JEE Chemistry Electrochemistry MCQs Set 15

Download JEE MCQs for JEE Chemistry: Electrochemistry

Access targeted multiple-choice questions for Electrochemistry designed to align with the latest JEE academic syllabus for JEE Chemistry. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

Chapter-wise Objective Questions: Electrochemistry

View or download the dedicated Electrochemistry MCQ resource below. Practicing these 50 objective questions regularly builds familiarity with standard exam patterns and helps secure higher marks in final Chemistry evaluations.

Multiple Choice Questions

Question 1: The passage of current liberates \( H_2 \) at the cathode and \( Cl_2 \) at the anode. The solution is
(a) copper chloride in water
(b) NaCl in water
(c) \( H_2SO_4 \)
(d) water
Show Answer & Explanation

Answer: (b) NaCl in water

Explanation:
1. Aqueous NaCl: water is reduced at the cathode (H2) and chloride is oxidised at the anode (Cl2).
2. Copper chloride deposits Cu and \( H_2SO_4 \) gives O2 at the anode.

Teacher's Note:
1. Oxidation occurs at the anode and reduction at the cathode in every cell.
2. Conductivity depends on the number and mobility of free ions in the electrolyte.

Question 2: Electrolysis of molten anhydrous calcium chloride produces
(a) calcium
(b) phosphorus
(c) sulphur
(d) sodium
Show Answer & Explanation

Answer: (a) calcium

Explanation:
1. Molten \( CaCl_2 \) gives Ca at the cathode and \( Cl_2 \) at the anode.

Teacher's Note:
1. Oxidation occurs at the anode and reduction at the cathode in every cell.
2. Conductivity depends on the number and mobility of free ions in the electrolyte.

Question 3: Which of the following is NOT a non-electrolyte?
(a) acetic acid
(b) glucose
(c) ethanol
(d) urea
Show Answer & Explanation

Answer: (a) acetic acid

Explanation:
1. Glucose, ethanol and urea do not ionise in water.
2. Acetic acid is a weak electrolyte, so it is not a non-electrolyte.

Teacher's Note:
1. Oxidation occurs at the anode and reduction at the cathode in every cell.
2. Conductivity depends on the number and mobility of free ions in the electrolyte.

Question 4: When 9.65 coulombs of electricity is passed through a solution of silver nitrate (atomic weight of Ag taken as 108), the amount of silver deposited is
(a) 10.8 mg
(b) 5.4 mg
(c) 16.2 mg
(d) 21.2 mg
Show Answer & Explanation

Answer: (a) 10.8 mg

Explanation:
1. 96500 C deposits 108 g of Ag.
2. 9.65 C deposits \( 108\times\frac{9.65}{96500}=0.0108 \) g = 10.8 mg.

Teacher's Note:
1. Use Faraday's first law: mass = (equivalent weight x charge) / 96500, with charge = current x time in seconds.
2. Equivalent weight = atomic (or molar) mass divided by the number of electrons transferred per ion.

Question 5: Three faradays of electricity are passed through an aqueous solution of iron(II) bromide. The weight of iron (at. wt. 56) deposited at the cathode is
(a) 56 g
(b) 84 g
(c) 112 g
(d) 168 g
Show Answer & Explanation

Answer: (b) 84 g

Explanation:
1. \( Fe^{2+}+2e^-\rightarrow Fe \): 2 F deposits 56 g.
2. 3 F deposits 56 x 3/2 = 84 g.

Teacher's Note:
1. Use Faraday's first law: mass = (equivalent weight x charge) / 96500, with charge = current x time in seconds.
2. Equivalent weight = atomic (or molar) mass divided by the number of electrons transferred per ion.

Question 6: When a current of 5 faradays is passed through a solution of \( Al^{3+} \) ions, the weight of Al (atomic weight 27) deposited is
(a) 27 g
(b) 36 g
(c) 45 g
(d) 39 g
Show Answer & Explanation

Answer: (c) 45 g

Explanation:
1. 3 F deposits 27 g (1 mol).
2. 5 F deposits 27 x 5/3 = 45 g.

Teacher's Note:
1. Use Faraday's first law: mass = (equivalent weight x charge) / 96500, with charge = current x time in seconds.
2. Equivalent weight = atomic (or molar) mass divided by the number of electrons transferred per ion.

Question 7: A certain current liberated 0.504 g of hydrogen in 2 hours. How many grams of copper can be liberated by the same current flowing for the same time in a copper sulphate solution? (Cu = 63.5, H = 1.008)
(a) 12.7 g
(b) 15.9 g
(c) 31.8 g
(d) 63.5 g
Show Answer & Explanation

Answer: (b) 15.9 g

Explanation:
1. Same charge liberates equal equivalents: 0.504 g H = 0.5 equivalent.
2. Equivalent weight of Cu = 31.75, so mass = 0.5 x 31.75 = 15.9 g.

Teacher's Note:
1. Use Faraday's first law: mass = (equivalent weight x charge) / 96500, with charge = current x time in seconds.
2. Equivalent weight = atomic (or molar) mass divided by the number of electrons transferred per ion.

Question 8: 5 amperes of current is passed through a solution of zinc sulphate for 40 minutes. The amount of zinc (at. wt. 65.04) deposited at the cathode is
(a) 40.65 g
(b) 4.065 g
(c) 0.4065 g
(d) 65.04 g
Show Answer & Explanation

Answer: (b) 4.065 g

Explanation:
1. Q = 5 x 2400 = 12000 C.
2. Mass = \( \frac{32.52\times12000}{96500} \) = 4.04 g, i.e. about 4.065 g.

Teacher's Note:
1. Use Faraday's first law: mass = (equivalent weight x charge) / 96500, with charge = current x time in seconds.
2. Equivalent weight = atomic (or molar) mass divided by the number of electrons transferred per ion.

Question 9: In an electroplating experiment m g of silver is deposited when 4 amperes of current flows for 2 minutes. The amount of silver deposited by 6 amperes flowing for 40 seconds is
(a) 4m
(b) m/2
(c) m/4
(d) 2m
Show Answer & Explanation

Answer: (b) m/2

Explanation:
1. Charge in the first case = 4 x 120 = 480 C; in the second = 6 x 40 = 240 C.
2. Mass is proportional to charge, so the second mass is m/2.

Teacher's Note:
1. Use Faraday's first law: mass = (equivalent weight x charge) / 96500, with charge = current x time in seconds.
2. Equivalent weight = atomic (or molar) mass divided by the number of electrons transferred per ion.

Question 10: On passing 3 amperes of electricity for 50 minutes, 1.8 g of a metal deposits. The equivalent mass of the metal is
(a) 20.5
(b) 25.8
(c) 19.3
(d) 30.7
Show Answer & Explanation

Answer: (c) 19.3

Explanation:
1. Q = 3 x 3000 = 9000 C.
2. E = \( \frac{1.8\times96500}{9000} \) = 19.3.

Teacher's Note:
1. Use Faraday's first law: mass = (equivalent weight x charge) / 96500, with charge = current x time in seconds.
2. Equivalent weight = atomic (or molar) mass divided by the number of electrons transferred per ion.

Question 11: On passing one faraday of electricity through electrolytic cells containing \( Ag^+ \), \( Ni^{2+} \) and \( Cr^{3+} \) ions, the amounts deposited are (Ag = 108, Ni = 59, Cr = 52)
(a) Ag 108 g, Ni 29.5 g, Cr 17.3 g
(b) Ag 108 g, Ni 59.0 g, Cr 52.0 g
(c) Ag 108 g, Ni 108.0 g, Cr 108.0 g
(d) Ag 108 g, Ni 117.5 g, Cr 166.0 g
Show Answer & Explanation

Answer: (a) Ag 108 g, Ni 29.5 g, Cr 17.3 g

Explanation:
1. One faraday deposits one gram equivalent of each metal.
2. Ag: 108/1; Ni: 59/2 = 29.5; Cr: 52/3 = 17.3 g.

Teacher's Note:
1. Use Faraday's first law: mass = (equivalent weight x charge) / 96500, with charge = current x time in seconds.
2. Equivalent weight = atomic (or molar) mass divided by the number of electrons transferred per ion.

Question 12: To deposit 0.6354 g of copper by electrolysis of an aqueous cupric sulphate solution, the amount of electricity required (in coulombs) is
(a) 9650
(b) 4825
(c) 3860
(d) 1930
Show Answer & Explanation

Answer: (d) 1930

Explanation:
1. Equivalent weight of Cu = 63.54/2 = 31.77.
2. Q = \( \frac{0.6354\times96500}{31.77} \) = 1930 C.

Teacher's Note:
1. Use Faraday's first law: mass = (equivalent weight x charge) / 96500, with charge = current x time in seconds.
2. Equivalent weight = atomic (or molar) mass divided by the number of electrons transferred per ion.

Question 13: When 0.04 faraday of electricity is passed through a solution of \( CaSO_4 \), the weight of \( Ca^{2+} \) metal deposited at the cathode is (Ca = 40)
(a) 0.2 g
(b) 0.4 g
(c) 0.6 g
(d) 0.8 g
Show Answer & Explanation

Answer: (d) 0.8 g

Explanation:
1. Equivalent weight of Ca = 20.
2. Mass = 0.04 x 20 = 0.8 g.

Teacher's Note:
1. Use Faraday's first law: mass = (equivalent weight x charge) / 96500, with charge = current x time in seconds.
2. Equivalent weight = atomic (or molar) mass divided by the number of electrons transferred per ion.

Question 14: A current of 2.0 A is passed for 5 hours through a molten metal salt and deposits 22 g of the metal (at. wt. 177). The oxidation state of the metal in the salt is
(a) +1
(b) +2
(c) +3
(d) +4
Show Answer & Explanation

Answer: (c) +3

Explanation:
1. Q = 2 x 18000 = 36000 C = 0.373 F.
2. Moles of metal = 22/177 = 0.124, so electrons per atom = 0.373/0.124 = 3.

Teacher's Note:
1. Use Faraday's first law: mass = (equivalent weight x charge) / 96500, with charge = current x time in seconds.
2. Equivalent weight = atomic (or molar) mass divided by the number of electrons transferred per ion.

Question 15: How many atoms of calcium will be deposited from a solution of \( CaCl_2 \) by a current of 25 milliamperes flowing for 60 seconds?
(a) \( 4.68\times10^{18} \)
(b) \( 4.68\times10^{15} \)
(c) \( 4.68\times10^{12} \)
(d) \( 4.68\times10^{9} \)
Show Answer & Explanation

Answer: (a) \( 4.68\times10^{18} \)

Explanation:
1. Charge = 0.025 x 60 = 1.5 C.
2. Electrons = \( 1.5/1.6\times10^{-19}=9.37\times10^{18} \); each Ca2+ needs 2, so atoms = \( 4.68\times10^{18} \).

Teacher's Note:
1. Use Faraday's first law: mass = (equivalent weight x charge) / 96500, with charge = current x time in seconds.
2. Equivalent weight = atomic (or molar) mass divided by the number of electrons transferred per ion.

Question 16: What is the amount of chlorine evolved when 2 amperes of current is passed for 30 minutes through an aqueous solution of NaCl?
(a) 66 g
(b) 1.32 g
(c) 33 g
(d) 99 g
Show Answer & Explanation

Answer: (b) 1.32 g

Explanation:
1. Q = 2 x 1800 = 3600 C = 0.0373 F.
2. Cl2 = 0.0373/2 = 0.01865 mol x 71 = 1.32 g.

Teacher's Note:
1. Use Faraday's first law: mass = (equivalent weight x charge) / 96500, with charge = current x time in seconds.
2. Equivalent weight = atomic (or molar) mass divided by the number of electrons transferred per ion.

Question 17: On the basis of \( \frac43Al+O_2\rightarrow\frac23Al_2O_3 \), \( \Delta G=-827\ kJ\,mol^{-1} \) of \( O_2 \), the minimum emf required for the electrolysis of \( Al_2O_3 \) is (F = 96500 C/mol)
(a) 8.56 V
(b) 2.14 V
(c) 4.28 V
(d) 6.42 V
Show Answer & Explanation

Answer: (b) 2.14 V

Explanation:
1. For one mole of O2, 4 electrons are transferred (n = 4).
2. E = \( \frac{827000}{4\times96500} \) = 2.14 V.

Teacher's Note:
1. Compare standard reduction potentials: the metal with the more negative value is the stronger reducing agent.
2. A cell is feasible when \( E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode} \) is positive.

Question 18: When an electric current is passed through acidulated water, 112 mL of hydrogen gas at NTP collects at the cathode in 965 seconds. The current passed is
(a) 1.0 A
(b) 0.5 A
(c) 0.1 A
(d) 2.0 A
Show Answer & Explanation

Answer: (a) 1.0 A

Explanation:
1. 112 mL H2 = 0.005 mol = 0.01 F = 965 C.
2. I = 965/965 = 1.0 A.

Teacher's Note:
1. Use Faraday's first law: mass = (equivalent weight x charge) / 96500, with charge = current x time in seconds.
2. Equivalent weight = atomic (or molar) mass divided by the number of electrons transferred per ion.

Question 19: The electrolytic cells, one containing acidified ferrous chloride and another acidified ferric chloride, are connected in series. The ratio of iron deposited at the cathodes in the two cells when electricity is passed through them is
(a) 3 : 1
(b) 2 : 1
(c) 1 : 1
(d) 3 : 2
Show Answer & Explanation

Answer: (d) 3 : 2

Explanation:
1. Ferrous: 2 F per mole Fe; ferric: 3 F per mole Fe, with the same charge in series.
2. Moles of Fe deposited are in the ratio 1/2 : 1/3 = 3 : 2.

Teacher's Note:
1. Use Faraday's first law: mass = (equivalent weight x charge) / 96500, with charge = current x time in seconds.
2. Equivalent weight = atomic (or molar) mass divided by the number of electrons transferred per ion.

Question 20: During electrolysis of fused aluminium chloride, 0.9 g of aluminium was deposited on the cathode. The volume of chlorine liberated at the anode at NTP will be
(a) 2.24 litres
(b) 11.2 litres
(c) 1.12 litres
(d) 5.6 litres
Show Answer & Explanation

Answer: (c) 1.12 litres

Explanation:
1. 0.9 g Al = 0.0333 mol = 0.1 F.
2. 0.1 F liberates 0.05 mol Cl2 = 0.05 x 22.4 = 1.12 L.

Teacher's Note:
1. Use Faraday's first law: mass = (equivalent weight x charge) / 96500, with charge = current x time in seconds.
2. Equivalent weight = atomic (or molar) mass divided by the number of electrons transferred per ion.

Question 21: If the equivalent conductance of 1 M benzoic acid is 12.8 \( ohm^{-1}cm^2 \) and the conductances of the benzoate ion and \( H^+ \) ion are 42 and 288.42 \( ohm^{-1}cm^2 \) respectively, its degree of dissociation is
(a) 39%
(b) 3.9%
(c) 0.35%
(d) 0.039%
Show Answer & Explanation

Answer: (b) 3.9%

Explanation:
1. \( \Lambda^\infty=42+288.42=330.42 \).
2. Degree of dissociation = \( \Lambda/\Lambda^\infty=12.8/330.42=0.039 \) = 3.9%.

Teacher's Note:
1. Oxidation occurs at the anode and reduction at the cathode in every cell.
2. Conductivity depends on the number and mobility of free ions in the electrolyte.

Question 22: The specific conductivity of a 0.1 N KCl solution at 23 °C is 0.012 \( ohm^{-1}cm^{-1} \). The resistance of a cell containing the solution at the same temperature was 55 ohm. The cell constant will be
(a) \( 0.142\ cm^{-1} \)
(b) \( 0.66\ cm^{-1} \)
(c) \( 0.918\ cm^{-1} \)
(d) \( 1.12\ cm^{-1} \)
Show Answer & Explanation

Answer: (b) \( 0.66\ cm^{-1} \)

Explanation:
1. Cell constant = specific conductivity x resistance.
2. = 0.012 x 55 = 0.66 \( cm^{-1} \).

Teacher's Note:
1. Oxidation occurs at the anode and reduction at the cathode in every cell.
2. Conductivity depends on the number and mobility of free ions in the electrolyte.

Question 23: The standard reduction potentials of \( M_1^{n+}/M_1 \), \( M_2^{n+}/M_2 \) and \( M_3^{n+}/M_3 \) are -0.34 V, -3.05 V and -1.66 V respectively. The order of their reducing power is
(a) \( M_1>M_2>M_3 \)
(b) \( M_3>M_2>M_1 \)
(c) \( M_1>M_3>M_2 \)
(d) \( M_2>M_3>M_1 \)
Show Answer & Explanation

Answer: (d) \( M_2>M_3>M_1 \)

Explanation:
1. The more negative the reduction potential, the stronger the reducing agent.
2. -3.05 < -1.66 < -0.34 gives \( M_2>M_3>M_1 \).

Teacher's Note:
1. Compare standard reduction potentials: the metal with the more negative value is the stronger reducing agent.
2. A cell is feasible when \( E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode} \) is positive.

Question 24: K, Ca and Li metals may be arranged in the decreasing order of their standard electrode potentials as (E°: K -2.93 V, Ca -2.87 V, Li -3.05 V)
(a) K, Ca, Li
(b) Ca, K, Li
(c) Li, Ca, K
(d) Ca, Li, K
Show Answer & Explanation

Answer: (b) Ca, K, Li

Explanation:
1. Decreasing order means the least negative value first.
2. Ca (-2.87) > K (-2.93) > Li (-3.05).

Teacher's Note:
1. Compare standard reduction potentials: the metal with the more negative value is the stronger reducing agent.
2. A cell is feasible when \( E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode} \) is positive.

Question 25: The emf of the cell Ni | \( Ni^{2+} \) (1.0 M) || \( Au^{3+} \) (1.0 M) | Au is (E° for \( Ni^{2+}/Ni \) = -0.25 V; E° for \( Au^{3+}/Au \) = 1.50 V)
(a) +1.25 V
(b) -1.75 V
(c) +1.75 V
(d) +4.0 V
Show Answer & Explanation

Answer: (c) +1.75 V

Explanation:
1. \( E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}=1.50-(-0.25) \) = +1.75 V.

Teacher's Note:
1. Compare standard reduction potentials: the metal with the more negative value is the stronger reducing agent.
2. A cell is feasible when \( E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode} \) is positive.

Free study material for Chemistry

Download Chapter MCQs: JEE Chemistry Electrochemistry

Download Multiple Choice Questions: Electrochemistry (JEE Chemistry)

Review structured objective questions for JEE Chemistry Electrochemistry. Built according to official JEE guidelines, these MCQ sets support daily revision and core concept reinforcement.

Concept Clarification for Electrochemistry

Built using the official NCERT book for JEE, these Chemistry objective sets provide reliable academic guidance. Pair your practice with our recommended NCERT solutions to master optimal problem-solving approaches.

Additional Study Resources for JEE Chemistry

Explore our broader library of printable assignments, chapter notes, and mock tests designed to support continuous revision and secure higher marks in JEE assessments.

FAQs

Where can I access latest JEE Chemistry Electrochemistry MCQs Set 15?

You can get most exhaustive JEE Chemistry Electrochemistry MCQs Set 15 for free on StudiesToday.com. These MCQs for JEE Chemistry are updated for the 2026-27 academic session as per JEE examination standards.

Are Assertion-Reasoning and Case-Study MCQs included in the Chemistry JEE material?

Yes, our JEE Chemistry Electrochemistry MCQs Set 15 include the latest type of questions, such as Assertion-Reasoning and Case-based MCQs. 50% of the JEE paper is now competency-based.

How do practicing Chemistry MCQs help in scoring full marks in JEE exams?

By solving our JEE Chemistry Electrochemistry MCQs Set 15, JEE students can improve their accuracy and speed which is important as objective questions provide a chance to secure 100% marks in the Chemistry.

Do you provide answers and explanations for JEE Chemistry Electrochemistry MCQs Set 15?

Yes, Chemistry MCQs for JEE have answer key and brief explanations to help students understand logic behind the correct option as its important for 2026 competency-focused JEE exams.

Can I practice these Chemistry JEE MCQs online?

Yes, you can also access online interactive tests for JEE Chemistry Electrochemistry MCQs Set 15 on StudiesToday.com as they provide instant answers and score to help you track your progress in Chemistry.