In the electrochemical cell: 
\(\mathrm{Z n   \left|\right. Z n S O_{4}   \left(\right. 0 . 01   M \left.\right)   \left|\right. \left|\right.   C u S O_{4} \left(\right. 1 . 0   M \left.\right)   \left|\right.   C u}, \)$$
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. \(\mathrm{E_{1} < E_{2}}\)

2. \(\mathrm{E_{1} > E_{2}}\)

3. \(\mathrm{E_{2} = 0 \neq E_{1}}\)

4. \(\mathrm{E_{1} = E_{2}}\)

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.

Consider the following cell reaction 

2Fe(s) + ${\mathrm{O}}_2$(g) + 4${\mathrm{H}}^{+}$(aq) $\to$ 2${\mathrm{Fe}}^{2+}$(aq) + 2${\mathrm{H}}_2\mathrm{O}$(l)

E° = 1.67 V, At [${\mathrm{Fe}}^{2+}$] = 10${}^{-3}$ M, ${\mathrm{P}}_{{\mathrm{O}}_2}$ = 0.1 atm and pH = 3, the cell potential at 25 °C is : 

1. 1.27 V

2. 1.77 V

3. 1.87 V

4. 1.57 V

\(Cu(s)|Cu^{+2}(10^{-3} \ M) \ || \ Ag^{+}(10^{-5} \ M)|Ag(s)\)
if \(E_{Cu^{+2}/Cu}^{o} \ = \ +0.34 \ V\), and \(E_{Ag^{+}/Ag}^{o} \ = \ +0.80 \ V\)
E_cell will be: 

1. 0.46 V
2. \(0.46-\frac{RT}{2F}ln10^{7}\)
3. \(0.46+\frac{RT}{2F}ln10^{7}\)
4. \(0.46-\frac{RT}{2F}ln10^{2}\)

Mg(s) + 2Ag⁺(0.0001M) → Mg²⁺(0.130M) + 2Ag(s)

If  E^Ɵ(cell) for the above mentioned cell is 3.17 V, then E(cell) value will be-

(log 13=1.1)

1. 2.87 V
2. 3.08 V
3. 2.96 V
4. 2.68 V