Faradays Law Of Electrolysis
High-Yield Trend
Questions 9 MCQs
The liberation of chlorine per min is:
| 1. | 17.6 mg | 2. | 21.3 mg |
| 3. | 24.3 mg | 4. | 13.6 mg |
A steady current of 1.5 A flows through a copper voltmeter for 10 min. If the electrochemical equivalent of copper is 30 \times 10-5 g C-1, the mass of copper deposited on the electrode will be:
1. 0.40 g
2. 0.50 g
3. 0.67 g
4. 0.27 g
Al2O3 is reduced by electrolysis at low potentials and high currents. If 4.0 x 104 A of current is passed through molten Al2O3 for 6 hours, the mass of aluminum produced is:
(Assume 100 % current efficiency, the atomic mass of Al = 27 g mol-1)
| 1. | 9.0 x 103 g | 2. | 8.1 x 104 g |
| 3. | 2.4 x 105 g | 4. | 1.3 x 104 g |
For the reduction of silver ions with copper metal, the standard cell potential was found to be +0.46 V at 25 °C. The value of standard Gibbs energy, ΔGo will be:
(F = 96500 C mol-1)
1. -89.0 kJ
2. -89.0 J
3. -44.5 kJ
4. -98.0 kJ
In producing chlorine by electrolysis, 100 kW power at 125 V is being consumed.
How much chlorine per minute is liberated:
(Given -ECE of chlorine is 0.367 X 10-6 kgC-1)
1.
2.
3.
4.
During the electrolysis of molten sodium chloride, the time required to produce 0.10 mol of chlorine gas using a current of 3 amperes is:
1. 55 minutes
2. 110 minutes
3. 220 minutes
4. 330 minutes
The number of electrons delivered at the cathode during electrolysis by a current of 1 ampere in 60 seconds is:
(Given: Charge on electron = 1.60 \times 10–19 C)
1.
2.
3.
4.
| List-I (Redox Conversion) | List-II (Number of Faraday required) | ||
| A. | 1 mol of H2O to O2 | I. | 3F |
| B. | 1 mol of to | II. | 2F |
| C. | 1.5 mol of from molten | III. | 1F |
| D. | 1 mol of FeO to Fe2O3 | IV. | 5F |
Choose the correct answer from the options given below:
| 1. | A - III, B - IV, C - I, D - II | 2. | A - II, B - III, C - I, D - IV |
| 3. | A - III, B - IV, C - II, D - I | 4. | A - II, B - IV, C - I, D - III |
| 1. | 0.315 g | 2. | 31.5 g |
| 3. | 0.0315 g | 4. | 3.15 g |