NEET Exam  >  NEET Tests  >  Chemistry Class 12  >  P. Bahadur Test: Redox Reactions & Electrochemistry - NEET MCQ

P. Bahadur Test: Redox Reactions & Electrochemistry - NEET MCQ


Test Description

25 Questions MCQ Test Chemistry Class 12 - P. Bahadur Test: Redox Reactions & Electrochemistry

P. Bahadur Test: Redox Reactions & Electrochemistry for NEET 2024 is part of Chemistry Class 12 preparation. The P. Bahadur Test: Redox Reactions & Electrochemistry questions and answers have been prepared according to the NEET exam syllabus.The P. Bahadur Test: Redox Reactions & Electrochemistry MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for P. Bahadur Test: Redox Reactions & Electrochemistry below.
Solutions of P. Bahadur Test: Redox Reactions & Electrochemistry questions in English are available as part of our Chemistry Class 12 for NEET & P. Bahadur Test: Redox Reactions & Electrochemistry solutions in Hindi for Chemistry Class 12 course. Download more important topics, notes, lectures and mock test series for NEET Exam by signing up for free. Attempt P. Bahadur Test: Redox Reactions & Electrochemistry | 25 questions in 35 minutes | Mock test for NEET preparation | Free important questions MCQ to study Chemistry Class 12 for NEET Exam | Download free PDF with solutions
P. Bahadur Test: Redox Reactions & Electrochemistry - Question 1

According to Faraday’s Second Law of Electrolysis the amounts of different substances liberated by the same quantity of electricity passing through the electrolytic solution are proportional:

Detailed Solution for P. Bahadur Test: Redox Reactions & Electrochemistry - Question 1

The amounts of different substances liberated by the same quantity of electricity passing through the electrolytic solution are proportional to their chemical equivalent weights.

P. Bahadur Test: Redox Reactions & Electrochemistry - Question 2

Charge on one mole of electrons i.e. one Faraday is approximately:

Detailed Solution for P. Bahadur Test: Redox Reactions & Electrochemistry - Question 2

Charge on one electron is 1.6 x 10-19 C. 
1 mole of electron 6.023 x 1023 electrons.
So charge on one mole of electron = 1.6 x 10-19 C x 6.023 x 1023 = 96500 C mol-1

1 Crore+ students have signed up on EduRev. Have you? Download the App
P. Bahadur Test: Redox Reactions & Electrochemistry - Question 3

In the primary batteries,

Detailed Solution for P. Bahadur Test: Redox Reactions & Electrochemistry - Question 3

Primary batteries cannot be recharged and reused.

P. Bahadur Test: Redox Reactions & Electrochemistry - Question 4

A secondary cell after use ________ recharged by passing current through it in the ________ so that it can/can’t be used again.

Detailed Solution for P. Bahadur Test: Redox Reactions & Electrochemistry - Question 4

Secondary cell can be recharged and reused.

P. Bahadur Test: Redox Reactions & Electrochemistry - Question 5

Fuel cells are _________ that are designed to convert the energy of combustion of fuels like hydrogen, methane, methanol, etc. directly into _________.

Detailed Solution for P. Bahadur Test: Redox Reactions & Electrochemistry - Question 5
  • fuel cell is an electrochemical cell that converts the chemical energy of a fuel (often hydrogen) and an oxidizing agent (often oxygen) directly into electrical energy through a pair of redox reactions. 
  • Fuel cells are different from most batteries in requiring a continuous source of fuel and oxygen (usually from air) to sustain the chemical reaction, whereas in a battery the chemical energy usually comes from metals and their ions or oxides that are commonly already present in the battery, except in flow batteries.
  • Fuel cells can produce electricity continuously for as long as fuel and oxygen are supplied.

Setup of a Fuel cell:

P. Bahadur Test: Redox Reactions & Electrochemistry - Question 6

In corrosion, a metal is _________ of electrons to oxygen and results in the formation of oxides.

Detailed Solution for P. Bahadur Test: Redox Reactions & Electrochemistry - Question 6

In Corrosion, a metal is oxidized by loss of electrons and results in the formation of oxides.

  • Corrosion is a natural process that converts a refined metal into a more chemically stable form such as oxide, hydroxide, or sulfide.
  • It is the gradual destruction of materials (usually a metal) by chemical and/or electrochemical reaction with their environment. 
  • Corrosion engineering is the field dedicated to controlling and preventing corrosion.
P. Bahadur Test: Redox Reactions & Electrochemistry - Question 7

One of the simplest methods of preventing corrosion is to prevent the surface of the metallic object to come in contact with atmosphere. This can be done:

Detailed Solution for P. Bahadur Test: Redox Reactions & Electrochemistry - Question 7

The best way to prevent corrosion is to prevent the surface of the metallic objects to come in contact with the atmosphere.

  • The rusting of iron can be prevented by painting, oiling, greasing or varnishing its surface.
  • Galvanization is another method of protecting iron from rusting by coating iron with a thin layer of zinc.
  • Corrosion of iron is prevented by coating iron with non-xcorrosive substance like carbon. This process is termed as alloying .
P. Bahadur Test: Redox Reactions & Electrochemistry - Question 8

An electrochemical method is to provide a sacrificial electrode of another metal (like Mg, Zn, etc.) which corrodes itself but saves the object. A typical example is

Detailed Solution for P. Bahadur Test: Redox Reactions & Electrochemistry - Question 8

Galvanization is a process in which iron is coated with reactive metal like Zn which get corroded hence preventing iron from corrosion.

Electro-Galvanization:

P. Bahadur Test: Redox Reactions & Electrochemistry - Question 9

Correct arrangement of Al, Cu, Fe, Mg and Zn in the order in which they displace each other from the solution of their salts is

Detailed Solution for P. Bahadur Test: Redox Reactions & Electrochemistry - Question 9

 A more reactive metal displaces a less reactive metal from the salt solution (refer to reactivity series).

The order in which metals displace each other from the solution of their salts can be given with the help of their standard electrode potential. Since magnesium has the least standard electrode potential so it is the most strong reducing agent.

So the required order we get is: Mg > Al > Zn > Fe > Cu

Reactivity Series:
What is the significance of the reactivity series of metals? - Quora

P. Bahadur Test: Redox Reactions & Electrochemistry - Question 10

The standard emf of galvanic cell involving 3 moles of electrons in its redox reaction is 0.59 V. The equilibrium constant for the reaction of the cell is

Detailed Solution for P. Bahadur Test: Redox Reactions & Electrochemistry - Question 10

The standard emf of a galvanic cell involving 3 moles of electrons in its redox reaction is 0.59V.

cell​ = 0.59V
n=3
The equilibrium constant for the reaction of the cell is given by the expression: ln K = RT nFEºcell
ln K = 8.314 × 298 × 3 × 96500 × 0.59 = 68.9
K≈1030

The equilibrium constant for the reaction of the cell is 1030.

P. Bahadur Test: Redox Reactions & Electrochemistry - Question 11

Emf of the cell 

Mg (s) | Mg2+ (0.001M) || Cu2+ (0.0001M) | Cu(s)   at 298 K is Cu Given : Eco2+, cu = +0.337 V, EºM22+ Me =- 2.37 V, F = 96500 C mol

Detailed Solution for P. Bahadur Test: Redox Reactions & Electrochemistry - Question 11

P. Bahadur Test: Redox Reactions & Electrochemistry - Question 12

Emf of the cell 

Fe (s) | Fe2+ (0.001 M)|| H+ (1M)|H2 (g) (1 bar) |Pt(s)   at 298 K is

(Given: Eocell ​= +0.44 V)

Detailed Solution for P. Bahadur Test: Redox Reactions & Electrochemistry - Question 12

⇒Ecell ​= 0.44 + 0.089V

⇒Ecell​ = 0.53V

P. Bahadur Test: Redox Reactions & Electrochemistry - Question 13

The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S  cm-1. Calculate its molar conductivity. 

Detailed Solution for P. Bahadur Test: Redox Reactions & Electrochemistry - Question 13

k (Conductivity) =0.0248 Scm−1, C (molarity) = 0.20 M, T (temperature) = 298 K
Λm​ = ? (molar conductivity) 

Λm​ = (1000×k​) / C = (1000×0.0248) / 0.20 ​= 124 S cm2 mol−1

The molar conductivity is 124 S cm2 mol−1

P. Bahadur Test: Redox Reactions & Electrochemistry - Question 14

The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500 Ω. What is the cell constant if conductivity of 0.001M KCl solution at 298 K is 0.146 × 10-3 S cm-1.

Detailed Solution for P. Bahadur Test: Redox Reactions & Electrochemistry - Question 14

Κ = G x cell constant and G = 1/R.

P. Bahadur Test: Redox Reactions & Electrochemistry - Question 15

How much charge is required for the reduction of 1 mol of Al3+ to Al?

Detailed Solution for P. Bahadur Test: Redox Reactions & Electrochemistry - Question 15

For reduction of 1 mol of Al3+ to Al, 3 mol of electrons are required so total charge will be 3F.

P. Bahadur Test: Redox Reactions & Electrochemistry - Question 16

How much electricity in terms of Faraday is required to produce 40.0 g of Al from molten  Al2O3?

Detailed Solution for P. Bahadur Test: Redox Reactions & Electrochemistry - Question 16

P. Bahadur Test: Redox Reactions & Electrochemistry - Question 17

Three electrolytic cells A, B, C containing solutions of ZnSO4, AgNO3 and CuSO4, respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell. How long did the current flow? What mass of copper and zinc were deposited?  

Detailed Solution for P. Bahadur Test: Redox Reactions & Electrochemistry - Question 17

And mass of Zn and Cu deposited will be in ratio of their equivalent mass.

P. Bahadur Test: Redox Reactions & Electrochemistry - Question 18

The standard emf of galvanic cell involving 3 moles of electrons in its redox reaction is 0.59 V. The equilibrium constant for the reaction of the cell is

Detailed Solution for P. Bahadur Test: Redox Reactions & Electrochemistry - Question 18

P. Bahadur Test: Redox Reactions & Electrochemistry - Question 19

An increase in equivalent conductance of a strong electrolyte with dilution is mainly due to

Detailed Solution for P. Bahadur Test: Redox Reactions & Electrochemistry - Question 19

Equivalent conductance increases on dilution for a strong electrolyte because of increase in mobility of ions.

P. Bahadur Test: Redox Reactions & Electrochemistry - Question 20

For the reduction of silver ions with copper metal the standard cell potential was found to be +0.46V at 25°C. The value of standard Gibbs energy, ΔG° wll be (F = 96500 C mol-1)

Detailed Solution for P. Bahadur Test: Redox Reactions & Electrochemistry - Question 20

P. Bahadur Test: Redox Reactions & Electrochemistry - Question 21

Which of the following electrolytic solutions has the least specific conductance?

Detailed Solution for P. Bahadur Test: Redox Reactions & Electrochemistry - Question 21

Specific conductance decreases with dilution because of decrease in the number of ions per unit volume.

P. Bahadur Test: Redox Reactions & Electrochemistry - Question 22

The highest electrical conductivity of the following aqueous solutions is of

Detailed Solution for P. Bahadur Test: Redox Reactions & Electrochemistry - Question 22

Acidity increases on attaching electron withdrawing group because of stability of conjugate base.

P. Bahadur Test: Redox Reactions & Electrochemistry - Question 23

The standard electrode potential is measured by

Detailed Solution for P. Bahadur Test: Redox Reactions & Electrochemistry - Question 23

Voltmeter measures the potential.

P. Bahadur Test: Redox Reactions & Electrochemistry - Question 24

When KMnO4 acts as an oxidizing agent and ultimately forms MnO42-, MnO2, Mn2O3, and Mn2+ then the number of electrons transferred in each case 

Detailed Solution for P. Bahadur Test: Redox Reactions & Electrochemistry - Question 24

No. of electrons = change in oxidation number of active elemnt.

P. Bahadur Test: Redox Reactions & Electrochemistry - Question 25

On the basis of the following E° values, the strongest oxidizing agent is: 

Detailed Solution for P. Bahadur Test: Redox Reactions & Electrochemistry - Question 25
  • The strongest oxidizing agent is one having more positive or less negative reduction potential.
  • For the strongest oxidizing agent, the oxidizing potential should be the least.

Here, the oxidizing potential of Fe+2 is less than that of [Fe(CN)6​]4−. Therefore, Fe+2 is a stronger oxidizing agent than [Fe(CN)6​]4−.

  • Also, the stronger oxidizing agent should easily reduce itself.

Here, Fe+3 is easily reduced than Fe+2. Therefore, among all the four, Fe+3 is the stronger oxidizing agent.

108 videos|286 docs|123 tests
Information about P. Bahadur Test: Redox Reactions & Electrochemistry Page
In this test you can find the Exam questions for P. Bahadur Test: Redox Reactions & Electrochemistry solved & explained in the simplest way possible. Besides giving Questions and answers for P. Bahadur Test: Redox Reactions & Electrochemistry, EduRev gives you an ample number of Online tests for practice

Top Courses for NEET

108 videos|286 docs|123 tests
Download as PDF

Top Courses for NEET