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.
Assertion : : Sodium reacts with water at room temperature, whereas magnesium decomposes only hot water.
Reason : Sodium is more electropositive than magnesium.
Assertion : Silver nitrate solution becomes bluish when copper rod is placed in it.
Reason : Copper is more electropositive than silver.
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
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%
Calculate the cell EMF in mV for
If values are at
(1) 456 mV
(2) 654 mV
(3) 546 mV
(4) None of these
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
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)
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 :