Calculate the emf of the given cell: 

Zn(s) | Zn⁺² (0.1M) || Sn⁺² (0.001M) | Sn(s)

(Given ${\mathrm{E}}_{{\mathrm{Zn}}^{+2}/\mathrm{Zn}}^{\mathrm{o}}=-0.76$ $\mathrm{V},$ ${\mathrm{E}}_{{\mathrm{Sn}}^{2+}/\mathrm{Sn}}^{\mathrm{o}}=-0.14$ $\mathrm{V)}$

1. 0.62 V

2. 0.56 V

3. 1.12 V

4. 0.31 V

The potential of a hydrogen electrode having a pH = 10 is : 

1. 0.59 V

2. –0.59 V

3.  0.0 V

4. –5.9 V

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. \(\mathrm{x-y+2z}\)       
2. \(\mathrm{x+y+z}\)          
3. \(\mathrm{x-y+z}\)       
4. \(\mathrm{{(x-y) \over 2}+z}\)          

$$
For the Cell \(Pt(s)|| Br^-(aq)(0.010M)| Br_2(l)|| H^+(aq) (0.0030M) |H_2(g) (1 bar) | Pt(s)\)

If the concentration of ${\mathrm{Br}}^{-}$ becomes 2 times and the concentration of ${\mathrm{H}}^{+}$ becomes half of the initial value, then emf of the cell

1.  Doubles

2.  Four times

3.  Eight times

4.  Remains the same

The specific conductance (K) of 0.02 M aqueous acetic acid solution at 298 K is $1.65 \times {10}^{-4}$ S ${\mathrm{cm}}^{-1}$. The degree of dissociation of acetic acid is [Given: Equivalent conductance at infinite dilution of ${\mathrm{H}}^{+}$ = 349.1 S ${\mathrm{cm}}^2$ ${\mathrm{mol}}^{-1}$ and ${\mathrm{CH}}_3{\mathrm{COO}}^{-}$ = 40.9 S ${\mathrm{cm}}^2$ ${\mathrm{mol}}^{-1}$)

1.  0.021

2.  0.21

3.  0.012

4.  0.12

The equilibrium constant of a 2 electron redox reaction at 298 K is 3.8 x ${10}^{-3}$. The cell potential E^o (in V) and the free energy change ∆G^o (in kJ mol^-1 ) for this equilibrium respectively, are -

The specific conductance of 0.01 M solution of a weak monobasic acid is 0.20 x 10^-3 S cm^-1. The dissociation constant of the acid is-

[Given  ${\Lambda }_{\mathrm{HA}}^{\infty }$ = 400 S ${\mathrm{cm}}^2$ ${\mathrm{mol}}^{-1}$]

Equivalent conductance of saturated ${\mathrm{BaSO}}_4$ solution is 400 ${\mathrm{ohm}}^{-1} {\mathrm{cm}}^{2 }{\mathrm{equivalent}}^{-1}$ and it's specific conductance is $8 \times {10}^{-5} {\mathrm{ohm}}^{-1} {\mathrm{cm}}^{-1}$; hence solubility product ${\mathrm{K}}_{\mathrm{sp}} \mathrm{of} {\mathrm{BaSO}}_4$ is : 

1.  $4 \times {10}^{-8} {\mathrm{M}}^2$

2.  $1 \times {10}^{-8} {\mathrm{M}}^2$

3.  $2 \times {10}^{-4} {\mathrm{M}}^2$

4.  $1 \times {10}^{-4} {\mathrm{M}}^2$

Aluminium oxide may be electrolysed at 1000 C to furnish aluminium metal (Atomic mass = 27 amu; 1 Faraday = 96,500 Coulombs). The cathode reaction is: ${\mathrm{Al}}^{3+}+3{\mathrm{e}}^{-}\to \mathrm{Al}$

To prepare 5.12 kg of aluminium metal by this method, would require : 

1. $5.49\times {10}^{\mathrm{7 }}C$ of electricity 

2. $1.83\times {10}^{\mathrm{7 }}C$ of electricity

3. $5.49\times {10}^{\mathrm{4 }}C$ of electricity 

4. $5.49\times {10}^{\mathrm{9 }}C$ of electricity

For a reaction A(s) + 2B⁺ $\to$ A²⁺ + 2B(s) ;  K_C has been found to be 10¹². The ${\mathrm{E}}_{\mathrm{cell}}^{°}$ is :