Chemistry 146 Lecture Problems
Zinc/Iron Electrochemical Cell
At Non-Standard Conditions
For the electrochemical system:
Zn | Zn2+ (1 M) || Fe2+ (1 M) | Fe
Anode (oxidation occurs at the anode, in the cell notation used above the anode is on the
left):
Bred -> Box + ne-
Oxidation Reaction
Zn(s) --> Zn2+ (1 M) + 2 e-
Reduction potential for Box
E anode
0.763 .volt
Cathode (reduction occurs at the cathode, in the cell notation used above the cathode is on
the right):
Reduction Reaction
Aox + ne - -> Ared
Fe2+ (1 M) + 2e- --> Fe (s)
Reduction potential for Aox
E cathode
0.409 . volt
Reaction: The overall electrochemical reaction has the form:
a Aox + b Bred <-> a Ared + b Box
For this problem
Zn(s) + Fe2+ (1 M)
--> Zn2+ (1 M) + Fe (s)
Eocell (Calculate the standard cell potential for this system):
E std_cell
E cathode
E anode
E std_cell = 0.354 volt
Zn_Fe_cell.mcd
S.E. Van Bramer
8/7/01
Ecell calculation with the Nernst Equation (This step is required any time the system is not
at the "standard state". In the "standard state" concentrations are 1 M, pressure is 1 atm,
and temperature is 298.15 K) In this problem you were given a series of different
concentrations to use for these calculations:
20 ) .K
Temperature
T
( 273.15
Faraday's Constant:
F
96485.309 .
coul
mole
Gas Law Constant:
R
8.314510 .
Number of electrons in balanced redox reaction:
n
2
Define Molarity
M
joule
mole .K
mole
liter
For each example you are given a concentration for Zn2+ and Fe2+. The following additional
calculations are required to solve:
Concentration anode:
C Zn2
1 .M
Concentration cathode:
C Fe2
1 .M
Equlibrium Quotient (Q)
Nernst Equation:
The Cell potential:
Zn_Fe_cell.mcd
Q
C Zn2
C Fe2
E cell
E std_cell
R .T .
ln( Q )
n .F
E cell = 0.354 volt
S.E. Van Bramer
8/7/01
Now Repeat for other concentrations
C Zn2
1 .M
E cell
E std_cell
C Fe2
2 .M
Q
C Zn2
C Fe2
R .T .
ln( Q )
n .F
E cell = 0.363 volt
Now Repeat for other concentrations
C Zn2
2 .M
E cell
E std_cell
C Fe2
1 .M
Q
C Zn2
C Fe2
R .T .
ln( Q )
n .F
E cell = 0.345 volt
Now Repeat for other concentrations
C Zn2
0.1 .M
E cell
E std_cell
C Fe2
0.1 . M
Q
C Zn2
C Fe2
R .T .
ln( Q )
n .F
E cell = 0.354 volt
Now Repeat for other concentrations
C Zn2
1.0 .M
E cell
E std_cell
C Fe2
0.01 . M
Q
C Zn2
C Fe2
R .T .
ln( Q )
n .F
E cell = 0.296 volt
Now Repeat for other concentrations
Zn_Fe_cell.mcd
C Zn2
0.01 . M
E cell
E std_cell
C Fe2
1 .M
Q
C Zn2
C Fe2
R .T .
ln( Q )
n .F
S.E. Van Bramer
E cell = 0.412 volt
8/7/01
Calculating the equlibrium concentrations for a set potential
For the electrochemical system:
Zn | Zn2+ (1 M) || Fe2+ (1 M) | Fe
Standard Potentials:
0.7628 .volt
E anode
E cathode
n
0.409 . volt
2
Standard Cell Potential:
E std_cell
E cathode
E anode
E std_cell = 0.354 volt
Applied Potential
E applied
0.3538 .volt
Since the applied potential controls the cell equlibrium:
E cell
E applied
Rearanges the Nernst Equation to find Q:
NOTE:
Q
exp
Q
reactants
E cell E std_cell
Q
2
products
Zn
2
Fe
R .T .
ln( Q )
n.F
E cell .n . F
E std_cell .n .F
( R .T )
Q=1
Zn_Fe_cell.mcd
S.E. Van Bramer
8/7/01
Repeat calculation for other applied potentials
E cell
0.3538 .volt
Q
exp
E cell
0.3530 .volt
Q
exp
E cell
0.3400 .volt
Q
exp
E cell
0.3550 .volt
Q
exp
E cell
0.3600 .volt
Q
exp
Zn_Fe_cell.mcd
E cell .n . F
E std_cell .n .F
( R .T )
E cell .n . F
E std_cell .n .F
( R .T )
E cell .n . F
E std_cell .n .F
( R .T )
E cell .n . F
E std_cell .n .F
( R .T )
E cell .n . F
E std_cell .n .F
( R .T )
S.E. Van Bramer
Q=1
Q = 1.065
Q = 2.982
Q = 0.909
Q = 0.612
8/7/01