A resistance of 50 is registered when two electrodes are suspended into a beaker containing a dilute solution of a strong electrolyte such that exactly half of the them are submerged into solution. If the solution is diluted by adding pure water (negligible conductivity) so as to just completely submerge the electrodes, the new resistance offered by the solution would be
(1) \(50 \Omega\)
(2) \(100 \Omega\)
(3) \(25 \Omega\)
(4) \(200 \Omega\)
The magnitude of the emf of the following cell \(at ~25^{\circ} \mathrm{C} \) is:
\(P t\left(H_2\right) \mid 0.1 M\text{ acetic acid+ 0.1 M sodium acetate} ~\\\| ~0.1 \mathrm{M}\text{ formic acid + 0.1 M}\text{ sodium formate} \mid P t\left(H_2\right) \)
\( \left(\text { Use } K_{a(\text { acetic acid })}=1.8 \times 10^{-5} \text { and } K_{a(\text { Formic acid })} \\=2.1 \times 10^{-4}\right)\)
1. 0.0315 volt
2. 0.0629 volt
3. 0.0455 volt
4. 0.0545 volt
Following cell has EMF 0.7995 V
.
If we add enough KCl to the Ag cell so that the final is 1M. Now the measured emf of the cell is 0.222 V. The of AgCl would be :
A flashlight cell has the cathodic reaction
If the flashlight cell is to give out 4.825 mA, how long could it run if initially 8.7 g of the
limiting reagent is present? [Mn = 55, O = 16]
(1)
(2)
(3)
(4)
An excess of granular zinc was added to 500 mL of 1M nickel nitrate till equilibrium was established. Find out the concentration of nickel at the equilibrium, if the standard electrode potential of are – 0.75 and – 0.24 V, respectively.
(1) 5.56 x 10-18 M
(2) 0.56 x 10-18 M
(3) 2.16 x 10-18 M
(4) 4.12 x 10-18 M
Calculate the cell EMF in mV for
If values are at
(1) 456 mV
(2) 654 mV
(3) 546 mV
(4) None of these
A current of 0.1A was passed for 2hr through a solution of cuprocyanide and 0.3745 g of
copper was deposited on the cathode. Calculate the current efficiency for the copper
deposition. (Cu – 63.5)
1. 79%
2. 39.5%
3. 63.25%
4. 63.5%
In acidic medium is an oxidising agent. ion concentration is doubled, electrode potential of the half cell will :
(1) Increase by 28.46 mV
(2) Decrease by 28.46 mV
(3) Increase by 14.23 mV
(4) Decrease by 142.30 mV
Calculate the EMF of the cell at 298 K
(1) – 0.604 V
(2) – 0.04 V
(3) 1.048 V
(4) emf depends on x and cannot be determined unless value of x is given
Assertion : Fluorine cannot be prepared from fluorides by chemical oxidation.
Reason : Fluorine is the strongest oxidizing agent due to its highly positive standard potential.