The molar conductance of an electrolyte increases with dilution according to the equation: 
\(\Lambda_{\mathrm{m}}=\Lambda_{\mathrm{m}}^{\circ}-\mathrm{A} \sqrt{\mathrm{c}} \)
Consider the following four statements:
A: This equation applies to both strong and weak electrolytes.
B: The value of the constant A depends upon the nature of the solvent. 
C: The value of constant A is the same for both \(BaCl_2\) and \(MgSO_4\)
D: The value of constant A is the same for both \(BaCl_2\) and \(Mg(OH)_2\)
Which of the above statements are correct? 
1. (A) and (B) only 
2. (A), (B), and (C) only 
3. (B) and (C) only 
4. (B) and (D) only 

Subtopic:  Conductance & Conductivity |
 55%
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The correct value of cell potential in volt for the reaction that occurs when the following two half cells are connected, is:
\(\mathrm{Fe}_{\text {(aq) }}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Fe}(\mathrm{s}), \mathrm{E}^{\circ}=-0.44 \mathrm{~V} \)
\( \mathrm{Cr}_2 \mathrm{O}_7^{2-}{ }_{\text {(aq) }}+14 \mathrm{H}^{+}+6 e^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O},\)
\( \mathrm{E}^{\circ}=+1.33 \mathrm{~V}\)
1. +1.77 V 2. +2.65 V 
3. +0.01 V  4. +0.89 V
Subtopic:  Electrode & Electrode Potential |
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The molar conductance of a solution, given its conductivity (0.248 S m–1) and concentration (0.2 mol m–3) is:
1. 0.124 S cm2 mol–1
2. 1.24 S mmol–1
3. 124 S cm2 mol–1
4. 124 S m2 mol–1
Subtopic:  Conductance & Conductivity |
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The conductivity of centimolar solution of KCl at 25°C is 0.0210 ohm–1 cm–1 and the resistance of the cell containing the solution at 25°C is 60 ohm. The value of the cell constant is: 
1. 3.34 cm–1 2. 1.34 cm–1
3. 3.28 cm–1 4. 1.26 cm–1
Subtopic:  Conductance & Conductivity |
 77%
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Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R):
Assertion (A): In equation \(\Delta_{\mathrm{r}} \mathrm{G}=-\mathrm{nFE} _{\text {cell }}, \) value \(\mathrm{\Delta_rG }\) depends on n. 
Reason (R): \(\mathrm{E_{cell} }\) is an intensive property and \(\mathrm{\Delta_rG }\)  is an extensive property. 
 
In the light of the above statements choose the correct answer from the options given below: 
 
1. (A) is False but (R) is True. 
2. Both (A) and (R) are True and (R) is the correct explanation of (A)
3. Both (A) and (R) are True and (R) is not the correct explanation of (A)
4. (A) is True but (R) is False.
Subtopic:  Relation between Emf, G, Kc & pH |
 57%
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Standard electrode potential for the cell with cell reaction
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
is 1.1 V. Calculate the standard Gibbs energy change for the cell reaction.
(Given F = 96487 C mol–1)
1. –200.27 kJ mol–1 2. –212.27 kJ mol–1
3. –212.27 J mol–1 4. –200.27 J mol–1
Subtopic:  Relation between Emf, G, Kc & pH |
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Two half cell reactions are given below:
\(\begin{aligned} &\mathrm{Co}^{3+}+e^{-} \rightarrow \mathrm{Co}^{2+}, \mathrm{E}_{\mathrm{Co}^{2+} / \mathrm{Co}^{3+}}^{\circ}=-1.81 \mathrm{~V} \\ &2 \mathrm{Al}^{3+}+6 e^{-} \rightarrow 2 \mathrm{Al}(\mathrm{s}), \mathrm{E}_{\mathrm{Al} / \mathrm{Al}^{3+}}^{\circ}=+1.66 \mathrm{~V} \end{aligned} \)
The standard EMF of a cell with feasible redox reaction will be:

1. +7.09 V 2. +0.15 V
3. +3.47 V 4. –3.47 V
Subtopic:  Electrode & Electrode Potential |
 66%
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\(\land^o_m\) for NaCl, HCl and \(\mathrm{CH_3COONa }\) are 126.4, 425.9, and 91.05 S cm2 mol–1 respectively. If the conductivity of 0.001028 mol L–1 acetic acid solution is \(4.95 \times 10^{-5} S ~cm^{-1} \), the degree of dissociation of the acetic acid solution is:

1. 0.01233 2. 1.00 
3. 0.1233  4. 1.233
Subtopic:  Conductance & Conductivity |
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The three cells with their \(E^\circ_{\text{(cell)}}\) values are given below:

Cells \(E^\circ_{\text{(cell)}}/V\)
(a) Fe|Fe2+||Fe3+|Fe 0.404
(b) Fe|Fe2+||Fe3+, Fe2+|Pt 1.211
(c) Fe|Fe3+||Fe3+, Fe2+|Pt 0.807
The standard Gibbs free energy change values for three cells are, respectively:
(F represents the charge on 1 mole of electrons.)
 
1. –1.212 F, –1.211 F, –0.807 F
2. +2.424 F, +2.422 F, +2.421 F
3. –0.808 F, –2.422 F, –2.421 F
4. –2.424 F, –2.422 F, –2.421 F
Subtopic:  Faraday’s Law of Electrolysis |
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Find the emf of the cell in which the following reaction takes place at 298 K:
\(\mathrm{Ni}(\mathrm{s})+2 \mathrm{Ag}^{+}(0.001 \mathrm{M}) \rightarrow \mathrm{Ni}^{2+}(0.001 \mathrm{M})+2 \mathrm{Ag}(\mathrm{s}) \)
\( \small{\text { (Given that } \mathrm{E}_{\text {cell }}^{\circ}=10.5 \mathrm{~V}, \frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059 \text { at } \ 298 \mathrm{~K})} \)
1. 1.05 V
2. 1.0385 V 
3. 1.385 V
4. 0.9615 V 

Subtopic:  Nernst Equation |
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