At a particular temperature, the ratio of equivalent conductance to specific conductance of a 0.01 N NaCl solution is
105 cm3
103 cm3
10 cm3
105 cm2
The number of moles of electrons required to deposit 1gm equivalent aluminium (atomic wt. = 27) from a solution of aluminium chloride will be
3
1
4
2
At temperature of 298K, the emf of the following electrochemical cell,
Ag (s) | Ag+ (0.1 M)|| Zn2+ (0.1 M) | Zn (s) will be (Given, E°cell = -1.562 V)
-1.532 V
-1.503 V
1.532 V
-3.06 V
Assertion: Potassium and caesium are used in photoelectric cells.
Reason: Potassium and caesium emit electrons on exposure to light.
If both assertion and reason are true and reason is a correct explanation of the assertion.
If both the assertion and reason are true but the reason is not a correct explanation of the assertion.
If the assertion is true but the reason is false.
If both the assertion and reason are false.
Li occupies a higher position in the electrochemical series of metals as compared to Cu since :
the standard reduction potential of Li+/Li is lower than that of Cu2+/Cu
the standard reduction potential of Cu2+/Cu is lower than that of Li+/Li
the standard oxidation potential of Li+/Li is lower than that of Cu2+/Cu
Li is smaller in size as compared to Cu
A hydrogen gas electrode is made by dipping platinum wire in a solution of HCl of pH=10 and by passing hydrogen gas around the platinum wire at 1 atm pressure. The oxidation potential of electrode would be
0.059 V
0.59 V
0.118V
1.18V
B.
0.59 V
For hydrogen electrode, oxidation half reaction is
At 25°C molar conductance of 0.1 molar aqueous solution of ammonium hydroxide is 9.54 ohm-1cm2 mol-1 and at infinite dilution its molar conductance is 238 ohm-1cm2 mol-1. The degree of ionisation of ammonium hydroxide at the same concentration and temperature is
2.080%
20.800%
4.008%
40.800%
The quantity of electricity needed to separately electrolyze 1 M solution of ZnSO4 , AlCl3 and AgNO3 completely is in the ratio of
2 : 3 : 1
2 : 1 : 1
2 : 1 : 3
2 : 2 : 1
On passing C ampere of current for the time 't' sec through 1 L of 2 (M) CuSO4 solution (atomic weight of Cu= 63. 5), the amount 'm' of Cu (in gram) deposited on cathode will be
m = Ct/(63.5 x 96500)
m = Ct/(31.25 X 96500)
m = (C x 96500)/(31.25 x t)
m = (31.25 X C x t)/96500