The specific conductance of a 0.1 M KCl solution at 23 is 0.012 .
The resistance of the cell containing the solution at the same temperature
was found to be 55 . 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
For the cell, Ti/Ti+(0.001M)||Cu2+(0.1M)|Cu, at
25 C is 0.83 V. Ecell 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:
1. | ohm-1 cm-1 | 2. | ohm cm |
3. | ohm cm-1 | 4. | ohm-1 cm |
The metal that cannot be produced upon reduction of its oxide by aluminium is :
1. | K | 2. | Mn |
3. | Cr | 4. | Fe |
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) + (g) + 4(aq) 2(aq) + 2(l)
E° = 1.67 V, At [] = 10 M, = 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 :
From the data given above, it can be concluded that \(\lambda_m^0 \) in (\(S\ cm^2\ mol^{-1}\)) for CH3COOH will be :
1. \(\mathrm{x-y+2z}\)
2. \(\mathrm{x+y+z}\)
3. \(\mathrm{x-y+z}\)
4. \(\mathrm{{(x-y) \over 2}+z}\)
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.
2.
3.
4. HBrO
On electrolysis of dilute sulphuric acid using Platinum (Pt) electrode, the product obtained at the anode will be:
1. Oxygen gas
2. gas
3. gas
4. Hydrogen gas
Calculate the Emf of the given cell:
Zn(s) | Zn+2 (0.1M) || Sn+2 (0.001M) | Sn(s)
(Given
1. 0.62 V
2. 0.56 V
3. 1.12 V
4. 0.31 V