If hydrogen electrodes dipped in two solutions of pH = 4 and pH = 6 are connected by a salt bridge, the emf of the resulting cell is -

1. 0.177 V               

2. 0.3 V               

3. 0.118 V             

4. 0.104 V

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

Salt with the highest electrolytic conductivity in solution is : 

1. K₂[PtCl₆] 

2. [Co(NH₃)₃(NO₂)₃]

3. K₄[Fe(CN)₆] 

4. [Co(NH₃)₄]SO₄

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 :

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 ${\mathrm{cm}}^2$ ${\mathrm{mol}}^{-1}$]

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

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

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
 \(\mathrm{2Fe^{3+}(aq) \ + \ 2I^{-}(aq)\rightarrow 2Fe^{2+}(aq) \ + \ I_{2}(aq)}\)

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

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

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