Chapter 9 - Electrochemistry

Example I1 - Pg 200

In [1]:
#Calculate the potential of the cell
#Initialization of variables
import math
Gr=-math.pow(10,5) #kJ/mol
v=1
F=9.6485*10000. #C/mol
#calculations
E=-Gr/(v*F)
#results
print '%s %d %s' %("potential of the cell =",E,"V")
potential of the cell = 1 V

Example I2 - Pg 202

In [2]:
#calculate the equilibrium constant of the reaction
#Initialization of variables
import math
V=1.1 #V
F=9.6485*10000. #C/mol
R=8.314 #J/K mol
T=298.15 #K
#calculations
lnK=2*F*V/(R*T)
k=math.pow(math.e,(lnK))
#results
print '%s %.1e' %("Equilibrium constant =",k)
Equilibrium constant = 1.5e+37

Example E1 - Pg 189

In [3]:
#Calculate the acidity constant of the acid
#Initialization of variables
import math
lw=34.96 #mS m^2 /mol
la=4.09 #mS m^2 /mol
C=0.010 #M
K=1.65 #mS m^2 /mol
#calculations
lmd=lw+la
alpha=K/lmd
Ka=C*alpha*alpha
pKa=-math.log10(Ka)
#results
print '%s %.2f' %("Acidity constant of the acid = ",pKa)
Acidity constant of the acid =  4.75

Example E6 - Pg 203

In [4]:
#Calculate if the reaction is favoring the products or not. Also calculate the E value
#Initialization of variables
ER=1.23 #V
EL=-0.44 #V
#calculations
E=ER-EL
#results
if(E>0):
    print '%s %.2f %s' %("The reaction is favouring products and E is",E,"V")
else:
    print '%s %.2f %s' %("The reaction is not favouring products and E is",E," V")
    
The reaction is favouring products and E is 1.67 V

Example E7 - Pg 203

In [5]:
#calculate the equilibrium constant 
#Initialization of variables
import math
ER=0.52 #V
EL=0.15 #V
#calculations
E=ER-EL
lnK=E*1000./(25.69)
K=math.exp(lnK)
#results
print '%s %.1e' %("Equilbrum constant K= ",K)
Equilbrum constant K=  1.8e+06

Example E8 - Pg 205

In [6]:
#calculate the biological standard potential
#Initialization of variables
import math
E0=-0.11 #V
H=math.pow(10,-7)
#calculations
pH=-math.log10(H)
E=E0-29.59*pH*math.pow(10,-3)
#results
print '%s %.2f %s' %("Biological standard potential =",E,"V")
Biological standard potential = -0.32 V

Example E9 - Pg 206

In [7]:
#calculate the equilibrium constant for the reaction
#Initialization of variables
import math
ER=-0.21 #V
EL=-0.6 #V
#calculations
E=ER-EL
lnK=2*E*1000./(25.69)
K=math.exp(lnK)
#results
print '%s %.1e' %("Equilibrium constant for the reaction = ",K)
Equilibrium constant for the reaction =  1.5e+13

Example E10 - Pg 209

In [8]:
#calculate the electric potential
#Initialization of variables
E1=2*(-0.340)
E2=-0.522 
#calculations
FE=-E1+E2
#results
print '%s %.3f %s' %("Electric potential =",FE,"V")
Electric potential = 0.158 V

Example E11 - Pg 210

In [9]:
#calculate the Gibbs enthalpy, Standard entropy and enthalpy of the process
#Initialization of variables
import math
v=2
F=9.6485*10000. #C/mol
E=0.2684 #V
V1=0.2699 #V
V2=0.2669 #V
T1=293. #K
T=298. #K
T2=303. #K
#calculations
Gr= -v*F*E/1000.
Sr=v*F*(V2-V1)/(T2-T1)
Hr=Gr+T*Sr/1000.
#results
print '%s %.2f %s' %("Gibbs enthalpy =",Gr,"kJ/mol")
print '%s %.1f %s' %("\n Standard Entropy =",Sr,"J /K mol")
print '%s %.1f %s' %("\n Enthalpy =",Hr,"kJ/mol")
Gibbs enthalpy = -51.79 kJ/mol

 Standard Entropy = -57.9 J /K mol

 Enthalpy = -69.0 kJ/mol