1
CHEM 1000 A, V
Midterm Test #3, February 5, 2010
Part A. Answer each question with a few sentences or equations where necessary. (5 Marks each)
1.
Name five things that involve electrochemical reactions. No reactions are required. Choose
examples from industrial processes, natural phenomena, household chemistry, etc. An example
would be “manufacturing aluminum metal from bauxite.”
Any five of: manufacturing F2, manufacturing Cl2 and NaOH, manufacturing aluminum,
electrowinning of metals, electrorefining of metals, corrosion, batteries, recharging batteries,
fuel cells, electrolysis of water to make H2, etc
2.
Write the balanced overall redox reaction for the electrochemical cell denoted by the following
shorthand notation: Pt(s) | C2O4 2–(aq) | CO2(g) || Ag+(aq) | Ag(s)
C2O4–2(aq) + 2 Ag+(aq) → 2 CO2(g) + 2 Ag(s)
3.
Name two reasons, related to the lead–acid battery, why it is more difficult to start your car on a
cold day.
– According to the Nernst equation, the potential of the battery will be lower when the
temperature is lower.
– The viscosity of the electrolyte increases at lower temperature, which lowers the current
through the electrolyte.
4.
In an electrolytic cell having two Platinum electrodes and ZnF2(aq) as an electrolyte, what is
produced at the anode and what is produced at the cathode?
At the anode there are two possibilities:
2 F– → F2 + 2 e–
(–2.87 V)
+
–
2 H2O → O2 + 4 H + 4 e
(–1.23 V)
Because the water oxidation potential is higher, O2 and H+ are produced at the anode.
the cathode there are two possibilities:
Zn+2 + 2 e– → Zn
(–0.76 V)
2 H2O + 2 e– → H2 + 2 OH– (–0.83 V)
Because the zinc reduction potential is higher, Zn metal is produced at the cathode.
5.
Arrange the following compounds in increasing order of boiling point and explain your answer
in terms of the intermolecular forces present:
2
H3C
O
CH3
Dimethyl ether
C2H5
O
C2H5
Diethyl ether
H3C C C O H
H2 H2
n–propanol
Lowest is dimethyl ether. These molecules interact via dipole–dipole forces.
In the middle is diethyl ether. These also experience dipole–dipole forces, but they are larger
than those in dimethyl ether because the diethyl ether molecules are larger.
n–propanol has the highest boiling point because the molecules interact via relatively strong H–
bonds.
6.
Explain why the viscosity of a liquid decreases at the temperature is increased.
At lower temperatures, the molecules in a liquid have less kinetic energy. Fewer of them have
enough kinetic energy to overcome the attractive intermolecular forces, and therefore the
molecules do not flow past one another as easily.
Part B. Answer any three of four questions B1-B4. If you answer all four, the best three will be used to
calculate your mark. (20 marks each)
B1.
A Galvanic electrochemical cell is constructed consisting of thallium (Tl) metal dipping in an
aqueous solution that is 0.050 M in Tl+, connected by a salt bridge to a 0.50 M aqueous solution
of HCl in contact with a Pt electrode over which H2 gas is bubbling at p = 0.90 bar.
(a) Write the balanced redox reaction [5 marks]
Tl(s) → Tl+(aq) + e–
2 H+(aq) + 2 e– → H2(g)
Eo = +0.34 V
Eo = 0.00 V
2 Tl(s) + 2 H+(aq) → 2 Tl+(aq) + H2(g)
Eocell = +0.34 V
(We know the reaction is in this direction because it is a Galvanic cell, which must have a
positive cell potential.)
(b) Determine the cell potential (in V) at 25oC. [10 marks]
3
 2
Q
[Tl ] p H2
 2

2
(0.05) (0.90)
 0.009
(0.50) 2
[H ]
RT
E  Eo 
ln(Q)
nF
8.314 J K 1mol 1 (298 K)
 0.34 V 
ln(0.009)
2(96487 C mol 1 )
 0.34 V  (0.06 V)
 0.40 V
(c) TlI is insoluble in water. What will happen to the cell potential as a solution NaI(aq) is
dripped into the anode compartment of this cell? Explain. [5 marks]
Since TlI is insoluble, adding NaI will result in the precipitation of Tl+ as TlI. This will reduce
[Tl+(aq)]. This will reduce Q, making ln(Q) more negative, hence making E more positive.
B2.
The first battery to find widespread commercial use was the carbon–zinc dry cell, in which the
cathode reaction is
2 MnO2(s) + Zn2+(aq) + 2 e– → ZnMn2O4(s)
In a flashlight, one of these batteries provides a current of 0.048 A. If the battery contains 4.0 g
of MnO2 and fails after 90% of its MnO2 is consumed, calculate the operating life of the
flashlight (in hours).
90%(4 g) = 3.6 g MnO2(s)
3.6 g
 0.0414 mol MnO 2
(54.9  32)g mol1
According to the balanced reaction above, 2 mol MnO2 requires 2 mol electrons. Thus, 0.0414
mol electrons are required (=n). [10 marks]
It  nF
nF
t
I
0.0414 mol(96487 C mol1 )
[10 marks]

0.048 C s 1
 8.3 104 s
 23.1 h
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B3. Octane (C8H18) has a vapour pressure of 0.462 atm at 100 °C, and an enthalpy of
vapourization of 50.1 kJ/mol. Calculate the normal boiling point of octane (oC).
 p  H vap  1 1 
ln  2  
  
R  T1 T2 
 p1 
p  1 1
R
[10 marks]
ln  2   
H vap  p1  T1 T2
1
 p 
R
T2   
ln  2  
 T1 H vap  p1  
1
Here, T1 = 100oC = 373 K
p1 = 0.462 atm
T2 = ?
p2 = 1.00 atm (since T2 is the normal boiling point, –10 for not knowing this)
ΔHvap = 50100 J mol–1
1
 1
8.314 J K 1mol1  1 atm  
T2  

ln 

50100 J mol1
 0.462 atm  
 373 K
[10 marks]
 (2.553 103 K 1 ) 1
 392 K
 119o C
B4. Referring to the phase diagram below for water, answer the following questions:
S
L
G
(a)
(b)
(c)
(d)
Label “S”, “L” and “G” on the diagram. [3 marks]
The normal melting point of water is ____0__ oC [1 mark]
The normal boiling point of water is _____100____ oC [1 mark]
The critical point of water is at ___374____ oC and ____218____ atm [2 marks]
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o
(e) The triple point of water is at ___0.01____ C and ____0.006__ atm [2 marks]
(f) Some H2O(g) at –1 oC and 0.001 atm is gradually pressurized to 300 atm, holding the
temperature constant at –1oC. Describe what happens, step by step. [5 marks]
Initially, only G is present
At the sublimation line, G Ý S
Between the sublimation line and the freezing (melting) line, there is only S
On the freezing (melting) line, S Ý L
Above the freezing (melting) line, there is only L
(g) H2O(g) at 25oC and 0.001 atm is cooled to –1oC, keeping the pressure constant at 0.001 atm.
Does any frost form? How do you know? [6 marks]
No frost forms. At no time is the sublimation curve crossed.