Name:
Chem 332
Analytical Chemistry
Exam I
1. A 100.0 mL solution of the ion V2+ at a concentration of 0.0500 M buffered to pH 9.00 was
titrated with .0500 M EDTA
a. What is the equivalence volume (Ve) in mLs?
100.00 mL
b. What is the concentration of free V2+ ion at V = ½ Ve?
.1(.025)/.150 = .01667M (.025 = .05/2)
c. What is concentration of EDTA in the Y4- form at this pH?
.05*á = .05 x .041 = .00205
d. Calculate the value of the conditional formation constant (Kf’) for the V2+@EDTA
complex at pH 9.00.
Kf *á = 2.05x1011
e. Calculate the concentration of V2+ at V = Ve
Kf’ = [EDTA@V2+]/[V2+][EDTA]
[EDTA@V2+] = .1(.05)/.2 = .025
[V2+]=[EDTA]=X
Kf’ = (.025-X)/X2
.025-X ~ .025
2
’
X =.025/Kf
X=sqrt(.025/2.01x1011) =1.22x10-11; X =3.5x106
2A.Make a diagram of a galvanic cell that uses the following two reactions to generate
electricity:
Al3+ + 3e- 6 Al(s)
Eo = -1.677
1+
Cu + e 6Cu(s)
Eo = .518
Make sure your diagram includes the following items:
Direction of electron flow. Anode cell, Cathode cell, salt bridge. Cell in which
oxidation occurs, cell in which reduction occurs. Chemical in solution in the
cells, metals used in electrodes of the cells, net reaction for the cell, Eo of the cell.
Cu metal electrode on right in a cell containing Cu+1 ion
This is where reduction occurs
This is cathode
This is + pole
Al metal electrode on left in a cell containing Al3+ ion
This is where oxidation occurs
This is anode
This is - pole
A wire connects the to metal electrodes
A salt bridge connects the two cells
Net reaction for cell
3 Cu1+ (aq) + Al(s) 63Cu(s) + Al3+(aq)
E0 = .518-(1.677)=2.195
2B. What is the E of the cell if you build it using .001M AlCl3 for one solution and .1M CuNO3
for the other solution?
ER = .518 - .05916/3 log(1/[Cu1+]3)
= .518 - .05916/3 log(1/.13)
= .518 - .05916/3 log(1/.001)
= .518 - .05916/3 x 3
= .518 - .05916
= .459
EL = -1.677 - .05916/3 log(1/.001)
= -1.677 - .05916/3 * 3
= -1.677 - .05916
= -1.736
Ecell = ER - EL = .459-(-1.736) = 2.195
No I did not deliberately make this equal to the previous answer. It just worked out that way!
3. Describe the construction of a Silver/Silver Chloride reference electrode. If I have a cell that
has a potential of -.1V vs a Silver/silver chloride electrode (saturated KCl), what would its
potential be when measured against a calomel electrode (also containing saturated Kcl)
Silver wire coated with AgCl
in a solution of either 1M KCl or saturated KCl,
Connect to external solution via a salt bridge
Potential of a saturated AgCl electrode is .197
Potential of a saturated Calomel electrode is .241
V = .097-.241 = -.144
4A.What is a junction potential?
A potential that arises at a junction between two solutions in which ions of one charge are
passing at a different rate than ions of another charge creating a charge imbalance at the interface
between the tow solutions.
4B. How does a junction potential arise in a solid-state ions selective electrode.
The most common solid state electrodes are for anions like F-,Cl-, Br- , I-, SCN-, CN- or
2S . In this kind of electrode a solid crystal acts as the interface between two solutions. This
crystal is composed of some insoluble salt of the ion which you are trying to detect. Since the
crystal is insoluble, it does not dissolve, but because the crystal contains the ions you are trying to
detect, these ions can diffuse into and out of the crystal. If these anions can diffuse into one side
of the crystal and out of the other side, we have created in interface that only allows a certain ion
to pass, and a junction potential forms across this interface as anions moved without their
corresponding cations.
4C. Use 4A and 4B to explain how a solid-state ions selective electrode works.
To detect the above junction potential you can place a reference electrode like a
silver/silver chloride electrode close to both sides of the interface. Since these reference
electrodes are in chemically different environments, the potential difference between the two
reference electrodes will reflect the potential difference across the two sides of the electrode, and
this potential difference will reflect the concentration difference of the ion that can cross the
crystal interface.
5. In theory you should now be able to make an electrode that would sense the presence of Fe3+
in our concentrated solution of sludge dissolved in acid. Describe the construction of such an
electrode and predict its potential.
This could be as simple as an iron wire placed in the solution coupled to some reference
electrode.
6. Another way we could determine the amount of iron in our sludge sample would be to do a
redox titration. If we were to use the two half reactions:
Fe3+ + e- 6Fe2+
Eo = .771
I3- + 2e- 6 3IEo =.535
to determine the amount of Fe3+ in a sample, answer the following questions
a. Write a balanced titration reaction.
2Fe3+ + 3I- 62Fe2+ + I3b. What is the Eo for this reaction
.771-.535 = .236
c. What is the K for this reaction
K = 10^(nE/.05916)
=10^[2(.236)/.05916]
=9.5x107
d. What is the ÄGo for this reaction?
= -nFE
= -2*96485*.236
= -45.5 kJ
Consistency check ÄG=-RTlnK
= -8.314(298)9.5x107
= -45.5 kJ
Both methods for ÄG agree!
e. What would we use for a titrant, I3- or I-?
If. If we had a 50 mL solution that was 1M Fe3+, and a titrant molarity of 1M, what
volume of titrant required to reach the equivalence point (Ve) ?
.05(1) X 3I-/2Fe2+ = .075 moles I1 = .075/X ; X = .075L = 75 mL
g. If we were to use an electrode to monitor the potential of the titration against a standard
hydrogen electrode, what potential would we measure at 0.5 @Ve?
.5 Ve = 37.5 mL
mMoles Fe3+ = 50*1 = 50 mmole
mMole I- = 37.5 *1 = 37.5 mmole
Reaction table
2Fe3+ +
3I- 6
2Fe2+ +
I3Initial
50
37.5
0
0
RXN
-37.5(2/3)
-37.5
+37.5(2/3)
+37.5(1/3)
-25
37.5
+25
+12.5
Net
25
0
25
12.5
Have finite amounts of Fe3+ and Fe2+ so will use Nernst of this reaction
E = E0 - .05912/n Log([Fe2+]/[Fe3+]
=.771 -.05912/1 log [25 mmole/(50+37.5)mL / 25 mole/(50+37.5)mL]
=.771 -.05912/1 log(1)
=.771
h. If we were to use an electrode to monitor the potential of the titration against a standard
hydrogen electrode, what potential would we measure at 1.0 @Ve ?
At equivalence point
E=( n1Eo1+n2Eo2)/(n1 + n2)
=[1(.771) + 2(.535)/(1+2)
= .614
i. Is there a redox indicator that could be used find the endpoint of this titration? If so,
what indicator would you use?
None really fit!
j. If we were to use an electrode to monitor the potential of the titration against a standard
hydrogen electrode, what potential would we measure at 1.1 @Ve?
1.1 Ve = 82.5 mL
mMoles Fe3+ = 50*1 = 50 mmole
mMole I- = 82.5 *1 = 82.5 mmole
Reaction table
2Fe3+ +
3I- 6
2Fe2+ +
I3Initial
50
82.5
0
0
RXN
-50
-50(3/2)
+50
+50(1/2)
-50
-75
+50
+25
Net
0
7.5
50
25
Have finite amounts of I- and I3- so will use Nernst of this reaction
E = E0 - .05912/n Log([I-]3/[I3-]
=.535 -.05912/2 log [(7.5 mmole/(50+82.5)mL)3 / 25 mmole/(50+82.5)mL]
=.535 -.05912/2 log [(7.5 mmole/(132.5)mL)3 / 25 mole/(132.5)mL]
=.535 -.05912/2 log [(.0566)3 / .189]
=.535 -.05912/2 log (.00096)
=.535 -.05912/2* (-3.02)
=.535 + .089
=.624