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:

1.	Y will oxidize X but not Z.	2.	Y will oxidize Z but not X.
3.	Y will oxidize both X and Z.	4.	Y will reduce both X and Z.

$∆\mathrm{G}°$ 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.

At $25°\mathrm{C}$, the standard emf of a cell reaction involving a two-electron exchange is found to be 0.295 V. The equilibrium constant of the reaction is approximately : 

1. $9.50\times {10}^9$

2. $9.51\times {10}^7$

3. 10

4. $1$ $\mathrm{x}$ ${10}^{10}$

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 : 

Pairs of reactants (R) and product (P) are given below. The pair which can be used in a fuel cell is :

1. $R\mathit{=}{H}_{\mathit{2}\mathit{\left(}\mathit{g}\mathit{\right)}}\mathit{,}{O}_{\mathit{2}\mathit{\left(}\mathit{g}\mathit{\right)}}\mathit{;}P\mathit{=}{H}_{\mathit{2}}{O}_{\mathit{2}\mathit{\left(}\mathit{g}\mathit{\right)}}$

2. $\mathrm{R}={\mathrm{H}}_{2\left(\mathrm{g}\right)},{\mathrm{O}}_{2\left(\mathrm{g}\right)};\mathrm{P}={\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{\ell }\right)}$

3. $\mathrm{R}={\mathrm{H}}_{2\left(\mathrm{g}\right)},{\mathrm{O}}_{2\left(\mathrm{g}\right)},{\mathrm{Cl}}_{2\left(\mathrm{g}\right)};\mathrm{P}={\mathrm{HClO}}_{4\left(\mathrm{aq}\right)}$

4. $\mathrm{R}={\mathrm{H}}_{2\left(\mathrm{g}\right)},{\mathrm{N}}_{2\left(\mathrm{g}\right)};\mathrm{P}={\mathrm{NH}}_{3\left(\mathrm{aq}\right)}$

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

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 negative sign in the expression E^o_Zn²⁺/Zn = -0.76 V indicates : 

1. Metal reactivity increases.

2. Metal reactivity decreases.

3. No effect on metal.

4. None of the above.