DEPARTMENT OF CHEMICAL ENGINEERING
COURSE: Electrochemical Engr.
COURSE NO.: CHEG 295-01; CHEG 320-02
INSTRUCTOR: Prof. J. M. Fenton
Faraday’s Law
The basis for electrochemical theory and technology are Faraday’s laws of
electrolysis which formed a crucial link between chemistry and electricity. These laws
are called in honor of Michael Faraday (1791-1867) who formulated them during 1833
and 1834:
1. The amount of chemical change produced by electrolysis is proportional to the
total amount of charge passed through the cell.
2. The amount of chemical change produced is proportional to the equivalent weight
of the substance undergoing chemical change.
W=
W
I
t
z
F
ItA
zF
= weight in grams of the compound undergoing electrolysis
= current in amperes; A = atomic weight
= elapsed time of electrolysis in seconds
= number of charges involved in the elementary reaction
= Faraday’s constant, 96,493.5 coulombs/equivalent
For engineering purposes, the value of the Faraday will be rounded off to 96,500
coulombs/equivalent.
Example Problem #1:
Calculate the daily aluminum production of a 150,000 ampere aluminum cell
which operates at a current efficiency of 89%. The cell reaction is
2Al 2 0 3 + 3 C = 4 Al + 3 CO 2
Answer:
 ItA 
W = εc

 zF 
(1.5 x 10 A )(24 x 60 x 60 sec)(26.985
= 0.89
(3 )(96,500 ) day
5
geq
gmol
A sec
geq
gm
gmole
)
= 1.08 x 10 6
gm
ton
or 1.18
day
day
Example Problem #2
The electrolysis of water at high pressure is used to produce hydrogen and oxygen
of very high purity. For a cell voltage of 1.78V, what is the power consumption in kwhrs per cubic meter of hydrogen gas at standard conditions of temperature and pressure?
Answer:
)(
(
lit
) / 22.414 gmol
1m 3 of H 2 @ STP = (1m 3 )(100 cm
10 3
m
3
Faradays Law →
(44.615 )(96,500 )(2 ) = 8.61 x 10
gmol
m3
Since Power Current X Voltage:
coul
geq
geq
gmol
(8.61 x 10
(3600
) = 44.615 gmol
6 coul
m3
)(1.78V )(1
)(10 )
6 coul
m3
sec
hr
cm 3
lit
3 watt
kW
waH⋅sec
coul⋅volt
)
= 4.26
kw − hr
m 3 @STP