Chemistry 222
Exam 4: Chapters 11, 13, 14
80 Points
Name__________________________________________
Fall 2015
Complete five (5) of the following problems. Each problem is worth 16 points. CLEARLY mark
the problem you do not want graded. You must show your work to receive credit for problems
requiring math. Report your answers with the appropriate number of significant figures.
1. Consider the half reactions below:
Reaction
Eo (volts)
MnO4-(aq) + 8H+(aq) + 5e- ⇌ Mn2+(aq) + 4H2O(l)
+1.507
Au+(aq) + e- ⇌ Au(s)
+1.690
a. Write the shorthand (line) notation for the galvanic cell that could be produced from the
two half reactions above. What is the Eo for this cell? (6 points)
Since we want the galvanic cell, we need to determine which have cell will need to be
the anode and which will need to be the cathode so that the overall cell reaction is
spontaneous (has a positive cell potential, Eocell).
Since Eocell = Eocathode - Eoanode, if the gold half cell is the cathode, the cell potential
becomes Eocell = +1.690V – (+1.507V) = +0.183 V. Since this potential is positive, this
must be the arrangement for the galvanic cell, with Eocell = +0.183 V
The shorthand notation is:
Pt(s)| MnO4- (aq), H+ (aq), Mn2+(aq) || Au+ (aq) | Au(s)
b. What is the cell potential if [MnO4-] = 0.140 M, [Mn2+] = 0.00200 M, [Au+] = 0.0100 M and
pH = 5.13? Is this more or less spontaneous than under standard conditions as in part a
above? (10 points)
The overall cell reaction is:
5Au+ + Mn2+ + 4H2O ⇌ 5Au + MnO4- + 8H+
We can use the Nernst equation to determine the cell potential:
Ecell =+0.183V-0.05916 Vlog[MnO4-][H+]8
5
[Au+]5[Mn2+]
Ecell =+0.183V-0.05916 Vlog [0.140][10-5.13]8 =+0.528V
5
[0.0100]5[0.0020]
Since this potential is more positive than that at standard conditions, the reaction is more
spontaneous under these conditions than under standard conditions.
You could also use the Nernst equation for each half reaction, then calculate a cell
potential:
[Mn2+]
= +1.0432 V
Eanode = +1.507V - 0.05916 V log
+ 8
5
[MnO4 ][H ]
Ecathode =
+1.690V -
0.05916 V
1
log
1
[Au+]
Ecell = Ecathode - Eanode = : +1.5716 V – (+1.0432 V) = +0.528 V
1
= +1.5716 V
2. Consider the titration of 20.0 mL of 0.0200 M Fe(NO3)2 with 0.0100 M EDTA in a solution
buffered at pH 8.00. (log Kf = 14.30 for the Fe2+-EDTA complex)
a. Calculate pFe2+ at two of the volumes below. (12 points)
 At the equivalence point
 11.2 mL before the equivalence point
 11.2 mL after the equivalence point.
First locate the equivalence point:
20.0 mL
x
0.0200 mol Fe2+
L
1 mol Y41 mol Fe2+
x
=
1L
0.0.0100 mol Y4-
=
40.0 mL
At the equivalence point: [FeY2-] = (20.0 mL x 0.0200 mol/L)/(60.0 mL) = 0.00667 M
Fe2+
+
Y4=
FeY2i
0
0
0.00667
c
+x
+x
-x
e
x
x
0.00667-x
Kf' = Y4-Kf =
= (0.0056)(1014.3) = 1.12x1012 = 0.00667-x
[FeY2-]
[Fe2+][Y4-]
(x)(x)
Solving for x yields [Fe2+] = 7.73 x 10-8M or pFe = 7.11
11.2 mL before the equivalence point: we will have consumed (40-11.2)/40 of the Fe2+, so
11.2/40 remains. Since K is large, we can assume that the dissociation of FeY2- is negligible.
20.0 mLx0.0200 mol Fe2+x11.2 mLx
1
=0.002295 M Fe2+ or, pFe = 2.64
L
40 mL 20 + 28.8 mL
11.2 mL after the equivalence point: we will have 11.2 mL unreacted EDTA in a total of 71.2
mL solution.
[FeY2-] = (20.0 mL x 0.0200 mol/L)/(71.2 mL) = 0.005617 M
[EDTA] = (11.2 mL x 0.0100 mol/L)/(71.2 mL) = 0.001573 M
Fe2+ +
Y4=
FeY2i
0
0.001573
0.005617
c +x
+x
-x
e
x
0.001573 + x
0.005617 -x
Kf' = Y4-Kf =
= (0.0056)(1014.3) =
0.005617 -x
[FeY2-]
2+
4[Fe ][Y ]
(x)(0.001573 + x)
Solving for x yields [Fe2+] = 3.196 x 10-12M or pFe = 11.50
b. Given that the fraction of the EDTA present as Y4- does not reach its maximum until pH
is over 12 or so, why would we choose to run the titration at pH 8.00 instead? (4 points)
Increasing the pH favors the formation of insoluble metal hydroxides, therefore, the
titration must be run at a higher pH or we must use an auxiliary complexing agent to
prevent hydroxide formation at higher pH.
2
3. You need to do a pH measurement and have a pH electrode, pH 4.00, 7.00, and 10.00
buffers, but no pH meter! You do, however, have access to a voltmeter capable of
measuring potential differences at the millivolt level. Using these materials, describe how
you could determine the pH of an unknown aqueous solution. Include a description of how
you would collect the data and how you would use the data to find the pH of your unknown.
In your description, you should address the following items.
1. Connect the leads of the pH electrode to the voltmeter.
2. Measure Ecell for each buffer. (You should briefly describe how this is done)
3. Plot Ecell vs pH, you should expect a straight line with a slope of ~59mV/pH unit. (Why
should this relationship be linear?)
4. Measure Ecell for the unknown and extract the unknown pH from your calibration curve.
3
4. In our magnesium determination in the laboratory, we added a solution made from ammonia
and ammonium chloride to each sample prior to titrating with EDTA. Explain the two primary
purposes served by the addition of this solution?
Your discussion should focus on the following two ideas:
1. In order for the titration to be effective, it is important to keep as much of the EDTA in its
fully deprotonated for as is reasonable. Keeping the pH high results in a larger fraction
of all EDTA present as Y4-. AS a result, the conditional formation constant, Kf’, will
remain large.
2. At high pH, many metal ions form insoluble metal hydroxides or hydrous oxides. Should
this happen, the analyte would precipitate out of solution and be unavailable for reaction
with EDTA. Ammonia serves as an auxiliary complexing agent in the titration by forming
a soluble magnesium complex. The tendency for ammonia to bind Mg2+ is greater than
that for hydroxide, preventing the formation of magnesium hydroxide. However, EDTA
binds more strongly than ammonia, allowing the titration reaction to occur.
4
5. Calculate the E at pH 11.00 for the half reaction Pd(OH)2 (s) + 2e- ⇌ Pd (s) + 2OH- given that
the Ksp for Pd(OH)2 is 3.0x10-28 and the Eo = +0.915 V for Pd2+ + 2e- ⇌ Pd (s).
There are several approaches to solving this problem, all start with finding Eo. Here’s one:
1
2
Net = 1 +2
Reaction
Pd2+ + 2e- = Pd (s)
Pd(OH)2 (s) = Pd2+ + 2OHPd(OH)2 (s) + 2e- = Pd (s) + 2OH-
Eo
+0.915 V
K
3.0x10-28
K1K2
Eo1 + Eo2
Go1
Go2
Go1
Go
= -nFEo
= -RTlnK
+ Go2
Since we can make our overall equation by summing the two components, we can
determine the Go for the reaction from the sum of the Gos for the component reactions.
Go1 = -nFEo = -(2 mol e-)(96485 coul/mole e-)(+0.915V) = -176,568 J
Go2 = -RTlnK = -(8.31441 J/mol K)(298K)ln(3.0x10-28) = 157,021 J
Gonet = Go1 + Go2 = 157021 J + (-176568 J) = -19,547 J
Eonet = Gonet/(-nF) = -19547 J/[-(2 mol e-)(96485 coul/mole e-)] = +0.1013 V
Another approach would be to convert everything to K’s and realize that Knet = K1 x K2:
For an equilibrium: -nFEo = -RTln K = -2.303RTlog K
So, for this equilibrium at standard temperature:
logK1 = (-nFEo1)/(-2.303RT) = 30.94
Knet = K1K2 = (1030.94)(3.0x10-28) = 2668
o
Then E net = (2.303RT/nF)log K = (0.05916V/2)log(2668) = +0.101V
Once you have the E0, use the Nernst equation to find the E at pH 11.00:
E = +0.101V - 0.05916 V
2
log [OH-]2
= +0.101V - 0.05916 V
2
log [10-3]2
= +0.278 V
Yet another approach is to use the Ksp expression to find the [Pd2+] in a saturated solution at
pH 11.0, then use the Nernst equation for the Pd2+ + 2e- =Pd reaction to find E.
Pd(OH)2 = Pd2+ + 2OH-, Ksp = [Pd2+][OH-]2
[Pd2+] = Ksp/[OH-]2 = 3.0 x 10-22 M
E=
+0.915V
-
0.05916 V
2
log
5
1
[Pd2+]
= +0.278 V
6. A 1.25 g sample of soil containing Cu2+ and Zn2+ and other nonmetals was digested in acid
and diluted to a final volume of 50.00 mL. This solution was treated with 25.0 mL of
0.0452 M EDTA to bind all the metal. The excess unreacted EDTA required 12.4 mL of
0.0123 M Mg2+ for complete reaction. An excess of the reagent 2,3-dimercapto-1-propanol
was then added to displace the EDTA from zinc only. Another 29.2 mL of Mg2+ were
required for reaction with the liberated EDTA. Calculate the percent by weight of Cu2+ and
percent by weight of Zn2+ in the original rock sample.
We need to figure out the mass of both metals in the sample.
25.0 mL x 0.0452 mol EDTA = 1.13 mmol EDTA delivered
L
12.4 mL x 0.0123 mol Mg x 1 mol EDTA = 0.1525 mmol EDTA in excess
L
1 mol Mg
So, (1.13 – 0.1525)mmol = 0.9775 mmol EDTA was used to bind all of the metal. Therefore,
mol Zn2+ + mol Cu2+ = 0.9775 mmol
We can determine the amount of zinc from the magnesium titration:
29.2 mL x 0.0123 mol Mg2+ x 1 mol EDTA x 1 mol Zn2+ = 0.3592 mmol Zn2+
L
1 mol Mg
1 mol EDTA
So, the difference between the total moles metal and the moles zinc must correspond to the
number of moles copper in the sample.
(0.9775 – 0.3592)mmol = 0.6183 mmol Cu2+
Now to determine percent composition:
0.3592 mmol Zn2+ x 65.37 g Zn x
1g
= 0.02348 g Zn
1 mol Zn
1000 mg
0.02348 g Zn x 100 % = 1.88 % Zn
1.25 g sample
0.6183 mmol Cu2+ x 63.54 g Cu x
1g
= 0.03929 g Cu
1 mol Cu
1000 mg
0.03929 g Cu x 100 % = 3.14 % Cu
1.25 g sample
6
Possibly Useful Information
Kw = 1.0 x 1014 = [H+][OH-]
E  Eo 
x
0.05916V
2.303RT
logQ
logQ  Eo 
n
nF
 b  b 2  4ac
2a
Go = -nFEo = -RTlnK
F = 96485 C mol-1
R = 8.31441 Jmol-1K-1
 0.05916V 
E  const.  
 log ion
n


m
y = mx + b ,
y
x
Values of y4- for EDTA at 20°C and  = 0.10 M
pH
y4-
y4-
pH
pH
y4-
3.7 x 10
-7
10
0.36
-5
11
0.85
1.3 x 10
-23
1
1.9 x 10
-18
6
2.3 x 10
2
3.3 x 10-14
7
5.0 x 10-4
12
0.98
3
2.6 x 10-11
8
5.6 x 10-3
13
1.00
4
-9
9
-2
14
1.00
0
3.8 x 10
5
5.4 x 10
7