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 :-

1. $R=H_{2\left(g\right)}, O_{2\left(g\right)}; P=H_2{O}_{2\left(l\right)}$

2. $R=H_{2\left(g\right)}, O_{2\left(g\right)}; P=H_2{O}_{\left(l\right)}$

3. $R=H_{2\left(g\right)}, O_{2\left(g\right)}, C{l}_2\left(g\right); P=HCl{O}_{4\left(aq\right)}$

4. $R=H_{2\left(g\right)}, N_{2\left(g\right)}; P=N{H}_{3\left(aq\right)}$

Limiting molar conductivities, for the given solutions, are :

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

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

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

From the data given above, it can be concluded that \(\lambda_m^0 \) in (\(S\ cm^2\ mol^{-1}\)) for CH₃COOH will be :

1. $\mathrm{x }- \mathrm{y }+ 2z$

2. $\mathrm{x }+ \mathrm{y }+ \mathrm{z }$

3. x - y + z

4. $\frac{(x-y)}{\mathrm{2 }}+ z$

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: $\left[\mathrm{Given} \mathrm{that} \frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059 \mathrm{V} \mathrm{at} \mathrm{T}=298 \mathrm{K}\right]$

1.  $1.0\times {10}^{30}$

2.  $1.0\times {10}^2$

3.  $1.0\times {10}^5$

4.  $1.0\times {10}^{10}$

For the cell reaction \(2Fe^{3+}(aq) \ + \ 2I^{-}(aq)\rightarrow 2Fe^{2+}(aq) \ + \ I_{2}(aq)\)

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

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

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 
\(\begin{aligned} \mathrm{Ni}(\mathrm{s})+2 \mathrm{Ag}^{+}(0.001 \mathrm{M}) \rightarrow \mathrm{Ni}^{2+}(0.001 \mathrm{M})+2 \mathrm{Ag}(\mathrm{s}) \\ \small{\text { (Given that } \mathrm{E}_{\text {cell }}^{\circ}=10.5 \mathrm{~V}, \frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059 \text { at } \ 298 \mathrm{~K})} \end{aligned}\)
1. 1.05 V 
2. 1.0385 V 
3. 1.385 V 
4. 0.9615 V