The number of Faradays (F) required to produce 20 g of calcium from molten CaCl2 (Atomic mass of Ca = 40 g mol–1) is:

1. 2

2. 3

3. 4

4. 1

Subtopic:  Faraday’s Law of Electrolysis |
 70%
From NCERT
NEET - 2020
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On electrolysis of dilute sulphuric acid using Platinum (Pt) electrode, the product obtained at the anode will be:

1. Oxygen gas

2. H2S gas

3. SO2 gas

4. Hydrogen gas 

Subtopic:  Electrolytic & Electrochemical Cell |
 66%
From NCERT
NEET - 2020
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In a typical fuel cell, the reactants (R) and products (P) are: 

1. R = H2(g), O2(g); P = H2O2(l)
2. R = H2(g), O2(g); P = H2O(l)
3. R = H2(g), O2(g), C l2(g); P = HClO4(aq)
4. R = H2(g), N2(g); P = NH3(aq)

Subtopic:  Batteries & Salt Bridge |
 75%
From NCERT
NEET - 2020
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Limiting molar conductivities, for the given solutions, are :

λm0(H2SO4)= x cm2 mol-1

λm0(K2SO4)= y cm2 mol-1

λm0(CH3COOK)= z cm2 mol-1

From the data given above, it can be concluded that \(\lambda_m^0 \) in (\(S\ cm^2\ mol^{-1}\)) for CH3COOH will be :
1. \(\mathrm{x-y+2z}\)       
2. \(\mathrm{x+y+z}\)          
3. \(\mathrm{x-y+z}\)       
4. \(\mathrm{{(x-y) \over 2}+z}\)          

Subtopic:  Conductance & Conductivity |
 70%
From NCERT
NEET - 2019
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The Gibb's energy for the decomposition of \(\mathrm{A l_{2} O_{3}}\) at \(\mathrm{500~ ^\circ C}\) is as follows: 

2/3Al2O3 → 4/3Al + O2 ; ∆rG = + 960 k J mol–1

The potential difference needed for the electrolytic reduction of aluminium oxide (Al2O3) at \(\mathrm{500~ ^\circ C}\) is at least,

1. 3.0 V 

2. 2.5 V 

3. 5.0 V 

4. 4.5 V 

Subtopic:  Relation between Emf, G, Kc & pH |
 62%
From NCERT
AIPMT - 2012
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Molar conductivities (°m) at infinite dilution of
NaCl, HCl, and CH3COONa are 126.4, 425.9, and 91.0 S cm2 mol–1 respectively.
 (°m)  for CH3COOH  will be: 

1. \(180.5~S~cm^2~mol^{-1}\) 2. \(290.8~S~cm^2~mol^{-1}\)
3. \(390.5~S~cm^2~mol^{-1}\) 4. \(425.5~S~cm^2~mol^{-1}\)
Subtopic:   Kohlrausch Law & Cell Constant |
 91%
From NCERT
AIPMT - 2012
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The correct expression that  represents the equivalent conductance at infinite dilution of Al2(SO4)3 is:

(Given that Al3+° and SO42-° are the equivalent conductances at infinite dilution of the respective ions)

1. Al3+° + SO42-°

2. Al3+° + SO42-°×6

3. 13Al3+° +12 SO42-°

4. 2Al3+° +3 SO42-°

Subtopic:  Conductance & Conductivity |
 63%
AIPMT - 2010
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Consider the following relations for emf of an electrochemical cell:

(a) emf of cell = (Oxidation potential of anode) – (Reduction potential of cathode)
(b) emf of cell = (Oxidation potential of anode) + (Reduction potential of cathode)
(c) emf of cell = (Reduction potential of anode) + (Reduction potential of cathode)
(d) emf of cell = (Oxidation potential of anode) – (Oxidation potential of cathode)


Which of the following combinations correctly represents the relation for the emf of the cell?

1. (a) and (b) 2. (c) and (d)
3. (b) and (d) 4. (c) and (a)
Subtopic:  Electrode & Electrode Potential |
 70%
From NCERT
AIPMT - 2010
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105 coloumb charge liberated 1 gm silver (Ag). If the charge is doubled then the amount of liberated Ag will be:

1. 1 gm 2. 2 gm
3. 3 gm 4. 4 gm
Subtopic:  Faraday’s Law of Electrolysis |
 89%
From NCERT
AIPMT - 1998
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The concentration of ZnCl2 solution will change when it is placed in a container which is made of:

1. Al 2. Cu
3. Ag 4. None
Subtopic:  Electrochemical Series |
 73%
From NCERT
AIPMT - 1998
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