A fuel cell develops an electrical potential from the combustion of butane at 1 bar and 298 K 

C₄H₁₀(g) + 6.5O₂(g) → 4 CO₂(g) + 5H₂O(l); E^o  of a cell is: 

(Given $∆G^o$ $=-\mathrm{2746 }kJ/mole$ )

1. 4.74 V

2. 0.547 V

3. 4.37 V

4. 1.09 V

A gas X at 1 atm is bubbled through a solution containing a mixture of 1 M Y^- and 1 M Z^- at 25 ^oC . If the reduction potential of Z > Y > X, then:

The standard Emf of a galvanic cell involving cell reaction with n=2 is found to be 0.295 V at 25$°$C. The equilibrium constant of the reaction would be : 
(Given F=96500 C mol^-1; R=8.314 JK^-1 mol^-1)

(1) 2.0$\times$10¹¹

(2) 4.0$\times$10¹²

(3) 1.0$\times$10²

(4) 1.0$\times$10¹⁰

$∆\mathrm{G}°$ for the reaction given below is: 

$\frac{1}{2}\underset{}{\mathrm{A}\left(\mathrm{g}\right) }+\frac{3}{2}\underset{}{\mathrm{B}\left(\mathrm{g}\right) }\rightleftharpoons \underset{}{\mathrm{C}\left(\mathrm{g}\right)}$

(${\mathrm{K}}_{\mathrm{eq}}=826 {\mathrm{atm}}^{-1} \mathrm{at} 298 \mathrm{K}$)

1. -8.32 kJ

2. 8.32 kJ

3. 16.64 kJ

4. -16.64 kJ

For the electrolysis of CuSO₄ solution the correct option is: 

1. Cathode reaction: Cu²⁺ + 2e^- $\to$ Cu using Cu electrode.

2. Anode reaction: Cu $\to$ Cu⁺ + e^- using Cu electrode.

3. Cathode reaction: 2H⁺ + 2e^- $\to$H₂ using Pt electrode.

4. Anode reaction: Cu $\to$ Cu²⁺ + 2e^- using Pt electrode.

Molten sodium chloride conducts electricity due to the presence of: 

1. Free ions.

2. Free molecules.

3. Free electrons.

4. Atoms of sodium and chlorine.

One liter of 0.5 M KCl solution is electrolyzed for one minute in a current of 1.608 mA. Considering 100 % efficiency, the pH of the resulting solution will be : 

The reaction from following having top position in EMF series (Std.red. potential) according to their electrode potential at 298 K is: 

1. ${\mathrm{Mg}}^{2+}+2{\mathrm{e}}^{-}\to {\mathrm{Mg}}_{\left(\mathrm{s}\right)}$

2. ${\mathrm{Fe}}^{2+}+2{\mathrm{e}}^{-}\to {\mathrm{Fe}}_{\left(\mathrm{s}\right)}$

3. ${\mathrm{Au}}^{3+}+3{\mathrm{e}}^{-}\to {\mathrm{Au}}_{\left(\mathrm{s}\right)}$

4. ${\mathrm{K}}^{+}+{\mathrm{le}}^{-}\to {\mathrm{K}}_{\left(\mathrm{s}\right)}$

On electrolysis of dilute sulphuric acid using Platinum (Pt) electrode, the product obtained at the anode will be:

1. Oxygen gas

2. ${\mathrm{H}}_2\mathrm{S}$ gas

3. ${\mathrm{SO}}_2$ gas

4. Hydrogen gas 

The number of Faradays (F) required to produce 20 g of calcium from molten CaCl₂ (Atomic mass of Ca = 40 g mol^–1) is:

1. 2

2. 3

3. 4

4. 1