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JEE Main 35 Year PYQs- Electrochemistry - Free MCQ Test with solutions


MCQ Practice Test & Solutions: Test: JEE Main 35 Year PYQs- Electrochemistry (35 Questions)

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Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 70 minutes
  • - Number of Questions: 35

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Test: JEE Main 35 Year PYQs- Electrochemistry - Question 1

Conductivity (unit Siemen’s S) is directly proportional toarea of the vessel and the concentration of the solution in itand is inversely proportional to the length of the vesselthen the unit of the constant of proportionality is                [2002]

Detailed Solution: Question 1

given :

Test: JEE Main 35 Year PYQs- Electrochemistry - Question 2

EMF of a cell in terms of reduction potential of its left andright electrodes is        [2002]

Detailed Solution: Question 2

Ecell = Reduction potential of cathode (right)
– Reduction potential of anode (left)
= Eright – Eleft.

Test: JEE Main 35 Year PYQs- Electrochemistry - Question 3

What will be the emf for the given cell [2002]
Pt | H2 (P1) | H+ (aq) | | H2 (P2) | Pt

Detailed Solution: Question 3

Oxidation half call:-

The net cell reaction

Test: JEE Main 35 Year PYQs- Electrochemistry - Question 4

Which of the following reaction is possible at anode?

Detailed Solution: Question 4

O.S. of Cr changes from +3 to +6 by loss of electrons. At anode oxidation takes place.

Test: JEE Main 35 Year PYQs- Electrochemistry - Question 5

When the sample of copper with zinc impurity is to be purified by electrolysis, the appropriate electrodes are

[2002]

Detailed Solution: Question 5

Pure metal always deposits at cathode

Test: JEE Main 35 Year PYQs- Electrochemistry - Question 6

Which of the following is a redox reaction? [2002]

Detailed Solution: Question 6

Test: JEE Main 35 Year PYQs- Electrochemistry - Question 7

For a cell reaction involving a two-electron change, thestandard e.m.f. of the cell is found to be 0.295 V at 25ºC. The equilibrium constant of the reaction at 25ºC will be               [2003

Detailed Solution: Question 7

The equilibrium constant is related to the standard emf of cell by the expression

Test: JEE Main 35 Year PYQs- Electrochemistry - Question 8

Standard reduction electrode potentials of three metals A, B& C are respectively + 0.5 V, – 3.0 V & –1.2 V. The reducingpowers of these metals are [2003]

Detailed Solution: Question 8

NOTE : The higher the negative value of reduction potential, the more is the reducing power.
Hence B > C > A.

Test: JEE Main 35 Year PYQs- Electrochemistry - Question 9

When during electrolysis of a solution of AgNO3 9650coulombs of charge pass through the electroplating bath,the mass of silver deposited on the cathode will be

Detailed Solution: Question 9

When 96500 coulomb of electricity is passed through the electroplating bath the amount of Ag deposited = 108g
∴ when 9650 coulomb of electricity is passed deposited Ag.

Test: JEE Main 35 Year PYQs- Electrochemistry - Question 10

For the redox reaction :                      
Zn(s) +Cu2+ (0.1M)→ Zn2+ (1M) + Cu(s) taking place in a cell,  is 1.10 volt. Ecell for the cell will  be                                [2003]

Detailed Solution: Question 10

Test: JEE Main 35 Year PYQs- Electrochemistry - Question 11

Several blocks of magnesium are fixed to the bottom of a ship to                   [2003]

Detailed Solution: Question 11

Magnesium, on account of its lightness, great affinity for oxygen and toughness is used in ship. Being a lighter element, magnesium makes the ship lighter when it is fixed to the bottom of the ship.

Test: JEE Main 35 Year PYQs- Electrochemistry - Question 12

In a hydrogen-oxygen fuel cell, combustion of hydrogenoccurs to               [2004]

Detailed Solution: Question 12

In H2 -O2 fuel cell, the combustion of H2 occurs to create potential difference between the two electrodes

Test: JEE Main 35 Year PYQs- Electrochemistry - Question 13

Consider the following Eº values

Under standard conditions the potential for the reaction

                               [2004]

Detailed Solution: Question 13

Fe3+ + e- → Fe2+ΔG° = -1x Fx0.77
Sn2+ + 2e- → Sn(s)ΔG° = -2 x F(-0.14)
for Sn(s) + 2Fe3+ (aq) → 2Fe2+ (aq) +Sn2+ (aq)
∴ Standard potential for the given reaction

∴  

Test: JEE Main 35 Year PYQs- Electrochemistry - Question 14

The standard e.m.f. of a cell involving one electron changeis found to be 0.591 V at 25ºC. The equilibrium constant ofthe reaction is (F = 96,500 C mol–1; R = 8.314 JK–1 mol–1)                             [2004]

Detailed Solution: Question 14

Test: JEE Main 35 Year PYQs- Electrochemistry - Question 15

The limiting molar conductivities Lº for NaCl, KBr and KClare 126, 152 and 150 S cm2 mol–1 respectively. The Lº forNaBr is [2004]

Detailed Solution: Question 15

operating (i) + (ii) - (iii)

= 126 +152 -150 = 128 S cm2 mol-1

Test: JEE Main 35 Year PYQs- Electrochemistry - Question 16

In a cell that utilises the reaction  Zn(s) + 2H+ (aq) → Zn2++ H2(g) addition of H2SOto cathode compartment, will          [2004]

Detailed Solution: Question 16

Test: JEE Main 35 Year PYQs- Electrochemistry - Question 17

The  values for Cr, Mn, Fe and Co are – 0.41, + 1.57, + 0.77 and + 1.97V respectively. For which one of these
metals the change in oxidation state from +2 to +3 is easiest?               [2004]

Detailed Solution: Question 17

The given values show that Cr has maximum oxidation potental, therefore its oxidation will be easiest. (Change the sign to get the oxidation values)

Test: JEE Main 35 Year PYQs- Electrochemistry - Question 18

For a spontaneous reaction the DG, equilibrium constant (K) and will be respectively [2005]

Detailed Solution: Question 18

NOTE : For spontaneous reaction DG should be negative. Equilibrium constant should be more than one

(ΔG = – 2.303 RT log Kc, If Kc = 1 then ΔG = 0; If Kc < 1

Test: JEE Main 35 Year PYQs- Electrochemistry - Question 19

The highest electrical conductivity of the following aqueoussolutions is of                     [2005]

Detailed Solution: Question 19

Thus difluoro acetic acid being strongest acid will furnish maximum number of ions showing highest electrical conductivity. The decreasing acidic strength of the carboxylic acids given is difluoro acetic acid > fluoro acetic acid > chloro acitic acid > acetic acid.

Test: JEE Main 35 Year PYQs- Electrochemistry - Question 20

Aluminium oxide may be electrolysed at 1000°C to furnishaluminium metal (At. Mass = 27 amu; 1 Faraday = 96,500Coulombs). The cathode reaction is– Al3+ + 3e- → Al° To prepare 5.12 kg of aluminium metal by this method werequire

Detailed Solution: Question 20

1 mole of e– = 1F = 96500 C
27g of Al is deposited by 3 × 96500 C
5120 g of Al will be deposited by

Test: JEE Main 35 Year PYQs- Electrochemistry - Question 21

Calculate  using appropriate molar conductances of the electrolytes listed above at infinite dilution in H2O
at 25°C

Detailed Solution: Question 21

Test: JEE Main 35 Year PYQs- Electrochemistry - Question 22

Which of the following chemical reactions depict the oxidizing beahviour of H2SO4?

Detailed Solution: Question 22

reaction oxidation number of S is decreasing from + 6 to +4 hence undergoing reduction and for HI oxidation

Number of I is increasing from –1 to 0 hence
underegoing oxidation therefore H2SO4 is acting as oxidising agent.

Test: JEE Main 35 Year PYQs- Electrochemistry - Question 23

The molar conductivities and at infinite dilution in water at 25ºC are 91.0 and 426.2 S cm2/mol
respectively. To calculate  , the additional value required is

Detailed Solution: Question 23

 is given by the following equation

Hence  is required.

Test: JEE Main 35 Year PYQs- Electrochemistry - Question 24

Resistance of a conductivity cell filled with a solution of an electrolyte of concentration 0.1 M is 100 W. The conductivity of this solution is 1.29 S m–1. Resistance of the same cell when filled with 0.2 M of the same solution is 520 W. The molar conductivity of 0.2 M solution of electrolyte will be                            [2006]

Detailed Solution: Question 24

Given, R = 520Ω, C = 0.2 M, μ (molar conductivity) = ?

now cell constant is known.)

Hence, 

Test: JEE Main 35 Year PYQs- Electrochemistry - Question 25

The equivalent conductances of two strong electrolytes at infinite dilution in H2O (where ions move freely through a
solution) at 25°C are given below :                              [2007]

What additional information/ quantity one needs to calculate   of an aqueous solution of acetic acid?

Detailed Solution: Question 25

NOTE : According to Kohlrausch’s law, molar
conductivity of weak electrolyte acetic acid (CH3COOH) can be calculated as follows:

∴ Value of   should also be known for
calculating value of

Test: JEE Main 35 Year PYQs- Electrochemistry - Question 26

The cell, Zn | Zn2+ (1 M) || Cu2+ (1 M) | Cu   was allowed to be completely discharged at 298 K. The relative
concentration of  is         [2007]   

Detailed Solution: Question 26

Ecell = 0; when cell is completely discharged.

Test: JEE Main 35 Year PYQs- Electrochemistry - Question 27

Given  The potential for the cell Cr|Cr3+ (0.1M)|| Fe2+ (0.01 M)| Fe is         [2008]

Detailed Solution: Question 27

From the given representation of the cell, Ecell can be found as follows.

Hence option (d) is correct answer.

Test: JEE Main 35 Year PYQs- Electrochemistry - Question 28

In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is

At 298 K standard Gibb’s energies of formation for CH3OH(l), H2O(l) and and CO2 (g) are –166.2,  –237.2 and –394.4 kJ
mol–1 respectively. If standard enthalpy of combustion of methonal is – 726 kJ mol–1, efficiency of the fuel cell will be:                   

[2009]

Detailed Solution: Question 28

= – 394.4 + 2 (–237.2) – (–166.2) – 0
= – 394.4 – 474.4 + 166.2 = – 702.6 k J
% efficiency = 

Test: JEE Main 35 Year PYQs- Electrochemistry - Question 29

Given: 

The value of standard electrode potential for the change, Fe3+ (aq) + e →Fe2+ (aq) will be:               

[2009]

Detailed Solution: Question 29

Given

we have to calculate

Fe3+ + e → Fe2+ … (iii)
As we know that ΔG = –nFE
Thus for reaction (iii)

ΔG = ΔG1 - ΔG ; – nFE° = – nFE1 – (–nFE2)
–nFE° = nFE2 – nFE1
–1FE° = 2× 0.439F – 3 × 0.036 F
–1 FE° = 0.770 F ∴ E° = – 0.770V

Test: JEE Main 35 Year PYQs- Electrochemistry - Question 30

The Gibbs energy for the decomposition of Al2O3 at 500°C is as follows :

The potential difference needed for electrolytic reductionof Al2O3 at 500°C is at least                      [2010]

Detailed Solution: Question 30

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