Practice CBSE Class 12 Chemistry Electrochemistry MCQs Set 07 provided below. The MCQ Questions for Class 12 Unit 2 Electrochemistry Chemistry with answers and follow the latest CBSE/ NCERT and KVS patterns. Refer to more Chapter-wise MCQs for CBSE Class 12 Chemistry and also download more latest study material for all subjects
MCQ for Class 12 Chemistry Unit 2 Electrochemistry
Class 12 Chemistry students should review the 50 questions and answers to strengthen understanding of core concepts in Unit 2 Electrochemistry
Unit 2 Electrochemistry MCQ Questions Class 12 Chemistry with Answers
MCQs
Question. Debye-Huckel Onsager equation for strong electrolytes: \( \wedge = \wedge_{\circ} - A\sqrt{C} \) Which of the following equality holds?
(a) \( \wedge = \wedge_{\circ} \) as \( C \to \sqrt{A} \)
(b) \( \wedge = \wedge_{\circ} \) as \( C \to \infty \)
(c) \( \wedge = \wedge_{\circ} \) as \( C \to 0 \)
(d) \( \wedge = \wedge_{\circ} \) as \( C \to 1 \)
Answer: (c) \( \wedge = \wedge_{\circ} \) as \( C \to 0 \)
Explanation: At infinite dilution, concentration approaches to zero and molar conductivity at infinite dilution is \( \wedge_{\circ} \), When \( C \to 0 \) Then \( \wedge = \wedge_{\circ} \)
Question. 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: (c) Conductivity does not depend upon solvation of ions present in solution.
Explanation: Conductivity depends upon the nature of the solute added, size of the ions produced, nature of solvent, viscosity of solvent and solvation of ions present in solution. Greater the solvation of ions of an electrolyte, lesser will be the electrical conductivity of the solution.
Question. When 0.1 mol \( \text{CoCl}_3(\text{NH}_3)_5 \) is treated with excess of \( \text{AgNO}_3 \), 0.2 mol of \( \text{AgCl} \) 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 0.1 mole of \( \text{CoCl}_3(\text{NH}_3)_5 \) was reacted with excess of \( \text{AgNO}_3 \), we get 0.2 moles of \( \text{AgCl} \). As, 1 mol \( \text{AgNO}_3 \) can give only 1 mole \( \text{AgCl} \). In the given reaction 1 mol \( \text{Co}(\text{NH}_3)_5\text{Cl}_3 \) will give 2 mol \( \text{Cl}^- \) ions, when \( 2\text{Cl}^- \) are not in coordinate sphere of complex. Thus, the formula for complex has to be \( [\text{Co}(\text{NH}_3)_5\text{Cl}]\text{Cl}_2 \).
\( [\text{Co}(\text{NH}_3)_5\text{Cl}]\text{Cl}_2 \to [\text{Co}(\text{NH}_3)_5\text{Cl}]^{2+} + 2\text{Cl}^- \)
Therefore, the conductivity of the solution will be 1: 2 electrolyte.
Question. 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 is directly proportional to distance between two conductive plates and inversely proportional to the cross section area. Thus, it remains constant for a particular cell.
Question. \( \Lambda^{\circ}_m [\text{NH}_4\text{OH}] \) is equal to ______ .
(a) \( \Lambda^{\circ}_m[\text{NH}_4\text{OH}] + \Lambda^{\circ}_m[\text{NH}_4\text{Cl}] - \Lambda^{\circ}_m[\text{HCl}] \)
(b) \( \Lambda^{\circ}_m[\text{NH}_4\text{Cl}] + \Lambda^{\circ}_m[\text{NaOH}] - \Lambda^{\circ}_m[\text{NaCl}] \)
(c) \( \Lambda^{\circ}_m[\text{NH}_4\text{Cl}] + \Lambda^{\circ}_m[\text{NaCl}] - \Lambda^{\circ}_m[\text{NaOH}] \)
(d) \( \Lambda^{\circ}_m[\text{NaOH}] + \Lambda^{\circ}_m[\text{NaCl}] - \Lambda^{\circ}_m[\text{NH}_4\text{Cl}] \)
Answer: (b) \( \Lambda^{\circ}_m[\text{NH}_4\text{Cl}] + \Lambda^{\circ}_m[\text{NaOH}] - \Lambda^{\circ}_m[\text{NaCl}] \)
Explanation:
\( \text{NH}_4\text{Cl} \rightleftharpoons \text{NH}_4^+ + \text{Cl}^- \) (i)
\( \text{NaCl} \rightleftharpoons \text{Na}^+ + \text{Cl}^- \) (ii)
\( \text{NaOH} \rightleftharpoons \text{Na}^+ + \text{OH}^- \) (iii)
\( \text{NH}_4\text{OH} \rightleftharpoons \text{NH}_4^+ + \text{OH}^- \) (iv)
To get equation (iv):
\( \Lambda^{\circ}_m(\text{NH}_4\text{Cl}) + \Lambda^{\circ}_m(\text{NaOH}) - \Lambda^{\circ}_m(\text{NaCl}) = \Lambda^{\circ}_m(\text{NH}_4\text{OH}) \)
Question. In the electrolysis of aqueous sodium chloride solution which of the half cell reaction will occur at anode?
(a) \( \text{Na}^+(\text{aq}) + e^- \to \text{Na}(\text{s}) \); \( E_{\text{Cell}}^{\circ} = -2.71 \text{ V} \)
(b) \( 2\text{H}_2\text{O}(\text{l}) \to \text{O}_2(\text{g}) + 4\text{H}^+(\text{aq}) + 4e^- \); \( E_{\text{Cell}}^{\circ} = 1.23 \text{ V} \)
(c) \( \text{H}^+(\text{aq}) + e^- \to \frac{1}{2}\text{H}_2(\text{g}) \); \( E_{\text{Cell}}^{\circ} = 0.00 \text{ V} \)
(d) \( \text{Cl}^-(\text{aq}) \to \frac{1}{2}\text{Cl}_2(\text{g}) + e^- \); \( E_{\text{Cell}}^{\circ} = 1.36 \text{ V} \)
Answer: (d) \( \text{Cl}^-(\text{aq}) \to \frac{1}{2}\text{Cl}_2(\text{g}) + e^- \); \( E_{\text{Cell}}^{\circ} = 1.36 \text{ V} \)
Explanation: During electrolysis:
\( \text{NaCl} \to \text{Na}^+ + \text{Cl}^- \)
\( \text{H}_2\text{O} \to \text{H}^+ + \text{OH}^- \)
\( \text{Na}^+ + e^- \to \text{Na} \) (\( E_{\text{Cell}}^{\circ} = -2.71 \text{ V} \))
\( \text{H}^+ + e^- \to \frac{1}{2}\text{H}_2 \) (\( E_{\text{Cell}}^{\circ} = 0.00 \text{ V} \))
At cathode:
\( \text{H}_2\text{O} + e^- \to \frac{1}{2}\text{H}_2 + \text{OH}^- \)
At anode, two reactions are possible where lower of \( E^{\circ} \) is preferred:
\( \text{Cl}^- \to \frac{1}{2}\text{Cl}_2 + e^- \); \( E_{\text{Cell}}^{\circ} = 1.36 \text{ V} \)
\( 2\text{H}_2\text{O} \to \text{O}_2 + 4\text{H}^+ + 4e^- \); \( E_{\text{Cell}}^{\circ} = 1.23 \text{ V} \)
But due to slow oxidation process of \( \text{O}_2 \) it needs high voltage. Thus, \( \text{Cl}_2 \) will form rather than \( \text{O}_2 \) as preference.
Question. Which of the following statement is correct?
(a) \( E_{\text{Cell}} \) and \( \Delta_r G \) of cell reaction both extensive properties.
(b) \( E_{\text{Cell}} \) and \( \Delta_r G \) of cell reaction both intensive properties.
(c) \( E_{\text{Cell}} \) is an intensive property while \( \Delta_r G \) of cell reaction is an extensive property.
(d) \( E_{\text{Cell}} \) is an extensive property while \( \Delta_r G \) of cell reaction is an intensive property.
Answer: (c) \( E_{\text{Cell}} \) is an intensive property while \( \Delta_r G \) of cell reaction is an extensive property.
Explanation: \( E_{\text{cell}} \) is an intensive property and it does not depend upon number of particles but \( \Delta_r G \) of the cell reaction is an extensive property because this depends upon number of particles.
Question. An electrochemical cell behaves like an electrolytic cell when:
(a) \( E_{\text{cell}} = E_{\text{external}} \)
(b) \( E_{\text{cell}} = 0 \)
(c) \( E_{\text{external}} > E_{\text{cell}} \)
(d) \( E_{\text{external}} < E_{\text{cell}} \)
Answer: (c) \( E_{\text{external}} > E_{\text{cell}} \)
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 flows through the cell and if there is any further increase in the external potential (\( E_{\text{external}} \)), then reaction starts functioning in opposite direction i.e., an electrochemical cell behaves like an electrolytic cell. \( E_{\text{external}} > E_{\text{cell}} \)
Question. 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 the options
Answer: (c) to complete the circuit so that current can flow
Explanation: In an electrochemical cell, a salt bridge is used to keep solution electrically neutral and allow the flow of ions from one cell to another so that reaction can not stop, otherwise due to accumulation of ions on cathode and anode can stop reactions.
Question. Following reactions occur at cathode during the electrolysis of aqueous silver chloride solution: \( \text{Ag}^+(\text{aq}) + e^- \to \text{Ag}(\text{s}) \); \( E^{\circ} = +0.80 \text{ V} \) and \( \text{H}^+(\text{aq}) + e^- \to \frac{1}{2}\text{H}_2(\text{g}) \); \( E^{\circ} = 0.00 \text{ V} \). On the basis of their standard reduction electrode potential (\( E^{\circ} \)) values, which reaction is feasible at the cathode?
(a) \( \text{Ag}^+(\text{aq}) + e^- \to \text{Ag}(\text{s}) \); \( E^{\circ} = +0.80 \text{ V} \)
(b) \( \text{H}^+(\text{aq}) + e^- \to \frac{1}{2}\text{H}_2(\text{g}) \); \( E^{\circ} = 0.00 \text{ V} \)
(c) Both reactions are feasible
(d) None of the options
Answer: (a) \( \text{Ag}^+(\text{aq}) + e^- \to \text{Ag}(\text{s}) \); \( E^{\circ} = +0.80 \text{ V} \)
Explanation: On the basis of their standard reduction potential (\( E^{\circ} \)) values, cathode reaction is given by the one with higher \( E^{\circ} \) values. Thus, \( \text{Ag}^+(\text{aq}) + e^- \to \text{Ag}(\text{s}) \) reaction will be more feasible at cathode.
Question. Consider the following reaction: \( \text{Cu}(\text{s}) + 2\text{Ag}^+(\text{aq}) \to 2\text{Ag}(\text{s}) + \text{Cu}^{2+}(\text{aq}) \). Depict the galvanic cell in which the given reaction takes place.
(a) \( \text{Cu}^{2+}(\text{aq})|\text{Cu}(\text{s}) || \text{Ag}^+(\text{aq})|\text{Ag}(\text{s}) \)
(b) \( \text{Cu}(\text{s}) | \text{Cu}^{2+}(\text{aq}) || \text{Ag}^+(\text{aq}) | \text{Ag}(\text{s}) \)
(c) \( \text{Ag}^+(\text{aq})|\text{Ag}(\text{s}) || \text{Cu}^{2+}(\text{aq})|\text{Cu}(\text{s}) \)
(d) \( \text{Ag}(\text{s})|\text{Ag}^+(\text{aq}) || \text{Cu}^{2+}(\text{aq})|\text{Cu}(\text{s}) \)
Answer: (b) \( \text{Cu}(\text{s}) | \text{Cu}^{2+}(\text{aq}) || \text{Ag}^+(\text{aq}) | \text{Ag}(\text{s}) \)
Explanation: Oxidation half reaction:
\( \text{Cu}(\text{s}) \to \text{Cu}^{2+}(\text{aq}) + 2e^- \)
Reduction half reaction:
\( \text{Ag}^+(\text{aq}) + e^- \to \text{Ag}(\text{s}) \)
\( \text{Cu}(\text{s}) | \text{Cu}^{2+}(\text{aq}) || \text{Ag}^+(\text{aq}) | \text{Ag}(\text{s}) \) (Anode/Oxidation - salt bridge - Cathode/Reduction)
Question. 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) \( \text{Mn}^{3+} \) and \( \text{Co}^{3+} \) are oxidising agents in aqueous solution.
(d) \( \text{Ti}^{2+} \) and \( \text{Cr}^{2+} \) are reducing agents in aqueous solution.
Answer: (a) Copper liberates hydrogen from acids.
Explanation: Copper does not liberate hydrogen from acids because copper lies below hydrogen in electrochemical series. So, copper does not have sufficient electrode potential to liberate elemental hydrogen from compounds in which oxidation state of hydrogen is +1.
Question. Calculate the emf of the following cell at 298 K: \( \text{Mg}(\text{s})|\text{Mg}^{2+}(0.1 \text{ M})||\text{Cu}^{2+}(1.0 \times 10^{-3} \text{ M})|\text{Cu}(\text{s}) \) [Given \( E^{\circ}_{\text{Cell}} = 2.71 \text{ V} \)]
(a) 1.426 V
(b) 2.503 V
(c) 2.651 V
(d) 1.8 V
Answer: (c) 2.651 V
Explanation: \( E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.059}{n} \log \frac{[\text{Mg}^{2+}]}{[\text{Cu}^{2+}]} \)
\( = 2.71 - \frac{0.059}{2} \log \frac{0.1}{0.001} \)
\( = 2.71 - \frac{0.059}{2} \log 10^2 \)
\( E_{\text{cell}} = 2.651 \text{ V} \)
Question. Reduction potentials of some ions are given below. Arrange them in decreasing order of oxidising power.
Ion: \( \text{ClO}_4^- \), \( \text{IO}_4^- \), \( \text{BrO}_4^- \)
Reduction potential \( E^{\circ} \text{/V} \): \( 1.19 \text{V} \), \( 1.65 \text{V} \), \( 1.74 \text{V} \)
(a) \( \text{ClO}_4^- > \text{IO}_4^- > \text{BrO}_4^- \)
(b) \( \text{IO}_4^- > \text{BrO}_4^- > \text{ClO}_4^- \)
(c) \( \text{BrO}_4^- > \text{IO}_4^- > \text{ClO}_4^- \)
(d) \( \text{BrO}_4^- > \text{ClO}_4^- > \text{IO}_4^- \)
Answer: (c) \( \text{BrO}_4^- > \text{IO}_4^- > \text{ClO}_4^- \)
Explanation: Higher the reduction potential, higher is its tendency to get reduced. Hence, the order of oxidising power is: \( \text{ClO}_4^- < \text{IO}_4^- < \text{BrO}_4^- \)
Question. Using the data given below find strongest reduction agent.
\( E^{\circ}_{\text{Cr}_2\text{O}_7^{2-}/\text{Cr}^{3+}} = 1.33 \text{ V} \), \( E^{\circ}_{\text{Cl}_2/\text{Cl}^-} = 1.36 \text{ V} \), \( E^{\circ}_{\text{MnO}_4^-/\text{Mn}^{2+}} = 1.51 \text{ V} \), \( E^{\circ}_{\text{Cr}^{3+}/\text{Cr}} = -0.74 \text{ V} \)
(a) \( \text{Cl}^- \)
(b) \( \text{Cr} \)
(c) \( \text{Cr}^{3+} \)
(d) \( \text{Mn}^{2+} \)
Answer: (b) \( \text{Cr} \)
Explanation: The negative value of standard reduction potential for \( \text{Cr}^{3+} \) to \( \text{Cr} \) means that the redox couple is a stronger reducing agent.
Question. What will happen during the electrolysis of aqueous solution of \( \text{CuSO}_4 \) 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: (a) and (c) are correct.
Explanation: At cathode: \( \text{Cu}^{2+} + 2e^- \to \text{Cu} \); \( E^{\circ}_{\text{cell}} = 0.34 \text{ V} \). Copper metal will be deposited at cathode due to higher reduction potential. At anode: \( 2\text{H}_2\text{O} \to \text{O}_2 + 4\text{H}^+ + 4e^- \); \( E^{\circ}_{\text{cell}} = 1.23 \text{ V} \). The reaction with lower value of \( E^{\circ} \) will be preferred at anode, hence \( \text{O}_2 \) is released at anode.
Question. What will happen during the electrolysis of aqueous solution of \( \text{CuSO}_4 \) 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) and (b) are correct.
Explanation: At anode, \( \text{SO}_4^{2-} \) and \( \text{OH}^- \) are stable, so \( \text{Cu} \) on anode oxidises to form \( \text{Cu}^{2+} \) ion and dissolves. At cathode, \( \text{Cu}^{2+} \) is reduced due to higher potential than \( \text{H}^+ \).
Question. Conductivity \( \kappa \), is equal to ______________.
(a) \( \Lambda_m \)
(b) \( \frac{G^*}{R} \)
(c) \( \frac{l}{A} \)
(d) All of the options
Answer: (b) \( \frac{G^*}{R} \)
Explanation: \( \kappa = \frac{1}{R} \cdot \frac{l}{A} \) or \( \frac{G^*}{R} \)
Assertion and Reason Based MCQs
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, but (R) is true.
Question. 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, but (R) is true.
Question. Assertion (A): \( \Lambda_m \) for weak electrolytes shows a sharp increase when the electrolytic solution is diluted.
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).
Question. 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.
Question. 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)
Question. Assertion (A): \( E_{\text{cell}} \) should have a positive value for the cell to function.
Reason (R): \( E_{\text{cathode}} < E_{\text{anode}} \)
Answer: (c) (A) is true, but (R) is false.
Question. Assertion (A): \( \text{Cu} \) is less reactive than hydrogen.
Reason (R): \( E^{\circ}_{\text{Cu}^{2+}/\text{Cu}} \) is negative.
Answer: (c) (A) is true, but (R) is false.
Question. Assertion (A): Copper sulphate can be stored in zinc vessel.
Reason (R): Zinc is more reactive than copper.
Answer: (d) (A) is false, but (R) is true.
Explanation: Zinc will get dissolved in \( \text{CuSO}_4 \) solution, since, zinc is more reactive than copper.
Question. Assertion (A): \( E_{\text{Ag}^+/\text{Ag}} \) increases with increase in concentration of \( \text{Ag}^+ \) ions.
Reason (R): \( E_{\text{Ag}^+/\text{Ag}} \) has a positive value.
Answer: (b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
Explanation: \( \text{Ag}^+ + e^- \to \text{Ag} \), \( E_{\text{Ag}^+/\text{Ag}} = E^{\circ}_{\text{Ag}^+/\text{Ag}} - \frac{RT}{nF} \ln \frac{1}{[\text{Ag}^+]} \). On increasing \( [\text{Ag}^+] \), \( E_{\text{Ag}^+/\text{Ag}} \) will increase and it has a positive value.
Question. Assertion (A): Electrolysis of NaCl solution gives chlorine at anode instead of \( \text{O}_2 \).
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).
Case-based MCQs
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 \( \Omega \). If the resistance of the same cell when filled with 0.02 M KCl solution is 420 \( \Omega \). (Conductivity of 0.1 M KCl solution is 1.29 S \( \text{m}^{-1} \).)
Question. What is the conductivity of 0.02 M KCl solution?
(a) 0.452 S \( \text{m}^{-1} \)
(b) 0.215 S \( \text{m}^{-1} \)
(c) 0.614 S \( \text{m}^{-1} \)
(d) 0.433 S \( \text{m}^{-1} \)
Answer: (c) 0.614 S \( \text{m}^{-1} \)
Explanation: Cell constant \( G^* = \kappa \times R = 1.29 \times 200 = 258 \text{ m}^{-1} \). For 0.02 M solution, \( \kappa = \frac{G^*}{R} = \frac{258}{420} = 0.614 \text{ Sm}^{-1} \)
Question. 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 because on dilution number of ions per unit volume that are charge carriers decreases.
Question. 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 remains constant for a cell as it only depends on distance between the electrodes and their cross-section area.
Question. SI unit for conductivity of a solution is:
(a) S \( \text{m}^{-1} \)
(b) S \( \text{m}^2\text{mol}^{-1} \)
(c) \( \text{mol cm}^{-3} \)
(d) S \( \text{cm}^2\text{mol}^{-1} \)
Answer: (a) S \( \text{m}^{-1} \)
Explanation: \( \kappa = C \frac{l}{A} \); unit of \( \kappa = \text{S} \frac{\text{m}}{\text{m}^2} = \text{S m}^{-1} \)
Question. 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 differs due to size, charge, concentration and ease with which ions move under potential gradient.
A galvanic cell consists of a metallic zinc plate immersed in 0.1 M \( \text{Zn}(\text{NO}_3)_2 \) solution and metallic plate of lead in 0.02 M \( \text{Pb}(\text{NO}_3)_2 \) solution.
Question. How will the cell be represented?
(a) \( \text{Zn}(\text{s})| \text{Zn}^{2+}(\text{aq})|| \text{Pb}^{2+}(\text{aq})|\text{Pb}(\text{s}) \)
(b) \( \text{Zn}^{2+}(\text{s})| \text{Zn}(\text{aq})|| \text{Pb}^{2+}(\text{aq})|\text{Pb}(\text{s}) \)
(c) \( \text{Pb}^{2+}(\text{aq})|\text{Pb}(\text{s})||\text{Zn}^{2+}(\text{s})| \text{Zn}(\text{aq}) \)
(d) \( \text{Pb}(\text{s})|\text{Pb}^{2+}(\text{aq})||\text{Zn}^{2+}(\text{s})| \text{Zn}(\text{aq}) \)
Answer: (a) \( \text{Zn}(\text{s})| \text{Zn}^{2+}(\text{aq})|| \text{Pb}^{2+}(\text{aq})|\text{Pb}(\text{s}) \)
Explanation: In galvanic cells, redox reactions that start from left corner follow:
Reduction : \( \text{Pb}^{+2} + 2e^- \to \text{Pb} \)
Oxidation : \( \text{Zn} \to \text{Zn}^{+2} + 2e^- \)
Complete cell reaction : \( \text{Zn} + \text{Pb}^{+2} \to \text{Zn}^{+2} + \text{Pb} \)
\( \text{Zn}(\text{s})|\text{Zn}^{2+}(\text{aq})||\text{Pb}^{2+}(\text{aq})|\text{Pb}(\text{s}) \)
Question. 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: \( E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.0591}{2} \log \frac{[\text{Zn}^{2+}]}{[\text{Pb}^{2+}]} \)
\( E_{\text{cell}} = [-0.13 - (-0.76)] - \frac{0.0591}{2} \log \frac{0.1}{0.02} \)
\( = 0.63 - 0.02955 \times \log 5 \)
\( = 0.63 - 0.02955 \times 0.6990 = 0.6094 \text{ V} \)
Question. What product is obtained at cathode?
(a) Zn
(b) Pb
(c) \( \text{Zn}^{2+} \)
(d) \( \text{Pb}^{2+} \)
Answer: (b) Pb
Explanation: Anode reaction: \( \text{Zn}(\text{s}) \to \text{Zn}^{2+}(\text{aq}) + 2e^- \)
Cathode reaction: \( \text{Pb}^{2+}(\text{aq}) + 2e^- \to \text{Pb}(\text{s}) \)
Question. 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: (d) It provides surface for redox reaction.
Explanation: Inert electrode does not participate in redox reaction and acts only as source or sink for electrons. It provides surface either for oxidation or for reduction reaction.
III. Read the passage given below and answer the following questions:
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 1 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. Assertion (A): The mass of copper and silver, deposited on the cathode are same.
Reason (R): Copper and silver have different equivalent masses.
Answer: (d) Assertion is wrong statement but reason is correct statement.
Question. Assertion (A): At equilibrium condition \( E_{\text{cell}} = 0 \) or \( \Delta_r G = 0 \).
Reason (R): \( E_{\text{cell}} \) is zero when both electrodes of the cell are of the same metal.
Answer: (b) Both Assertion and reason both are correct statements but reason is NOT correct explanation for assertion
Question. Assertion (A): The negative sign in the expression \( E_{\text{Zn}^{2+}/\text{Zn}} = -0.76 \text{V} \) means \( \text{Zn}^{2+} \) cannot be oxidised to Zn.
Reason (R): Zn is more reactive than hydrogen and Zn will be oxidised and \( \text{H}^+ \) will get reduced.
Answer: (a) Both Assertion and reason both are correct statements and reason is correct explanation for assertion.
Question. 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) Both Assertion and reason both are correct statements and reason is correct explanation for assertion
Question. Assertion (A): Zinc sulphate cannot be stored in copper vessel.
Reason (R): Zinc is more reactive than copper.
Answer: (d) Assertion is wrong statement but reason is correct statement.
MCQs for Unit 2 Electrochemistry Chemistry Class 12
Students can use these MCQs for Unit 2 Electrochemistry to quickly test their knowledge of the chapter. These multiple-choice questions have been designed as per the latest syllabus for Class 12 Chemistry released by CBSE. Our expert teachers suggest that you should practice daily and solving these objective questions of Unit 2 Electrochemistry to understand the important concepts and better marks in your school tests.
Unit 2 Electrochemistry NCERT Based Objective Questions
Our expert teachers have designed these Chemistry MCQs based on the official NCERT book for Class 12. We have identified all questions from the most important topics that are always asked in exams. After solving these, please compare your choices with our provided answers. For better understanding of Unit 2 Electrochemistry, you should also refer to our NCERT solutions for Class 12 Chemistry created by our team.
Online Practice and Revision for Unit 2 Electrochemistry Chemistry
To prepare for your exams you should also take the Class 12 Chemistry MCQ Test for this chapter on our website. This will help you improve your speed and accuracy and its also free for you. Regular revision of these Chemistry topics will make you an expert in all important chapters of your course.
You can get most exhaustive CBSE Class 12 Chemistry Electrochemistry MCQs Set 07 for free on StudiesToday.com. These MCQs for Class 12 Chemistry are updated for the 2025-26 academic session as per CBSE examination standards.
Yes, our CBSE Class 12 Chemistry Electrochemistry MCQs Set 07 include the latest type of questions, such as Assertion-Reasoning and Case-based MCQs. 50% of the CBSE paper is now competency-based.
By solving our CBSE Class 12 Chemistry Electrochemistry MCQs Set 07, Class 12 students can improve their accuracy and speed which is important as objective questions provide a chance to secure 100% marks in the Chemistry.
Yes, Chemistry MCQs for Class 12 have answer key and brief explanations to help students understand logic behind the correct option as its important for 2026 competency-focused CBSE exams.
Yes, you can also access online interactive tests for CBSE Class 12 Chemistry Electrochemistry MCQs Set 07 on StudiesToday.com as they provide instant answers and score to help you track your progress in Chemistry.