Electrochemical Cell – Part I: Voltaic Cell
Introduction
An electrochemical cell – voltaic cell produces a voltage, i.e. electromotive force (EMF),
that is generated by an oxidation-reduction (or redox) reaction and a transfer of electrons.
A electrochemical cell functions only when a complete electric circuit occurs. An
external wire provides a path for electrons to travel from the anode to the cathode. The
circuit is completed when the solutions surrounding these electrodes are in contact. Ions
can then pass from one solution to the other. Because several different ways of achieving
this contact have been devised, there are several designs for voltaic cells. A salt bridge, a
porous pot. a porous plug, or membrane can be used to separate the solutions while
maintaining electrical contact. The size of the voltage produced depends on the nature of
oxidizing and reducing agents and their relative concentrations.
It is conceptually possible to divide any redox reaction into two half-reactions. The
oxidation half-reaction involves giving up electrons and occurs at the anode. The
reduction half-reaction involves absorption of electrons and occurs at the cathode. For
example, the redox reaction:
Pb(s) + Cu2+ → Cu(s) + Pb2+
shows lead being oxidized while the copper ion is reduced. Hence, a cell maybe thought
of as two half-cells, an anode and a cathode each comprised of a metal electrode inserted
in container filled with a solution known as the electrolyte. Each half-cell has a
standardized voltage (Eº) determined by measuring it with respect to a hydrogen half-cell
at standard temperature and pressure (STP) using 1.0 M solutions.
Cu2+ + 2e- → Cu(s)
Pb(s) → Pb2+ + 2e-
Eº = +0.345 v
Eº = -0.126 v
Eºcell = 0.471 v
The two half-cells corresponding to the above redox reaction are described by the
following notation:
Pb(s)/ Pb(NO3)2 (aq) // Cu(NO3)2 (aq) /Cu(s)
These electrode/electrolyte pairs must be chemically separated but at the same time
electrically connected. In other words, a barrier must prevent mixing while permit
conduction of ions. In this experiment, a test tube with a small hole in the bottom has
been dipped into a hot agar and kNO3 solution that forms a conductive plug when cooled.
This test tube becomes one of the half-cells by filling it with an electrolyte and inserting
an electrode. The other half-cell is composed of a beaker also filled with an electrolyte.
By placing another electrode along side the test tube and inserting them into the beaker, a
complete cell is formed. When a piece of wire connects the two electrodes, the redox
reaction occurs with metal forming on the anode and ions being released from the
cathode.
In this experiment, a voltage is applied that is equal but opposite to the voltage generated
by the cell. This reverse potential holds the redox reactions in equilibrium and enables us
to measure the cell’s output voltage.
Table of Standard Electrode (Reduction) Potentials in Aqueous Solution at 25ºC.
(for chemicals used in this experiment)
Cathode (Reduction) Half-Reaction
Zn2+(aq) + 2e- ↔ Zn(s)
Fe2+(aq) +2e- ↔ Fe(s)
Fe3+(aq) + e- ↔ Fe2+(aq)
Pb2+(aq) + 2e- ↔ Pb(s)
Cu2+(aq) + 2e- ↔ Cu(s)
Ag+(aq) + e- ↔ Ag(s)
Standard Potential, Eº (Volts)
-0.76
-0.41
0.77
-0.13
0.34
0.80
Electrochemical Cell – Part I: Voltaic Cell
Procedure
**NOTE: Use instructions provide by the instructor – DO NOT USE INSTRUCTIONS
IN CHEMLAB! To remove the instructions on the screen, and free-up more working
area, perform the following operation: click on the OPTIONS tab; then click on LAB
ONLY. The instructions “disappear” and all of the area is now lab space.**
1. Select an Electrochemical Cell from the Equipment menu.
2. Choose Copper for the Left Electrode and choose Lead for the Right Electrode
from the drop-down lists in the dialog box. (You can verify or change these by
selecting the Half-Cell Assembly and re-selecting Electrochemical Cell from the
Equipment menu.)
3. Select the Half-Cell Assembly and add 4 ml of 1.0 M Lead nitrate to the test tube
from the Chemical menu.
4. Obtain a 100 ml beaker and add 40 ml of 1.0 M Copper nitrate from the Chemical
menu.
5. Place the Half-Cell Assembly within the beaker by selecting both and combining
them.
6. Read the voltage from the meter. Record the value and note the positive electrode
(i.e. which metal) in Observations.
7. Repeat steps (1.) to (6.) for each of the six half-cell combinations found in the
table below:
Pb/ 1.0 M Pb(NO3)2 // 1.0 M Cu(NO3)2 /Cu
Pb/ 1.0 M Pb(NO3)2 // 1.0 M Ag(NO3) /Ag
Pb/ 1.0 M Pb(NO3)2 // 1.0 M Fe2(SO4) 3/ 2.0 M Fe(SO4),Pt*
Cu/ 0.1 M Cu(NO3)2 // 1.0 M Ag(NO3) /Ag
Cu/ 1.0 M Cu(NO3)2 // 1.0 M Fe2(SO4) 3/ 2.0 M Fe(SO4),Pt*
Ag/ 1.0 M Ag(NO3) // 1.0 M Fe2(SO4) 3/ 2.0 M Fe(SO4),Pt*
* Mix equal amounts, i.e. 2ml or 20ml each, of 1.0 M Fe2(SO4) 3 and 2.0 M Fe(SO4) in
the electrolyte solution to maintain equal ion concentrations. Since Iron corrodes in
water, it cannot be used as an electrode. Hence, Platinum is used.
Electrochemical Cell
Part I: Voltaic Cell
Name:__________
Sect:____________
Observations
Data
Note: All observations taken at STP (25oC and 100 kPa)
Reaction Equation
+ve Electrode
1.0 M Voltages
Pb(s) + Cu2+ → Cu(s) + Pb2+
V
Pb(s) + 2Ag2+ → 2Ag(s) + Pb2+
V
Pb(s) + 2Fe3+ → 2Fe2+ + Pb2+
V
Cu(s) + 2Ag+ → 2Ag(s) + Cu2+
V
Cu(s) + 2Fe3+ → 2Fe2+ + Cu2+
V
Fe3+ + Ag(s) → Ag+ + Fe2+
V
Questions
1. According to the pattern of half-cell potentials, what seems to determine the polarity
of a cell?
2. Provide a definition for Anode:
3. Proved a definition for Cathode
4. Proved a definition for Voltaic Cell
5. Proved a definition for emf
6. Identify the half-cell reactions for the electrodes that you studied in this experiment.
Identify the anode and cathode for each half reaction.
Half-cell reactions ANODE
Half-cell reactions CATHODE
7. Calculate Eºcell use the Table of Standard Electrode (Reduction) Potentials in
Aqueous Solution at 25ºC. This is given in the introduction. Fill in the table below
using your results from every electrochemical cell evaluated. Show calculations on
separate piece of paper:
Over All Reaction in Voltaic Cell
Calculated
Eºcell (V)
Pb(s) + Cu2+ → Cu(s) + Pb2+
V
Pb(s) + 2Ag+ → 2Ag(s) + Pb2+
V
Pb(s) + 2Fe3+ → 2Fe2+ + Pb2+
V
Cu(s) + 2Ag+ → 2Ag(s) + Cu2+
V
Cu(s) + 2Fe3+ → 2Fe2+ + Cu2+
V
Fe3+ + Ag(s) → Ag+ + Fe2+
V
8. Compare calculated Eºcell* with measured Ecell. Why are some of the values
different?
9. Show that your experimental results are in accord with the statement that Pb will
reduce Cu2+ ions, but Pb2+ ions are not reduced by Cu.
Discussion