Chapter 10 Electmotive Force

Example 10.1 , Page no:235

In [1]:
import math
from __future__ import division

#initialisation of variables
T= 25 #C
M= 0.08 #m
P= 1 #atm
F= 96500 #coloumbs
R= 8.31 #J/mol K

#CALCULATIONS
E= -R*(273+T)*2.3*math.log10(M)/F

#RESULTS
print"oxidation potential of hydrogen elctrode =",round(E,3),"v";

oxidation potential of hydrogen elctrode = 0.065 v

Example 10.2 , Page no:236

In [2]:
import math
from __future__ import division

#initialisation of variables
E= -0.337 #v
R= 8.31 #J/mol K
T= 25 #C
F= 96500 #coloums
M= 0.12 #m

#CALCULATIONS
E1= E-(R*(273+T)*2.3*math.log10(M)/(2*F))

#RESULTS
print"oxidation potential of copper elctrode =",round(E1,3),"v";

oxidation potential of copper elctrode = -0.31 v

Example 10.3 , Page no:236

In [3]:
import math
from __future__ import division

#initialisation of variables
E= -0.771 #v
R= 8.31 #J/mol K
T= 25 #C
F= 96500 #coloums
M= 0.02 #m
M1= 0.1 #m

#CALCULATIONS
E1= E-(R*(273+T)*2.3*math.log10(M/M1)/F)

#RESULTS
print"oxidation potential of copper elctrode =",round(E1,2),"v";

oxidation potential of copper elctrode = -0.73 v

Example 10.4 , Page no:238

In [4]:
import math
from __future__ import division

#initialisation of variables
E= 0.763 #v
R= 8.31 #J/mol K
T= 25 #C
F= 96500 #coloums
M= 0.1 #m
M1= 0.01 #m

#CALCULATIONS
E1= E-(R*(273+T)*2.3*math.log10(M)/(2*F))+R*(273+T)*2.3*math.log10(M1)/F

#RESULTS
print"oxidation potential of copper elctrode =",round(E1,2),"v";

oxidation potential of copper elctrode = 0.67 v

Example 10.5 , Page no:238

In [5]:
import math
from __future__ import division

#initialisation of variables
E1= 0.126 #v
E2= -1.360 #v
M= 0.02 #m
M1= 1/0.1 #m
R= 8.31 #J/mol K
T= 25 #C
F= 96500 #coloums

#CALCULATIONS
E= (E1-R*(273+T)*2.3*math.log10(M)/(2*F))-(E2-R*(273+T)*2.3*math.log10(M1)/(F))

#RESULTS
print"oxidation potential of copper elctrode =",round(E,3),"v";

oxidation potential of copper elctrode = 1.595 v

Example 10.6 , Page no:239

In [6]:
import math
from __future__ import division

#initialisation of variables
E1= 0.763 #v
c= 0.1 #mol/lit
c1= 0.01 #mol/lit
R= 8.31 #J/mol K
T= 25 #C
F= 96500 #coloums
c2= 1 #molar
c3= 1 #molar

#CALCULATIONS
E= E1-(math.log10(c*c2/(c1**2*c3))*R*(273+T)*2.3/(2*F))

#RESULTS
print"potential of the cell =",round(E,3),"v";

potential of the cell = 0.674 v

Example 10.7 , Page no:239

In [7]:
import math
from __future__ import division

#initialisation of variables
R= 8.31 #J/mol K
T= 25 #C
F= 96500 #coloums
c= 0.02 #molar
c1= 0.1 #molar
c2= 1 #molar
c3= 1 #molar
E1= 1.486 #v

#CALCULATIONS
E= E1-R*(273+T)*2.3*math.log10(c*c1**2/(c2*c3))/(2*F)

#RESULTS
print"potential of the cell =",round(E,3),"v";

potential of the cell = 1.595 v

Example 10.8 , Page no:241

In [8]:
import math
from __future__ import division

#initialisation of variables
R= 8.31 #J/mol K
T= 25 #C
F= 96500 #coloums
c= 0.08 #molar
c1= 0.04 #molar

#CALCULATIONS
E= R*(T+273)*math.log(c/c1)/(2*F)
E1= 2*E
#RESULTS
print"potential of the cell =",round(E,3),"v";
print"potential of the cell =",round(E1,3),"v";

potential of the cell = 0.009 v
potential of the cell = 0.018 v