# Chapter 15 - Electrochemistry¶

## Example 1 - pg 384¶

In [1]:
#calculate the Weight of copper leaving
#Initialization of variables
I=0.5 #amp
t=55  #min
we=31.77
#calculations
Q=I*t*60
n=Q/96496.
w=n*we
#results
print '%s %.3f %s' %("Weight of copper leaving =",w," g")

Weight of copper leaving = 0.543  g


## Example 2 - pg 386¶

In [2]:
#calculate the values of tplus and tminus
#Initialization of variables
w1=0.7532 #g
w2=0.9972 #g
wdep=0.4 #g
we=31.77 #g
#calculations
dn=w2/we - w1/we
t=dn/(wdep/we)
dne=wdep/we
dnmig=dn-dne
tplus=-dnmig/dne
tminus=1-tplus
#results
print '%s %.3f' %("tplus =",tplus)
print '%s %.3f' %("\n tminus=",tminus)

tplus = 0.390

tminus= 0.610


## Example 3 - pg 393¶

In [3]:
#calculate the Equivalent conductance
#Initialization of variables
R1=312 #ohms
R2=1043 #ohms
c=0.01 #N
kdash=0.002768 #ohm^-1cm^-1
#calculations
k=kdash*R1
kdash2=k/R2
ambda=kdash2/(c/1000.)
#results
print '%s %.1f %s' %("Equivalent conductance =",ambda,"ohm^-1 cm^2 equiv^-1")

Equivalent conductance = 82.8 ohm^-1 cm^2 equiv^-1


## Example 4 - pg 393¶

In [4]:
#calculate the Conductance for acetic acid
#Initialization of variables
l1=349.8
l2=40.9
#calculations
l=l1+l2
#results
print '%s %.1f %s' %("Conductance for acetic acid =",l," ohm^-1 cm^2")

Conductance for acetic acid = 390.7  ohm^-1 cm^2


## Example 5 - pg 395¶

In [5]:
#calculate the Specific conductanc
#Initialization of variables
l1=63.6
l2=79.8
n=1 #mg/lt
we=116.7 #g/equiv
#calculations
l=l1+l2
c=n*10**-3 /we
k=c*l/1000.
#results
print '%s %.2e %s' %("Specific conductance =",k," ohm^-1 cm^-1")

Specific conductance = 1.23e-06  ohm^-1 cm^-1


## Example 6 - pg 402¶

In [6]:
#calculate the SEP of the cell
#Initialization of variables
e1=0.763 #volt
e2=0.337 #volt
#calculations
e0=e1+e2
#results
print '%s %.3f %s' %("Standard electrode potential of the cell =",e0,"volts")

Standard electrode potential of the cell = 1.100 volts


## Example 7 - pg 403¶

In [7]:
#calculate the Emf of the cell
#Initialization of variables
import math
aZn=0.1
aCu=0.01
e1=0.763 #volt
e2=0.337 #volt
#calculations
e0=e1+e2
Q=aZn/aCu
E=e0- 0.05915*math.log10(Q) /2
#results
print '%s %.3f %s' %("Emf of the cell =",E," volts")

Emf of the cell = 1.070  volts


## Example 8 - pg 410¶

In [8]:
#calculate the Decomposition potential
#Initialization of variables
e1=1.2 #volts
e2=0.15 #volts
e3=0.45 #volts
#calculations
E=e1+e2+e3
#results
print '%s %.1f %s' %("Decomposition potential =",E," volt")

Decomposition potential = 1.8  volt