Chapter8, Electrolytic Processes

Ex1 : Page 414

In [1]:
from __future__ import division
#To determine the quantity of current

#Parameters of the Copper Plating
TSA=0.3716  #Total Surface Area 
t=0.0254*(10**-3)  #Thickness of the plating
density=8.883*(10**3)  #Density of Copper

W=TSA*t*density*1000  #Weight of copper deposited in gms
ECEcu=0.329  #Electrochemical Equivalent of copper in gm per coulomb
C=W*1000/ECEcu  #Quantity of Current required.

F=96500  #One Farad Charge

Q=C/F  #Quantity of Current in Faradays

print 'The Quantity of Current Required is %0.3f F'%Q

The Quantity of Current Required is 2.641 F

Ex2 : Page 420

In [2]:
#To detemine the Electrochemical Equivalent of Silver

I=20  #Current Passed
W=26.84  #amount of silver deposited in gms
t=20*60  #Time Period in seconds

TC=I*t  #Total Charge in the given time period

ECEAg=W*1000/TC  #Electrochemical Equivalent of Silver

print 'The Electrochemical Equivalent of Silver is %0.3f mg/C'%ECEAg 

The Electrochemical Equivalent of Silver is 1.118 mg/C

Ex3 : Page 420

In [3]:
#To Determine the Annual output of copper

I=2000  #Current Passed
NW=52  #Number of weeks in a Year
T=100*3600  #Number of seconds per week
TC=NW*T*I  #Total Charge supplied all over the year.
ECu=31.8  #Equivalent Weight of Copper in grams
F=96500  #One Farad of Charge

# 1 F of charge gives 31.8 gms of copper

W=(TC/F)*ECu/(1000*1000)  #Weight of copper in tonnes

print 'The Annual Output of Copper is %0.1f Tonnes'%W 

The Annual Output of Copper is 12.3 Tonnes

Ex4 : Page 420

In [4]:
#To Calculate the Energy usedd in producing chemical action developed in an electrolytic cell

T=15/60  #Time Period of Operation in Hours
I=100  #Current in Amperes
V=15  #Potential Difference
R=0.05  #Resistance of Solution

NetEP=V-(I*R)  #Net Electrode Potential

#Energy Equation, E = VIt.
E=NetEP*I*T/1000  # Energy in KiloWatts
print 'The Energy used in producing the chemical action developed in an electrolytic Cell for 15 mins is %g kWhr'%E

The Energy used in producing the chemical action developed in an electrolytic Cell for 15 mins is 0.25 kWhr

Ex5 : Page 420

In [5]:
#To determine the weights of Nickel and Silver Deposited

F=96500  #One Farad of Charge
#Equivalent Weights of the Following metals
EWAg= 108  #silver
EWNi= 58.6/2  #Nickel

I=20  #Current Passed
T=1*60*60  #Time for which the current is passed
TC=T*I  #Total Charge produced

#One Equivalent of Metal requires 1 F of charge

Q=TC/F  # Total Charge in Farad

#Amount of metal deposited
WAg=Q*EWAg  #Silver
WNi=Q*EWNi  #Nickel

print 'The Weight of Nickel And Silver deposited by 20A for an hour is %0.2f gm and %0.2f gm respectively.'%(WNi,WAg)

The Weight of Nickel And Silver deposited by 20A for an hour is 21.86 gm and 80.58 gm respectively.