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

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 :

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

The specific conductance of a 0.1 M KCl solution at 23$°\mathrm{C}$ is 0.012 ${\mathrm{\Omega }}^{-1}{\mathrm{cm}}^{-1}$.
The resistance of the cell containing the solution at the same temperature
was found to be 55 $\mathrm{\Omega }$. The cell constant will be:

1. 0.142 cm^-1
2. 0.66 cm^-1
3. 0.918 cm^-1
4. 1.12 cm^-1

For the cell, Ti/Ti⁺(0.001M)||Cu²⁺(0.1M)|Cu, ${\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}$ $$at

25 $°$C is 0.83 V. E_cell can be increased :

1. By increasing [Cu²⁺]

2. By increasing [Ti⁺]

3. By decreasing [Cu²⁺]

4. None of the above.

The unit of specific conductance is:

The metal that cannot be produced upon reduction of its oxide by aluminium is :

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

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

\(E_{cell}^{o} \ = \ 0.24 \ V\) 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. $46.32$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$

If ${\mathrm{E}}_{{\mathrm{Fe}}^{2+}/\mathrm{Fe}}^{\mathrm{o}}$ = -0.441 V and  ${\mathrm{E}}_{{\mathrm{Fe}}^{3+}/{\mathrm{Fe}}^{2+}}^{\mathrm{o}}$ = 0.771 V, the standard emf of the reaction: 

Fe + 2Fe³⁺→ 3Fe²⁺ will be: