Harris Quantitative Analysis, Quantitative Analysis

1. Need to Know
What is oxidation/reduction
What is oxidizing or reducing agent
Oxidation Numbers
Balancing redox reactions
Focus: Electrochemistry as a Chemical Analysis Tool
MnO4
-
+ NO2
-
 Mn2+
+ NO3
-
What is reduced in the following equation
a. Mn
b. N
c. O

2. Split into two reactions
Ce4+
+ e- = Ce3+
(Reduction)
Fe2+
= Fe3+
+ e- (Oxidation)
Electrochemical Cells
Galvanic cell - cell in which the reactions occur
spontaneously to produce electrical energy
Electrolytic cell – electrical energy used to
force a non-spontaneous reaction to occur
Cathode: where reduction takes place
Anode: where oxidation takes place
memorize

3. Two ways to make an electrochemical cell
1- all in same container
2- separate containers : half cells
connect with bridge
Copper – oxidized
soln blue
Silver – reduced
Cu turns grey
Piece of Cu in
a soln of AgNO3

4. 2Ag+
+ Cus  2Ags + Cu2+
A “silver” xmas tree
Reaction: spontaneous

5. 2Ag+
+ 2e-
 2AgsCds  Cd2+
+ 2e-
anode cathode
Balanced rxn = ?
∆G0
= - nF∆E0
∆E0
= E0
cath – E0
anode
E0
= ?

6. Reaction proceeds until
“chemical equilibrium”
Cu + 2Ag+
 Cu2+
+ 2Ag
E0
= ?
From skoog, west, holler etal

7. Can not measure the potential of any half-cell rxn
Can measure Potential Difference
Establish a reference electrode
tabulate wrt that half cell
Half cell Ref: SHE or NHE
1. Make SHE the anode
2. Make the other cell the cathode
3. Measure potential difference
2H+
+ 2e-  H2(g) E0
= 0.00 V
memorize

8. E0
cell = E0
cathode - E0
anode
If E0
anode is 0
then E 0
cell = E 0
cathode
All reactions must be written as
reduction rxns

9. SHE
anode
E0
cell = ?

11. What is E0
for
5Fe3+
+ 5e- = 5Fe2+
given
Fe3+
+ e- = Fe2+
E0
= 0.771 V
a. 0.771
b. 3.86
c. 0.154
d. 0

12. MnO4
-
+ NO2
-
 Mn2+
+ NO3
-
What is oxidizing agent in the following equation
a. MnO4
-
b. NO2
-
c. Mn

13. Important Points:
Simplified Cell Schematic
Anode ll Cathode
E0
cell = E0
cathode – E0
anode
All rxns written as reduction
ΔG0
= - nFE0
memorize
memorize
ΔG0
is negative: spontaneous
ΔG0
is positive: non-spontaneous

14. Which one of the following would be the best
oxidizing agent
a. Ag+
b. H+
c. Cd2+
d. Zn2+
Ag+
+ e- = Ags E0
= 0.799 V
2H+
+ 2e- = H2(g) E0
= 0.00V
Cd2+
+ 2e- = Cds E0
= -0.403
Zn2+
+ 2e- = Zns E0
= -0.763

15. Write a simplified cell schematic for the following
unbalanced redox reaction (std conditions)
Ni2+
+ Crs  Cr3+
+ Nis
If E0
(Ni2+
/Ni) is -0.25 V and E0
(Cr3+
/Cr) is -0.74,
what is E0
cell?
Ni2+
+ Crs  Cr3+
+ Nis

16. Cell Potential is Concentration Dependent
Nernst Equation
ba
dc
BA
DC
nF
RT
EE
][][
][][
ln0
−=
Nernst Equation

17. Nernst Equation:
at 250
C
ba
dc
BA
DC
n
EE
][][
][][
log
0592.00
−=
At Equilibrium
0592.0
0
10
cellnE
K =
E0
cell: units of V
n: total #e-
memorize
memorize

18. What is the total number of electrons transferred in
the following unbalanced redox rxn
MnO4
-
+ NO2
-
= Mn2+
+ NO3
-
a. 2
b. 4
c. 5
d. 7
e. 10

19. Given the following cell, calculate E0
cell
Ag l Ag+
(a = 0.02 M) ll Cu2+
(a = 0.02 M) l Cu

20. Calculate the equilibrium constant for the following
redox reaction given E0
(MnO4
-
/Mn2+
) is 1.51 V and E0
(Fe3+
/Fe2+
) is 0.771 V.
MnO4
-
+ 5Fe2+
+ 8H+
 Mn2+
+ 5Fe3+
+ 4H2O

21. What is the concentration of aluminum ions if the cell potential
Is 0.47 V and the Mn2+
concentration is 0.49 M.
E0
(Mn2+
/Mn) is -1.18 V and E0
(Al3+
/Al) is -1.66 V.
2Als + 3Mn2+
 2Al3+
+ 3Mns