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


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


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This mock test of Revisal Problems (Past 13 Year) JEE Advanced (Electrochemistry) for JEE helps you for every JEE entrance exam. This contains 30 Multiple Choice Questions for JEE Revisal Problems (Past 13 Year) JEE Advanced (Electrochemistry) (mcq) to study with solutions a complete question bank. The solved questions answers in this Revisal Problems (Past 13 Year) JEE Advanced (Electrochemistry) quiz give you a good mix of easy questions and tough questions. JEE students definitely take this Revisal Problems (Past 13 Year) JEE Advanced (Electrochemistry) exercise for a better result in the exam. You can find other Revisal Problems (Past 13 Year) JEE Advanced (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 12 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE is correct.

Q. 

For the following potentiometric titration,


EMF at the equivalent point is given by the graph.
What is the value of (log Keq)?

Solution:

From potentiometric curve, EMF at the equivalent point = 2,0 V.

QUESTION: 2

Given at 298 K,

Thus, ΔG° (Gibbs free energy) for the reaction,

Solution:




(ΔG° > 0, reaction is non-spontaneous)

QUESTION: 3

One molal solution of AgNO3 containing 1 kg of water is electrolysed in two electrolytic cells containing Ag and Pt electrodes.2 A current is passed for 2 h in each of them.After electrolysis AgNO3 is

Solution:

 With Ag electrodes

Ag electrode dissolves at anode and equivalent amount is deposited at the cathode. Thus, there is no change in concentration of AgNO3.
With Pt electrodes

Ag is deposited at the cathode and thus , concentration of AgNO3 decreases.


= 16.12 g = 0.15 mol
AgNO3 left = 1.0 - 0.15 = 0.85 mol in 1 kg water
Molality = 0.85 mol kg-1
= 0.85 molal

QUESTION: 4

Following two electrolytic cells are taken and current is passed as indicated at 298 K.  

After current is passed,they were connected using a salt-bridge.Change in emf of the cell is 

Solution:

In I, Ag+ + e → Ag
Ag deposited by 1F = 1mol
Ag deposited by 0.5F = 0.5 mol
Thus, [Ag+]1 = 1.0 - 0.5 = 0.5 mol L-1
In II, Ag deposited by 1F = 1mol


QUESTION: 5

When temperature of the following cell Cu | Cu2+(1 M) || Ag+(1 M) | Ag is increased by 10°, E°cell becomes 0.48 V.

Given, E° Cu2+ /Cu = 0.34 V, E°Ag+ /Ag = 0.80 V. Thus, entropy change is

Solution:

Initially



QUESTION: 6

Select the correct statement based on following half-cell reactions.  

 

Solution:



cell > 0, thus spontaneous
Thus, Fe3+ is reduced to Fe2+ by iron nails.
Then (b) is correct.

cell > 0, thus spontaneous.
Hence, corrosion of iron takes place in air.

QUESTION: 7

For the following cell reaction,

 

Thus Ecell is  

Solution:




QUESTION: 8

Given, limiting equivalent conductance (S cm2 equiv-1).
BaCI2 = x1, H2SO4 = x2, HCI = x3

Specific conductance of saturated BaSO4 = y S cm-1
Thus, Ksp (solubility product) of BaSO4 is

Solution:



QUESTION: 9

 

 1 x 10-7 M AgNO3 solution is added to 1 L saturated solution of AgBr: Then specific conductance of AgBr solution is  

Solution:

Let solubility of AgBr in presence of Ag NO3 = x





QUESTION: 10

In the electrolysis of aqueous sodium chloride solution,following reactions may take place at cathode(reduction):

 

Select the correct option(s). 

Solution:



If amalgam is formed, I is preferred thus (a), (b) both are correct.

QUESTION: 11

For the following half-cell reaction,

Potential at the equivalence point is

Solution:



QUESTION: 12

For a thermocell, 

operating between 800 K and 1200 K,

Thus, emf in the operating temperature 850 K and 1050 K is

Solution:

QUESTION: 13

Statement Type

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

Q.

Statement I : In the conductometric titration of CH3COOH with NaOH,conductance first increases slowly and after complete reaction of CH3COOH, it increases rapidly as shown

                                        

Statement II : Initial increase in conductance is due to hydrolysis of salt  formed.

Solution:


As CH3COONa is formed, it is hydrolysed at the same time forming CH^COOH (weak acid) and NaOH (strong base). Due to formation of strong base, conductance increases at every addition of NaOH. After CH3COOH is completely neutralised further addition of NaOH causes rapid increase in conductance (no CH3COOH at this stage)

Thus, both Statements I and II are correct and Statement II is the correct explanation of Statement I.

QUESTION: 14

Statement I : Electrolysis of NaCl solution gives chlorine at anode instead of O2.

Statement II: Formation of oxygen at anode requires over voltage.

Solution:


In concentrate d NaCI, Cl2 is favoured and in dilute NaCI. O2 is favoured. However, product is Cl2. It is due to over-voltage. Experiments indicate that the applied voltage required is always greater than the voltage calculated from the standard potentials. The additional voltage required is called over-voltage. Thus, Statements I and II are correct and Statement II is the correct explanation of Statement I.

*Multiple options can be correct
QUESTION: 15

One or More than One Options Correct Type

This section contains 5 multiple type questions. Each question has 4 choices (a), (b), (c) and (d), out of which ONE or MORE THAN ONE are correct.

Q.

Standard electrode potential data are useful for understanding the suitability of an oxidant in a redox titration. Some half-cell reactions and their standard potentials are given below:  

Identify correct statement regarding the quantitative estimation of aqueous Fe(NO3)2  

Solution:


This cell reaction is spontaneous since, E°cell > 0.
Thus, MnO4- cannot be used in aqueous HCI.

Thus  is not affected in HCI medium. Thus, HCI can be used in    solution.
(c) In I, source of H+ is H2SO4, is not further oxidised by MnO4-hence, H2SOcan be used.
(d) same as (c).

*Multiple options can be correct
QUESTION: 16

For a given KCl solution,molar conductance L(conductance) = 10 L (H2O), (L in S) of solution.

For pure of solution water k (specific conductance = 5.7 x 10-6 S m-1).Thus,molar conductance of KCl solution is 

Solution:

k (specific conductance of water) = 5.7 x 10-6 Sm-1
k = L (conductance) x cell constant

∴ Conductance of pure KCI solution = L (solution) - L (water)

∴ Specific conductivity of solution = L x cell constant


*Multiple options can be correct
QUESTION: 17

Electrolytic conductance of an electrolyte is decreased due to

Solution:

(a) Due to dilution, ionisation increases hence, conduction increases.
(b) Increase in temperature, increases ionisation thus, conduction increases.
(c) In higher concentration range, ions are closer to one another thus interionic attraction increases. Hence, conduction decreases.

*Multiple options can be correct
QUESTION: 18

For the reduction of NO3in an aqueous solution E° is 0.96 V. Values of E° for some metals are given below: 

The pair(s) of metals that is (are) oxidised by NO3- ion in aqueous solution is (are)  

Solution:

(N)red is the reduction product of NO3-.




Since, E°cell < 0, hence Au is not oxidised to Au3+ by NO3- ion.
Thus, (a) V and Hg, (b) Hg and Fe, (d) Fe and V
pairs are oxidised by NO3- in aqueous medium.

*Multiple options can be correct
QUESTION: 19

For the reaction, 2F3+ + Fe → 3Fe2+, E° = 1.21 V. Thus,

Solution:

(a) Fe3+ is reduced to Fe2+ by Fe which is oxidised thus, reaction can take place even in the absence of second half-cell or a current-correct.

Since, oxidised species and reduced species are ions, hence Pt cathode is used, Fe3+, Fe2+ | Pt-correct.
(c) Fe as cathode would directly react with Fe3+ thus, short-circuiting the celi-correct.
(d) Cell set up is

Thus, given cell is incorrect.

QUESTION: 20

Matching List Type

Choices for the correct combination of elements from Column I and Column II are given as options (a), (b), (c) and (d), out of which one is correct

Q.

1 L buffer of 1.0 M NaH2PO4 and  1.0 M Na2HPOare taken in two electrolytic vessel separately.Electrolysis was carried using platinum electrodes for 201 min with constant current of 1.25 A.Assume the electrolysis of water takes place.

                                

Match the parameters given in Column I with their values in Column II and select the answer from the codes given below list. 

Solution:

(i) Before current is passed, mixture of and is a buffer of equimolar value.







0.156g L-1 = 0.156 Mat anode
Thus, OH- formed = 0.156 M at cathode.
Thus (iv) → (q) 

QUESTION: 21

For H2O at 298 K, 

Match the parameters  in Column I (in terms of -log(value) i.e pvalue ) with their values in Column II 

Solution:




QUESTION: 22

Ka(CH3COOH)= 1.8 x 10-5, [CH3COOH] = 0.040 M
Cell constant of the cell used = 0.206 cm-1

Match the parameters given in Column I with their values in Column II and select the answer from the codes given below list.

Solution:




QUESTION: 23

Comprehension Type

This section contains a passage describing theory, experiments, data, etc. Two questions related to the paragraph have been given. Each question has only one correct answer out of the given 4 options (a), (b), (c) and (d)

                                                   Passage I

During the discharge of a lead storage battery,the density of sulphuric acid fell from 1.294 g/mL (39% H2SOby weight) to 1.139 g/ML (20% H2SOby weight).The battery holds 3.5 L of the acid and the volume remains practically constant during discharge.

Q.

Total H2SO4 used is   

Solution:



Thus, H2SO4 consumed per litre = (505 - 228) g = 277 g
Total acid in 3.5 L = 969.5 g
H2SO4 consumed = 9.89 mol
Electrons consumed = 2 mol for 2 mol H2SO4 
= 1 mol for 1 mol H2SO4

QUESTION: 24

Passage I

During the discharge of a lead storage battery,the density of sulphuric acid fell from 1.294 g/mL (39% H2SOby weight) to 1.139 g/ML (20% H2SOby weight).The battery holds 3.5 L of the acid and the volume remains practically constant during discharge.

Q.

Total number of ampere-hour for which battery must have been used is

Solution:



Thus, H2SO4 consumed per litre = (505 - 228) g = 277 g
Total acid in 3.5 L = 969.5 g
H2SO4 consumed = 9.89 mol
Electrons consumed = 2 mol for 2 mol H2SO4 
= 1 mol for 1 mol H2SO4

QUESTION: 25

Passage II

Variation of molar conductancewith molar concentration C is given following Debye-Huckel Onsager equation. 

With A and B as Debye-Huckel constants andis the molar conductance of infinitely diluted(≈ zero concentration) solution.

Q.

Select the correct option.

Solution:



I. As solution is diluted, C decreases and vice-versa. Increase in concentration, increases interionic attraction between ions, hence Am decreases as shown for strong electrolytes.

II. By Ostwald's dilution law

Thus, in a very dilute solution of weak salt, ionisation is very rapid (as compared to strong salt.) In such case cannot be obtained by extrapolation to zero concentration,
Ill. For strong electrolytes, is uniform at all concentration range, hence can be obtained by extrapolation of graph between to zero concentration.

Thus, I, II and III are correct. 

QUESTION: 26

Passage II

Variation of molar conductancewith molar concentration C is given following Debye-Huckel Onsager equation. 

With A and B as Debye-Huckel constants andis the molar conductance of infinitely diluted(≈ zero concentration) solution.

Q. 

Following are some facts about electrolytic conduction :
I. decreases with due to increase in interionic attraction between ions.
II. For weak electrolytes, there is rapid increase in the degree of ionisation with dilution, hence, interionic attraction decreases and increases rapidly in lower concentration range.
III. can be obtained by extrapolation of graph of to zero concentration for strong electrolytes.

Correct facts are  

Solution:



I. As solution is diluted, C decreases and vice-versa. Increase in concentration, increases interionic attraction between ions, hence Am decreases as shown for strong electrolytes.

II. By Ostwald's dilution law

Thus, in a very dilute solution of weak salt, ionisation is very rapid (as compared to strong salt.) In such case cannot be obtained by extrapolation to zero concentration,
Ill. For strong electrolytes, is uniform at all concentration range, hence can be obtained by extrapolation of graph between to zero concentration.

Thus, I, II and III are correct. 

*Answer can only contain numeric values
QUESTION: 27

One Integer Value Correct Type

This section contains 4 questions, when worked out will result in an integer value from 0 to 9 (both inclusive)

Q.

A cell, at 298 K is set up Ag | Ag+ (1 M) || Cu2+(1 M) | Cu

A current of 9.65 A is passed for 1 h. What is change in EMF (in centivolt)?


Solution:



Thus, [Cu2+] decreases and that of [Ag+] increases.
Reaction quotient


Thus, [Ag+] formed = 0.36 equivalent = 0.36 mol
[Ag+] after current is passed = 1.36 M
[Cu2+] reduced = 0.36 equivalent = 0.36/2 = 0.18 mol
∴ [Cu2+] after current is passed = 1.0 - 0. 18 = 0. 82 M


*Answer can only contain numeric values
QUESTION: 28

An acidic solution of Cu2+ salt containing 0.4 g of Cu2+ is electrolysed untill all the copper is deposited.Calculate the volume of gas liberated at the other electrode(Express result in centilitre at STP)


Solution:

In electrolysis, cation goes to cathode and anion to anode 


*Answer can only contain numeric values
QUESTION: 29

At 298 K,the EMF of the cell  

                                              

If liquid junction potential is zero,the pH of the solution x is


Solution:


*Answer can only contain numeric values
QUESTION: 30

Silver is electro-deposited on a metaliic vessel surface of area 800 cm2 by passing a current of 0.20 A for 3 h.Thickness of the silver deposited was X x 10-4 cm.What is the value of X? (Density of Ag = 10.20 g cm-3)


Solution: