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. ${\mathrm{H}}_2\mathrm{S}$ gas

3. ${\mathrm{SO}}_2$ gas

4. Hydrogen gas 

In a typical fuel cell, the reactants (R) and products (P) are: 

Limiting molar conductivities, for the given solutions, are :

${\lambda }_m^0\left(H_{2}S{O}_4\right)=\hspace{0.17em}x$ $\mathrm{S }c{m}^2$ $mo{l}^{-1}$

${\lambda }_m^0\left(K_{2}S{O}_4\right)=\hspace{0.17em}y$ $\mathrm{S }c{m}^2$ $mo{l}^{-1}$

${\lambda }_m^0\left(C{H}_{3}COOK\right)=\hspace{0.17em}z$ $\mathrm{S }c{m}^2$ $mo{l}^{-1}$

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. $1.75\times {10}^{-5}$ mol L^–1 

2. $2.50\times {10}^{-5}$ mol L^–1 

3. $1.75\times {10}^{-4}$ mol L^–1 

4. $2.50\times {10}^{-4}$ mol L^–1 

For a cell involving one electron ${\mathrm{E}}_{\mathrm{cell}}^{⊝}=0.\mathrm{59 }\mathrm{V}$ at 298 K. The equilibrium constant for the cell reaction is :
\(\mathrm{[Given~ that~ \frac {2.303 ~RT}{F} = 0.059 ~V~ at~ T = 298 K]}\)

Given the following cell reaction:
 \(\mathrm{2Fe^{3+}(aq) \ + \ 2I^{-}(aq)\rightarrow 2Fe^{2+}(aq) \ + \ I_{2}(aq)}\)

\(E_{cell}^{o} \ = \ 0.24 \ V\) at $298$ $\mathrm{K}$.
The standard Gibbs energy ∆_rG^⊝ of the cell reaction is:

[Given: $\mathrm{F }=\hspace{0.17em}96500$ $\mathrm{C}$ ${\mathrm{mol}}^{-1}$]

1. $23.16$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$

2. $-46.32$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$

3. $-23.16$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$

4. $46.32$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$

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