Example 1.
Find the
charge in coulomb on 1 g-ion of N3-.
Solution:
Charge on
one ion of N3-
= 3 × 1.6
× 10-19 coulomb
Thus,
charge on one g-ion of N3-
= 3 × 1.6 10-19 × 6.02 × 1023
= 2.89 × 105 coulomb
= 3 × 1.6 10-19 × 6.02 × 1023
= 2.89 × 105 coulomb
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How much
charge is required to reduce (a) 1 mole of Al3+ to Al and (b)1 mole of to Mn2+ ?
Solution:
(a) The
reduction reaction is
Al3+ + 3e- → Al
Al3+ + 3e- → Al
Thus, 3
mole of electrons are needed to reduce 1 mole of Al3+
Q = 3 × F = 3 × 96500 = 289500 coulomb
Q = 3 × F = 3 × 96500 = 289500 coulomb
(b) The
reduction is
Mn4-+ 8H+ 5e- → Mn2+ + 4H2O
1 mole 5 mole
Q = 5 × F = 5 × 96500 = 48500 coulomb
1 mole 5 mole
Q = 5 × F = 5 × 96500 = 48500 coulomb
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Example 3.
How much
electric charge is required to oxidise (a) 1 mole of H2O to O2 and (b)1 mole of FeO to Fe2O3?
Solution:
(a) The
oxidation reaction is
H2O → 1/2 O2 + 2H+ + 2e-
Q = 2 × F = 2 × 96500 =193000 coulomb
Q = 2 × F = 2 × 96500 =193000 coulomb
(b) The
oxidation reaction is
FeO + 1/2
H2O → 1/2 Fe2O3 + H+ + e-
Q = F = 96500 coulomb
Q = F = 96500 coulomb
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Example 4.
Exactly
0.4 faraday electric charge is passed through three electrolytic cells
in series, first containing AgNO3, second CuSO4 and third FeCl3 solution.
How many gram of rach metal will be deposited assuming only cathodic
reaction in each cell?
Solution:
The
cathodic reactions in the cells are respectively.
Ag+ + e- → Ag
Cu2+ + 2e- → >Cu
and Fe3+ + 3e- → Fe
Ag+ + e- → Ag
Cu2+ + 2e- → >Cu
and Fe3+ + 3e- → Fe
Hence, Ag
deposited = 108 × 0.4 = 43.2 g
Cu
deposited = 63.5/2×0.4=12.7 g
and Fe
deposited = 56/3 ×0.4=7.47 g
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Example 5.
An
electric current of 100 ampere is passed through a molten liquid of
sodium chloride for 5 hours. Calculate the volume of chlorine gas
liberated at the electrode at NTP.
Solution:
The
reaction taking place at anode is
2Cl- → Cl2 + 2e-
Q = I × t = 100 × 5 × 600 coulomb
The amount of chlorine liberated by passing 100 × 5 × 60 × 60 coulomb of electric charge. =1/(2×96500)×100×5×60×60=9.3264 mole
Volume of Cl2 liberated at NTP = 9.3264 × 22.4 = 208.91 L
2Cl- → Cl2 + 2e-
Q = I × t = 100 × 5 × 600 coulomb
The amount of chlorine liberated by passing 100 × 5 × 60 × 60 coulomb of electric charge. =1/(2×96500)×100×5×60×60=9.3264 mole
Volume of Cl2 liberated at NTP = 9.3264 × 22.4 = 208.91 L
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Example 6.
A 100
watt, 100 volt incandescent lamp is connected in series with an
electrolytic cell containing cadmium sulphate solution. What mass of
cadmium will be deposited by the current flowing for 10 hours?
Solution:
We know
that
Watt =
ampere × volt
100 =
ampere × 110
Ampere =
100/110
Quantity of charge = ampere × second = 100/110×10×60×60 coulomb
Quantity of charge = ampere × second = 100/110×10×60×60 coulomb
The
cathodic reaction is
Cd2+ + 2e- → Cd
Mass of cadmium deposited by passing 100/110×10×60×60
Coulomb charge = 112.4/(2×96500)×100/110×10×60×60=19.0598 g
Cd2+ + 2e- → Cd
Mass of cadmium deposited by passing 100/110×10×60×60
Coulomb charge = 112.4/(2×96500)×100/110×10×60×60=19.0598 g
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Example 7.
In an
electrolysis experiment, a current was passed for 5 hours through two cells
connected in series. The first cell contains a solution gold salt and the
second cell contains copper sulphate solution. 9.85 g of gold was
deposited in the first cell. If the oxidation number of gold is +3, find
the amount of copper deposited on the cathode in the second cell. Also
calculate
the magnitude of the current in ampere.
the magnitude of the current in ampere.
We know
that
(Mass of Au deposited)/(Mass f Cu deposited)=(Eq.mass of Au)/(Eq.Mass of Cu)
Eq. mass of Au = 197/3;
(Mass of Au deposited)/(Mass f Cu deposited)=(Eq.mass of Au)/(Eq.Mass of Cu)
Eq. mass of Au = 197/3;
Eq. mass
of Cu 63.5/2
Mass of copper deposited = 9.85 × 63.5/2 x 3/197 g = 4.7625 g
Let Z be the electrochemical equivalent of Cu.
E = Z × 96500
or Z =E/96500=63.5/(2×96500)
Applying W = Z × I × t
T = 5 hour = 5 × 3600 second
4.7625 = 63.5/(2×96500) × I × 5 × 3600
or I = (4.7625 × 2 × 96500)/(63.5 × 5 × 3600)=0.0804 ampere
Mass of copper deposited = 9.85 × 63.5/2 x 3/197 g = 4.7625 g
Let Z be the electrochemical equivalent of Cu.
E = Z × 96500
or Z =E/96500=63.5/(2×96500)
Applying W = Z × I × t
T = 5 hour = 5 × 3600 second
4.7625 = 63.5/(2×96500) × I × 5 × 3600
or I = (4.7625 × 2 × 96500)/(63.5 × 5 × 3600)=0.0804 ampere
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Example 8.
How long
has a current of 3 ampere to be applied through a solution of silver nitrate
to coat a metal surface of 80 cm2 with 0.005 cm thick layer? Density of silver
is 10.5
g/cm3.
g/cm3.
Solution:
Mass of
silver to be deposited = Volume × density = Area ×thickness ×
density
Given: Area = 80 cm2
thickness = 0.0005 cm and density = 10.5 g/cm3
Mass of silver to be deposited = 80 × 0.0005 × 10.5 = 0.42 g
Applying to silver E = Z × 96500
Z = 108/96500 g
Let the current be passed for r seconds.
We know that
W = Z × I × t
So, 0.42 = 108/96500 x 3 x t
or t = (0.42 × 96500)/(108×3)=125.09 second
Given: Area = 80 cm2
thickness = 0.0005 cm and density = 10.5 g/cm3
Mass of silver to be deposited = 80 × 0.0005 × 10.5 = 0.42 g
Applying to silver E = Z × 96500
Z = 108/96500 g
Let the current be passed for r seconds.
We know that
W = Z × I × t
So, 0.42 = 108/96500 x 3 x t
or t = (0.42 × 96500)/(108×3)=125.09 second
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Example 9.
What
current strength in ampere will be required to liberate 10 g of chlorine
from sodium chloride solution in one hour?
Solution:
Applying E
= Z × 96500 (E for chlorine = 35.5)
35.5 = Z × 96500
or Z = 35.5/96500 g
Now, applying the formula
35.5 = Z × 96500
or Z = 35.5/96500 g
Now, applying the formula
W = Z × I
× t
Where W =
10 g, Z= 35.5/96500 t = 60×60 =3600 second
I =
10x96500/35.5x96500 = 7.55 ampere
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Example 10.
0.2964 g
of copper was deposited on passage of a current of 0.5 ampere for
30 minutes through a solution of copper sulphate. Calculate the atomic
mass of copper. (1 faraday = 96500 coulomb)
Quantity
of charge passed
0.5 × 30 ×
60 = 900 coulomb
900 coulomb deposit copper = 0.2964 g
96500 coulomb deposit copper = 0.2964/900×96500=31.78 g
Thus, 31.78 is the equivalent mass of copper.
At. mass = Eq. mass × Valency = 31.78 × 2 = 63.56
900 coulomb deposit copper = 0.2964 g
96500 coulomb deposit copper = 0.2964/900×96500=31.78 g
Thus, 31.78 is the equivalent mass of copper.
At. mass = Eq. mass × Valency = 31.78 × 2 = 63.56
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Example 11.
19 g of
molten SnCI2 is electrolysed for some time using inert electrodes until 0.119 g
of Sn is deposited at the cathode. No substance is lost during electrolysis.
Find the ratio of the masses of SnCI2 : SnCI4 after electrolysis.
Solution:
The
chemical reaction occurring during electrolysis is
2SnCl2 → SnCl4 + Sn
2×190 g 261 g 119 g
2SnCl2 → SnCl4 + Sn
2×190 g 261 g 119 g
119 g of
Sn is deposited by the decomposition of 380 g of SnCl2
So, 0.119
g of SnCl2 of Sn is deposited by the decomposition
of
380/119×0.119=0.380 g of SnCl2
Remaining amount of SnCl2 = (19-0.380) = 18.62 g
380/119×0.119=0.380 g of SnCl2
Remaining amount of SnCl2 = (19-0.380) = 18.62 g
380 g of
SnCl2 produce = 261 g of SnCl4
So 0.380 g of SnCl2 produce = 261/380×0.380=0.261 g of SnCl
Thus, the ratio SnCl2 : SnCl4 =18.2/0.261 , i.e., 71.34 : 1
So 0.380 g of SnCl2 produce = 261/380×0.380=0.261 g of SnCl
Thus, the ratio SnCl2 : SnCl4 =18.2/0.261 , i.e., 71.34 : 1
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Example 12.
A current
of 2.68 ampere is passed for one hour through an aqueous solution
of copper sulphate using copper electrodes. Calculate the change in mass
of cathode and that of the anode. (At. mass of copper = 63.5).
Solution:
The
electrode reactions are:
Cu2+ + 2e- → Cu (Cathode)
1 mole 2 × 96500 C
Cu → Cu2+ + 2e-
(Anode)
Thus, cathode increases in mass as copper is deposited on it and the anode decreases in mass as copper from it dissolves.
Charge passed through cell = 2.68 × 60 × 60 coulomb
Copper deposited or dissolved = 63,5/(2×96500)×2.68×60×60 =3.174 g
Increase in mass of cathode = Decrease in mass of anode = 3.174 g
Cu2+ + 2e- → Cu (Cathode)
1 mole 2 × 96500 C
Cu → Cu2+ + 2e-
(Anode)
Thus, cathode increases in mass as copper is deposited on it and the anode decreases in mass as copper from it dissolves.
Charge passed through cell = 2.68 × 60 × 60 coulomb
Copper deposited or dissolved = 63,5/(2×96500)×2.68×60×60 =3.174 g
Increase in mass of cathode = Decrease in mass of anode = 3.174 g
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Example 13.
An ammeter
and a copper voltameter are connected in series through which a constant
current flows. The ammeter shows 0.52 ampere. If 0.635 g of copper is deposited
in one hour, what is the percentage error of the ammeter? (At. mass of
copper = 63.5)
Solution :
The
electrode reaction is:
Cu2+ + 2e → Cu
Cu2+ + 2e → Cu
1 mole 2 ×
96500 C
63.5 g of copper deposited by passing charge = 2 × 96500 Coulomb
0.635 g of copper deposited by passing charge =(2×96500)/63.5×0.653 coulomb = 2 × 965 coulomb = 1930 coulomb
63.5 g of copper deposited by passing charge = 2 × 96500 Coulomb
0.635 g of copper deposited by passing charge =(2×96500)/63.5×0.653 coulomb = 2 × 965 coulomb = 1930 coulomb
We know
that
Q = l × t
1930 = I × 60 × 60
I= 1930/3600=0.536 ampere
Percentage error = ((0.536-0.52))/0.536×100=2.985
Q = l × t
1930 = I × 60 × 60
I= 1930/3600=0.536 ampere
Percentage error = ((0.536-0.52))/0.536×100=2.985
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Example 14.
A current
of 3.7 ampere is passed for 6 hours between platinum electrodes in 0.5
litre of a 2 M solution of Ni(NO3)2. What will be the molarity of the solution
at the end of electrolysis?
What will
be the molarity of solution if nickel electrodes are used? (1 F = 96500
coulomb; Ni = 58.7)
Solution:
The
electrode reaction is
Ni2+ + 2e- → Ni
1 mole 2 ×
96500 C
Quantity
of electric charge passed = 3.7 × 6 × 60 × 60 coulomb = 79920
coulomb
Number of moles of Ni(NO3)2 decomposed or nickel deposited = (1.0 - 0.4140) = 0.586
Since 0.586 moles are present in 0.5 litre,
Molarity of the solution = 2 × 0.586 = 1.72 M
When nickel electrodes are used, anodic nickel will dissolve and get deposited at the cathode.
The molarity of the solution will, thus, remain unaffected
Number of moles of Ni(NO3)2 decomposed or nickel deposited = (1.0 - 0.4140) = 0.586
Since 0.586 moles are present in 0.5 litre,
Molarity of the solution = 2 × 0.586 = 1.72 M
When nickel electrodes are used, anodic nickel will dissolve and get deposited at the cathode.
The molarity of the solution will, thus, remain unaffected
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Example 15:
An acidic
solution of Cu2+ salt containing 0.4 g of Cu2+ is electrolysed until all the copper is
deposited. The electrolysis is continued for seven more minutes with volume of
solution kept at 100 mL and the current at 1.2 amp. Calculate the gases
evolved at NTP during the entire electrolysis.
0.4 g of
Cu2+ = 0.4/31.75 = 0.0126 g equivalent
At the same time, the oxygen deposited at anode = 0.0126 g equivalent = 8/32 × 0.0126 = 0.00315 g mol
After the complete deposited of copper, the electrolysis will discharge hydrogen at cathode
and oxygen at anode. The amount of charge passed = 1.2 × 7 × 60 = 504 coulomb
So, Oxygen liberated = 1/96500 × 504 = 0.00523 g equivalent
= 8/32 × 0.00523 = 0.001307 g mole
At the same time, the oxygen deposited at anode = 0.0126 g equivalent = 8/32 × 0.0126 = 0.00315 g mol
After the complete deposited of copper, the electrolysis will discharge hydrogen at cathode
and oxygen at anode. The amount of charge passed = 1.2 × 7 × 60 = 504 coulomb
So, Oxygen liberated = 1/96500 × 504 = 0.00523 g equivalent
= 8/32 × 0.00523 = 0.001307 g mole
Hydrogen
liberated = 0.00523 g equivalent
= 1/2 × 0.00523 = 0.00261 g mole
Total gases evolved = (0.00315 + 0.001307 + 0.00261) g mole
= 0.007067 g mole
Volume of gases evolved at NTP = 22400 × 0.007067 mL = 158.3 mL
= 1/2 × 0.00523 = 0.00261 g mole
Total gases evolved = (0.00315 + 0.001307 + 0.00261) g mole
= 0.007067 g mole
Volume of gases evolved at NTP = 22400 × 0.007067 mL = 158.3 mL
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Example 16:
Consider
the reaction,
2Ag+ + Cd → 2Ag + Cd2+
The
standard electrode potentials for Ag+ --> Ag and Cd2+ --> Cd couples are 0.80 volt and
-0.40 volt, respectively.
(i) What
is the standard potential Eo for this reaction?
(ii) For
the electrochemical cell in which this reaction takes place which electrode is
negative electrode?
Solution:
(i) The
half reactions are:
2Ag+ + 2e- → 2Ag.
Reduction
Cathode)
EoAg+/Ag =0.80 volt
(Reduction potential)
Cd → Cd2+ + 2e-,
Oxidation
(Anode)
EoCd+/Cd = -0.40 volt
(Reduction potential)
or
EoCd+/Cd2 = +0.40 volt
Eo = EoCd+/Cd2 + EoAg+/Ag = 0.40+0.80 = 1.20 volt
(ii) The
negative electrode is always the electrode whose reduction potential has
smaller value or the electrode where oxidation occurs. Thus, Cd electrode is
the negative electrode.
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Example 17:
Calculate
the electricity that would be required to reduce 12.3 g of nitrobenzene
to aniline, if the current efficiency for the process is 50 per cent. If
the potential drop across the cell is 3.0 volt, how much energy will be
consumed?
Solution:
The
reduction reaction is
C6H5NO2 + 3H2 C6H5NH2 + 2H2O
Hydrogen required for reduction of 12.3/123 or 0.1 mole of nitrobenzene = 0.1 × 3 = 0.3 mole
Amount of charge required for liberation of 0.3 mole of hydrogen = 2 × 96500 × 0.3 = 57900 coulomb
C6H5NO2 + 3H2 C6H5NH2 + 2H2O
Hydrogen required for reduction of 12.3/123 or 0.1 mole of nitrobenzene = 0.1 × 3 = 0.3 mole
Amount of charge required for liberation of 0.3 mole of hydrogen = 2 × 96500 × 0.3 = 57900 coulomb
Actual
amount of charge required as efficiency is 50% = 2 × 57900 = 115800
coulomb
Energy
consumed = 115800 × 3.0 = 347400 J = 347.4 kJ
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Example 18:
After
electrolysis of a sodium chloride solution with inert electrodes for
a certain period of time, 600 mL of the solution was left which was found
to be 1 N in NaOH. During the same period 31.75 g of copper was deposited
in the copper voltameter in series with the electrolytic cell. Calculate
the percentage theoretical yield of NaOH obtained.
Solution:
Equivalent
mass of NaOH = 40/1000 × 600 = 24 g
Amount of NaOH formed = 40/1000 × 600 = 24 g
31.75 g of Cu = 1 g equivalent of Cu.
During the same period, 1 g equivalent of NaOH should have been formed.
1 g equivalent of NaOH = 40 g
% yield = 24/40 × 100 = 60
Amount of NaOH formed = 40/1000 × 600 = 24 g
31.75 g of Cu = 1 g equivalent of Cu.
During the same period, 1 g equivalent of NaOH should have been formed.
1 g equivalent of NaOH = 40 g
% yield = 24/40 × 100 = 60
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Example 19:
To find
the standard potential of M3+/M electrode, the following cell is
constituted:
Pt|M|M3+(0.0018 mol-1L)||Ag+(0.01 mol-1L)|Ag
The emf of
this cell is found to be 0.42 volt. Calculate the standard potential of the
half reaction M3+ + 3e- M3+. = 0.80 volt.
Solution:
The cell
reaction is
M + 3Ag+ → 3Ag + M3+
Applying
Nernst equation,
Ecell = Ecello - 0.0591/n log(Mg2+)/[Ag+]3
0.42
= Ecello - 0.0591/n log (0.0018)/(0.01)3 = Ecello - 0.064
Ecello =(0.042+0.064)= 0.484 volt
Eocell = Eocathode - Eoanode
or Eoanode = Eocathode - Eocell = (0.80-0.484) = 0.32 volt
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Example 20:
Cadmium
amalgam is prepared by electrolysis of a solution of CdCl2 using a mercury c0thode. Find how long a
current of 5 ampere should be passed in order to prepare 12% Cd-Hg
amalgam on a cathode of 2 g mercury. At mass of Cd = 112.40.
Solution:
2 g Hg require
Cd to prepare 12% amalgam = 12/88 × 2 = 0.273 g
Cd2+ + 2e- → Cd
1 mole 2 × 96500C
112.40g
Charge required to deposit 0.273 g of Cd = 2*96500/112.40 × 0.273 coulomb
Charge = ampere × second
Second = 2*96500*0.273/112.40*5 = 93.75
Cd2+ + 2e- → Cd
1 mole 2 × 96500C
112.40g
Charge required to deposit 0.273 g of Cd = 2*96500/112.40 × 0.273 coulomb
Charge = ampere × second
Second = 2*96500*0.273/112.40*5 = 93.75
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