Mass in grams of copper deposited by passing 9.6487 A current through a voltmeter containing copper sulphate for 100 seconds is:
(Given : Molar mass of Cu: \(63~ g mol^{-1}\); 1F = 96487 C)
1. 0.315 g  2. 31.5 g 
3. 0.0315 g  4. 3.15 g 
Subtopic:  Faraday’s Law of Electrolysis |
 68%
Level 2: 60%+
NEET - 2024
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Match the redox conversions in List-I with the corresponding number of Faradays required in List-II.
List-I
(Redox Conversion)
List-II
(Number of Faraday required)
A. 1 mol of H2O to O2 I. 3F
B. 1 mol of \(MnO^-_4\) to \(Mn^{2+}\) II. 2F
C. 1.5 mol of \(Ca\) from molten \(CaCl_2\) 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
Subtopic:  Faraday’s Law of Electrolysis |
 66%
Level 2: 60%+
NEET - 2024
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The electricity required in coulombs for the oxidation of 1 mole of FeO to \(Fe_{2} O_{3 }\) is:

1. 964.87 C 2. 96487 C
3. 96.487 C 4. 9.6487 C
Subtopic:  Faraday’s Law of Electrolysis |
 89%
Level 1: 80%+
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The amount of charge required for the reduction of 1 mol of MnO4- to Mn2+ is:

1. 4.8 × 105 C
2. 3.2 × 106 C
3. 1.8 × 105 C
4. 4.1 × 104 C

Subtopic:  Faraday’s Law of Electrolysis |
 84%
Level 1: 80%+
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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 × 10–19 C)

1.  6×1023

2.  6×1020

3.  3.75×1020

4.  7.48×1020

Subtopic:  Faraday’s Law of Electrolysis |
 81%
Level 1: 80%+
NEET - 2016
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When 0.1 mol MnO42– is oxidized, the quantity of electricity required to completely oxidise MnO42– to MnO4– is:

1. 96500 C 2. 2 × 96500 C
3. 9650 C 4. 96.50 C
Subtopic:  Faraday’s Law of Electrolysis |
 78%
Level 2: 60%+
AIPMT - 2014
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The number of Faradays required to produce 20.0 g of Ca from molten CaCl2 is-

1. 2F

2. 1F

3. 4F

4. 3F

Subtopic:  Faraday’s Law of Electrolysis |
 78%
Level 2: 60%+
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A steady current of 1.5 A flows through a copper voltmeter for 10 min. If the electrochemical equivalent of copper is 30 × 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

Subtopic:  Faraday’s Law of Electrolysis |
 78%
Level 2: 60%+
AIPMT - 2007
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For the cell, Ti/Ti+(0.001M)||Cu2+(0.1M)|Cu, Ecello at

25 °C is 0.83 V. Ecell can be increased :

1. By increasing [Cu2+]

2. By increasing [Ti+]

3. By decreasing [Cu2+]

4. None of the above.

Subtopic:  Nernst Equation | Faraday’s Law of Electrolysis |
 73%
Level 2: 60%+
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The  number of electrons involved in the deposition of 63.5 g of Cu from a solution of CuSO4 is:

1. 6.022 × 1023 

2. 3.011 × 1023 

3. 12.044 × 1023 

4. 6.022 × 1022

Subtopic:  Faraday’s Law of Electrolysis |
 71%
Level 2: 60%+
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