Electrochemistry MCQ Chemistry Chapter 3 Question 1.

Debye-Huckel Onsager equation for strong electrolytes: \(\wedge=\wedge_{0}-\mathrm{A} \sqrt{\mathrm{C}}\) Which of the following equality holds?

(A) \(\wedge=\wedge_{\mathrm{o}} \text { as } C \longrightarrow \sqrt{A}\)
(B) \(\wedge=\wedge_{0} \text { as } \mathrm{C} \longrightarrow \infty\)
(C) \(\wedge=\wedge_{\mathrm{o}} \text { as } \mathrm{C} \longrightarrow 0\)
(D) \(\wedge=\wedge_{o} \text { as } \mathrm{C} \longrightarrow 1\)
Answer:
(B) \(\wedge=\wedge_{0} \text { as } \mathrm{C} \longrightarrow \infty\)

Explanation:
When c → ∞
Then \(\wedge=\wedge_{o}\)

Electrochemistry Class 12 MCQ Chapter 3 Question 2. 

Which of the following option will be the limiting molar conductivity of CH₃COOH if the limiting molar conductivity of CH₃COONa is 91 Scm² mol^-1? Limiting molar conductivity for individual ions are given in the following table.

(A) 350 Scm² mol^-1
(B) 375.3 Scm² mol^-1
(C) 390.5 Scm² mol^-1
(D) 340.4 Scm² mol ^-1
Answer:
(C) 390.5 Scm² mol^-1

Explanation:
The limiting molar conductivity (A$ for strong and weak electrolyte can be determined by using Kohlrausch’s law which states that “the limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte.”

\(\wedge\) CH₃COONa = \(\wedge\) CH₃COO“ + ANS1
91 Scm² mol^-1 = \(\wedge\) CH₃COO + 50.1 Scm² mol^-1
=> \(\wedge\)CH3COO” = 40.9 Scm² mol^-1
For acetic acid,
\(\wedge\) CH₃COOH = \(\wedge\)CH₃COO^– + \(\wedge\)H⁺
= 40.9 Scm² mol^-1 + 349.6 Scm² mol^-1 = 390.5 Scm² mol^-1

MCQ On Electrochemistry Chemistry Chapter 3 Question 3.

Which of the statements about solutions of electrolytes is not correct ?

(A) Conductivity of solution depends upon size of ions.
(B) Conductivity depends upon viscosity of solution.
(C) Conductivity does not depend upon solvation of ions present in solution.
(D) Conductivity of solution increases with temperature.
Answer:
(B) Conductivity depends upon viscosity of solution.

Explanation:
Conductivity depends upon solvation of ions present in solution. Greater the solvation of ions of an electrolyte, lesser will be the electrical conductivity of the solution.

Electrochemistry MCQ Questions Chapter 3 Question 4.

When 0.1 mol COCl₃(NH₃)₅ is treated with excess of AgNO₃, 0.2 mol of AgCl are obtained. The conductivity of solution will correspond to

(A) 1 : 3 electrolyte
(B) 1 : 2 electrolyte
(C) 1 : 1 electrolyte
(D) 3 : 1 electrolyte
Answer:
(B) 1:2 electrolyte

Explanation:
When 0.L mole of COCl₃(NH₃)₅ was reacted with excess of Ag.NO₃, we gel 0.2 moles of Agt’l. So, there are two chloride ions that are free and not part of the complex. The formula for 1 complex has to be [CO(NH₃)₅Cl]Cl₂ [CO(NH₃)Cl]Cl₂ → [CO(.NH₃)₅Cl]²⁺ + 2Cl^– Therefore, the conductivity of the solution will be 1: 2 electrolyte.

MCQ Of Electrochemistry Class 12 Chapter 3 Question 5.

The cell constant of a conductivity cell

(A) Changes with change of electrolyte.
(B) Changes with change of concentration of electrolyte.
(C) Changes with temperature of electrolyte.
(D) Remains constant for a cell.
Answer:
(D) Remains constant for a cell.

Explanation:
The cell constant of a conductivity cell remains constant for a cell.

Electrochemistry MCQ Class 12 Chemistry Question 6.

\(\Lambda^{0} m\left[\mathrm{NH}_{4} \mathrm{OH}\right]\) is equal to …………..

(A) \(\Lambda_{\mathrm{m}\left[\mathrm{NH}_{4} \mathrm{OH}\right]}^{0}+\Lambda_{\mathrm{m}\left[\mathrm{NH}_{4} \mathrm{Cl}\right]}^{0}-\Lambda_{[\mathrm{HCl}]}^{0}\)
(B) \(\Lambda_{\mathrm{m}\left[\mathrm{NH}_{4} \mathrm{Cl}\right]}^{0}+\Lambda_{\mathrm{m}[\mathrm{Na} \mathrm{OH}]}^{0}-\Lambda_{[\mathrm{NaCl}]}^{0}\)
(C) \(\Lambda_{\mathrm{m}\left[\mathrm{NH}_{4} \mathrm{Cl}\right]}^{0}+\Lambda_{\mathrm{m}[\mathrm{NaCl}]}^{0}-\Lambda_{[\mathrm{NaOH}]}^{0}\)
(D)\(\Lambda_{\mathrm{m}[\mathrm{NaOH}]}^{0}+\Lambda_{\mathrm{m}[\mathrm{NaCl}}^{0}-\Lambda_{\left[\mathrm{NH}_{\mathrm{C}} \mathrm{Cl}\right]}^{0}\)
Answer:
(B) \(\Lambda_{\mathrm{m}\left[\mathrm{NH}_{4} \mathrm{Cl}\right]}^{0}+\Lambda_{\mathrm{m}[\mathrm{Na} \mathrm{OH}]}^{0}-\Lambda_{[\mathrm{NaCl}]}^{0}\)

Explanation:
NH₄Cl NH₄Cl ⇌ NH⁺₄Cl^–
NaCl ⇌ Na⁺ + Cl^– _{(ii)
NaOH ⇌ Na⁺ OH^– (iii)
NH4OH ⇌ NH⁺4 + 0H^– (iv)
To get equation (iv)
\(\mathrm{A}_{\mathrm{m}}^{\circ}\left(\mathrm{NH}_{4} \mathrm{Cl}\right)+(\mathrm{NaOH})^{-} \Lambda_{\mathrm{m}}^{\circ}(\mathrm{NaCl})=\Lambda^{\circ}{ }_{\mathrm{m}}\left(\mathrm{NH}_{4} \mathrm{OH}\right)\)}

Class 12 Chemistry Chapter 3 MCQ Chemistry Question 7.

In the electrolysis of aqueous sodium chloride solution which of the half cell reaction will occur at anode?

(A) Na⁺ (aq) + e^– → Na(s); E^Θ_cell= 2.71 V
(B) 2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e^– E^Θ_cell = 1.23V
(C) H⁺ (aq) + e^– → \(\frac {1}{2}\) H₂(g); E^Θ_cell = 0.00 V
(D)C^– → \(\frac {1}{2}\) Cl₂(g) + e; E^Θ_cell = 1.36V
Answer:
(B) 2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e^– E^Θ_cell = 1.23V

Explanation:
During electrolysis
NaCl →Na⁺ + Cl^–
H₂O →H⁺ + OH^–
Na^– + e^– → Na(E^Θ_cell = – 2.71V)
H⁺ + e^– \(\frac {1}{2}\) 4H₂ (E^Θ_cell = 0.00V)
Atcathode,
H₂O + e^– → \(\frac {1}{2}\)H₂ + OH^–
At anode, two reactions are possible.
Cl^– → \(\frac {1}{2}\)Cl₂ + e^–; E^Θ_cell = 1.36 V
2H₂O → O₂ + 4H + 4e^–; E^Θ_cell = 1.23 V

MCQ On Electrochemistry Class 12 With Answers Question 8.

Which of the following statement is corred?

(A) E_cell and ∆_rG of cell reaction both extensive properhes.
(B) E_cell and ∆_rG of cell reaction both intensive properties.
(C) E_cell is an intensive property while ∆_rG of cell reaction is an extensive property.
(D) E_cell is an extensive property while ∆_rG of cell reaction is an intensive property.
Answer:
(C) E_cell is an intensive property while ∆_rG of cell reaction is an extensive property.

Explanation:
Eu is an intensive property and it does not depend upon number of partides but ∆_rG of the cell reaction is an extensive property because this depends upon number of particles.

Electrochemistry MCQs Chemistry Chapter 3 Question 9.

An elecftochemical cell behaves like an electrolytic cell when

(A) E_cell = E_external
(B) E_external = 0
(C) E_external > E_cell
(D) E_external < E_cell
Answer:
(C) E_external > E_cell

Explanation:
If an external opposite potential is applied on the electrochemical cell, the reaction continues to take place till the opposite voltage reaches the value 1.1 V. At this stage, no current flow through the cell and if there is any further increase in the external potentia1(EeJ, then reaction starts functioning in opposite direction i.e. an electrochemical cell behaves like an electrolytic cell.
E_external < E_cell

MCQ Electrochemistry Class 12 Chapter 3 Question 10.

ELectrode potential for Mg electrode varies according to the equation:
\(\mathrm{E}_{\mathrm{Mg}^{2+} \mathrm{Mg}}=\mathrm{E}_{\mathrm{Mg}^{2+} \mathrm{Mg}}^{\circ}-\frac{0.059}{2} \log \frac{1}{\left[\mathrm{Mg}^{2+}\right]}\)
The graph of \(\mathrm{E}_{\mathrm{Mg}^{2+} / \mathrm{Mg}}\) , vs. log [Mg²⁺] is

Answer:
Option (B) is correct.

Explanation:
\(\mathrm{E}_{\mathrm{Mg}^{2+} \mathrm{Mg}}=\mathrm{E}_{\mathrm{Mg}^{2+} \mathrm{Mg}}^{\circ}-\frac{0.059}{2} \log \frac{1}{\left[\mathrm{Mg}^{2+}\right]}\)
Compare this equation with the equation of straight line y = mx + c.
The graph of \(\mathrm{E}_{\mathrm{Mg}^{2+} / \mathrm{Mg}}\) vs. log [Mg²⁺] is a straight line with a positive slope and intercept \(\mathrm{E}_{\mathrm{Mg}^{2+} / \mathrm{Mg}}\)

MCQ Of Electrochemistry Chemistry Chapter 3 Question 11.

In an electrochemical process, a salt bridge is used

(A) as a reducing agent
(B) as an oxidizing agent
(C) to complete the circuit so that current can flow
(D) None of these
Answer:
(C) to complete the circuit so that current can flow

Explanation:
In an electrochemical cell, a salt bridge is used to complete the circuit so that current can flow.

Class 12 Electrochemistry MCQ Chemistry Question 12.

Following reactions occur at cathode during the electrolysis of aqueous silver chloride solution:
\(\mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{e}^{-} \rightarrow \mathrm{Ag}(\mathrm{s})\) E⁰ = +0.80 V
\(\mathrm{H}^{+}(\mathrm{aq})+\mathrm{e}^{-} \rightarrow \frac{1}{2} \mathrm{H}_{2}(\mathrm{~g})\) E⁰ = +0.80 V
On the basis of their standard reduction electrode potential (E°) vaLues, which reaction is feasible at the cathode?

(A) Ag⁺(aq) + e^– → Ag(s) E⁰ = + 0.80 V
(B) H⁺(aq) + e^– → \(\frac {1}{2}\) H₂(g) E⁰ = 0.00 V
(C) Both reactions are feasible
(D) None of the above
Answer:
(A) Ag⁺(aq) + e^– → Ag(s) E⁰ = + 0.80 V

Explanation:
Ag⁺(aq) + e^– → Ag(s) E⁰ = + 0.80 V
H⁺(aq) + e^– → \(\frac {1}{2}\) H₂(g) E⁰ = 0.00 V
On the basis of their standard reduction potential (E⁰) values, cathode reaction is given by the one with higher E⁰ values. Thus, Ag⁺(aq) + e^– → Ag(s) reaction will be more feasible at cathode.

MCQ Questions For Class 12 Chemistry Chapter 3 Question 13.

Consider the following reaction: Cu(s) + 2Ag⁺(aq) → 2Ag(s) + Cu²⁺(aq) Depict the galvanic cell in which the given reaction takes place.

(A) Cu²⁺ (aq)|Cu(s) ||Ag⁺(aq)|Ag(s)
(B) Cu(s) | Cu²⁺(aq) || Ag⁺ (aq)|Ag(s)
(C) Ag⁺(aq)|Ag(s)||Cu²⁺(aq)|Cu(s)
(D)Ag⁺(s)|Ag⁺(aq)||Cu²⁺(aq)|Cu(s)
Answer:
(B) Cu(s) | Cu²⁺(aq) || Ag⁺ (aq)|Ag(s)

Explanation:
Oxidation half reaction
Cu(s) + Cu²⁺(aq) + 2e^– .
Reduction half reaction

Chapter 3 Chemistry Class 12 MCQ Question 14.

Which of the following statements is not correct?

(A) Copper liberates hydrogen from acids.
(B) In its higher oxidation states, manganese forms stable compounds with oxygen and fluorine.
(C) Mn³⁺ and CO³⁺ are oxidising agents in aqueous solution.
(D) Ti³⁺ and Cr²⁺ are reducing agents in aqueous solution.
Answer:
(A) Copper liberates hydrogen from acids.

Explanation:
Copper does not liberate hydrogen from adds because copper lies below hydrogen in electrochemical series. So, copper does not have sufficient electrode potential to liberate elemental hydrogen form compounds in which oxidation state of hydrogen is +1.

MCQ On Electrochemistry Class 12 Pdf Question 15.

Calculate the emf of the following cell at 298 K: Mg(s)|Mg2 (0.1 M)||Cu²⁺ (1.0 x 10^-3M)|Cu(s) [Given E⁰_cell = 2.71 V]

(A) 1.426 V
(B) 2.503 V
(C) 2.651 V
(D) 1.8 V
Answer:
(C) 2.651 V

Explanation:
E_cell = \(E_{\text {Cell }}^{0}-\frac{0.059}{n} \log \frac{\left[\mathrm{Mg}^{2+}\right]}{\left[\mathrm{Cu}^{2+}\right]}\)
= \(2.71 \mathrm{~V}-\frac{0.059}{2} \log \frac{0.1}{0.001}\)
= \(2.71 \mathrm{~V}-\frac{0.059}{2} \log 10^{2}\)
E_cell = 2.651V

MCQs On Electrochemistry Chemistry Chapter 3 Question 16.

Reduction potentials of some ions are given below. Arrange them in decreasing order of oxidising power.

(A) ClO^–₄> IO^–₄> BrO^–₄
(B) IO^–₄> BrO^–₄> ClO^–₄
(C) BrO^–₄> IO^–₄> ClO^–₄
(D) BrO^–₄> ClO^–₄> IO^–₄
Answer:
(C) BrO^–₄> IO^–₄> ClO^–₄

Explanation:
Higher the reduction pøtential, higher is its tendency to get reduced. Hence, the order of oxidising power is: ClO^–₄> IO^–₄> BrO^–₄

Electrochemistry Class 12 MCQ Questions Question 17.

Using the data given below find strongest reduction agent.

\(\mathrm{E}_{\mathrm{Cr}_{2} \mathrm{O}_{7}^{2}-\mathrm{Cr}^{3+}}^{-1}=1.33 \mathrm{~V}, \mathrm{E}_{\mathrm{Cl}_{2} / \mathrm{Cl}^{-}}\) = 1.36 V
\(\mathrm{E}_{\mathrm{MnO}_{4} / \mathrm{Mn}^{2+}}^{-}=1.51 \mathrm{~V}, \mathrm{E}_{\mathrm{Cr}^{3} / \mathrm{Cr}}^{-}\) = – 0.74 V
(A) Cl^–
(B) Cr
(C) Cr³⁺
(D) Mn²⁺
Answer:
(B) Cr

Explanation:
The negative value of standard reduction potential for Cr to Cr means that the redox couple is a stronger reducing agent.

Chemistry Class 12 Chapter 3 MCQ Question 18.

What will happen during the electrolysis of aqueous solution of CuSO₄ by using platinum electrodes?

(A) Copper will deposit at cathode.
(B) Copper will deposit at anode.
(C) Oxygen will be released at anode.
(D) Copper will dissolve at anode.
Answer:
(C) Oxygen will be released at anode.

Explanation:
CuSO₄ ⇌ Cu²⁺ + SO^2-₄
H₂O = H⁺ + OH^–
At cathode, .
Cu²⁺ + 2e^2- → Cu; E^Θ_cell = 0.34 V
H⁺ + e^– → \(\frac {1}{2}\); H₂E^Θ_cell= 0.00 V
This reaction will take place due to higher reduction potential.
At anode,
2SO^2- + 2e^– → S₂O^2- + 2e⁸ E^Θ_cell = 1. 96 V
2H₂O → O₂ + 4H⁺ + 4e^– E^Θ_cell = 1.23 V

The reaction with lower value of E⁰ will be preferred a t anode, hence O₂ is released at anode.

Class 12 Chemistry Ch 3 MCQ Question 19.

What will happen during the electrolysis of aqueous solution of CuSO₄ in the presence of Cu electrodes?

(A) Copper will deposit at cathode.
(B) Copper will dissolve at anode.
(C) Oxygen will be released at anode.
(D) Copper will deposit at anode.
Answer:
(A) Copper will deposit at cathode.

Explanation:
Electrolysis of CuSO₄ can be represented by two half-cell reactions these occurring at cathode and anode, respectively, as given below:
At cathode: Cu²⁺ + 2e^2- → Cu(s)
At anode : Cu(s) →Cu²⁺ + 2e^–
In above reaction Cu will deposit at cathode while copper will dissolve at anode. Hence, (a) and (b) are the correct options.

Ch 3 Chemistry Class 12 MCQ Question 20.

Conductivity K, is equal to …………..

(A) \(\wedge_{m}\)
(B) \(\frac {G}{R}\)
(C) \(\frac {1}{A}\)
(D) All of the above
Answer:
(B) \(\frac {G}{R}\)

Explanation:
k = \(\frac{1}{R} \cdot \frac{1}{A} \text { or } \frac{G}{R}\)

Assertion And Reason Based MCQs (1 Mark each)

Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false and R is True

Question 1.

Assertion (A): Conductivity of an electrolyte increases with decrease in concentration.
Reason (R): Number of ions per unit volume decreases on dilution.

Answer:
(D) A is false and R is True

Explanation:
Conductivity of an electrolyte I decreases with decrease in concentration because of ions per unit volume decreases on dilution.

Question 2.

Assertion (A): \(\Lambda_{\mathrm{m}}\) for weak electrolytes shows a sharp increase when the electrolytic solution is diluted. The reaction with lower value of E⁰ will be preferred at anode, hence 0₂ is released at anode.
Reason (R): For weak electrolytes degree of dissociation increases with dilution of solution.

Answer:
(A) Both A and R are true and R is the correct explanation of A

Explanation:
Weak electrolytes dissociate partially in concentrated solution. On dilution, their degree of dissociation increases hence, their A. increases sharply.

Question 3.

Assertion (A): Electrolytic conduction increases with increase in temperature.
Reason (R): Increase in temperature cause the electronic movement more rapid

Answer:
(C) A is true but R is false

Explanation:
As the temperature of electrolytic solution is increased, the kinetic energy of the ion increases. This results in the increase of electrical conductance of electrolytic solutions.

Question 4.

Assertion (A): Molar Conductivity of an ionic solution depends on temperature.
Reason (R): Molar Conductivity of an ionic solution depends on the concentration of electrolytes in the solution.

Answer:
(B) Both A and R are true but R is NOT the correct explanation of A

Explanation:
Molar Conductivity of .an ionic J solution depends on (he temperature as well E as on the i omentiation of the electrolytes ini Ihe solution.

Question 5.

Assertion (A): EceU should have a positive value for the cell to function.
Reason (R): E_cathode < E_anode

Answer:
(C) A is true but R is false

Explanation:
E_cell = E_cathode – E_anode To have positive value of E_cell, E_Cathode should be
greater than E_anode

Question 6.

Assertion (A): Cu is less reactive than hydrogen.
Reason (R): \(\mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\Theta}\) is negative.

Answer:
(C) A is true but R is false.

Explanation:
Cu is less reactive than hydrogen because \(\mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\Theta}\) is positive.

Question 7.

Assertion (A): Copper sulphate can be stored in zinc vessel.
Reason (R): Zinc is more reactive than copper.

Answer:
(D) A is false and R is True

Explanation:
Zinc will get dissolved in CuSO₄ solution, since, zinc is more reactive than copper.

Question 8.

Assertion (A): \(\mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}\) increases with increase in concentration of Ag⁺ ions.
Reason (R): \(\mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}\) has a positive value.

Answer:
(B) Both A and R are true but R is NOT the correct explanation of A

Explanation:
Ag + e^– → Ag
\(\mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}=\mathrm{E}^{\Theta}{\mathrm{Ag}^{\prime} / \mathrm{Ag}}-\frac{\mathrm{RT}}{\mathrm{nF}} \log \frac{1}{\left[\mathrm{Ag}^{+}\right]}\)
On increasing [Ag4], EAg+/ Ag will increase and it will increase and it has a positive value.

Question 9.

Assertion (A): Electrolysis of NaCl solution gives chlorine at anode instead of O₂.
Reason (R): Formation of oxygen at anode requires over voltage.

Answer:
(A) Both A and R are true and R is the correct explanation of A

Explanation:
Formation of oxygen has lower I value of I:° than formation of chlorine even then it is not formed because it requires over voltage.

Case-Based MCQS’

I. Read the passage given below and answer the following questions:
The cell constant is usually determined by measuring the resistance of the cell containing a solution whose conductivity is already known. For this purpose, we generally use KCl solutions whose conductivity is known accurately at various concentrations and at different temperatures. Consider the resistance of a conductivity cell filled with 0.1 M KCl solution is 200 Ohm. If the resistance of the same cell when filled with 0.02 M KCl solution is 420 Ohm. (Conductivity of 0.1 M KCl solution is 1.29 Sm^-1.) The following questions are Multiple Choice

Questions. Choose the most appropriate answer:

Question 1.

What is the conductivity of 0.02 M KCl solution?

(A) 0.452 S m^-1
(B) 0.215 S m^-1
(C) 0.614 S m^-1
(D) 0.433 S m^-1
Answer:
(C) 0.614 S m^-1

Explanation:
Conductivity of 0.02 mol L^-1 KCl Solution = Cell constant resistance \(\frac {258}{420}\) = 0.614 Sm^-1

Question 2.

What will happen to the conductivity of the cell with the dilution ?

(A) First decreases then increases
(B) Increases
(C) First increases then decreases
(D) Decreases
Answer:
(D) Decreases

Explanation:
The conductivity decreases with dilution.

Question 3.

The cell constant of a conductivity cell …………

(A) Changes with change of electrolyte.
(B) Changes with change of concentration of electrolyte.
(C) Changes with temperature of electrolyte.
(D) Remains constant for a cell.
Answer:
(D) Remains constant for a cell.

Explanation:
The cell constant of a conductivity cell remains constant for a cell.

Question 4.

SI unit for conductivity of a solution is ………….

(A) S m^-1
(B) Sm² mol^-1
(C) mol cm^-3
(D) S cm² mol^-1
Answer:
(A) S m^-1

Explanation:
SI unit for conductivity of a solution is Sm^-1

OR

Which of the following is not true? The conductivity of solutions of different electrolytes in the same solvent and at a given temperature differs due to

(A) size of the ions in which they dissociate
(B) concentration of ions
(C) charge of the ions in which they dissociate
(D) is independent of ions movement under a potential gradient
Answer:
(D) is independent of ions movement under a potential gradient

Explanation:
The conductivity of solutions of different electrolytes in the same solvent and at a given temperature differs due to size and charge of the ions in which they dissociate, concentration of ions, ease with which the ions move under a potential gradient.

II. Read the passage given below and answer the following questions:
A galvanic cell consists of a metallic zinc plate immersed in 0.1 M Zn(NO₃)₂ solution and metallic plate of lead in 0.02 M Pb(NO₃)₂ solution. The following questions are multiple choice questions. Choose the most appropriate answer:

Question 5.

How will the cell be represented ?

(A) Zn(s) | Zn²⁺(aq)| | Pb²⁺(aq)|Pb(s)
(B) Zn²⁺(s)| Zn(aq) | | Pb²⁺(aq)|Pb(s)
(C) Pb²⁺(aq)|Pb(s)| |Zn²⁺(s)| Zn(aq)
(D) Pb(s)|Pb²⁺(aq)| |Zn²⁺(s)| Zn(aq)
Answer:
(A) Zn(s) | Zn²⁺(aq)| | Pb²⁺(aq)|Pb(s)

Explanation:
Cell representation:
Zn(s) | Zn²⁺ (aq) | | Pb²⁺ (aq),|Pb(s)

Question 6.

Calculate the emf of the cell.

(A) 6.01 V
(B) 0.412 V
(C) 0.609 V
(D) 4.12 V
Answer:
(C) 0.609 V

Explanation:
According to Xernst equation:
\(\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}-\frac{0.0591}{2} \log \frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Pb}^{2+}\right]}\)
\(\mathrm{E}_{\text {cell }}=[-0.13-(-0.76)]-\frac{0.0591}{2} \log \frac{0.1}{0.02}\)
= 0.63 – 0.02955 x log 5
= 0.63 – 0.02955 x 0.6990
= 0.63 – 0.0206 – 0.6094 V

Commonly Made Errors :

• The cell representation is given incorrectly by manv candidates.
• The calculation of emf of the cell by using Xernst equation is incorrect, in some cases.

Answering Tip :

• Do more practice of cell representation and numerical based on Nernst equation.

Question 7.

What product is obtained at cathode?

(A) Zn
(B) Pb
(C) Zn²⁺
(D) Pb²⁺
Answer:
(B) Pb

Explanation:
Anode reaction: Zn(s) → Zn²⁺ (aq) + 2e^–
Cathode reaction: Pb²⁺(aq) + 2e ^– → Pb(s)

Question 8.

Which of the following statement is not correct about an inert electrode in a cell ?

(a) It does not participate in the cell reaction.
(b) It provides surface either for oxidation or for reduction reaction.
(c) It provides surface for conduction of electrons.
(d) It provides surface for redox reaction.
Answer:
(a) It does not participate in the cell reaction.

Explanation:
Inert electrode dot’s not participate in redox reaction and acts only as source or sink tor electrons. It provides surface either fori oxidation or for reduction reaction.

III. Products of electrolysis depend on the nature of material being electrolysed and the type of electrodes being used. If the electrode is inert (e.g., platinum or gold), it does not participate in the chemical reaction and acts only as source or sink for electrons. On the other hand, if the electrode is reactive, it participates in the electrode reaction. Thus, the products of electrolysis may be different for reactive and inert electrodes. Aqueous copper sulphate solution and aqueous silver nitrate solution are electrolysed by ampere current for 10 minutes in separate electrolytic cells. In these questions, a statement of assertion followed by a statement of reason. Choose the correct answer out of the following choices.

(A) Assertion and reason both are correct statements and reason is correct explanation for assertion.
(B) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
(C) Assertion is correct statement but reason is wrong statement.
(D) Assertion is wrong statement but reason is correct statement.

Question 1.

Assertion (A): The mass of copper and silver, deposited on the cathode be same.
Reason (R): Copper and silver have different equivalent masses.

Answer:
(D) Assertion is wrong statement but reason is correct statement.

Explanation:
W = itE/96300 = 1 x 10 x 60 x 31.8/96500 for copper. It will be different for silver since the equivalent weight of silver is different.

Question 2.

Assertion (A): At equilibrium condition E_cell = 0 or ∆_rG = 0.
Reason (R): E_cell is zero when both electrodes of the cell are of the same metal.

Answer:
(B) Assertion and reason both are correct statements but reason is not correct explanation for assertion.

Explanation:
At equilibrium, condition of E_cell = 0, ∆_rG = 0

Question 3.

Assertion (A): The negative sign in the expression \(\mathrm{E}_{\mathrm{Zn}^{2}+/ \mathrm{Zn}}\) = – 0.76V means Zn²⁺cannot be oxidised to Zn.
Reason (R): Zn is more reactive than hydrogen & Zn will oxidised, & H⁺ will get reduced.

Answer:
(A) Assertion and reason both are correct statements and reason is correct explanation for assertion.

Explanation:
It shows that the reduced form of (Zn²⁺) is not stable. It is difficult to reduce Zn²⁺ to Zn. Rather the reverse reaction i.e Zn can get oxidised to Zn²⁺ and H~ will get reduced as it is stabler among both the reduced species.

Question 4

Assertion (A): In a galvanic cell, chemical energy is converted into electrical energy.
Reason (R): Redox reactions provide the chemical energy to the cell.

Answer:
(A) Assertion and reason both are correct statements and reason is correct explanation for assertion.

Explanation:
The redox reactions provide the chemical energy’ to the galvanic cell which is converted into electrical energy.

OR

Assertion (A): Copper sulphate cannot be stored in zinc vessel.
Reason (R): Zinc is less reactive than copper.

Answer:
(C) Assertion is correct statement but reason is wrong statement.

Explanation:
Copper sulphate cannot be stored in zinc vessel as zinc is more reactive than copper.

MCQ Questions for Class 12 Chemistry with Answers