Introduction to
Electrochemical Cells
Thanks to Eric Edens ‘07
Bollens, Rachel
A
Pitkin, Julia
D
Calhoun, Corinne
A
Liu, Melinda
D
Clute-Reinig, Nicholas
A
Takahashi, Alexis
D
Crawford, Catherine
A
Tarhan, Leyla
D
Cyr, Christina
B
Valdez, Stephanie
E
Evarkiou-Kaku, Anatolia
B
Vega, Alejandra
E
Ferguson, Luke
B
Vega, Marvin
E
Fernald, Samuel
B
Wang, Kevin
E
Folse, Katherine
C
Washington, Shannon
F
Forrester, William
C
Winninghoff, Hayley
F
Ho, Vanessa
C
Zahedi, Leila
F
Kuo, Linus
C
Experimental Goals
• Reinforce topics from last semester
– Provide experience working with electrochemical cells
– Investigate dependence of cell potential on
concentration
• Nernst Equation
– Using cell potential to determine equilibrium constants
• Introduce related material
– Concentration Cells
– Ionic Strength
Two Types of Electrochemical Cells
e-
2+
Zn(aq)
Zinc anode
Silver cathode
+
Ag(aq)
Information on Procedure
• Working with a partner
• Check out kit from stockroom
• Using small volumes for measurements
– Take only what you need, a few mLs
– Avoid making cells that form a precipitate,
such as silver with halides
• If precipitates form, combine half-cell of
interest with different half-cell.
– Why?
Part A: Potential Measurements
• Use filter crucible and beaker to construct
electrochemical cells
– Red lead is Cathode
– Black lead is Anode
– Make sure that your multimeters are set to
measure VDC
• All half-cells will be expressed vs Ag+|Ag
half-reaction in the end
• Need to make measurements that allow this
comparison
Example
Suppose you have data for the following two
reactions:
2Ag+ + Cu  2Ag + Cu2+ ΔE1 = 0.46V
Zn2+ + Cu  Zn + Cu2+
ΔE2 = -1.10V
You can then calculate ΔE for
2Ag+ + Zn  2Ag + Zn2+ ΔE3 = ?
Under standard conditions 1atm pressure, each metal 1.0M
Example
2Ag+ + Cu  2Ag + Cu2+ ΔE1 = 0.46V
Zn + Cu2+  Zn2+ + Cu - ΔE2 = 1.10V
2Ag+ + Zn  2Ag + Zn2+ ΔE3 = ΔE1+ (- ΔE2)
ΔE3 = 0.46 + 1.10 = 1.56V
Part A - Important Notes
• For Fe3+/Fe2+ half-reaction, you will need
to combine equal volumes of the two
solutions immediately prior to making your
measurements
• All measurements should be made three
times to ensure that you have obtained the
correct values
• I2/I- half-reaction
– What is really going on?
Remember the Nernst Eq?
Gr  Gr  RT ln Q

Gr
Gr   nFE
  nFE

 nFE   nFE  RT ln Q
 RT 
E  E  
 ln Q
 nF 

Demonstrates cell voltage (ΔE) depends on concentration (Q)

Part B: Concentration Dependence
• Exploring the Nernst Equation with real cells
• Since E°cell=E°cathode-E°anode=0.00V if both Cu+2
Experimentally, Cu/Cu concentration cell is hard to
measure, so we will vary the copper concentration
(crucible) and measure it against a zinc half-cell
(beaker).
Part B: Ionic Strength
• When salt concentrations are large
enough, we begin to see deviations from
Nernstian behavior
• Maintain the same ionic strength for all
measurements
• Make dilutions with stock, water, &
KNO3
Calculating Ionic Strength
1 2
I   zi ci
2
For the stock copper(II) nitrate solution
( NO3 )
( Cu 2 )


1
2
2
I  (2) 0.1M  (1) 0.2M 
2 

I  .3
For the 0.01M copper(II) nitrate solution
I  .03

The difference in ionic strengths is .3-.03 =.27
How much KNO3 needs to be
added to your dilutions?
All dilutions made in 100ml volumetric flasks (in kits).
How much 0.4M KNO3 is required to make up for the .27
difference?
M1V1=M2V2
0.4M*V1=0.27M*100ml
V1=67.5 ml
Each time a dilution is made, the ionic strength is maintained.
Parts C - Equilibrium Constant
• Using concentration cells, determine the
value of Q.
• Use Q to determine the unknown
concentration
Part C: Ksp for AgCl
• Starting with ~50 ml of 0.1M KCl in a beaker, add AgNO3
dropwise until a precipitate forms.
• Place inside this half-cell, a crucible containing 0.1M
AgNO3
• The resulting cell is:
Ag|Ag+(unknown concentration)|| Ag+(0.1 M)|Ag
Use E = E°- (RT/nF)ln([Ag+unknown]/[.1M Ag+])
then calculate Ksp = [Ag+unknown]x[Cl-]
Safety and Waste Disposal
• Two separate waste bottles in the lab
– One for silver nitrate
– One for all other solutions (halides)
• Silver nitrate stains hands, so be careful!
Before Starting
• Find a partner
• Check out kit from stockroom
• Have fun!
ALSO, make sure to read your experiment for
next week for the round robin…
Part D Formation Constant for
[Cu(NH3)4]2+
• The overall reaction is:
2
Cu(aq)
 4NH3(aq)  CuNH3 4 2
• Use the same type of arrangement as in
Part C, replacing Ag with Cu in the
cathode half-cell
Table of Changes Using Moles**
Cu2+
NH3
Cu(NH3)42+
Initial
0.03L*0.1M
0.003L*6M
0.000
Change
-X
-4X
+X
Equilibrium
moles
[Cu2+]unk*.033L (0.018-4X)
** Assumes 30ml Cu(NO3)2 solution and 3 ml NH3
For equilibrium concentrations divide by 0.033L
Qe

nF
Ecell
RT
[Cu 2 ]unk

0.1M
X
Helpful hints
Recall : Gr   nFE   RT ln K
so : RT ln K  nFE
RT
E 
ln K
nF
 RT 
E  E  
 ln Q
 nF 
RT
 RT 

ln K  
 ln Q
nF
 nF 
RT K

ln
nF Q

What happens when Q  1?