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 Gibb's energy for the decomposition of $A{l}_2{O}_3$ at $500°C$ 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 $500°C$ is at least,

1. 3.0 V 

2. 2.5 V 

3. 5.0 V 

4. 4.5 V 

Molar conductivities $(\wedge {°}_m)$ at infinite dilution of NaCl, HCl and $C{H}_{3}COONa$ are 126.4, 425.9 and 91.0 S cm² mol^-1 respectively. $(\wedge {°}_m)$  for $C{H}_{3}COOH$  will be:

1.  $180.5 S c{m}^{\mathrm{2 }}mo{l}^{-1}$

2.  $290.8 S c{m}^{\mathrm{2 }}mo{l}^{-1}$

3.  $390.5 S c{m}^{\mathrm{2 }}mo{l}^{-1}$

4.  $425.5 S c{m}^{\mathrm{2 }}mo{l}^{-1}$

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. $\left({\wedge }_{A{l}^{3+}}^{°} + {\wedge }_{S{O}_4^{2-}}^{°}\right)\times 6$

3. $\frac{1}{3}{\wedge }_{A{l}^{3+}}^{°} +\frac{1}{2} {\wedge }_{S{O}_4^{2-}}^{°}$

4. $2{\wedge }_{A{l}^{3+}}^{°} +3 {\wedge }_{S{O}_4^{2-}}^{°}$

Consider the following relations for emf of an electrochemical cell:

The correct relation among the given options is : 

1. (a) and (b)

2. (c) and (d)

3. (b) and (d)

4. (c) and (a)

${10}^5$coloumb charge liberated 1 gm silver (Ag). If now charge is doubled then the amount of liberated Ag will be :
1.  1 gm
2.  2 gm
3.  3 gm
4.  4 gm

The concentration of $ZnC{l}_2$ solution will change when it is placed in a container which is made of :
1. Al
2. Cu
3. Ag
4. None