1
CHEM 1000 A, V
Midterm Test #3, February 6, 2004 Calculator allowed.
This test should have six pages. A periodic table and a table of reduction potentials are on the last page.
Part A. Answer each question with a few sentences or equations where necessary. (5 Marks each)
1. [5]
2. [5]
In an electrolytic cell containing two platinum electrodes and an aqueous solution of PbI2, what will be produced at
the anode and what will be produced at the cathode?
Possible oxid’n reactions: 2 I-(aq)  I2(s) + 2 e2 H2O(l)  O2(g) + 4 H+(aq) + 4 e-
Eo = -0.54 V
Eo = -1.23 V Thus, I2(s) is produced at the anode
Possible red’n reactions: Pb+2(aq) + 2 e-  Pb(s)
2 H2O(l) + 2 e-  H2(g) + 2 OH-
Eo = -0.13 V
Eo = -0.83 V Thus, Pb(s) is produced at the cathode
Write the oxidation and reduction reactions in the cell having the shorthand notation:
Cu(s) | Cu+2(aq) || O2(g) | H+(aq) | H2O(l) | Pt(s)
Oxidation:
Reduction:
3. [5]
Cu(s)  Cu+2(aq) + 2 e4 H+(aq) + O2(g) + 4 e-  2 H2O(l)
Why does connecting a piece of Mg(s) to a piece of iron metal prevent the iron from corroding?
Mg(s)  Mg+2(aq) + 2 e- occurs at a higher oxidation potential (+2.37 V) than that of Fe (s)  Fe+2(aq) + 2 e- (+0.45
V). The oxidation of Mg is therefore preferred. Thus, the reduction of iron is preferred and the iron stays as solid
metal rather than being corroded.
4. [5]
Name two advantages of a lithium ion battery over a dry cell.
1. These batteries are light weight since lithium is a low density metal. 2. Lithium ion batteries have a high
potential, since the oxidation potential of lithium is so high. 3. Lithium ion batteries are rechargeable, dry cells are
not. 4. Lithium ion batteries last longer, since the potential is higher. (any two of these reasons)
5. [5]
Explain why water forms little droplets on a waxed surface.
Water has a high surface tension. This means that the water molecules have greater forces of attraction to one
another than to the molecules in the wax.
6. [5]
Which of diemethyl ether (H3C-O-CH3) and methyl alcohol (H3C-O-H) has the higher boiling point? Why?
Methyl alcohol has the higher boiling point since its intermolecular forces (H-bonds) are stringer than those of the
ether (dispersion forces). These forces must be overcome in order to vaporize molecules from the liquid into the
gas phase.
2
Part B. Answer any three questions. If you answer all four, the best three will be used to calculate your
mark. (20 marks each)
1.
Referring to the phase diagram for water below,
(a) Name all phases present at points A, B, C, D, E, F and G.
(b) Name points E and H.
(c)Explain why the melting point line has a negative slope.
(d)What is the region above and to the right of point H called?
(e)Explain what will be observed if, starting at point J, the pressure is increased at constant
temperature.
F
Pressure
.J
.
.B
.A .
.D E
.H
.G
.C
Temperature
(a) [7] A = solid
B = liquid
C = gas
D = solid, gas
E = solid, liquid, gas F = solid, liquid
G = liquid, gas
(b) [2] E is the triple point
H is the critical point
(c) [5] Solid water (ice) has a lower density than liquid water. Thus, when pressure is increased, this
increases the density, i.e. tend to convert solid into liquid. This can only happen if the line has a
negative slope.
(d) [1] Supercritical fluid
(e) [5] Initially, at point J, there is only gas. At the S-G equilibrium line, solid will appear. Above this
line, the gas will disappear, leaving only solid. Then at the S-L line, liquid will appear. Above
this, the solid will disappear, leaving only liquid.
2.
The reactions in a dry cell battery are:
Anode: Zn(s)  Zn+2(aq) + 2eCathode: 2 MnO2(s) + 2 NH4+(aq) + 2e-  Mn2O3(s) + 2 NH3(aq) + H2O(l)
[4]
E0 = +0.76 V
E0 = +0.75 V
(a) Calculate the standard free energy change for the overall reaction (kJ).
The overall reaction is Zn(s) + 2 MnO2(s) + 2 NH4+(aq)  Zn+2(aq) + Mn2O3(s) + 2 NH3(aq) + H2O(l)
Eo = Eocathode + Eoanode = 0.75 + 0.76 V = 1.51 V
Go = -nFEo = -2 mol e- (96487 C/mol e-)(1.51 J/C) = -291391 J = -291 kJ
3
[8]
(b) Calculate the battery potential at 25°C when the battery is fully charged ([NH4
= 0.01 M, [Zn+2(aq)] = 0.01 M)
+
(aq)
= 1.0 M, [NH3(aq)]
Zn(s) + 2 MnO2(s) + 2 NH4+(aq)  Zn+2(aq) + Mn2O3(s) + 2 NH3(aq) + H2O(l)
For this reaction, Q 
[Zn 2 (aq) ][NH 3 ]2
[NH 4  ]2

(0.01)(0.01) 2
 1.0 106
(1.0) 2
RT
ln(Q)
nF
8.314 JK 1mol1 (298 K)
 1.51 J C 1 
ln(1.0 106 )

 1
2 mol e (96487 C (mol e ))
E  E0 
 1.68 J C 1
 1.68 V
[8]
(c) Calculate the battery potential at 10°C when the battery is almost discharged ([NH4+(aq) = 0.01 M,
[NH3(aq)] = 1.0 M, [Zn+2(aq)] = 0.5 M)
For these conditions, Q 
[Zn 2 (aq) ][NH3 ]2
[NH 4  ]2

(0.5)(1.0) 2
 5000
(0.01) 2
RT
ln(Q)
nF
8.314 JK 1mol 1 (263K)
 1.51 J C 1 
ln(5000)
2 mol e  (96487 C (mol e  )) 1
E  E0 
 1.41J C 1
 1.41V
3. [20] A surface is to be electroplated with copper (Cu) metal by electrolyzing an aqueous solution of
Cu(NO3)2(aq). If a 1.0 cm2 surface is made the cathode in an electrolytic cell, how thick (in cm) will the
copper layer be after passing a current of 0.05 A through the solution for 10 minutes? The density of
copper metal is 8.92 g/cm3.
q = it = nf
[10]
60s 

0.05Cs 1 10.0 min

it
min 

n 
F
96487 C (mol e  ) 1
 3.11104 mol e 
In water, Cu(NO3)2(aq) is really Cu+2(aq) + 2 NO3 (aq).
Thus, the reduction reaction is Cu+2(aq) + 2 e  Cu(s). The quantity of copper reduced is therefore
[5]
3.11104 mol e  
1mol Cu
 1.55 104 mol Cu

2 mol e
1.55 x 10-4 mol Cu x 65.3 g mol-1 = 0.00987 g Cu reduced
4
0.00987 g Cu
 0.0011cm3 Cu
3
8.92 g cm
[5]
4.
0.0111cm3 Cu
 0.0011cm thickness
1cm 2 Cu
Some data for the vapor pressure of water are presented in the table below:
Temperature, oC
5.0
10.0
90.0
95.0
[8]
Vapor Pressure, mm Hg
6.54
9.21
525.8
633.9
(a) Calculate the enthalpy of vaporization, Hvap (kJ mol-1) at low temperature, i.e. use the data at 5.0oC
and 10.0oC.
ln(p 2 )  ln(p1 ) 
thus, H vap
H vap  1 1 
  
R  T1 T2 


 ln(p )  ln(p ) 
2
1

 R 
1
1





T
T
2
1




 ln(9.21)  ln(6.54) 
  44,800 J  44.8 kJ
 8.314 JK 1mol 1 
1
1


 283K  278 K 


[8]
(b) Calculate the enthalpy of vaporization, Hvap (kJ mol-1) at high temperature, i.e. use the data at
90.0oC and 95.0oC.
H vap


 ln(p )  ln(p ) 
2
1

 R 
1
1





T
T
2
1




 ln(633.9)  ln(525.8) 
  41,500 J  41.5 kJ
 8.314 JK 1mol 1 
1
1


 368 K  363K 


[4]
(c) You should have found that Hvap is smaller at the higher temperature. Why is this so?
This is because at higher temperature, the water molecules already have higher energy, so less is
required to vaporize them from the liquid into the gas phase.