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35 Year JEE Previous Year Questions: Electrochemistry

35 Questions MCQ Test Class 12 Chemistry 35 Years JEE Mains &Advance Past yr Paper | 35 Year JEE Previous Year Questions: Electrochemistry

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Attempt 35 Year JEE Previous Year Questions: Electrochemistry | 35 questions in 70 minutes | Mock test for JEE preparation | Free important questions MCQ to study Class 12 Chemistry 35 Years JEE Mains &Advance Past yr Paper for JEE Exam | Download free PDF with solutions

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]

A.
Sm mol^–1
B.
Sm² mol^–1
C.
S^–2m² mol
D.
S²m² mol^–2.

Solution:

given :

QUESTION: 2

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

A.
E = E_left - E_right
B.
E = E_left + E_right
C.
E = E_right - E_left
D.
E = – (E_right + E_left).

Solution:

E_cell = Reduction potential of cathode (right)
– Reduction potential of anode (left)
= E_right – E_left.

QUESTION: 3

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

A.
B.
C.
D.
None of these.

Solution:

Oxidation half call:-

The net cell reaction

QUESTION: 4

Which of the following reaction is possible at anode?

A.
B.
F₂→ 2F^-
C.
(1/2) O₂ + 2H⁺ → H₂O
D.
None of these.

Solution:

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

QUESTION: 5

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

[2002]

A.
a
B.
b
C.
c
D.
d

Solution:

Pure metal always deposits at cathode

QUESTION: 6

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

A.
NaCl + KNO₃ → NaNO₃ + KCl
B.
CaC₂O₄ + 2HCl →CaCl₂ + H₂C₂O₄
C.
Mg(OH)₂ + 2NH₄Cl →MgCl₂ + 2NH₄OH
D.
Zn + 2AgCN → 2Ag + Zn(CN)₂.

Solution:

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

A.
29.5 × 10^–2
B.
10]
C.
1 × 10¹⁰
D.
1 × 10^–10

Solution:

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

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]

A.
A > B > C
B.
C > B > A
C.
A > C > B
D.
B > C > A

Solution:

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

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

A.
10.8 g
B.
21.6 g [2003]
C.
108 g
D.
1.08 g

Solution:

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.

QUESTION: 10

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

A.
1.80 volt
B.
1.07 volt
C.
0.82 volt
D.
2.14 volt

Solution:

QUESTION: 11

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

A.
make the ship lighter
B.
prevent action of water and salt
C.
prevent puncturing by under-sea rocks
D.
keep away the sharks

Solution:

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.

QUESTION: 12

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

A.
produce high purity water
B.
create potential difference between two electrodes
C.
generate heat
D.
remove adsorbed oxygen from elctrode surfaces

Solution:

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

QUESTION: 13

Consider the following Eº values

Under standard conditions the potential for the reaction

[2004]

A.
0.91 V
B.
1.40 V
C.
1.68 V
D.
0.63 V

Solution:

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

∴

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]

A.
1.0 × 10¹⁰
B.
1.0 × 10⁵
C.
1.0 × 10¹
D.
1.0 × 10³⁰

Solution:

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]

A.
278 S cm² mol^–1
B.
176 S cm² mol^–1
C.
128 S cm² mol^–1
D.
302 S cm² mol^–1

Solution:

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

= 126 +152 -150 = 128 S cm² mol^-1

QUESTION: 16

In a cell that utilises the reaction Zn_(s) + 2H⁺ (aq) → Zn²⁺+ H_2(g) addition of H₂SO₄to cathode compartment, will [2004]

A.
increase the E and shift equilibrium to the right
B.
lower the E and shift equilibrium to the right
C.
lower the E and shift equlibrium to the left
D.
increase the E and shift equilibrium to the left

Solution:

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]

A.
Fe
B.
Mn
C.
Cr
D.
Co

Solution:

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

QUESTION: 18

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

A.
–ve, >1, –ve
B.
–ve, <1, –ve
C.
+ve, >1, –ve
D.
–ve, >1, +ve

Solution:

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

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

QUESTION: 19

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

A.
0.1 M difluoroacetic acid
B.
0.1 M fluoroacetic acid
C.
0.1 M chloroacetic acid
D.
0.1 M acetic acid

Solution:

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.

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

A.
5.49 × 10¹ C of electricity
B.
5.49 × 10⁴ C of electricity
C.
1.83 × 10⁷ C of electricity
D.
5.49 × 10⁷ C of electricity

Solution:

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

QUESTION: 21

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

A.
217.5
B.
390.7
C.
552.7
D.
517.2

Solution:

QUESTION: 22

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

A.
NaCl + H₂SO₄ → NaHSO₄ + HCl
B.
2PCl₅ + H₂SO₄ → 2POCl₃ + 2HCl + SO₂Cl₂
C.
2HI + H₂SO₄ → I₂ + SO₂ + 2H₂O
D.
Ca(OH)₂ + H₂SO₄ → CaSO₄ + 2H₂O

Solution:

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.

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

Solution:

is given by the following equation

Hence is required.

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]

A.
1.24 × 10^–4 S m₂ mol^–1
B.
12.4 × 10^–4 S m² mol^–1
C.
124 × 10^–4 S m² mol^–1
D.
1240 × 10^–4 S m² mol^–1

Solution:

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

now cell constant is known.)

Hence,

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?

A.
of chloroacetic acid (ClCH₂COOH)
B.
of NaCl
C.
of CH₃COOK
D.
the limiting equivalent coductance of

Solution:

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

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]

A.
9.65 × 10⁴
B.
antilog (24.08)
C.
37.3
D.
10^37.3

Solution:

E_cell = 0; when cell is completely discharged.

QUESTION: 27

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

A.
0.22 V
B.
0.336 V
C.
– 0.339
D.
0.26 V

Solution:

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

Hence option (d) is correct answer.

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]

A.
87%
B.
90%
C.
97%
D.
80%

Solution:

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

QUESTION: 29

Given:

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

[2009]

A.
0.385 V
B.
0.770 V
C.
–0.270 V
D.
–0.072 V

Solution:

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

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]

A.
4.5 V
B.
3.0 V
C.
- 2.5 V
D.
5.0 V

Solution:

QUESTION: 31

The correct order of values with negative sign for the four successive elements Cr, Mn, Fe and Co is [2010]

A.
Mn > Cr > Fe > Co
B.
Cr < Fe > Mn > Co
C.
Fe > Mn > Cr > Co
D.
Cr > Mn > Fe > Co

Solution:

The value of for given metal ions are

The correct order of values without
considering negative sign would be

Mn²⁺ > Cr²⁺ > Fe²⁺ > Co²⁺.

QUESTION: 32

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

[2011]

A.
p(H₂) = 1 atm and [H⁺] = 2.0 M
B.
p(H₂) = 1 atm and [H⁺] = 1.0 M
C.
p(H₂) = 2 atm and [H⁺] = 1.0 M
D.
p(H₂) = 2 atm and [H⁺] = 2.0 M

Solution:

Now if = 2 atm and [H⁺] = 1M

QUESTION: 33

The standard reduction potentials for Zn²⁺/Zn, Ni²⁺/Ni and Fe²⁺/Fe are –0.76,–0.23 and –0.44 V respectively.
The reaction X+Y²⁺ →X²⁺+Y will be spontaneous when :

A.
X = Ni, Y = Fe
B.
X = Ni, Y = Zn
C.
X= Fe, Y = Zn
D.
X= Zn, Y = Ni

Solution:

For a spontaneous reaction DG must be –ve
Since ΔG = – nFE°
Hence for ΔG to be -ve ΔE° has to be positive. Which is possible when X = Zn, Y = Ni
Zn + Ni⁺⁺ → Zn⁺⁺ + Ni

QUESTION: 34

Given :

Based on the data given above, strongest oxidising agentwill be : [JEE M 2013]

A.
Cl
B.
Cr³⁺
C.
Mn²⁺
D.

Solution:

higher the value of standard reduction potential
stronger will be the oxidising agent, hence is the strongest oxidising agent.

QUESTION: 35

Resistance of 0.2 M solution of an electrolyte is 50 Ω.The specific conductance of the solution is 1.4 S m^–1.The resistance of 0.5 M solution of the same electrolyte is 280 Ω. The molar conductivity of 0.5 M solution of the electrolyte in S m² mol^–1 is:

[JEE M 2014]