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.

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

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 

Consider the change in the oxidation state of Bromine corresponding to different emf values as shown in the diagram below:

\(\small{BrO_4^-\ \overset{1.82V}{\longrightarrow}\ BrO_3^-\ \overset{1.5V}{\longrightarrow} HBrO\ \overset{1.595V}{\longrightarrow}\ Br_2 \overset{1.0652V}{\longrightarrow}\ Br^-}\) 
Then the species undergoing disproportionation is:-

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

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

The resistance of a cell containing 0.001 M KCl solution at 298 K is 1500 Ω. The conductivity is 0.146 × 10^–3 S cm^–1. The cell constant would be-

$1.$ $0.12$ ${\mathrm{cm}}^{-1}$

$2.$ $0.56$ ${\mathrm{cm}}^{-1}$

$3.$ $0.22$ ${\mathrm{cm}}^{-1}$

$4.$ $1.36$ ${\mathrm{cm}}^{-1}$

The amount of charge required for the reduction of 1 mol of ${\mathrm{MnO}}_4^{-}$ to ${\mathrm{Mn}}^{2+}$ is -

$1.$ $4.8$ $\times$ ${10}^5$ $\mathrm{C}$
$2.$ $3.2$ $\times$ ${10}^6$ $\mathrm{C}$
$3.$ $1.8$ $\times$ ${10}^5$ $\mathrm{C}$
$4.$ $4.1$ $\times$ ${10}^4$ $\mathrm{C}$

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.