The electrode potential for Mg electrode varies according to the equation

\(E_{Mg^{2+}/Mg}\ = \ E_{Mg^{2+}/Mg}^{o} \ - \ \frac{0.059}{2}log\frac{1}{[Mg^{2+}]}\) 

The graph of E_{Mg²⁺ / Mg} vs log [Mg²⁺] among the following is:

1.		2.
3.		4.

In the electrochemical cell:

Zn|ZnSO₄(0.01 M) || CuSO₄(1.0M),Cu, 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. The relationship between E₁ and E₂ is : 
( Given, $\frac{RT}{\mathrm{F }}$= 0.059)

1. E₁ = E₂

2. E₁ < E₂

3. E₁ > E₂

4. E₂ = 0 $\ne$E₁

The change in reduction potential of a hydrogen electrode when its solution initialy at pH = 0 is neutralised to pH = 7, is a/an-

The voltage of the cell given below increases with: 

Cell: Sn(s) + 2Ag⁺(aq) → Sn²⁺(aq) + 2Ag(s)

1. Increase in size of the silver rod.

2. Increase in the concentration of Sn²⁺ ions.

3. Increase in the concentration of Ag⁺ ions.

4. None of the above.

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

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.

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 

The electrode potential of Cu electrode dipped in 0.025 M CuSO₄ solution at 298 K is:

(standard reduction potential of Cu = 0.34 V)

1. 0.047 V

2. 0.293 V

3. 0.35 V

4. 0.387 V