# The correct expression that  represents the equivalent conductance at infinite dilution of $A{l}_{2}{\left(S{O}_{4}\right)}_{3}$ is: (Given that ${\wedge }_{A{l}^{3+}}^{°}$ $and$ ${\wedge }_{S{O}_{4}^{2-}}^{°}$ are the equivalent conductances at infinite dilution of the respective ions) 1. ${\wedge }_{A{l}^{3+}}^{°}$ $+$ ${\wedge }_{S{O}_{4}^{2-}}^{°}$ 2. $\left({\wedge }_{A{l}^{3+}}^{°}\right)$ $+$ ${\wedge }_{S{O}_{4}^{2-}}^{°}×6$ 3. $\frac{1}{3}{\wedge }_{A{l}^{3+}}^{°}$ $+\frac{1}{2}$ ${\wedge }_{S{O}_{4}^{2-}}^{°}$ 4. $2{\wedge }_{A{l}^{3+}}^{°}$ $+3$ ${\wedge }_{S{O}_{4}^{2-}}^{°}$

Subtopic:  Conductance & Conductivity |
60%
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)

The correct relation among the given options is:

 1 (a) and (b) 2 (c) and (d) 3 (b) and (d) 4 (c) and (a)
Subtopic:  Electrode & Electrode Potential |
68%
From NCERT
AIPMT - 2010
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Molar conductivities $\left(\wedge {°}_{m}\right)$ at infinite dilution of
NaCl, HCl, and $C{H}_{3}COONa$ are 126.4, 425.9, and 91.0 S cm2 mol–1 respectively.
$\left(\wedge {°}_{m}\right)$  for $C{H}_{3}COOH$  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 |
90%
From NCERT
AIPMT - 2012
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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 |
61%
From NCERT
AIPMT - 2012
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Consider the half-cell reduction reaction:
$$\text{Mn}^{2+}+2e^-\rightarrow \text{Mn},\ E^{0}= -1.18~ \text V$$
$$\text{Mn}^{2+}\rightarrow \text{Mn}^{3+}+e^-,\ E^{0}= -1.5~ \text V$$

The $$E^{0}$$ for the reaction $$\mathrm{3\ Mn^{2+}\rightarrow Mn^{0}+2Mn^{3+} }$$ and possibility of the forward reaction are respectively:

 1 –4.18 V and Yes 2 +0.33 V and Yes 3 +2.69 V and No 4 –2.69 V and No
Subtopic:  Electrode & Electrode Potential |
53%
From NCERT
NEET - 2013
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How many grams of cobalt metal will be deposited when a solution of cobalt (II) chloride is electrolyzed with a current of 10 amperes for 109 minutes?
(1 Faraday = 96,500 C; Atomic mass of Co = 59 u)

 1 4 2 20 3 40 4 0.66
Subtopic:  Faraday’s Law of Electrolysis |
71%
From NCERT
NEET - 2013
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A hypothetical electrochemical cell is shown below.
A|A+(x M) || B+(y M)|B
The Emf measured is +0.20 V. The cell reaction is:

1. A+ + B → A + B+

2.  A+ + e- → A ; B+ + e- → B

3. The cell reaction cannot be predicted.

4. A + B+ → A+ + B

Subtopic:  Electrochemical Series | Nernst Equation |
75%
From NCERT
AIPMT - 2006
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If ${\mathrm{E}}_{{\mathrm{Fe}}^{2+}/\mathrm{Fe}}^{\mathrm{o}}$ = -0.441 V and  ${\mathrm{E}}_{{\mathrm{Fe}}^{3+}/{\mathrm{Fe}}^{2+}}^{\mathrm{o}}$ = 0.771 V, the standard emf of the reaction:

Fe + 2Fe3+→ 3Fe2+ will be:

 1 0.330 V 2 1.653 V 3 1.212 V 4 0.111 V
Subtopic:  Electrode & Electrode Potential |
72%
From NCERT
AIPMT - 2006
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In producing chlorine through electrolysis, 100 W power at 125 V is being consumed.
The liberation of chlorine per min is:
(ECE of chlorine is 0.367×10-6 kg/C)
 1 17.6 mg 2 21.3 mg 3 24.3 mg 4 13.6 mg
Subtopic:  Faraday’s Law of Electrolysis |
63%
From NCERT
AIPMT - 2006
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The efficiency of a fuel cell is given by:

1. $\frac{∆H}{∆G}$

2. $\frac{∆G}{∆S}$

3. $\frac{∆G}{∆H}$

4. $\frac{∆S}{∆G}$

Subtopic:  Batteries & Salt Bridge |
65%
AIPMT - 2007
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