Find the emf of the cell in which the following reaction takes place at 298 K:
\(\mathrm{Ni}(\mathrm{s})+2 \mathrm{Ag}^{+}(0.001 \mathrm{M}) \rightarrow \mathrm{Ni}^{2+}(0.001 \mathrm{M})+2 \mathrm{Ag}(\mathrm{s}) \)
\( \small{\text { (Given that } \mathrm{E}_{\text {cell }}^{\circ}=10.5 \mathrm{~V}, \frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059 \text { at } \ 298 \mathrm{~K})} \)
1. 1.05 V
2. 1.0385 V 
3. 1.385 V
4. 0.9615 V 

In the electrochemical cell: 
  $Zn | ZnS{O}_4 (0.01 M) || CuS{O}_4(1.0 M) | C\mathrm{u, }$the emf of this Daniel cell is E₁. When the concentration of ZnSO₄ is changed to 1.0 M and that of CuSO₄ is changed to 0.01 M, the emf changes to E₂. From the following, which one is the relationship between E₁ and E₂ ? 
(Given, \(\frac{RT}{F}\) = 0.059)

1. $E_1<E_2$

2. $E_1>E_2$

3. $E_2=0\ne E_1$

4. $E_1=E_2$

The pressure of H₂ required to make the potential of H₂ - electrode zero in pure water at 298 K is:

For a given reaction, 
Cu(s) + 2Ag⁺ (aq) → Cu²⁺(aq)+2Ag(s); 
E⁰=0.46 V at 298 K . The equilibrium constant will be :

1. $2.4\times {10}^{10}$

2. $2.0\times {10}^{10}$

3. $4.0\times {10}^{10}$

4. $4.0\times {10}^{15}$

A hypothetical electrochemical cell is shown below.
A|A⁺(x M) || B⁺(y M)|B
The Emf measured is +0.20 V. The cell reaction is:

1. A⁺ + B → A + B⁺

2.  A⁺ + e^- → A ; B⁺ + e^- → B

3. The cell reaction cannot be predicted.

4. A + B⁺ → A⁺ + B