Revisal Problems (Past 13 Year) JEE Main (Electrochemistry)


40 Questions MCQ Test Chemistry for JEE Advanced | Revisal Problems (Past 13 Year) JEE Main (Electrochemistry)


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This mock test of Revisal Problems (Past 13 Year) JEE Main (Electrochemistry) for JEE helps you for every JEE entrance exam. This contains 40 Multiple Choice Questions for JEE Revisal Problems (Past 13 Year) JEE Main (Electrochemistry) (mcq) to study with solutions a complete question bank. The solved questions answers in this Revisal Problems (Past 13 Year) JEE Main (Electrochemistry) quiz give you a good mix of easy questions and tough questions. JEE students definitely take this Revisal Problems (Past 13 Year) JEE Main (Electrochemistry) exercise for a better result in the exam. You can find other Revisal Problems (Past 13 Year) JEE Main (Electrochemistry) extra questions, long questions & short questions for JEE on EduRev as well by searching above.
QUESTION: 1

Only One Option Correct Type

This section contains 34 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE is correct.

Q.

Some of the batteries are rechargeable :

I. Dry cell,
II. Lead-storage, 
III. Nickel-Cadmium,
IV. Lithium,
V. Fuel cell

Select the batteries which can be recharged 

Solution:

II. Lead-storage, III. Nickel- Cadmium and IV. Lithium batteries are rechargeable.

QUESTION: 2

Consider the following facts about the electrolysis of CuSO4 solution:

I. Copper will deposit at cathode.
II. Copper will deposit at anode.
III. Oxygen will be released at anode.
IV. Copper will dissolve at anode.

Thus, with platinum electrodes observed facts are 

Solution:

With platinum electrodes

With copper electrodes

Note When inert electrodes are used, principle is used in electroplating. Blue colour of CuSO4 gradually fades. With copper electrodes, Cu dissolves from the anode and deposits at cathode. Concentration of copper solution remains unchanged. This makes the basis of purification of metals called refining.

QUESTION: 3

In a galvanic cell,the salt-bridge

Solution:

In a galvanic cell, for example

two half-cells are joined by salt-bridge. It does not take part in chemical reaction

QUESTION: 4

For aqueous NH3(NH4OH), at 298 K
Molar conductance at infinite dilution = 2.38 x 10-2 S m2 mol-1.
Kb (ionisation constant) = 1.6x 10-5

What is the molar conductance at 0.01 M NH4OH?

Solution:

 = molar condu cta nce at given concentration.
= molar conductance at infinite dilution (= zero concentration)
Thus,

By Ostwald’s dilution law

QUESTION: 5

Electrode potential for Mg electrode varies according to the equation at 298 K

Solution:


This represents a straight line, y = c + mx
Thus, y increases as x increases with intercept on Y-axis. When [Mg2+] = 1 M

QUESTION: 6

(Ecell - E°cell) has maximum value for spontaneous reaction in

Solution:

(a)





QUESTION: 7

For strong electrolytes,the values of molar conductivity at infinite dilution are given below:

Solution:

Ba(OH)2 = BaCI2 + 2NaOH - 2NaCI

QUESTION: 8

The reduction potential of hydrogen half-cell will be negative if

Solution:





QUESTION: 9

At 298 K,given  

Thus, stability constant of the complex [Cu(NH3)4]2+ is   

Solution:


When complex is formed at equilibrium,
Keq = K (stability constant)

QUESTION: 10

For the half-cell

 

At pH = 2,electrode potential is   

Solution:

This is a quinhydrone electrode in which [A] = [S] where,

QUESTION: 11

The hydrazine fuel cell is based on the reaction,

Solution:


Change in oxidation number (ON) = 4
ΔG° = - nF E°cell = - 4 x 96500 x 1.56
= -602160 J 
= - 602.160 kJ
ΔG° = 2 ΔG°(H2O) - ΔG°(N2H4)
- 602.160 = - 2 x 237.2 - ΔG° (N2H4)
ΔG° (N2H4) = - 474.40 + 602.16
= 127.76 kJ mol-1

QUESTION: 12

The reduction of oxygen gas in 1 M aqueous acid is represented by

The potential for this reaction at physiological pH where [H+] = 1 x 10-7 M,and the partial pressure of Ois 1 atm,will be closest to which value?  

Solution:



QUESTION: 13

Ksp of AI(OH)3 = 1.0 x 10-33 and E°A|3+/A| = -1.66 V

Reduction potential of the above couple at pH = 14 is

Solution:

For the reduction potential 



QUESTION: 14

Consider the half-cell reactions :

 

If Fe2+ and H2O2 are mixed ,Which reaction is more likely?  

Solution:

 If H2O2 is a reducing agent,

Since, E°cell < 0, hence
H2O2 does not reduce Fe2+ to Fe.
if H2O2 is an oxidising agent,

Since, E°cell > 0, hence Fe2+ changes to Fe3+

QUESTION: 15

If for the half-cell reactions, E° values are known

Predict which exists in aqueous solution? 

Solution:


cell > 0, thus Cu+ disproportionates to Cu2+ and Cu. Thus, copper (li) sulphate exists. Copper (I) sulphate is unstable.

QUESTION: 16

For the half-cell

Hence, E° for the disproportionation reaction 2Cu+→ Cu2+ + Cu would be

Solution:

QUESTION: 17

Consider the following half-cell reactions :

What combination of two half-cells would result in a cell with the largest potential?   

Solution:

D2+/D = - 1.12 V
Its value being most negative, D is oxidised to D2+ most easily and is thus is the best reducing agent. Hence, it is anodic part.

Its value being most positive, A is reduced to A- by D

QUESTION: 18

In a cell that utilises the reaction
Mg(s)+ 2H+(aq) → Mg2+(aq)+ H2(g)

addition of dil. H2SO4 to cathode compartment will

Solution:

Oxidation at anode

Reduction at cathode

Complete cell is


On adding H+, [H+] increases, Q decreases.
Equilibrium is displaced towards right and Ecell increased.

QUESTION: 19

The standard reduction potentials for Zn2+ /Zn, Ni2+/Ni and Fe2+ /Fe are -0.76,-0.23 V and -0.44 V respectively.The reaction

 

will be spontaneous when     

Solution:

For spontaneous reaction , E°cell > 0.



QUESTION: 20

The cell Zn| Zn2+(1 M) ||Cu2+(1 M) | Cu, E°cell = 1.10 V  was allowed to be discharged at 298K. Thus at a time when,Ecell = 0.8634 V,log [Zn2+] /[Cu2+] is  

Solution:



During discharging, [Zn2+] increases, [Cu2+] decreases. Thus, [Zn2+] / [Cu2+] increases. Ecell  decreases.

QUESTION: 21

Given the data at 298 K,

Thus, solubility product of Agl(s) is (given, 

Solution:

When Agl(s). a sparingly soluble salt is dissolved, following equilibrium is set up


QUESTION: 22

The conductivity of a solution of 0.001 mol dm-3 Na2SOis 2.6 x 10-2 S m-1.Given that the molar conductivity of Na+ is 5.0 x 10-3 S m2 mol-1 ,thus molar conductivity of sulphate ion is (In S m2 mol-1)

Solution:


QUESTION: 23

At 25°C,limiting equivalent conductances (in Ω-1 cmequiv-1) of BaCl2 ,H2SO4 and HCl respectively are 100,250 and 150.Specific conductance (conductivity) of saturated aqueous BaSO4 solution is 1 x 10-6 Ω-1 cm-1.Thus solubility product of BaSO4 solution is

Solution:

In saturated solution of sparingly soluble salt,


QUESTION: 24

Cell constant (C) dependent of resistance (R) between two electrodes is shown by which of the following graph in terms of logR and logC?

Solution:


is called cell constant (C) and k is called specific resistance log (R) = log k + log(C) - a straight line (y = constant + mx)
Thus, graph between log R and logC is of the type
Slope = 1 = tan 45°
OA = log K

QUESTION: 25

The limiting molar conductivities for NaCI, KBr and KCI are 126,152 and 150 S cm2 mol-1, respectively. The for NaBr is (in S cm2 mol-1

Solution:

By Kohlrausch's law at infinite dilution,

QUESTION: 26

Following solutions having equal ionisation have been provided

I. 0.01 M HCI,
II. 0.01 MH2SO4,
III. 0.01 M H3PO4

Maximum value of molar conductanceand equivalent conductanceare of  

Solution:


Molarity is same hence, equal molar conductance.

Normality = Basicity x Normality
0.01 M HCI = 0.01 N HCI 
0.01 M H2SO4 = 0.02 N H2SO4
0.01 M H3PO4 = 0.03 N H3PO4 

Thus, Aequiv of HCI is maximum.

QUESTION: 27

The resistivity of aluminium is 2.824 x 10-8 Ωm. Thus,conductance across a piece of aluminium wire,that is 2.0 mm in diameter and 1.00 m long is (assume current= 1.25 A)

Solution:


Area of cross-section and resistance (R) is related to the resistivity as,
(Specific resistance), the length and area by

QUESTION: 28

Electrolysis products in gaseous state at cathode and anode are same when reactant is

Solution:


At anode and cathode both, H2 is evolved.

QUESTION: 29

Aluminium oxide may be electrolysed at 1270 K to form Al metal (atomic mass = 27 u, 1F = 96500 C)  

To prepare 5.12 kg of Al metal by this method,it would require

Solution:

By Faraday’s first law,
w(amount) = zit = zQ
where, it = Q = quantity of electricity
z = electrochemical equivalent 

QUESTION: 30

For an electrosynthesis of a substance, sign of ΔG and Ecell should be

Solution:

In the electrosynthesis (as electrolysis), ions are discharged at the electrodes thus, reverse reaction takes place.

QUESTION: 31

96500 coulombs of electricity deposits 108 g of silver.If electronic charge is 1.6 x 10-19 C,the Avogadro's number is

Solution:


Number of Ag atoms in 1 mole Ag = N0 atoms
Total charge carried by N0 atoms = N0 x 1.6 x 10-19

QUESTION: 32

Cost of electricity for the production of XLH2 gas at STP at cathode is rupess x then cost of electricity for the production of XLO2 gas at STP at the anode is (assume 1 mole of electrons as one unit of electricity).

Solution:


QUESTION: 33

Given below are the half-cell reactions

Thus, E° for the reaction

will be

Solution:


Mn2+ is oxidised as well as reduced. Thus, it is a disproportionation reaction.

Since, E° < 0, hence reaction will not occur.

QUESTION: 34

The formation of rust on the surface of iron occurs through the reaction(s)

Solution:

Fe is active metal and is oxidised toFe2+; Fe2+ is dissolved in O2 in acidic medium (or alkaline medium) forming Fe2O3(s)- rust.

QUESTION: 35

Statement Type

This section is based on Statement I and Statement II. Select the correct anser from the codes given below

Statement II : Spontaneous change in a concentration cell always occurs in the direction in which the concentration solution becomes more dilute and the dilute solution becomes more concentrated.   

Solution:



Ecell > 0 and spontaneous. Thus, Statement I is correct.
Zn2+ ion flows from 0.1 M (concentrated) to 0.01 M (dilute). Thus, concentrated solution becomes more dilute and dilute becomes more concentrated. Thus,Statements I and II both are correct and Statement II is the correct explanation of Statement I.

QUESTION: 36

Statement I : Iron can be protected from rusting by connecting it to the more active metals, as zinc.

Statement II : It is a type of galvanisation.

Solution:

Iron materials as electricity poles can be protected from rusting by connecting it to active metals as zinc. This is a type of cathodic protection and zinc which is oxidised is the sacrified anode. It is not the galvanisation. Thus, Statement I is correct but Statement II is incorrect.

QUESTION: 37

Statement I : In an electrochemical cell,two half-cells are connected with a salt-bridge that allows cations and anions to move between two half-cells

Statement II : Electrolyte used in salt-bridge can be NaCl or KCl with silver electrodes.

Solution:

Salt-bridge connects two half-cells and it allows cations and anions to move between two half-cells.
Electrolyte is generally NaNO3, NH4NO3, KNO3. NaCI, KCI cannot be used with silver electrodes as it react with silver forming AgCI. Thus, Statement I is correct and Statement II is incorrect.

QUESTION: 38

Statement I CuSO4 solution can be stored in a vessel made of zinc.

Statement II

Solution:


Since, E°cell > 0, thus CuSO4 is reduced to Cu.
Thus, CuSO4 cannot be stored in a vessel made of zinc. Thus, Statement I is incorrect and Statement II is correct (Standard E°red values)

QUESTION: 39

Statement I :  If

 

then,

Statement II : ΔG3 = | ΔG1|  + | ΔG2|

Solution:

If different number of electrons are involved,

Hence, Statement I is incorrect and Statement II is correct.

QUESTION: 40

Only One Option Correct Type

This section contains a multiple choice question. A question has four choices (a), (b), (c) and (d), out of which ONLY ONE is correct

9.65 C of electric current is passed through fused anhydrous MgCl2.The magnesium metal thus obtained is completely converted into a Grignard's reagent(R-Mg-X).The number of moles of Grignard's reagent obtained is

Solution: