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.

The unit of specific conductance is:

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

The specific conductance of a 0.1 M KCl solution at 23$°\mathrm{C}$ is 0.012 ${\mathrm{\Omega }}^{-1}{\mathrm{cm}}^{-1}$.
The resistance of the cell containing the solution at the same temperature was found to be 55 $\mathrm{\Omega }$. The cell constant will be:

1. 0.142 cm^–1
2. 0.66 cm^–1
3. 0.918 cm^–1
4. 1.12 cm^–1

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

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

Limiting molar conductivities, for the given solutions, are:

\(\lambda_{m}^{0} \left(\right. H_{2} S O_{4} \left.\right) = x\) \(S  c m^{2}\) \(m o l^{- 1}\)

\(\lambda_{m}^{0} \left(\right. K_{2} S O_{4} \left.\right) = y\) \(S  c m^{2}\) \(m o l^{- 1}\)

\(\lambda_{m}^{0} \left(\right. C H_{3} C O O K \left.\right) = z\) \(S  c m^{2}\) \(m o l^{- 1}\)

From the data given above, it can be concluded that \(\lambda_m^0 \) in (\(S\ cm^2\ mol^{-1}\)) for CH₃COOH will be:

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:-

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 

Calculate the emf of the given cell: 

Zn(s) | Zn⁺² (0.1M) || Sn⁺² (0.001M) | Sn(s)

(Given ${\mathrm{E}}_{{\mathrm{Zn}}^{+2}/\mathrm{Zn}}^{\mathrm{o}}=-0.76$ $\mathrm{V},$ ${\mathrm{E}}_{{\mathrm{Sn}}^{2+}/\mathrm{Sn}}^{\mathrm{o}}=-0.14$ $\mathrm{V)}$

1. 0.62 V

2. 0.56 V

3. 1.12 V

4. 0.31 V