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:-

1. ${\mathrm{BrO}}_3^{-}$

2. ${\mathrm{BrO}}_4^{-}$

3. ${\mathrm{Br}}_2$

4. HBrO

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 

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.

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

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

The metal that cannot be produced upon reduction of its oxide by aluminium is :

The unit of specific conductance 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.