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


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30 Questions MCQ Test - Revisal Problems (Past 13 Year) JEE Main (Electrochemistry)

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

Detailed Solution for Revisal Problems (Past 13 Year) JEE Main (Electrochemistry) - Question 1

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

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

Detailed Solution for Revisal Problems (Past 13 Year) JEE Main (Electrochemistry) - Question 2

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.

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Revisal Problems (Past 13 Year) JEE Main (Electrochemistry) - Question 3

In a galvanic cell,the salt-bridge

Detailed Solution for Revisal Problems (Past 13 Year) JEE Main (Electrochemistry) - Question 3

In a galvanic cell, for example

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

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

Detailed Solution for Revisal Problems (Past 13 Year) JEE Main (Electrochemistry) - Question 4

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

By Ostwald’s dilution law

Revisal Problems (Past 13 Year) JEE Main (Electrochemistry) - Question 5

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

Detailed Solution for Revisal Problems (Past 13 Year) JEE Main (Electrochemistry) - Question 5


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

Revisal Problems (Past 13 Year) JEE Main (Electrochemistry) - Question 6

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

Detailed Solution for Revisal Problems (Past 13 Year) JEE Main (Electrochemistry) - Question 6

(a)





Revisal Problems (Past 13 Year) JEE Main (Electrochemistry) - Question 7

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

Detailed Solution for Revisal Problems (Past 13 Year) JEE Main (Electrochemistry) - Question 7

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

Revisal Problems (Past 13 Year) JEE Main (Electrochemistry) - Question 8

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

Detailed Solution for Revisal Problems (Past 13 Year) JEE Main (Electrochemistry) - Question 8





Revisal Problems (Past 13 Year) JEE Main (Electrochemistry) - Question 9

At 298 K,given  

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

Detailed Solution for Revisal Problems (Past 13 Year) JEE Main (Electrochemistry) - Question 9


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

Revisal Problems (Past 13 Year) JEE Main (Electrochemistry) - Question 10

For the half-cell

 

At pH = 2,electrode potential is   

Detailed Solution for Revisal Problems (Past 13 Year) JEE Main (Electrochemistry) - Question 10

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

Revisal Problems (Past 13 Year) JEE Main (Electrochemistry) - Question 11

The hydrazine fuel cell is based on the reaction,

Detailed Solution for Revisal Problems (Past 13 Year) JEE Main (Electrochemistry) - Question 11


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

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

Detailed Solution for Revisal Problems (Past 13 Year) JEE Main (Electrochemistry) - Question 12



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

Detailed Solution for Revisal Problems (Past 13 Year) JEE Main (Electrochemistry) - Question 13

For the reduction potential 



Revisal Problems (Past 13 Year) JEE Main (Electrochemistry) - Question 14

Consider the half-cell reactions :

 

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

Detailed Solution for Revisal Problems (Past 13 Year) JEE Main (Electrochemistry) - Question 14

 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+

Revisal Problems (Past 13 Year) JEE Main (Electrochemistry) - Question 15

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

Predict which exists in aqueous solution? 

Detailed Solution for Revisal Problems (Past 13 Year) JEE Main (Electrochemistry) - Question 15


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

Revisal Problems (Past 13 Year) JEE Main (Electrochemistry) - Question 16

For the half-cell

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

Detailed Solution for Revisal Problems (Past 13 Year) JEE Main (Electrochemistry) - Question 16

Revisal Problems (Past 13 Year) JEE Main (Electrochemistry) - Question 17

Consider the following half-cell reactions :

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

Detailed Solution for Revisal Problems (Past 13 Year) JEE Main (Electrochemistry) - Question 17

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

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

Detailed Solution for Revisal Problems (Past 13 Year) JEE Main (Electrochemistry) - Question 18

Oxidation at anode

Reduction at cathode

Complete cell is


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

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

Detailed Solution for Revisal Problems (Past 13 Year) JEE Main (Electrochemistry) - Question 19

For spontaneous reaction , E°cell > 0.



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

Detailed Solution for Revisal Problems (Past 13 Year) JEE Main (Electrochemistry) - Question 20



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

Revisal Problems (Past 13 Year) JEE Main (Electrochemistry) - Question 21

Given the data at 298 K,

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

Detailed Solution for Revisal Problems (Past 13 Year) JEE Main (Electrochemistry) - Question 21

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


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

Detailed Solution for Revisal Problems (Past 13 Year) JEE Main (Electrochemistry) - Question 22


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

Detailed Solution for Revisal Problems (Past 13 Year) JEE Main (Electrochemistry) - Question 23

In saturated solution of sparingly soluble salt,


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

Detailed Solution for Revisal Problems (Past 13 Year) JEE Main (Electrochemistry) - Question 24


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

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

Detailed Solution for Revisal Problems (Past 13 Year) JEE Main (Electrochemistry) - Question 25

By Kohlrausch's law at infinite dilution,

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

Detailed Solution for Revisal Problems (Past 13 Year) JEE Main (Electrochemistry) - Question 26


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.

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

Detailed Solution for Revisal Problems (Past 13 Year) JEE Main (Electrochemistry) - Question 27


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

Revisal Problems (Past 13 Year) JEE Main (Electrochemistry) - Question 28

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

Detailed Solution for Revisal Problems (Past 13 Year) JEE Main (Electrochemistry) - Question 28


At anode and cathode both, H2 is evolved.

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

Detailed Solution for Revisal Problems (Past 13 Year) JEE Main (Electrochemistry) - Question 29

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

Revisal Problems (Past 13 Year) JEE Main (Electrochemistry) - Question 30

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

Detailed Solution for Revisal Problems (Past 13 Year) JEE Main (Electrochemistry) - Question 30

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

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