Chapter 9 Chemical effects of current

Example 9.1 Page no 285

In [5]:
#Given
m=1*10**-3                         #kg
I=2                                  #A
z=3.3*10**-7                          #kg/C

#Calculation
t=m/(z*I)

#Result
print"Time required is", round(t,1),"s"

Time required is 1515.2 s

Example 9.2 Page no 285

In [10]:
#Given
m=0.15*10**-3                      #Kg
z=3.3*10**-7                         #Kg/C
t=900                                  #S
I1=0.6                                #A

#Calculation
I=m/(z*t)
I2=I-I1

#Result
print"Correction required for the ammeter reading is", round(I2,1),"A"

Correction required for the ammeter reading is -0.1 A

Example 9.3 Page no 285

In [17]:
#Given
t=0.002                         #m
A=72                             #cm**2
d=8.9                             #g/cm**3
z=33*10**-5                        #g/C
I=5                                  #A

#Calculation
V=t*A
m=V*d
t1=m/(z*I)

#Result
print"Time required is", round(t1,0),"S"

Time required is 777.0 S

Example 9.4 Page no 286

In [25]:
#Given
m1=2                            #g
m2=1                            #g
t=1800                          #s
z1=1118*10**-6                  
z2=3294*10**-7 
a=6

#Calculation
l1=m1/(z1*t)
l2=m2/(z2*t)
l=l1+l2
p=l*a

#Result
print"Power of the current is",round(p,3),"W"

Power of the current is 16.082 W

Example 9.5 Page no 286

In [51]:
#Given
m1=0.403*10**-3                    #Kg
z1=1.12*10**-6
z2=3.3*10**-7
t=900                               #s
e=12                                #V

#Calculation
m2=(m1*z2)/z1
d=(e*m1)/z1

#Result
print"(a) Mass of the copper deposited is",round(m2*10**3,3),"10**-3"
print"(b) The energy supplied by the battery is",round(d*10**-3,2),"10**3","J"

(a) Mass of the copper deposited is 0.119 10**-3
(b) The energy supplied by the battery is 4.32 10**3 J

Example 9.6 Page no 286

In [64]:
#Given
r=100                                #ohm
t=600                                #s
z=3.3*10**-7
m=10**-4

#Calculation
I=m/(z*t)
Q=I**2*r*t

#Result
print"Heat produced in the resistance coil is",round(Q,1),"J"

Heat produced in the resistance coil is 15304.6 J

Example 9.7 Page no 286

In [104]:
#Given
a=1.008
v=1
m1=1.05*10**-8
a1=63.54
v1=2
m2=3.29*10**-7
a2=107.9
v2=1
m3=1.12*10**-6
a3=55.85
v3=3

#Calculation
#For water voltameter
E=a/v
s=m1/E

#For copper voltameter
E2=a1/v1
s1=m2/E2

#For silver voltameter
E3=a2/v2
s2=m3/E3

#For iron voltameter
E4=a3/v3
m4=(s/a)*E4

#Result
print"Mass of hydrogen liberated is",round(s*10**8,4)*10**-8
print"Mass of copper deposited is",round(s1*10**8,4)*10**-8
print"Mass of silver deposited is",round(s2*10**8,4)*10**-8
print"Mass of iron deposited is", round(m4*10**7,2)*10**-7,"Kg/C"

Mass of hydrogen liberated is 1.0417e-08
Mass of copper deposited is 1.0356e-08
Mass of silver deposited is 1.038e-08
Mass of iron deposited is 1.92e-07 Kg/C

Example 9.8 Page no 287

In [110]:
#Given
m=2.5                       #g
I=10                        #A
F=96500                     #C/mol
m1=63.5
n=2.0

#Calculation
E=m1/n
t=m*F/(E*I)

#Result
print"Time required is",round(t,1),"S"

Time required is 759.8 S

Example 9.9 Page no 287

In [119]:
#Given
m=16.43                            #g
t=4000                             #s
F=96485                            #C/mol
m1=63.54
n=2.0
I1=12.6                            #A

#Calculation
E=m1/n
I=m*F/(E*t)
I2=I1-I

#Result
print"Error in the ammeter reading is", round(I2,3),"A"

Error in the ammeter reading is 0.126 A

Example 9.10 Page no 287

In [123]:
#Given
t=600                              #S
m=5.92                             #g
F=96500                             #C/mol
V1=1.62                             #V
V2=1.34
m1=63.5
n=2.0

#Calculation
V=V1-V2
E=m1/n
I=m*F/(E*t)
R=V/I

#Result
print"Resistance of the voltmeter is",round(R*10**3,2),"m ohm"

Resistance of the voltmeter is 9.34 m ohm

Example 9.11 Page no 287

In [133]:
#Given
E=8                  #V
r=1                     #ohm
R=15                    #ohm
E1=120
t=300                    #s

#Calculation
I=(E1-E)/(R+r)
V=E+(I*r)
E12=E*I*t

#Result
print"(a) Current in the circuit is", I,"A"
print"(b) Terminal voltage across the battery is",V,"V"
print"(c) Chemical energy stored in the battery is",E12,"J"

(a) Current in the circuit is 7 A
(b) Terminal voltage across the battery is 15 V
(c) Chemical energy stored in the battery is 16800 J

Example 9.12 Page no 288

In [136]:
#Given
I=10                          #A
t=300                         #S
m=2.016
n=2.0
n1=1.0
m1=1.008
F=96500
V=22.4

#Calculation
q=I*t
M=m/n
M2=m1/n1
q1=F*m/m1
V1=V*q/q1

#Result
print"Volume of hydrogen is",round(V1,4),"litre"

Volume of hydrogen is 0.3482 litre

Example 9.13 Page no 288

In [143]:
#Given
w=107.9                           #g/mol
r=2                               #ohm
E=12                              #V
V=10
F=96500
t=1800

#Calculation
R=r*(V/(E-V))
I=E/(R+r)
E=w/I
z=E/F
m=z*I*t

#Result
print"Silver deposited at the cathode is", round(m,2),"g"

Silver deposited at the cathode is 2.01 g

Example 9.14 Page no 288

In [153]:
#Given
I=5.0                      #A
a=0.5                      #mole
n=1
F=96500
I1=10                       #A
t1=9650*2
n1=2.0
m=63.54
m2=55.85
n2=3.0

#Calculation
q=F*a
t=q/I
E=m/n1
m1=(E*I1*t1)/F
E3=m2/n2
m3=(E3*I1*t1)/F

#Result
print"Molar mass of copper is", m1,"equal to its atomic mass i.e 1 mole of copper is liberated."
print"Molar mass of iron is",round(m3,3),"Hence 2/3 mole of iron will be deposited."

Molar mass of copper is 63.54 equal to its atomic mass i.e 1 mole of copper is liberated.
Molar mass of iron is 37.233 Hence 2/3 mole of iron will be deposited.