Numericals on faraday's law of electrolysis
Answers
- 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
- 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
Thus, 3 mole of electrons are needed to reduce 1 mole of Al3+
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
- 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
(b) The oxidation reaction is
FeO + 1/2 H2O → 1/2 Fe2O3 + H+ + e-
Q = F = 96500 coulomb
- 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
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
- 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
- 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
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
- 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.
Solution:
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;
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
- 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.
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
- 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
W = Z × I × t
Where W = 10 g, Z= 35.5/96500 t = 60×60 =3600 second
I = 10x96500/35.5x96500 = 7.55 ampere
- 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)
Solution:
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