COMEDK UGET 2025 Electrochemistry Practice Questions with Solutions
COMEDK UGET 2025 practice questions with solutions are focused on key topics such as EMF of a cell, Standard electrode potential, Nernst equation etc. Solving COMEDK Electrochemistry sample questions will acquaint you with question format, exam structure etc.
COMEDK UGET 2025 Electrochemistry Practice Questions with Solutions: The chapter on Electrochemistry holds decent weightage in the COMEDK UGET syllabus. COMEDK UGET 2025 Electrochemistry practice questions will cover key concepts such as Redox Reactions, EMF of a Cell, Nernst Equation, Conductance, Electrolysis, Fuel Cells, and Corrosion. With an estimated weightage of 3-4 questions in COMEDK UGET 2025 exam , Electrochemistry can play a crucial role in boosting your overall Chemistry score. Since COMEDK UGET 2025 syllabus covers topics from both 11th and 12th standards, it is essential to not just memorize concepts but also practice questions regularly. By solving COMEDK Electrochemistry sample questions, you will get familiar with the question types, difficulty level, and exam pattern, improve problem-solving speed and accuracy, understand concepts and numerical applications effectively, and identify important and frequently asked questions. In this article, we have provided COMEDK UGET 2025 Electrochemistry practice questions with solutions to help you prepare efficiently and maximize your chances of scoring well in the exam.
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COMEDK Chapter Wise PYQs for Chemistry
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COMEDK UGET 2025 Electrochemistry Important Topics
With several sub-topics included in COMEDK UGET Electrochemistry, such as Redox reactions, Lead accumulator, Nerst equation and its application to chemical cells, this chapter plays an important role in the exam preparation. If you wish to sit for the exam, then you must make sure to be well-acquainted with all the topics of the Electrochemistry chapter. You can find more details related to COMEDK UGET Electrochemistry sub-topics for exam preparation below:-
Chapter | Topics |
Electrochemistry |
|
Also Check - Do or Die Chapters for COMEDK UGET 2025 Chemistry
COMEDK UGET 2025 Electrochemistry Expected Weightage
Electrochemistry chpater holds approximately 5% of the total weightage in COMEDK UGET 2025 Chemistry syllabus . While doing your preparation/ revision, you are advised to have a good knowledge of the expected weightage as well as the number of questions that will be asked from the said chapter. The table below entails the crux related to the information of COMEDK UGET 2025 Electrochemistry expected weightage:-
Topic | Expected Number of Questions | Expected Weightage |
Electrochemistry | 3-4 | 1-5% |
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COMEDK UGET 2025 Electrochemistry Practice Questions with Solutions
Regularly practicing COMEDK Electrochemistry sample questions can significantly enhance your exam preparation. Working through COMEDK Electrochemistry practice questions with solutions allows you to familiarize yourself with potential exam questions, verify correct answers, and understand step-by-step solutions. Below are some essential COMEDK UGET 2025 Electrochemistry practice questions, carefully selected from past exam papers:-
Q1. What is the quantity of charge, in Faraday units, required for the reduction of 3.5 moles of Cr 2 O 2- 7 in acid medium?
a) 6.0
b) 10.5
c) 21.0
d) 3.0
Correct answer - c) 21.0
Solution: To determine the quantity of charge, in Faraday units, required for the reduction of 3.5 moles of Cr 2 O 2- 7 in acid medium, we need to look at the half reaction for the reduction of Cr 2 O 2- 7 to Cr 3+ :
Cr 2 O 2- 7 + 14H + + 6e- ⇾ 2Cr 3 + 7H 2 O
The balanced equation shows that 6 moles of electrons (e- ) are needed to reduce 1 mole of Cr 2 O 2- 7 . Therefore, for 3.5 moles of Cr 2 O 2- 7
The total charge required in moles of electrons is calculated as:
Total charge = 3.5 moles\ of CR2O 2- 7 x 6 moles of e-/mole\ofCr 2 O 2- 7
So:
Total charge = 3.5 x 6 = 21 Faradays
Thus, the quantity of charge required for the reduction of 3.5 moles of Cr 2 O 2- 7 in acid medium is 21.0 Faradays.
The correct answer is Option C: 21.0
Q2. A dry cell consists of a moist paste of NH 4 Cl and ZnCl 2 contained in a Zn casing which encloses a Carbon rod surrounded by black Mno2 paste. What is the role of ZnCl 2 in it?
a) It prevents pressure being developed in the cell due to NH3 gas formation.
b) It serves as cathode thus permitting Carbon rod to act as anode.
c) It acts as the anode while Carbon rod acts as the cathode.
d) It keeps the contents dry and prevents leakage of electrolyte.
Correct answer - a) It prevents pressure being developed in the cell due to gas formation
Solution: In a dry cell, the following reactions occur:
At the anode (Zinc casing):
Zn(s)→Zn 2 +(aq)+2e -
At the cathode (Carbon rod):
2MnO 2 (S) + 2NH + 4 (aq)+2e - →Mn2O 3 (s) + 2NH3(aq) + H 2 O(l)
The ammonia (NH3) produced at the cathode is a gas. If it's not controlled, it can build up pressure inside the cell, potentially causing it to leak or even explode.
The role of ZnCl 2 is to react with the ammonia to form a complex ion:
Zn 2+ (aq) + 4NH 3 (aq)⇌[Zn(NH 3 ) 4 ] 2+ (aq)
This reaction effectively removes the ammonia gas from the system, preventing the buildup of pressure and ensuring the stability of the cell.
Let's look at why the other options are incorrect:
Option B: The cathode is the carbon rod where reduction occurs, and the anode is the zinc casing where oxidation takes place. The Zinc doesn't directly affect this role.
Option C: Zinc acts as the anode, not the cathode. The carbon rod acts as the cathode.
Option D: ZnCl 2 doesn't prevent the leakage of the electrolyte. The cell's design and the paste's consistency are more important for preventing leaks.
Q3. Identify the incorrect statement among the following.
a) Electrolytic conductance increases with increase in temperature
b) Conductivity does not depend on the viscosity of the solution
c) Conductivity depends on the size of the ions
d) Conductivity depends on the solvation of the ions
Correct answer:
b) Conductivity does not depend on the viscosity of the solution
Solution: Let's analyze each option:
Option A: "Electrolytic conductance increases with increase in temperature."
As temperature increases, ions move faster due to decreased viscosity and increased kinetic energy. This leads to higher conductance. Therefore, Option A is correct.
Option B: "Conductivity does not depend on the viscosity of the solution."
Viscosity affects the mobility of ions. According to models like the Stokes-Einstein relation, higher viscosity means lower ion mobility, which decreases conductivity. Hence, conductivity does indeed depend on viscosity, making Option B incorrect.
Option C: "Conductivity depends on the size of the ions."
Smaller ions typically move faster than larger one, leading to better conductivity. Thus, the size of the ions has an effect on conductivity, making Option C correct.
Option D: "Conductivity depends on the solvation of the ions."
Solvation affects the effective size and mobility of the ions. Strong solvation increases the effective radius and may reduce mobility, thereby affecting conductivity. So, Option D is also correct.
Based on the analysis, the incorrect statement is:
Option B
Q4. When Lead Storage battery is in the process of getting charged which one of the following reactions takes place?
a) PbO 2(s) is reduced.
b) PbSO 4(s) gets converted to Pb(s) and PbO2 (s)
c) SO 3 gas gets liberated and PbO2(s) is formed
d) Pb
(s)
gets confused
Correct answer:
b) PbSO
4(s)
gets converted to Pb(s) and PbO2
(s)
Solution:
The lead-acid battery charging process involves reversing the chemical reactions that occur during discharging. In a lead-acid battery, there are two electrodes: a lead dioxide (PbO2) cathode and a lead (Pb) anode, both immersed in an electrolyte of sulfuric acid (H
2
SO
4
).
During the discharge process, at the anode (negative electrode), lead (Pb) is oxidized to lead sulfate (PbSO4):
Pb+SO 2- 4 →PbSO 4 +2e -
At the cathode (positive electrode), lead dioxide (PbO2) is reduced to lead sulfate (PbSO4) as well using the electrons that flow through the external circuit:
PbO 2 + 4H + + SO 2- 4 + 2e - →PbSO 4 + 2H 2 O
When the battery is being charged, these reactions are reversed. At the anode:
PbSO 4 + 2e - →Pb + SO 2- 4
This is where lead sulfate is converted back to metallic lead. And at the cathode:
PbSO 4 + 2H 2 O → PbO 2 + 4H + + SO 2- 4 + 2e -
Here, lead sulfate is converted back to lead dioxide.
In this context, Option B is the correct answer. During the charging process, PbSO4 indeed gets converted to Pb and PBO2 and according to the reactions described. To be specific, these reactions take place at the two different electrodes of the battery (lead sulfate at both electrodes being converted into lead and lead dioxide, respectively).
Q5. For a cell reaction, A(s) + B 2+ (aq) → A 2+ (aq) + B(s); the standard emf of the cell is 0.295V at 25°C. The equilibrium constant at 25°C will be:
a) 1 x 10 10
b) 10
c) 2.95 x 10 -2
d) 2.95 x 10 -10
Correct answer: a) 1 x 10 10
Solution:
For this given reaction, n = 2. At 25°C,
K = antilog [nE°/0.059] = antilog [2x(-0.295/0.059]
K = 1x10 10
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