ELECTROCHEMISTRY
OBJECTIVE:
In this unit we shall define certain terms used in electrochemistry and then discuss the
voltaic cell, the electrolytic cell, electrolysis, and Faraday's law of electrolysis.
SECTION I
The students should be familiar with the following terms:
Coulomb: A coulomb is defined as the quantity of electricity required to deposit
.001118g of silver from a solution of silver ion.
Equation I. Coulombs = amperes x seconds
Ampere: Is a unit used to measure the rate of flow of electrons.
Ohm: Ohm is defined as the resistance offered by a uniform thread of mercury 106.3
cm long and containing 14.45g of Hg at O°C.
Volt: Volt is defined as the electrical potential required to cause the flow of coul/sec
through a uniform thread of mercury 106.3 cm long and containing 14.45g of mercury
at O°C.
Equation II: Volts = amperes x ohms
Equation II is known as Ohm's Law.
Faraday is equal to 96,500 coulombs and this quantity of electricity represents the
passage of Avogadro's number of electrons 1 F = 96,500 coul. = 6.02 x 10 23 electrons.
SECTION II
A voltaic cell is defined as a cell in which a chemical reaction produces an electrical current. By definition,
the electrode at which oxidation takes place is called the anode and the electrode at which reduction takes
place is called the anode and the electrode at which reduction takes place is called the cathode.
Reaction I
Zn
Reaction II
Zn2+ + 2e
Zn2+ + 2e
Zn
Note the following: In order for the voltaic cell to deliver a current, the voltage has to be positive. The
negative sign of any electrode indicated that the electron density of the electrode is greater than the electron
density of the hydrogen electrode. The positive sign of any electrode indicates that the electron density of
the electrode is less than the electron density of the hydrogen electrode.
By convention; to represent a voltaic cell the anode is written first, and the cathode last, the anode and
cathode are separated by two vertical lines indicating salt bridge. To express the anode, first write the sold
phase, followed by the electrolyte. The two are separated by a semicolon.
Zn ; Zn2+
Pt ; H2 ; H+
Please Study. The general form of the voltaic cell:
Anode (oxidation)
Cathode (reduction)
The cell reaction is the algebraic sum of the two half-cell reactions and the voltage of the cell is the
algebraic sum of the potential of the half-cell reactions.
Example I
A voltaic cell made up of Zn and Cu electrodes. The cell delivers an electric current spontaneously.
a.
Express half-cell reaction for each electrode.
b.
Calculate the voltage of the cell.
c.
Identify the anode and the cathode
d.
Represent the cell in the conventional fashion
1st half-cell reaction Zn2+ + 2e
Zn
-
0.76 v
2nd half-cell reaction Cu2+ + 2e
Cu
-
0.34 v
Zn2+ + 2e
Zn
Cu
Cu2+ + 2e
cell reaction Cu + Zn2+
Zn
Zn2+ + 2e
2+
Cu + 2e
Cu
Zn + Cu2+
Zn 2+ + Cu
Zn + Cu2+
0.76 v
0.34
-1.1. . . . . . . (non-spontaneous)
.76 v
.34 v
1.1 v. . . . . (spontaneous)
Oxidation occurred at the Zn electrode, there fore, the Zn electrode is the anode and reduction at the Cu
electrode, thus copper is the cathode.
The conventional method of expressing the cell is also shown.
Zn ; Zn2+ || Cu2+ ; Cu
It will be helpful to remember some of the important uses of the electromotive series. (Standard reduction
potential)
1 st
Metals with high negative potentials are good reducing agents in the free state.
2nd
Elements with positive potentials are good oxidizing agents when they are in oxidized state i.e.,
when they are in the form of ions.
3rd
The reduced form of any element will reduce the oxidized form of any element under it in the
series.
4th
There is an increasing potential to gain e’s from T to B in the chart.
SECTION III
1.
2.
Calculate the voltage of a voltaic cell composed of the following electrodes:
(a)
Co and Ni
(b)
H2 and Ag
(c)
Au and Pb
(d)
Zn and Ca
In each of the above cells indicate the anode and the cathode, the direction of current, the reaction
of the cell, and represent each cell in the conventional fashion.
Answers for Section III
1.
(a) .03 V
(b) .80 V
(c) 1.63 V
(d) 2.11 V
2.
ANODE
CATHODE
(a)
Co
Ni
Co ; Co2+ || Ni2+ ; Ni
(b)
H2
Ag
Pt; H2 ; H+ || Ag+ ; Ag
(c)
Pb
Au
Pb ; Pb2+ || Au3+ ; Au
(d)
Ca
An
Ca ; Ca2+ || Zn2+ ; Zn
In each case the current flows from anode to cathode.
CELL
SECTION IV
An electrolytic cell is defined as a cell in which an electric current produces a chemical change. A typical
electrolytic cell is shown in your text. Note that in the electrolytic cell the anode is the positive electrode
while the cathode is the negative electrode. Electroplating is one of the most important industrial uses of
the electrolytic cell.
Faraday’s Law of electrolysis states that 1 Faraday or 96,500 coul. of electricity (6.02  1023 electrons) are
required to deposit an equivalent weight of any element.
Remember that one mole of any element contains 6.02  1023 ions therefore 6.02  1023 electrons are
required to deposit 1 mole of silver which weighs 107.87.
Example 2
3 amp current is allowed to pass through a solution of Ag+ for 30 minutes. How many grams of Ag
are deposited?
Solution
3 amps for 30 minutes is equal to
3 30  60 = 5,400 coul. of electricity pass through the solution
We know that
96,500 coul ~~ 107.87 g of Ag
107.9  5,400
5,400 coul ~~
?
= 6.06 g of Ag
96,500
Example 3
3 g of Ca were deposited by an electric current in 1 hour. How many amps were required?
Solution
1 mole of Ca2+ (which weighs 40g) requires 2 x 6.02 x 10 23 electrons, i.e. requires 2F or 2 x 96,500
coul.
193,000 coul  40g of Ca
?
 3g of Ca
193,000  3
= 14,475 coul.
40
Remember that:
coul = amp x sec
14,475 coul = amp x 60 x 60
14 ,475
= 4.02 amp
3,600
SECTION V
1. Calculate the number of Faradays of electricity required to produce the following:
a. 40 g of Ca
b. 1 mole of Pb from Pb2+ solution
c. 1/3 mole of Al from Al3+ solution
d. 2 moles of Pb from Pb4+ solution
2. How long will it take a current of 6 amp. to produce 10 grams of each of the following:
a. H2 from H+
b. Pb from Pb2+
c. Ag from Ag+
3. A current of 3 amp was passed through 3 cell connected in series for 1 hour. How many grams of
Cu (Cu+ solution) Ag (Ag+ solution) and Au ( Au3+ solution) will be deposited in each cell.
Answers for Section V
1.
(a) 2F
(b) 2F
(c) 1F
(d) 8F
2.
(a) 44.67 hours
(b) 25.90 minutes
(c) 24.82 minutes
3.
(a)
7.11 g of Cu
(b) 12.09 g of Ag
(c)
7.35 g of Au