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Important Questions for Class 12 Chemistry Chapter 3 - Electrochemistry

Class 12 chemistry important questions with answers are provided here for chapter 3 Electrochemistry. These important questions are based on CBSE board curriculum and correspond to the most recent Class 12 chemistry syllabus. By practicing these Class 12 important questions, students will be able to quickly review all of the ideas covered in the chapter and prepare for the Class 12 Annual examinations as well as other entrance exams such as NEET and JEE.

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Class 12 Electrochemistry Important Questions with Answers

Short Answer Type Questions

1. Can absolute electrode potential of an electrode be measured?

Solution: No, the absolute potential of an electrode can not be measured because the half-cell containing a single electrode cannot work on its own. It can work only in combination with another half cell.

2. Can E0cell or ΔrG0 for a cell reaction ever be equal to zero?

Solution: No, it cannot be equal to zero for a cell reaction proceeding in a particular direction (forward or backward direction)

3. Under what conditions is E0cell = 0 and ΔrG0 = 0 ?

Solution: At the condition of equilibrium, E0cell = 0 and ΔrG0 = 0

4. What does the negative sign in the expression E0Zn2+ /Zn = – 0.76 V means?

Solution: It implies that Zn is more reactive than hydrogen or it is a stronger reducing agent. In a cell containing zinc electrode and standard hydrogen electrode present in two half cells, zinc will be oxidised to Zn2+ ions while H+ ions will get reduced to hydrogen.

5. Aqueous copper sulphate solution and aqueous silver nitrate solution are electrolysed by 1 ampere current for 10 minutes in separate electrolytic cells. Will the mass of copper and silver deposited on the cathode be same or different ? Explain your answer.

Solution: The mass of copper and silver deposited on the cathode in the two electrolytic cells will not be the same.

According to second law of Electrolysis.

(Mass of Cu deposited /mass of Ag deposited) = (equivalent mass of Cu/equivalent mass of Ag)

= 31.75/ 108.

6. Depict the galvanic cell in which the cell reaction is : Cu + 2Ag+ → 2Ag + Cu2+.

Solution: In the cell reaction: Cu + 2Ag+ → 2Ag + Cu2+

At anode : Cu → Cu2+ + 2e

At Cathode : 2 Ag+ + 2e → 2Ag

The representation of cell = Cu(s) | Cu2+(aq) || Ag+(aq) | Ag (s)

7. Value of standard electrode potential for the oxidation of Clions is more positive than that of water, even then in the electrolysis of aqueous sodium chloride, why are Cl ions oxidised at anode instead of water ?

Solution: The oxidation reactions taking place at anode is:

2Cl (aq) → Cl2 (g) + 2e E0(oxid) = -1.36 V

2H2O (l) → O2 (g) + 4H+ (aq) + 4e E0(oxid) = -1.23 V

Due to the overvoltage of oxygen (O₂), its liberation is kinetically slower than that of CI ions. Therefore, Cl ions are oxidised to Cl₂ gas.

8. What is electrode potential ?

Solution: Electrode potential appears at the interface between an electrode and electrolyte due to the transfer of charged species across the interface.

9. Why is alternating current used for measuring resistance of an electrolytic solution?

Solution: Alternate current is used for measuring the resistance of an electrolytic solution because direct current will cause the electrolysis of the electrolyte. As a result Concentration of the ions in the solution will change.

10. How will the pH of brine (NaCl solution) be affected when it is electrolysed?

Solution: NaCl is a salt of strong acid and strong base. It is neutral in nature. pH of NaCl solution is 7. However, upon electrolysis of the aqueous solution, Na+ and OH ions are formed. The solution becomes basic and its pH rises.

11. Solutions of two electrolytes ‘A’ and ‘B’ are diluted. The Λm of ‘B’ increases 1.5 times while that of A increases 25 times. Which of the two is a strong electrolyte? Justify your answer.

Solution: Dilution of the electrolyte mainly decreases interionic attractive force, therefore Λm increases. In case of strong electrolyte it dissociated completely in solution so on dilution Λm increases to a small extent. On the other hand, the electrolyte weak is ionised only to a small extent. Upon dilution, its ionisation or dissociation increases to a large extent. That is why, Λm for weak electrolyte is more higher on dilution.

From the available information A is a weak electrolytes and B is a strong electrolyte.

12. In an aqueous solution how does specific conductivity of electrolytes change with addition of water?

Solution: The specific conductivity or conductivity (🇰) of electrolytes decreases with the addition of water or upon dilution because the number of ions per unit volume decrease.

13. Which reference electrode is used to measure the electrode potential of other electrodes ?

Solution: Standard hydrogen electrode (SHE) is used as reference electrode to measure the electrode potential of other electrodes. Its electrode potential is assumed as zero. The electrode potential of any electrode as measured with respect to a standard hydrogen electrode is known as its standard electrode potential (E).

14. Consider a cell given below

Cu |Cu²+ || Cl- | Cl₂, Pt

Write the reactions that occur at anode and cathode.

Solution: The reaction taking place at the two electrodes are:

At anode : Cu(s) → Cu2+(aq) + 2e

At cathode: Cl2 (g) + 2e → 2 Cl

15. Write the Nernst equation for the cell reaction in the Daniell cell. How will the Ecell be affected when concentration of Zn2+ ions is increased ?

Solution: Daniell cell involves the redox reaction :

Zn(s) + CuSO4(aq) → ZnSO4 (aq) + Cu(s)

Thus in the cell Zn(s) is oxidised to Zn²+ (aq) ions in the oxidation half cell while Cu²+ (aq) ions are reduced to Cu(s) in the reduction half cell.

According to Nernst equation,

Ecell = E0cell – (0.059/2) {log[Zn2+(aq)] / [Cu2+(aq)]

Hence Ecell decreases as the [Zn2+(aq)] increases.

16. What advantage do the fuel cells have over primary and secondary batteries?

Solution: Primary batteries or cells contain a limited amount of reactants and are discharged when the reactants are consumed. Secondary batteries or cells can be recharged but the process takes a long time. Fuel cells work continuously as long the reactants (i.e. fuel) are supplied.

17. Write the cell reaction of a lead storage battery when it is discharged. How does the density of the electrolyte change when the battery is discharged?

Solution: The cell reaction of a lead storage battery when it is discharged, may be given as:

Pb(s) + 4H+(aq) + 2SO42- (aq) → PbO₂(s) + 2PbSO4 (s) + 2H₂O(l)

Density of the electrolyte i.e. conc.H2SO4 solution decreases because of the dilution of electrolyte taking place since water is formed as one of the products.

18. Why on dilution the Λm of CH3COOH increases drastically, while that of CH3COONa increases gradually?

Solution: CH3COOH is a weak electrolyte and is dissociated to small extent. With dilution, its degree of dissociation increases i.e. more ions are released in solution. Therefore, Λm of CH3COOH increases drastically. On the other hand, CH3COONa is a strong electrolyte and is almost completely dissociated in aqueous solution. With dilution, only interionic forces of attraction increase resulting in only small increase in the value of Λm .

Matching Types Questions

1. Match the terms given in Column I with the units given in Column II.

Column I Column II
(a) Λm (i) S cm-1
(b) Ecell (ii) m-1
(c) 🇰 (iii) S cm2 mol-1
(d) G* (iv) V

Solution: (a) – (iii); (b) – (iv); (c) – (i); (d) – (ii)

2. Match the items of Column I and Column II.

Column I Column II
(a) Lead storage battery (i) maximum efficiency
(b) Mercury cell (ii) prevented by galvanisation
(c) Fuel cell (iii) gives steady potential
(d) Rusting (iv) Pb is anode, PbO₂ is cathode

Solution: (a) – (iv); (b) – (iii); (c) – (i); (d) – (ii)

3. Match the items of Column I and Column II.

Column I Column II
(a) 🇰 (i) I x t
(b) Λm (ii) Λm / Λ0m
(c) α (iii) 🇰/c
(d) Q (iv) G*/R

Solution: (a) – (iv); (b) – (iii); (c) – (ii); (d) – (i)

4. Match the items of Column I and Column II.

Column I Column II
(a) Lechlanche cell (i) cell reaction 2H₂ + O₂ → 2H₂O
(b) Ni-Cd cell (ii) does not involve any iron in solution and is used in hearing aids.
(c) Fuel cell (iii) rechargeable
(d) Mercury cell (iv) reaction at anode, Zn → Zn2+ + 2e
(v) converts energy of combustion into electrical energy.

Solution: (a) – (iv); (b) – (iii); (c) – (i),(v); (d) – (ii)

Assertion-Reason Type Questions

Note: In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices:

(a) Both assertion and reason are true and the reason is correct explanation for assertion.

(b) Both assertion and reason are true and the reason is not correct explanation for assertion.

(c) Assertion is true but the reason is false.

(d) Both assertion and reason are false.

(e) Assertion is false but reason is true.

1. Assertion : Cu is less reactive than hydrogen.

Reason : E0Cu2+/Cu is negative.

Solution: (c) Assertion is true but the reason is false.

2. Assertion: Ecell should have a positive value for the cell to function.

Reason: Ecathode < Eanode

Solution:(c) Assertion is true but the reason is false.

3. Assertion : Conductivity of all electrolytes decreases on dilution.

Reason : On dilution number of ions per unit volume decreases.

Solution: (a) Both assertion and reason are true and the reason is correct explanation for assertion.

4. Assertion : Λm for weak electrolytes shows a sharp increase when the electrolytic solution is diluted.

Reason : For weak electrolytes degree of dissociation increases with dilution of solution.

Solution: (a) Both assertion and reason are true and the reason is correct explanation for assertion.

5. Assertion :Mercury cell does not give steady potential.

Reason: In the cell reaction, ions are not involved in solution.

Solution:(e) Assertion is false but reason is true.

Long Answer Types Questions

1. Define the terms specific conductance, molar conductance and equivalent conductance. Derive the relationship between molar conductance and equivalent conductance.

Solution:

(a) Specific conductance:

The reciprocal of specific resistance is termed as specific conductance or it is the conductance of one centimeter cube of a conductor.

It is denoted by the symbol 🇰 (kappa).

🇰 = 1/ρ or 🇰 = (l/a) x C

Specific conductance is also called conductivity.

(b) Molar conductance:

Molar conductance is defined as the conductance of all the ions produced by ionisation of 1 g- mole of an electrolyte when present in per litre of solution. It is denoted by Λm.

Molar conductance Λm = k x 1000/M

(c) Equivalent conductance:

Equivalent conductance is defined as the conductance of all ions produced by one gram-equivalent of an electrolyte present per litre of solution. It is denoted by Λeq.

Equivalent conductance Λeq = k x 1000/N.

Relationship between molar conductance and equivalent conductance.

Λm = k x 1000/M , Λeq = k x 1000/N

∴ Λm / Λeq = n

Where n= valence factor.

2. State Kohlrausch’s law. How does the law help in calculating Λ0 CH3COOH ?

Solution:

At infinite dilution, when dissociation is complete, each ion makes a definite contribution towards equivalent conductance of the electrolyte irrespective of the nature of the ion with which it is associated and the value of equivalent conductance at infinite dilution for any electrolyte is the sum of contribution of its constituent ions”, i. e. anions and cations.

Λ = λa + λc

The λa and λc are called the ionic conductances of cation and a anion at infinite dilution respectively.

The molar conductivity of an strong electrolyte at infinite dilution is the sum of the ionic conductivity of the cations and anions.

In case of weak electrolytes, it is not possible to determine the value of limiting molar conductance at infinite dilution. However this can be determine by using Kohlrausch’s law.

Example CH3COOH is a weak electrolyte.

Λm (CH3COOH) = λm (CH3COO-) + λm (H+)

=[ λm (CH3COO-) + λm (Na+)] + [ λm (H+) + λm (Cl)] – [ λm (Na+) + λm (Cl)]

= Λm (CH3COONa) + Λm (HCl) – Λm (NaCl)

3. What is the relationship between Gibbs free energy of the cell reaction in a galvanic cell and the emf or Ecell ? When will the maximum work be obtained from a galvanic cell ?

Solution:

The relation may be given as :

ΔG = -nF Ecell

If the redox reaction is under standard conditions, then

ΔG° = -nFEcell

Here nF is the quantity of charge passed. In case we want to obtain maximum work from the galvanic charge has to be passed reversibly. As pointed above, the reversible work done by the galvanic cell is equal decrease in its Gibbs free energy.

4. What is Nernst equation ? write the Nernst equation for the reaction.

2Cr(s) + 3Cd2+ (aq) → 2Cr3+ (aq) + 3Cd (s)

Solution:

Nernst equation is an equation relating the capacity of an atom/ion to take up one or more electrons (reduction potential) measured at any conditions to that measured at standard conditions (standard reduction potentials) of 298K and one molar or one atmospheric pressure.

Ecell = E0 – [RT/nF] ln Q

Where,

Ecell = cell potential of the cell

E0 = cell potential under standard conditions

R = universal gas constant

T = temperature

n = number of electrons transferred in the redox reaction

F = Faraday constant

Q = reaction quotient

In the equation : 2Cr(s) + 3Cd2+ (aq) → 2Cr3+ (aq) + 3Cd (s)

Ecell = E0 – [RT/nF] ln Q

Ecell = E0 – [RT/nF] ln {[Cr3+]2 /[Cd2+]3 }

5. State Faraday’s first law of electrolysis. How much charge in terms of Faraday is required for reduction of 1 mole of Cu2+ ions to Cu?

Solution:

When an electric current is passed through an electrolyte, the amount of substance deposited is proportional to the quantity of electric charge passed through the electrolyte.

If W be the mass of the substance deposited by passing Q coulomb of charge, then according to the law, we have the relation:

W ∝ Q

A coulomb is the quantity of charge when a current of one ampere is passed for one second. Thus, amount of charge in coulombs,

Q = current in amperes x time in seconds

Q = I x t

W ∝ I x t

W = Z x I x t

where, Z is a constant, known as electrochemical equivalent, and is characteristic of the substance deposited.

When a current of one ampere is passed for one second,i.e one coulomb (Q = 1), then

W = Z

The electrode reaction is Cu2+ + 2e−→Cu

∴ quantity of charge required for reduction of 1 mole of Cu2+ =2F

Cu2+=2F =2×96500=193000C

Test your Knowledge on Important questions class 12 chemistry chapter 3 electrochemistry!

1 Comment

  1. Awesome ..It’s very useful.

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