Electrochemical cells

Electrochemical cells
Sudha Madhugiri D.Chem.
Collin College
Department of Chemistry
Have you used a battery before for some purpose? I bet you have. The type of chemistry that is
used in batteries is always happening around us. Tarnishing of silver and restoring the shine is another
example of the same phenomenon. The chemistry we are talking about here is the process of oxidationreduction (also known as redox). Loss of electrons is oxidation, and gain of electrons is reduction.
Though these reactions always occur together, they can be physically separated and studied. The
substance that undergoes oxidation is called a reducing agent and the substance that undergoes reduction
is called an oxidizing agent.
Redox Reactions and Voltaic Cells:
In a redox reaction, the electrons lost by the reducing agent are gained by the oxidizing agent. If
these two reactions are made to occur in separate regions connected by a metal wire that allows the flow
of electrons between the two, it forces the electrons to travel through the external wire from the oxidizing
agent to the reducing agent compartments, or half-cells. The natural tendency for the electrons to flow
through the external wire is called the electromotive force (EMF), or potential. This flow creates
electricity. Electricity is typically viewed as the flow of electrons or charge through a medium. This
principle is used in building an apparatus called a voltaic cell. In a voltaic cell, a spontaneous reaction
(negative Go) is used to create electricity. The magnitude of the potential is a direct measure of the
thermodynamic driving force of the reaction. If a voltmeter is connected between the two electrodes (the
conducting metal or substance) immersed in each of the oxidation and reduction reaction chambers, the
voltage registered is the sum of oxidation and reduction potentials.
Eocell = Eooxidation + Eoreduction
Standard conditions for redox reactions are 1 M concentration for solutions and 1 atm pressure for
gases and a temperature of 298 K. Although not always the case, redox reactions are typically carried out
at room temperature, 298K. By convention, all the half cell potentials are tabulated as standard reduction
potentials. It is for you to determine which electrode undergoes oxidation and which undergoes
reduction. The electrode with more positive reduction potential undergoes reduction and the one with less
positive reduction potential undergoes oxidation. That means for oxidation, the half reaction is flipped,
which also means the standard reduction potential carries an opposite sign for oxidation potential. As a
voltaic cell is spontaneous, it always has a positive cell potential (Eocell).
In general, in a voltaic cell there are two half cell compartments, an anode (assigned the - sign)
and a cathode (assigned the + sign) compartment. In a half cell, an electrode is dipped in an electrolytic
solution, which is typically a salt solution of the metal electrode. If the electrode is used to make the
connection and if it is also part of the reaction, it is known as active electrode. However, if the reactant is
an aqueous solution, or a liquid or a gas, an electrode made of either a graphite rod or a platinum wire is
used to make the connection. These are called inactive electrodes. Oxidation (loss of electrons) always
occurs at the anode and reduction (gain of electrons) always occurs at the cathode. A shorthand form of
representing a cell, called line notation, is shown below:
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The electron flow which occurs when the anode and cathode are connected produces electricity in a
spontaneous cell.
As the cell is discharging (the redox reaction is producing electron flow), the half cells end up
electrically charged as the anode compartment loses electrons and becomes more positive while the
cathode compartment gains electrons to become more negatively charged. This electrical charge hinders
the flow of electrons in the redox reaction. To overcome this obstacle, one can use either a porous cup for
one of the compartments or a salt bridge. Both work on the same principle. The salt bridge contains a
saturated solution of an inert salt. As the charge begins to build up in the compartments, the ions present
in the salt bridge will naturally flow from the salt bridge into each compartment to make it electrically
neutral. The porous cup allows ions to flow through the pores to accomplish electrical neutrality in both
Electrolytic Cell:
Contrary to a voltaic cell that produces electricity from a spontaneous chemical reaction, an
electrolytic cell uses electricity to drive a non-spontaneous chemical reaction (positive Go). A power
source, usually a battery, causes oxidation, still at the anode (now assigned a + sign), and reduction, still
at the cathode (now assigned a - sign), where as the overall reaction is non-spontaneous. Note the reversal
in the signs for the electrodes as compared to voltaic cells.
Electrolysis (electro means electricity, lysis means breaking up) has great industrial importance as
this is the main process used in recovering metals from their ores. Electrolysis of brine (NaCl solution) is
extensively used to produce chlorine gas, sodium hydroxide and hydrogen gas. In electrolysis of aqueous
salt solutions, two oxidation reactions and two reduction reactions are possible.
At the cathode (-): The metal cation itself can be reduced or water can be reduced based on the following
reduction half reactions.
M+ (aq) + e-  M (s)
2H2O (l) + 2e-  H2 (g) + 2 OH- (aq) E = -0.83 V
In general, water gets reduced at the cathode.
At the anode (+): The non-metal anion can be oxidized or water can be oxidized based on the following
two half reactions:
X- (aq)  X (s) + e2H2O (l)  O2 (g) + 4H+ (aq) + 4 e- E = -1.23 V
The non-metal normally gets oxidized at the anode.
However, these results depend on the individual half cell potentials. The reaction with less negative
reduction potential will be reduced (for the most part it is water) at the cathode and the reaction with less
positive reduction potential will be oxidized (which is normally the metal) at the anode (M. Silberberg
Copyright© 2011 Sudha Madhugiri
From the reactions at the anode and cathode, it can be deduced that more the # of moles of electrons
transferred, greater will be the product formed at a given electrode. This relationship was first put forth
by Michael Faraday and is known as Faraday’s law of electrolysis. This law states that "the amount of
substance produced at each electrode is directly proportional to the quantity of charge flowing through the
cell.” 1
The charge in coulombs (C) can be calculated by:
Charge (C) = Current (amps) * time (seconds)
Using Faraday’s constant, it takes 96,500 C of charge to transport 1 mole of electrons.
Using Faraday’s law, the current transferred in amperes (amps) can also be calculated:
1 amp = 1 coulomb/second
This simple relationship combined with a balanced half reaction can be used to calculate the required
parameters involving stoichiometry of electrolysis.
Electrolysis has a variety of applications as mentioned earlier and one is electroplating. Electroplating is
a technique that uses electric current to coat a layer of a desired metal by reducing cations of that metal
from a solution. This process is performed to bestow properties that are not intrinsic to the material being
coated. For example, chrome plating is done on a variety of materials including parts of automobiles to
provide corrosion resistance, ease of cleaning, and aesthetic appeal. Gold or silver plating can be done on
pieces of jewelry and to keep expensive electronic material from tarnishing. The material that needs to be
coated can be made a cathode and using an electroplating solution, the cathode can be coated with the
desired metal using electrochemical reduction.
PART I: Voltaic Cell
You are provided with the following materials:
Strips of zinc , copper , iron and 1M aqueous solutions of ZnSO4 , CuSO4, FeSO4, and KBr
A porous container, graphite rods, a voltmeter, and a few wires with alligator clips.
Copyright© 2011 Sudha Madhugiri
You are trying to design a voltaic cell with the maximum voltage to be used as a power source for your
experiment. Use Figure 1 as a guide to construct 4 different voltaic cells using the above materials.
Cathode (+)
Anode (-)
Porous cup acting as a
cathode compartment
Figure 1. Schematic of a voltaic cell
NOTE: If you are using a salt solution for more than one cell, you can reuse it before carefully disposing
of it in the designated waste container. Make sure you don’t contaminate your solutions through the trials
as this will affect your results.
If your half cell uses an active electrode, be sure to place the metal strip (which functions as an active
electrode) in its respective salt solution (i.e., copper rod needs to be placed in CuSO 4 solution).
If your half cell uses an inactive electrode, use a graphite rod as your inactive electrode to make the
Use a 100 mL beaker for one half cell and the porous cup for the other half cell.
Connect the electrode you decide to be the anode to the negative voltmeter terminal and the cathode to the
positive voltmeter terminal and record the voltage. While using KBr solution, use a graphite rod as an
inactive electrode to make the contact.
Be sure to discuss and decide the possible anode and cathode for each cell among your group members
before you put the voltaic cells together.
Copyright© 2011 Sudha Madhugiri
Prelab Questions for Part I:
1. A voltaic cell made of a Cr electrode in a solution of 1.0 M in Cr 3+ and a gold electrode in a
solution that is 1.0 M in Au3+.
a) Provide the half cell reactions for each of the electrodes.
b) Which is the anode and which is the cathode? How did you decide?
c) Give an overall balanced equation for the cell.
d) Calculate the Eocell at 25oC.
2. Is it possible to create a cell that would follow the following cell reaction?
Fe2+ (aq) + Ni(s)  Fe(s) + Ni2+ (aq)
If yes, calculate Eocell at 25oC.
If it is not possible, explain why.
Critical Data/Discussion to Include in Your Lab Report for Part I:
For each cell, which electrode was the anode? Cathode?
The anode and cathode half reactions
Balanced equation for each of the four cells you successfully constructed
Both calculated and measured Eocell
Line notation for the four cells
Discussion of the four cells; which gave the most voltage? The least? Do your observations agree with
the calculated values? If not, discuss why not?
What is the use of porous cup in the set up?
References cited in the format specified by your instructor
Part II: Electrolytic Cell
Construct an electrolytic cell based on the schematic provided in Figure 2 and coat a piece of copper strip
with nickel metal to understand the process of electroplating. Using the data collected by electroplating
nickel, calculate the current supplied in amperes (amps) in this process.
Current (amps)
and time (s)
Calculate charge
(coulombs) 
 of
moles of
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Mass of
product (g)
Anode (+)
Cathode (-)
Figure 2. Schematic of the cell for electroplating
Set up of the electrolytic cell: Attach a brightly sanded, weighed, copper strip to the negative pole of a 6
V battery source. The positive pole of the battery source should be connected to a graphite rod using a
lead wire. Place the electrodes in a 50 mL beaker containing 40 mL of “nickel plating” solution. You
should be noticing a layer of shiny metal getting deposited on the copper strip. Allow this process to
continue for 10 minutes. Disassemble the electrolytic cell, pat dry and weigh the copper strip and dispose
the chemicals in appropriate waste containers. Remove the nickel coating on the copper strip using sand
paper before returning and perform the required calculations.
Prelab Questions for Part II:
1. Calculate the amount of copper produced in 1.0 h when aqueous CuBr 2 solution was electrolyzed
by using a current of 4.50 A.
2. In another electroplating experiment, if electric current was passed for 3 hours and 2.00 g of silver
was deposited from a AgNO3 solution, what was the current used in amperes?
Critical Data/Discussion to Include in Your Lab Report for Part II:
1. Construct a table using Microsoft Word containing the following data/information
Process (oxidation or reduction) occurring at the cathode
The half reaction for the reaction at cathode
Observations at the graphite electrode
The exact time for which the current was passed through solution
Mass of the copper strip before electroplating
Mass of the copper strip after electroplating
Mass of the metal deposited
2. Calculations for the charge (in coulombs) passed through the solution in this process
3. Calculations for the current supplied (in amps)
4. Discussion of results, things learned and ways to improve the experiment
Copyright© 2011 Sudha Madhugiri
5. References cited in the format specified by your instructor
6. If you were to plate silver onto an object you would need to use either AgNO 3 in a basic medium
along with ammonium hydroxide or silver cyanide (AgCN). This is matter of concern with
respect to safety and the cost involved in electroplating silver onto an object.
Perform a cost analysis on using AgNO3 if you were to use 1M AgNO3 solution in one
electroplating experiment and 1M CuSO4•5H2O solution for copper plating in an another
experiment. (500 g of CuSO4•5H2O costs $19 and 500 g of AgNO3 costs $540). Show your
Silberberg, Martin. (2007). Principles of general chemistry. 2007.
Copyright© 2011 Sudha Madhugiri