Q. No.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
On electrolysis of dilute sulphuric acid using Platinum (Pt) electrode, the product obtained at the anode will be:
1. Oxygen gas
2. gas
3. gas
4. Hydrogen gas
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) |
Limiting molar conductivities, for the given solutions, are :
\(\lambda_{m}^{0} \left(\right. H_{2} S O_{4} \left.\right) = x\) \(S c m^{2}\) \(m o l^{- 1}\)
\(\lambda_{m}^{0} \left(\right. K_{2} S O_{4} \left.\right) = y\) \(S c m^{2}\) \(m o l^{- 1}\)
\(\lambda_{m}^{0} \left(\right. C H_{3} C O O K \left.\right) = z\) \(S c m^{2}\) \(m o l^{- 1}\)
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. \(\mathrm{{(x-y) \over 2}+z}\)
The molar conductance of NaCl, HCI, and CH3COONa at infinite dilution are 126.45, 426.16, and 91.0 S cm mol–1 respectively. The molar conductance of CH3COOH at infinite dilution will be:
1. 698.28 S cm2 mol–1
2. 540.48 S cm2 mol–1
3. 201.28 S cm2 mol–1
4. 390.71 S cm2 mol–1
The molar conductivity of 0.007 M acetic acid is 20 S cm2 mol–1. The dissociation constant of acetic acid is :
(\(\mathrm{\Lambda_{H^{+}}^{o} \ = \ 350 \ S \ cm^{2} \ mol^{-1} }\))
(\(\mathrm{\mathrm{\Lambda_{CH_{3}COO^{-}}^{o} \ = \ 50 \ S \ cm^{2} \ mol^{-1} }}\))
1. mol L–1
2. mol L–1
3. mol L–1
4. mol L–1
For a cell involving one electron \(E_{cell}^{\ominus} = 0 . 59 V\) at 298 K. The equilibrium constant for the cell reaction is :
\(\mathrm{[Given~ that~ \frac {2.303 ~RT}{F} = 0.059 ~V~ at~ T = 298 K]}\)
| 1. | \(1 . 0 \times \left(10\right)^{30}\) | 2. | \(1 . 0 \times \left(10\right)^{2}\) |
| 3. | \(1 . 0 \times \left(10\right)^{5}\) | 4. | \(1 . 0 \times \left(10\right)^{10}\)
|
Given the following cell reaction:
\(\mathrm{2Fe^{3+}(aq) \ + \ 2I^{-}(aq)\rightarrow 2Fe^{2+}(aq) \ + \ I_{2}(aq)}\)
The standard Gibbs energy ∆rG⊝ of the cell reaction is:
[Given: \(F = 96500\) \(C\) \(mol^{- 1}\)]
1. \(23 . 16\) \(kJ\) \(mol^{- 1}\)
2. \(- 46 . 32\) \(kJ\) \(mol^{- 1}\)
3. \(- 23 . 16\) \(kJ\) \(mol^{- 1}\)
4. \(46 . 32\) \(kJ\) \(mol^{- 1}\)
\(\text{MnO}_{4}^{-}+8 \text{H}^{+}+5 \text{e}^{-} \rightarrow \text{Mn}^{2+}+4 \text{H}_{2} \text{O}, \)
\( \text{E}_{\text{Mn}^{2+}}^{\circ} / \text{MnO}_{4}^{-}=-1.510 \text{ V} \)
\( \frac{1}{2} \text{O}_{2}+2 \text{H}^{+}+2 \text{e}^{-} \rightarrow \text{H}_{2} \text{O}, \)
\( \text{E}_{\text{O}_{2} / \text{H}_{2} \text{O}}^{\circ}=+1.223 \text{ V}\)
Will the permanganate ion, \(\text{MnO}_{4}^{-}\) , liberate \(\text{O}_{2}\) from water in the presence of an acid?
1. No, because \(\text{E}_{\text {cell }}^{\circ}=-2.733 \text{ V}\)
2. Yes, because \(\text{E}_{\text {cell }}^{\circ}=+0.287 \text{ V}\)
3. No, because \(\text{E}_{\text {cell }}^{\circ}=-0.287 \text{ V}\)
4. Yes, because \(\text{E}_{\text {cell }}^{\circ}=+2.733 \text{ V}\)
Find the emf of the cell in which the following reaction takes place at 298 K:
\(\mathrm{Ni}(\mathrm{s})+2 \mathrm{Ag}^{+}(0.001 \mathrm{M}) \rightarrow \mathrm{Ni}^{2+}(0.001 \mathrm{M})+2 \mathrm{Ag}(\mathrm{s}) \)
\(\small{\text { (Given that } \mathrm{E}_{\text {cell }}^{\circ}=1.05 \mathrm{~V}; \dfrac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059} )\)
1. 1.05 V
2. 1.0385 V
3. 1.385 V
4. 0.9615 V
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