2H5OH) burns in oxygen it forms CO2 and water. If the heat of combustion
is -1237.7 kJ/mol, the heat of formation of water is -241.8 kJ/mol, and the heat of formation of
carbon dioxide is -393.5 kJ/mol, what is the heat of formation of C2H5OH?
C2H5OH + 3 O2 → 2 CO2 + 3 H2O
ΔH° = ΔH°prod - ΔH°react
-1237.7 kJ/mol= [2(-393.5kJ/mol) + 3(-241.8kJ/mol)] – [ΔH°C2H5OH + 3(0kJ/mol)]
ΔH°C2H5OH = -274.7 kJ/mol
4) Given a block of ice at 0̊C how much of the ice will melt if you put a 100 gram block of
aluminum on it that is initially at 100̊C? Cp(Al)=24.1 J/mol-K, Cp(H2O)=75.2 J/mol-K, Heat of
Fusion = 6.01 kJ/mol for ice, FWT(Al) = 26.98 g/mol, FWT (water) = 18 g/mol
-nCpTAl = nfus
100g
-
26.98
ቀ24.1
J
ቁ൫100°C – 0°C൯=n ቀ6010
molK
J
ቁ
molK
n = 1.4863 mol ice x 18g/mol = 26.75 g ice will melt
5) 10 grams of diethyl ether was placed into a bomb calorimeter and then combusted in excess
oxygen. The balanced combustion reaction was as follows;
C4H10O + 6 O2 -----> 4 CO2 + 5 H2O
The heat released by this reaction heated 2500 mL of water 8.6̊C. What is the heat of formation
of the diethyl ether? Cp(H2O) = 75.2 J/mol-K, ΔH̊ f(CO2) = -393.5 kJ/mol, ΔH̊ f(H2O) = -241.8
kJ/mol, FWT (Diethyl ether) = 74 g/mol, FWT (water) = 18 g/mol.
-nCpTH2O = rxn note: a negative sign was added because energy was given off.
2500g
-
18
ቀ75.2
-89,822 J
భబ ೒
೒ ቇ
ళర
೘ ೚೗
ቆ
J
ቁ(8.6°C )= -89,822 J for 10 grams
molK
= -664,683 J/mol
ΔH° = ΔH°prod - ΔH°react
-664,683 kJ/mol= [4(-393.5kJ/mol) + 5(-241.8kJ/mol)] – [ΔH°Ether + 6(0kJ/mol)]
ΔH°Ether = -2118.3 kJ/mol
6) If a 25 gram block of copper at 45̊C is added to 50 grams of liquid ammonia at -60̊C,
calculate the final temperature of the system. For NH3: FWT(NH3) = 17 g/mol, ΔHvap = 23.4
kJ/mol, Cp(liq)= 35.1 J/mol.K and for Cu : Cp = 24.4 J/mol.K, FWT(Cu) = 63.55 g/mol
-nCpTCu = nCpTNH3
25g
J
- 63.55 ቀ24.4 molKቁ൫Tf – 45°C൯=
50g
17
J
ቀ35.1 molKቁ(Tf –(-60°C))
-9.9587(Tf – 45°C) = 103.235(Tf + 60°C)
-Tf + 45 = 10.3663(Tf + 60) = 10.3663Tf + 621.98
-11.3663Tf = 666.98
Tf = -50.76°C
7) You have 50 grams of ice at -2.5 C sitting in a dixie cup. If you heat 5 nickels and throw them
into the cup and 8 grams of the ice melts, how hot were the nickels? Each nickel weighs 5 grams
and has a heat capacity of 24.3 J/mol-K. Cpice = 37.7 J/molK. Assume that nickels are pure
nickel with an atomic mass of 57.81 g/mol.
Note: You must heat all of the ice but melt only 8 grams of it.
nCpTNi = nCpTice + nfus
25 g
57.81
ቀ24.3
J
ቁ൫Tf – 0°C൯=
molK
଼௚
ଵ଼
ቀ6010
Tf – 0°C = 279.1°C so Tf = 279.1°C
J
ቁ+
molK
50g
18
ቀ37.7
J
ቁ(2.5°C)
molK
8) What is the ΔG and ΔS for the dilution of 4.5 M H2SO4 to 0.03 M H2SO4 at 25°C? The
enthalpy for this reaction is - 210 kJ/mol.
G = G° + RT ln Q = 0 + (8.314 J/molK)(298K) ln (0.03 M/ 4.5M)
G = -12,414 J/mol
∆ீି∆ு
G = H - TS => − ቀ
்
಻
ା
೘ ೚೗
ିଵଶ,ସଵସ
ቁ = ∆ܵ = − ቆ
ଶଵ଴,଴଴଴
ଶଽ଼ ௄
಻
೘ ೚೗
ቇ = −663 ‫ܬ‬/݉ ‫ܭ݈݋‬
9) Calculate the ΔG for the making of 6M NaOH from solid NaOH. Saturated solutions of
NaOH have a concentration of 19.25 M.
There are two processes going on,
1. Solid NaOH ↔ Sat’d NaOH
2. Sat’d NaOH → 6 M NaOH
The first process is an equilibrium so G = 0 for the first process.
For the second process, since all solutions have a standard state of 1M, the standard state
for Sat’d NaOH and 6 M NaOH are both 1 M and the G° = 0 since there is no difference
between the G°’s for these two solutions. So,
G1 = 0
G2 = G° + RT ln Q
G2 = 0 + (8.314 J/molK)(298K) ln (6 M/ 19.25M) = -2888.2 J/mol
G1+ G2 = 0 + (- 2888.2 J/mol) = - 2888.2 J/mol
10) Mathematically we write the first law of thermodynamics as U = q + w. The work, w, is
pressure-volume work. What is the sign on this work (+ or -) and why does it have this sign?
Work = PV
A chemist thinks of himself as being inside the system pushing out (like expanding a
balloon). To a chemist, pushing out (expanding the balloon) is negative work. Negative
work is work that you must do as opposed positive work with is work being done for you.
In mathematics, a change in volume is always the Final Volume – Initial Volume. In the
case of an expanding balloon, the final volume will be larger than the initial volume so
Final Volume – Initial Volume will produce a positive value (Final Volume – Initial
Volume = + change in volume). Since PV = negative for an expanding balloon, and
both P and V are positive, we must add a negative sign to the work expression to make
this work overall negative. So, chemist define pressure-volume work as,
Work = - PV (Chemist)
It should be noted that engineers see the world opposite from chemists. Engineers view
themselves as being outside of the system pushing in. Work for an engineer is
compressing the balloon, making it smaller. Since V = Vfinal – V initial, and the Vfinal
is smaller than Vinitial, the V is negative. So, for an engineer,
Work = PV (Engineer)
This value is still negative, like the chemist, but we see the world differently from an
engineer.
11) Please give me a definition of a heat capacity.
It is the amount of energy needed to raise the temperature of one mole of substance 1°C.
12) What is the ΔG° for the following reaction at 25°C?
BaBr2(aq) + CuSO4(aq) ----> BaSO4(s) + CuBr2(s)
ΔH°
BaBr2
CuSO4
BaSO4
CuBr2
S°
-186.47 kJ/mol
-201.51 kJ/mol
-345.57 kJ/mol
-25.1 kJ/mol
42.0 J/mol K
19.5 J/mol K
7.0 J/mol K
21.9 J/mol K
ΔH° = ΔH°prod - ΔH°react and ΔS° = S°prod - S°react
ΔH° = ΔH°prod - ΔH°react
ΔH° = [(-345.57 kJ/mol) + (-25.1 kJ/mol)] – [(-186.47 kJ/mol) + (-201.51 kj/mol)]
ΔH° = 17.31 kJ/mol
ΔS° = S°prod - S°react
ΔS° = [(7.0 J/molK) + (21.9 J/molK)] – [(42.0 J/molK) + (21.9 J/molK)]
ΔS° = -32.6 kJ/mol
G° = H° - TS°
G° = 17,310 J/mol + (298 K) (-32.6 J/molK)
G° = 27,024.8 J/mol
Circle all that apply about this reaction,
spontaneous
exothermic
entropy increases
the reaction occurs quickly
not spontaneous
endothermic
entropy decreases
the reaction occurs slowly
What is the equilibrium constant for this reaction?
G° = - RT ln Keq
∆ృ°
Keq = eି ౎౐ = e
మళ,బమర.ఴ
ି (ఴ.యభర)(మవఴ)
= 1.832x10ିହ
13) Given the following set of thermodynamic data calculate the ΔG°, ΔH°, ΔS°, and Keq for the
following reaction at 25C (Enthalpies are in kJ/mol and entropies are in J/mol-k)
CoCl2(s) ↔ Co2+ + 2 ClCoCl2
Co2+
Cl-
ΔH°
-325.5
-67.36
-167.4
S°
106.3
-155.2
55.1
ΔH° = ΔH°prod - ΔH°react and ΔS° = S°prod - S°react
ΔH° = ΔH°prod - ΔH°react
ΔH° = [(-67.36 kJ/mol) + 2(-167.4 kJ/mol)] – [(-325.5 kj/mol)]
ΔH° = -76.66 kJ/mol (exothermic)
ΔS° = S°prod - S°react
ΔS° = [(-155.2 J/molK) + 2(55.1 J/molK)] – [(106.3 J/molK)]
ΔS° = -151.3 J/molK (entropy decreases)
G° = H° - TS°
G° = -76,660 J/mol - (298 K) (-151.3 J/molK)
G° = -31,572.6 J/mol (reaction is spontaneous as written)
G° = - RT ln Keq
∆ృ°
Keq = eି ౎౐ = e
ష యభ,ఱళమ.ల
ି (ఴ.యభర)(మవఴ)
= 3.423x10ହ
Electrochemical Cells
14) Please balance each of the following redox reactions;
MnO4- + Fe2+ —> Fe3+ + Mn2+
Fe2+ → Fe3+ + e5e- + 8 H+ + MnO4- → Mn2+ + 4 H2O
CrO42- + I- —> I2 + Cr3+
3 e- + 8 H+ + CrO42- → Cr3+ + 4 H2O
2 I- → I2 + 2 eCo2+ + NO3- —> Co2O3 + NO
3 H2O + 2 Co2+ → Co2O3 + 6 H+ + 2e3 e- + 4 H+ + NO3- → NO + 2 H2O
I3- + S2O32- —> I- + S4O622 e- + I3- → 3 IS2O32- → S4O62- + 2 eH2S + Cr2O72- —> Cr3+ + S
H2S → S + 2 H+ + 2 e6 e- + 14 H+ + Cr2O72- → 2 Cr3+ + 7 H2O
Fe2+ + NiOOH —> Fe3+ + Ni(OH)2
Fe2+ → Fe3+ + ee- + H+ + NiOOH → Ni(OH)2
15) Consider the following electrochemical reaction,
PbI2 + 2 e- → Pb + 2 IPb → Pb2+ + 2 ePbI2 → Pb2+ + 2 IHow would the voltage change if you,
a) Add KI(s) to the oxidation cell
b) Add AgNO3 to the reduction cell
c) Add heat
d) Add water to the reduction cell?
e) Add NaNO3(s) to the oxidation cell
f) Enlarge the oxidation electrode?
Inc.
Inc.
Inc.
Inc.
Inc.
Inc.
N.C.
N.C.
N.C.
N.C.
N.C.
N.C.
Dec.
Dec.
Dec.
Dec.
Dec.
Dec.
In electrochemical cells all concentrations are 1 M. If all concentrations are 1 M, the reverse
reaction is going to occur. Therefore, the real reaction going on is,
Pb + 2 I- → PbI2 + 2 e- (oxidation)
Pb2+ + 2 e- → Pb
(reduction)
2+
Pb + 2 I → PbI2
To understand what would happen, you need to write down the Nernst Equation,
 = ° - RT/n ln 1/[Pb2+][I-]2
a) When you put KI in the oxidation cell you are increasing the I- concentration. This
causes 1/[Pb2+][I-]2 to become smaller. The natural log of a smaller number is a negative
number. The negative of a negative is a positive so the voltage will increase.
b) Nothing happens. There is nothing for the silver to react with.
c) Adding heat increases T in the equation. If everything else remains the same, the
- RT/n ln 1/ [Pb2+][I-]2 becomes more negative and the voltage decreases.
d) Adding water is a dilution. Diluting the reduction cell lowers the concentration of I-.
If I- gets smaller, the 1/[Pb2+][I-]2 gets larger. The natural log will become larger but
there is a negative so it becomes more negative, so the voltage goes down.
e) NaNO3 does not participate in this reaction. No change.
f) Changing the size of the electrodes does not change the voltage. The electrodes are
not a part of the reaction. They are not in the Nernst Equation.
16) Consider the following electrochemical reaction,
Au → Au3+ + 3eAuCl4- + 3e- → Au + 4 Cl-----------------------------AuCl4- → Au3+ + 4 ClHow would the voltage change if you,
a) Enlarge the Au electrodes?
b) Add KCl(s)?
c) Add AgNO3?
d) Add water to the reduction cell?
e) Add NaNO3(s)
f) Cool the reaction vessel?
Inc.
Inc.
Inc.
Inc.
Inc.
Inc.
N.C.
N.C.
N.C.
N.C.
N.C.
N.C.
Dec.
Dec.
Dec.
Dec.
Dec.
Dec.
Note: This is NOT a Ksp problem since AuCl4- is not a solid (solids are not charged) so
we will leave the reaction as is.
 = ° - RT/n ln [Au3+][Cl-]4/[ AuCl4-]
a) No change. Changing the size of the electrode makes no difference.
b) Adding Cl- makes the natural log large which decreases the voltage
c) The Ag+ will react with the Cl- to make AgCl. This removes Cl-, making the natural
log smaller (more negative). So you will be subtracting a smaller number so the voltage
will increase.
d) Diluting the reduction cell lowers the concentration of both AuCl4- and Cl- but the Clis raised to the 4th power so it reduces faster. Reducing Cl- makes the natural log smaller
(more negative so you are subtracting a smaller (more negative) number so voltage
increases.
e) No change.
f) If T is smaller, then you are subtracting a smaller number so voltage increases.
17) Given the following half-reactions draw an electrochemical cell, calculate the voltage of the
cell and label the anode and cathode. You may use a carbon electrode for the I- solution.
Fe3+ + e- → Fe2+
+0.771 eV
2 I- → I2 + 2 e-0.535 eV
2 Fe3+ + 2 I- → 2 Fe2+ + I2 0.236 eV
e-
Anode
Negative
Oxidation
C
Pt
I2
1M FeCl3
1M FeCl2
1M KI
18) What is the value of the half reaction,
Cu2+ + e- ----> Cu+
given that,
Cu2+ + 2e- → Cu ° = +0.3402
Cu+ + e- → Cu ° = +0.522
Cu2+ + 2e- → Cu ° = +0.3402
Cu → Cu+ + e- ° = -0.522
Cu2+ + e- → Cu+
° = 2(0.3402) + 1(-0.522) = 0.1584 volts
1
Cathode
Positive
Reduction
19) Given the following half-reactions draw an electrochemical cell that would work. Calculate
the voltage of the cell and label the anode and cathode, tell which electrode is positive and which
is negative, and where the oxidation and reduction reactions are occurring. In addition indicate
the direction of electron flow, and the concentration of any ionic species in solution.
Co3+ + e- → Co2+ Δ° = + 1.842 eV
Pb2+ + 2e- → Pb Δ° = - 0.126 eV
e-
Anode
Negative
Oxidation
Pb
Pt
1M
Pb(NO3)2
1M CoCl3
1M CoCl2
Cathode
Positive
Reduction
Co3+ + e- → Co2+ Δ° = + 1.842 eV
Pb → Pb2+ + 2e- Δ° = + 0.126 eV
3 Pb + 2 Co3+ → 3 Pb2+ + 2 Co
Δ° = 1.968 V
20) Given the following half-reactions, properly set up a WORKING electrochemical cell.
AgI + e----> Ag + IΔ° = -0.1519 V
PbI2 + 2e ---> Pb + 2 I Δ° = 0.3580 V
Label the anode, the cathode, the direction of electron flow, and indicate the concentrations of all
ionic species. Indicate the composition of the electrodes.
-
e
Ag
Anode
Negative
Oxidation
AgI(s)
Pb
1M KI
Cathode
Positive
Reduction
PbI2(s)
20b) Draw the cell diagram for the above cell.
Ag/AgI(s)/1 M KI/PbI2(s)/Pb
20c) What is the Keq for the above cell?
 = 0.5099 volts = 0.0592/2 log Keq
Keq = 1.684x1017
20d) Look very carefully at the overall reaction, and explain whether your answer for 20c) is
reasonable or not.
The problem is that everything is a solid except for the 1 M KI. The equilibrium constant must be,
Keq 
[ I  ]Pb
[ I  ] Ag
So the voltage is completely dependent on the difference between the amount of I- in equilibrium with the
Pb and the I- in equilibrium with the Ag. This makes what is known as a concentration cell, that is, a cell
whose voltage is due only to differences in concentration on the two sides of the cell.
21) Balance the following half-reactions in a BASE and then use these reactions to set up a
WORKING electrochemical cell.
6 OH- + 2 Bi(s) → Bi2O3(s) + 3 H2O + 6 e2 e- + H2O + Hg2O(s) → 2 Hg(l) + 2 OH-
0.460 Volts
-0.123 Volts
Label the anode, the cathode, the direction of electron flow, and indicate the concentrations of all
ionic species. Indicate the composition of the electrodes.
e
Anode
Negative
Oxidation
-
Bi
Cathode
Positive
Reduction
1M NaOH
Bi2O3(s)
Hg2O
Pt
21b)
Draw the cell diagram for the above cell.
Bi / Bi2O3(s) / 1 M NaOH / Hg2O(s) / Hg / Pt
21c) What is the Keq for the above cell?
 = 0.337 volts = 0.0592/6 log Keq
Keq = 1.43x1034
22) How long will it take to deposit 1 ounce of gold (31.1 g) at a constant current flow of 10
amps? MW (Au) = 197.8
31.1 g/197.8 g/mol = 0.1572 mol Au has a 3+ charge, so multiply by 3.
3 mole e-/mol Au x 0.15723 mol Au = 0.4717 mol e0.4717 mol e- x 96,487 C/mol e- = 45,511.8 C
Amp x sec = Coulombs (C)
(10 amps)(sec) = 45,511.9 C
sec = 4,551.18 sec => 75.85 minutes
23) How long will it take to deposit 1 lb. of Palladium (453.6 g) from a solution of PdCl4 at a
constant current of 25 amps? MW (Pd) = 106.4
453.6 g/106.4 g/mol = 4.263 mol Pd, but Pd has a 4+ charge, so multiply by 4.
4 mole e-/mol Pd x 4.263 mol Pd = 17.053 mol e17.053 mol e- x 96,487 C/mol e- = 1,645,357 C
Amp x sec = Coulombs (C)
(25 amps)(sec) = 1,645,357 C
sec = 65,814 sec => 18.28 hours
Chemistry 121
Third Exam
Name____________________
May 15, 2007
CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for full
credit. You may use a calculator.
Question
Credit
1 ( 18)
2 (40)
3(12)
4(15)
5(15)
Total
R = 8.314 J/mol-K
 =  - 0.0592/n log Q
ΔG = -n
 = 96486 C/mol eΔG = ΔG + RT ln Q
PV = nRT
C = amp x sec
ΔG= -RTlnKeq
R = 0.08205 L-atm/mol-K
1) Balance each of the following oxidation-reduction half-reactions;
IO3- —> I3- (in acid)
Mn(OH)2 —> Mn2O3 (in base)
2) Given the following set of half-reactions, set up a WORKING electrochemical cell.
Hg2Cl2(s) + 2 e- —> 2Hg + 2Cl2 H+ + 2 e- —> H2
 = 0.27 V
 = 0.000 V
Label the anode, the cathode, the direction of electron flow, and indicate the concentrations of all
ionic species. Indicate the composition of the electrodes.
2b) Draw the cell diagram for the above cell.
2c) What is the voltage of the cell when all ionic species have a concentration of 0.10 M?
3) How long will it take to galvanize (coat with Zinc) a nail with 2 grams of zinc using a current
(amperage) of 0.8 amps? FWT of Zn2+ = 65.41 g/mol
4) You are given 100 mL of water at 25̊C. You add a 20 gram ice cube at 0̊C and 50 mL of
boiling water at 100̊C. What is the final temperature of the system?
5) What is the ΔG and ΔS for the dilution of 4.5 M H2SO4 to 0.03 M H2SO4 at 25̊C? The ΔH = 210 kJ/mol.for this reaction.
Chemistry 121
Third Exam
Name__Answer Key____
May 15, 2007
CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for full
credit. You may use a calculator.
Question
Credit
1 ( 18)
2 (40)
3(12)
4(15)
5(15)
Total
R = 8.314 J/mol-K
 =  - 0.0592/n log Q
ΔG = -n
 = 96486 C/mol eΔG = ΔG + RT ln Q
PV = nRT
C = amp x sec
ΔG= -RTlnKeq
R = 0.08205 L-atm/mol-K
1) Balance each of the following oxidation-reduction half-reactions;
IO3- —> I3- (in acid)
16 e- + 18 H+ + 3 IO3- —> I3- + 9 H2O
Mn(OH)2 —> Mn2O3 (in base)
2 Mn(OH)2 —> Mn2O3 + H2O + 2 H+ + 2e2 OH- —> 2 OH2 Mn(OH)2 + 2 OH- —> Mn2O3 + 3 H2O + 2e-
2) Given the following set of half-reactions, set up a WORKING electrochemical cell.
Hg2Cl2(s) + 2 e- —> 2Hg + 2Cl2 H+ + 2 e- —> H2
e-
H2 1atm
Anode
Negative
Oxidation
 = 0.27 V
 = 0.000 V
Pt
1M HCl
Cathode
Positive
Reduction
1M HCl
Hg2Cl2
Pt
Label the anode, the cathode, the direction of electron flow, and indicate the concentrations of all
ionic species. Indicate the composition of the electrodes.
2b) Draw the cell diagram for the above cell. (Note: The solution in both cells is HCl so no salt
bridge is required).
Pt/H2 (1atm) / 1M HCl / Hg2Cl2(s) / Hg(l)/ Pt
2c) What is the voltage of the cell when all ionic species have a concentration of 0.10 M?
The overall reaction is,
Hg2Cl2(s) + 2 e- —> 2Hg + 2Cl = 0.27 V
H2 —> 2 H+ + 2 e= 0.000 V
Hg2Cl2(s) + H2 –> 2 Hg + 2 H+ + 2 Cl= 0.27 V
So the voltage is,
 = 0.27 - 0.0592/2 log [H+]2 [Cl-]2 = 0.27 - 0.0591/2 log [0.10]2 [0.10]2 = 0.3882 Volts
3) How long will it take to galvanize (coat with Zinc) a nail with 2 grams of zinc using a current
(amperage) of 0.8 amps? FWT of Zn2+ = 65.41 g/mol
2 grams/65.41g/mol = 0.03058 mole of Zn
0.03058 mole of Zn x 2 mole e-/mol Zn = 0.06115 mole e0.06115 mole e- x 96,486 C/mol e- = 5900.4 Coulombs
1 C = 1 amp x sec
5900.4 C = 0.8 amps x sec
5900.4 C/0.8 amps = 7375.47 seconds
4) You are given 100 mL of water at 25°C. You add a 20 gram ice cube at 0°C and 50 mL of
boiling water at 100°C. What is the final temperature of the system? ΔHfus = 6.01 kJ/mol and
Cp(H2O) = 75.2 J/mol K, FWT H2O = 18g/mol (These were not given in the original problem).
The easiest way of doing this is by mixing the hot and cold water and then adding the ice cube.
First add 100 mL of 25°C water to 50 mL of 100°C water,
nCpΔT25°C = - nCpΔT100°C
100g/18g/mol (75.2 J/molK)(Tf - 25) = -50g/18g/mol(75.2 J/molK)(Tf-100)
The 18g/mol and the 75.2 J/molK cancel on both sides so,
100(Tf-25) = -50(Tf-100)
Tf = 50°C
So now we have 150 mL of water at 50̊C and we add the 20 gram ice cube. I assume that the ice
cube will melt and then waters will mix to get to some final temperature so,
nΔHfus + nCpΔT0°C = - nCpΔT50°C
20g/18g/mol(6010 J/mol) + 20g/18g/mol (75.2)(Tf - 0) = - 150g/18g/mol(75.2)(Tf - 50)
All the 18 g/mol cancel so,
120,200 + 1504Tf = - 11280Tf + 564000
12784Tf = 443,800
Tf = 34.71°C
5) What is the ΔG and ΔS for the dilution of 4.5 M H2SO4 to 0.03 M H2SO4 at 25°C? The ΔH =
- 210 kJ/mol.for this reaction.
ΔG = ΔG + RT ln Q
But ΔG = 0 since both concentrations have the same standard state of 1M. So,
ΔG = 0 + RT ln 0.03M/4.5M
ΔG = -5391.42 J/mol
Now, since ΔG = ΔH - TΔS, we can solve for ΔS,
-5391.42 J/mol = -210,000 J/mol - (298)ΔS
ΔS = (-5391.42 J/mol + 210,000 J/mol )/298 = 686.61 J/mol K
Chemistry 121
Third Exam
Name____________________
May 14, 2009
CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for full credit.
You may use a calculator.
Question
Credit
1 (18)
2 (28)
3(10)
4(20)
Total
R = 8.314 J/mol-K
 =  - 0.0592/n log Q
ΔG = -n
 = 96486 C/mol eΔG = ΔG + RT ln Q
PV = nRT
C = amp x sec
ΔG= -RTlnKeq
R = 0.08205 L-atm/mol-K
1a) Please balance the following half-reaction in a base.
BH3 —> B2O3
1b) What is the voltage of the following half-cell reactions when added together?
Cu2+ + e- —> Cu+
Cu+ + e- —> Cu
Cu2+ + 2 e- —> Cu
̊ = 1.29
= 1.68
= ?
2) In lab, you were asked to experimentally determine the Ksp of PbI2. Given the information
below (found in the CRC), what voltage should you have obtained for your cell, and what should
the Ksp have been?
PbI2(s) ↔ Pb2+ + 2 IΔH
S
PbI2 -174.1 kJ/mol
176.98 J/mol-K
Pb2+
1.63 kJ/mol
21.34 J/mol-K
I-55.94 kJ/mol
109.37 J/mol-K
The reaction is (Circle all that apply).
Spontaneous
Exothermic
Entropy increases
Occur Fast
Not Spontaneous
Endothermic
Entropy Decreases
Occurs Slow
2c) Like you did in lab, draw the electrochemical cell that was used to determine the Ksp of PbI2.
Label the anode, the cathode, the direction of electron flow, and indicate the concentrations of all
ionic species. Indicate the composition of the electrodes.
2c) Draw the cell diagram for the above cell.
3) It took 35 minutes and 10 amps to electroplate 10.615 grams of platinum (FWT = 195.08
g/mol). What was the charge on the platinum?
4) A 20 gram block of ice initally at -10.0̊C was dropped into 250 mL of water at 25̊C. What is
the final temperature of the system? ΔHfus = 6.01 kJ/mol, Cp(H2O) = 75.2 J/mol-K, Cp(ice) =
35.2 J/mol-K, FWT (H2O)= 18 g/mol
Chemistry 121
Third Exam
Name__Answer Key___
May 14, 2009
CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for full credit.
You may use a calculator.
Question
Credit
1 (18)
2 (28)
3(10)
4(20)
Total
R = 8.314 J/mol-K
 =  - 0.0592/n log Q
ΔG = -n
 = 96486 C/mol eΔG = ΔG + RT ln Q
PV = nRT
C = amp x sec
ΔG= -RTlnKeq
R = 0.08205 L-atm/mol-K
1a) Please balance the following half-reaction in a base.
BH3 —> B2O3
3 H2O + 2 BH3 —> B2O3 + 12 H+ + 12 e12 OH- —> 12 OH12 OH- + 2 BH3 —> B2O3 + 9 H2O + 12 e-
1b) What is the voltage of the following half-cell reactions when added together?
Cu2+ + e- —> Cu+
Cu+ + e- —> Cu
Cu2+ + 2 e- —> Cu
 = 1.29
 = 1.68
 = (1.29(1) + 1.68 (1))/2 = 1.485 Volts
2) In lab, you were asked to experimentally determine the Ksp of PbI2. Given the information
below (found in the CRC), what voltage should you have obtained for your cell, and what should
the Ksp have been?
PbI2(s) <--> Pb2+ + 2 IPbI2
Pb2+
I-
ΔH
-174.1 kJ/mol
1.63 kJ/mol
-55.94 kJ/mol
S
176.98 J/mol-K
21.34 J/mol-K
109.37 J/mol-K
ΔH̊ = [1.63 + 2(-55.94)] - [-174.1] = 63.85 kJ/mol
Δ S = [21.34 + 2(109.37)] - [176.98] = 63.1 J/molK
ΔG = 63,850 - 298(63.1) = 45,046.2 J/mol
45,046.2 J/mol = -8.314(298)ln(Ksp)
Ksp = 1.27x10-8
The reaction is (Circle all that apply).
Spontaneous
Exothermic
Entropy increases
Occurs Fast
Not Spontaneous
Endothermic
Entropy Decreases
Occurs Slow
2c) Like you did in lab, draw the electrochemical cell that was used to determine the Ksp of PbI2.
Label the anode, the cathode, the direction of electron flow, and indicate the concentrations of all
ionic species. Indicate the composition of the electrodes.
e-
Cathode
Positive
Reduction
Pb
Pb
Anode
Negative
Oxidation
PbI2(s)
1M
Pb(NO3)2
1M KI
2c) Draw the cell diagram for the above cell.
Pb / PbI2(s) / 1 M KI // 1 M Pb(NO3)2 / Pb
3) It took 35 minutes and 10 amps to electroplate 10.615 grams of platinum (FWT = 195.08
g/mol). What was the charge on the platinum?
10.615 g/195.08 g/mol = 0.05441 mole of Pt
0.05441 mole of Pt x Charge on Pt = moles of eAlso,
amp x sec/ 96,486 C/mol e- = mole of eSo,
0.05441 mole of Pt x Charge on Pt = amp x sec/ 96,486 C/mol e0.05441 mole of Pt x Charge on Pt = 10 amps x 35 min x 60 sec/min / 96,486 C/mol eCharge on Pt = [10 amps x 35 min x 60 sec/min / 96,486 C/mol e-]/0.05441 mole of Pt
Charge on Pt = +4
4) A 20 gram block of ice initally at -10.0°C was dropped into 250 mL of water at 25°C. What is
the final temperature of the system? ΔHfus = 6.01 kJ/mol, Cp(H2O) = 75.2 J/mol-K, Cp(ice) =
35.2 J/mol-K, FWT (H2O)= 18 g/mol
nCpΔTice + nΔHfus + nCpΔT0°C H2O = -nCpΔT25°C H2O
(20/18)35.2(10) + (20/18)6010 + (20/18)75.2(Tf-0) = -(250/18)75.2(Tf-25)
All the 18's cancel so,
7040 + 120200 + 1504Tf = -18800Tf + 470000
20304Tf = 342760
Tf = 16.88°C