Pratcice NEET questions from all capters from huge question bank for free. All MCQs are based on NCERT syllabus. To practice from a specific subject and chapter, select a subject below. Login to practice in a structured way with explanations, bookmarks, lists, notes etc. Click here to Login or Sign up for free.
How much time is required for the complete decomposition of 2 moles of water using a current of 2 ampere-
(B)
A certain current liberates 0.504 g of hydrogen in 2 hours. How many gram of copper can be liberated by the same current flowing for the same time in aqueous solution :
(B)
Calculate the maximum work that can be obtained from the Daniell cell given below -
Given that
(A) Cell reaction is :
Here n=2
(On the basis of reduction potential)
= + 0.34 – (– 0.76) = 1.10 V
We know that :
The metal that cannot be produced on reduction of its oxide by aluminium is :
(A)
We have taken a saturated solution of of AgBr is If mole of are added to 1 litre of this solution then the conductivity of this solution in terms of units will be
(A). The solubility of AgBr in presence of molar
Therefore
Therefore
The standard reduction potentials, , for the half reactions are as and V; the e.m.f. for the cell reaction,
(B). Since oxidation potential of Zn is higher than Fe so it will act as anode simultaneously Fe will act as cathode so EMF of cell will be Eº (ox.) of Zn + Eº (red) of Fe.
E = 0.76 + ( - 0.41) = + 0.35 V
The calomel electrode is reversible with respect to-
(D). Since half cell reaction;
Zn amalgam is prepared by electrolysis of aqueous using Hg cathode (9 gm). How much current is to be passed through solution for 1000 seconds to prepare a Zn Amalgam with 25% Zn by wt. (Zn = 65.4)
C). Let x gm of Zn deposit on 9 gm of Hg
% of Zn in Amalgam
Eq. of Zn=
The standard oxidation potentials of and respectively. The standard electrode potential of would be :
(B). Reactions
Adding, we get
Acidified water is electrolysed using an inert electrode. The volume of gases liberated at STP is 0.168L. The quantity of charge passed through the acidified water would be:
(C)
The amount of electricity passed = 0.01F = 965C
(B). Since EMF of the cell is negative i.e. Free energy change will be positive so cell reaction will not be feasible.
How much charge should be supplied to a cell for the electrolytic production of 245 gm from if the anode efficiency for the required reaction is 60%?
(A)
Number of equivalents of
No. of Faradays
During an electrolysis of conc. , perdisulphuric acid form in equimolar amount. The amount of that will form simultaneously will be
(A) Anode
Cathode
Hence ratio of and is 1 : 3.
The specific conductivity of solution depends upon :
(C). Specific conductance is the conductance per c.c. solution
Value of for in water at from the following data
Conc. (mol/It) 0.25 1
260 250
(A).