JEE Chemistry Electrochemistry MCQs Set 20

Chemistry Objective Questions and Answers: 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.

Download Electrochemistry MCQs with Answers

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: E(Cu2+/Cu) = 0.337 V and E(Sn2+/Sn) = -0.136 V. Which statement is correct?
(a) Cu can be oxidised by H+
(b) Sn2+ ions can be reduced by H2(g)
(c) Cu can reduce Sn2+
(d) Cu2+ ions can be reduced by H2(g)
Show Answer & Explanation

Answer: (d) Cu2+ ions can be reduced by H2(g)

Explanation:
1. E(Cu2+/Cu) is above the hydrogen electrode (0), so H2 reduces Cu2+
2. Cu cannot be oxidised by H+ and H2 cannot reduce Sn2+, which has a negative E

Teacher's Note:
1. Check the sign of the potential and the number of electrons before substituting.

Question 2: Given Fe2+ + 2e- gives Fe, E = -0.44 V; Cu2+ + 2e- gives Cu, E = +0.34 V; Ag+ + e- gives Ag, E = +0.80 V. Which statements are correct? I Copper displaces iron from FeSO4. II Iron displaces copper from CuSO4. III Silver displaces Cu from CuSO4. IV Iron displaces silver from AgNO3.
(a) I and IV
(b) II and III
(c) I and II
(d) II and IV
Show Answer & Explanation

Answer: (d) II and IV

Explanation:
1. A metal displaces another from its salt if its reduction potential is lower
2. Fe (-0.44) displaces Cu2+ and Ag+
3. Cu and Ag cannot displace Fe, and Ag cannot displace Cu

Teacher's Note:
1. Check the sign of the potential and the number of electrons before substituting.

Question 3: Red hot carbon removes oxygen from AO and BO but not from MO, while B removes oxygen from AO. The activity of metals A, B and M in decreasing order is:
(a) B > A > M
(b) A > B > M
(c) M > A > B
(d) M > B > A
Show Answer & Explanation

Answer: (d) M > B > A

Explanation:
1. Carbon cannot reduce MO, so M is the most active
2. B reduces AO, so B is more active than A
3. The order is M > B > A

Teacher's Note:
1. Check the sign of the potential and the number of electrons before substituting.

Question 4: Which statement is correct?
(a) in H2O2 + O3 giving 2H2O + 2O2, H2O2 is oxidised to H2O
(b) by IUPAC convention the standard electrode potential refers to oxidation only
(c) the absolute value of an electrode potential cannot be determined
(d) in the SHE the pressure of dihydrogen is low and the pH is zero
Show Answer & Explanation

Answer: (c) the absolute value of an electrode potential cannot be determined

Explanation:
1. Only potential differences can be measured, so the SHE is used as a reference with H2 at 1 atm and [H+] = 1 M
2. In the given reaction H2O2 is oxidised to O2
3. IUPAC potentials are reduction potentials

Teacher's Note:
1. Check the sign of the potential and the number of electrons before substituting.

Question 5: A galvanic cell has two hydrogen electrodes, one of which is standard. In which solution should the other electrode be immersed to get the maximum emf? Ka(CH3COOH) = 2 x 10^-5, Ka(H3PO4) = 10^-3.
(a) 0.1 M HCl
(b) 0.1 M CH3COOH
(c) 0.1 M H3PO4
(d) 0.1 M H2SO4
Show Answer & Explanation

Answer: (b) 0.1 M CH3COOH

Explanation:
1. The emf is largest when [H+] in the second solution is the smallest
2. 0.1 M acetic acid has [H+] = sqrt(2 x 10^-5 x 0.1) = 1.4 x 10^-3 M, much less than the others

Teacher's Note:
1. Check the sign of the potential and the number of electrons before substituting.

Question 6: Two weak acid solutions HA1 and HA2 of the same concentration with pKa values 3 and 5 are each in contact with a hydrogen electrode (1 atm, 25 C) and connected through a salt bridge. The emf of the cell is:
(a) 0.21 V
(b) 0.059 V
(c) 0.021 V
(d) 0.018 V
Show Answer & Explanation

Answer: (b) 0.059 V

Explanation:
1. pH = (pKa - log c)/2 for a weak acid, so the pH difference = (5 - 3)/2 = 1
2. E = 0.059 x 1 = 0.059 V

Teacher's Note:
1. Check the sign of the potential and the number of electrons before substituting.

Question 7: What is the emf at 25 C of the cell Ag, AgBr(s), Br- (a = 0.34) || Fe3+ (a = 0.1), Fe2+ (a = 0.02) | Pt? E(AgBr/Ag) = +0.0713 V and E(Fe3+/Fe2+) = +0.770 V.
(a) 0.529 V
(b) 0.713 V
(c) 0.474 V
(d) 0.356 V
Show Answer & Explanation

Answer: (b) 0.713 V

Explanation:
1. Cell reaction: Ag + Br- + Fe3+ gives AgBr + Fe2+, with E = 0.770 - 0.0713 = 0.6987 V
2. Q = 0.02/(0.1 x 0.34) = 0.588
3. E = 0.6987 - 0.059 log 0.588 = 0.713 V

Teacher's Note:
1. Check the sign of the potential and the number of electrons before substituting.

Question 8: To purify a sample of copper containing zinc impurity by electrolysis, the appropriate electrodes are:
(a) pure zinc as cathode and pure copper as anode
(b) pure copper as cathode and impure sample as anode
(c) impure sample as cathode and pure copper as anode
(d) impure zinc as cathode and impure sample as anode
Show Answer & Explanation

Answer: (b) pure copper as cathode and impure sample as anode

Explanation:
1. Impure copper is the anode and dissolves, giving Cu2+ and Zn2+
2. Pure copper at the cathode collects only Cu, because Zn2+ stays in solution

Teacher's Note:
1. Check the sign of the potential and the number of electrons before substituting.

Question 9: Four moles of electrons are transferred in the electrolysis of water. The total volume of the two gases (dry, at STP) produced is approximately:
(a) 67.2 L
(b) 22.4 L
(c) 44.8 L
(d) 89.4 L
Show Answer & Explanation

Answer: (a) 67.2 L

Explanation:
1. 4 mol e- gives 2 mol H2 (44.8 L) and 1 mol O2 (22.4 L)
2. The total is 67.2 L

Teacher's Note:
1. Check the sign of the potential and the number of electrons before substituting.

Question 10: Passing a current of 27 A for 24 hours in the electrolysis of MnSO4 in aqueous sulphuric acid gives 1 kg of MnO2. The current efficiency is:
(a) 82.951 percent
(b) 80 percent
(c) 100 percent
(d) 95.185 percent
Show Answer & Explanation

Answer: (d) 95.185 percent

Explanation:
1. 1000 g MnO2 = 11.50 mol, which needs 2 e- each: 23.0 F = 2.22 x 10^6 C
2. Total charge passed = 27 x 86400 = 2.333 x 10^6 C
3. Efficiency = 95.2 percent

Teacher's Note:
1. Check the sign of the potential and the number of electrons before substituting.

Question 11: In the preparation of H2S2O8, O2 is released at the anode as a by-product. When 9.72 L of H2 is released at the cathode and 2.35 L of O2 at the anode (STP), the mass of H2S2O8 produced is about:
(a) 83.42 g
(b) 43.56 g
(c) 51.74 g
(d) 87.12 g
Show Answer & Explanation

Answer: (b) 43.56 g

Explanation:
1. Total electrons = 2 x (9.72/22.4) = 0.868 mol
2. Electrons used for O2 = 4 x (2.35/22.4) = 0.420 mol
3. Remaining 0.448 mol e- gives 0.224 mol H2S2O8 (2 e- each) = 0.224 x 194 = 43.5 g

Teacher's Note:
1. Check the sign of the potential and the number of electrons before substituting.

Question 12: When iron rusts, it is:
(a) evaporated
(b) decomposed
(c) reduced
(d) oxidised
Show Answer & Explanation

Answer: (d) oxidised

Explanation:
1. Fe is converted to Fe2+ and then Fe3+ in hydrated Fe2O3, so it is oxidised

Teacher's Note:
1. Check the sign of the potential and the number of electrons before substituting.

Question 13: A Daniell cell has 1.0 M Zn2+ and originally 1.0 M Cu2+. Enough NH3 is added to the cathode compartment to make [NH3] = 2.0 M at equilibrium. Kf for [Cu(NH3)4]2+ = 1 x 10^12 and E(cell) = 1.1 V for Zn + Cu2+ giving Zn2+ + Cu. Taking 2.303RT/F = 0.06 and log 6.25 = 0.8, the cell potential is about:
(a) 0.70 V
(b) 1.10 V
(c) 0.40 V
(d) 0.96 V
Show Answer & Explanation

Answer: (a) 0.70 V

Explanation:
1. [Cu2+] = 1/(Kf [NH3]^4) = 1/(10^12 x 16) = 6.25 x 10^-14 M
2. E = 1.1 - (0.06/2) log(1/6.25 x 10^-14) = 1.1 - 0.03 x 13.2
3. = 0.70 V

Teacher's Note:
1. Check the sign of the potential and the number of electrons before substituting.

Question 14: A hydrogen gas electrode is made by dipping platinum wire into NaOH solution of pH = 10 with H2 at 1 atm. The oxidation potential of the electrode (2.303RT/F = 0.059) is:
(a) +0.59 V
(b) -0.59 V
(c) -0.30 V
(d) +0.30 V
Show Answer & Explanation

Answer: (a) +0.59 V

Explanation:
1. For H2 giving 2H+ + 2e-, E(oxidation) = -(0.059/2) log [H+]^2 = 0.059 x pH
2. = 0.059 x 10 = +0.59 V

Teacher's Note:
1. Check the sign of the potential and the number of electrons before substituting.

Question 15: The molar conductivity of 0.04 M MgCl2 at 298 K is 200 S cm2 mol-1. The cell has electrodes of area 4 cm2 at a distance of 8 cm, with a potential difference of 10 V. The current is:
(a) 0.4 mA
(b) 4 mA
(c) 40 mA
(d) 400 mA
Show Answer & Explanation

Answer: (c) 40 mA

Explanation:
1. kappa = 200 x 0.04/1000 = 0.008 S/cm
2. G = kappa x A/l = 0.008 x 4/8 = 0.004 S
3. I = G V = 0.04 A = 40 mA

Teacher's Note:
1. Check the sign of the potential and the number of electrons before substituting.

Question 16: The conductivity of a solution that is 0.1 M in Ba(NO3)2 and 0.2 M in AgNO3 is 5.3 S/m. If lambda(Ag+) = 6 x 10^-3 and lambda(Ba2+) = 13 x 10^-3 S m2/mol, lambda(NO3-) in the same unit is:
(a) 7 x 10^-3
(b) 14 x 10^-3
(c) 3.5 x 10^-3
(d) 9 x 10^-3
Show Answer & Explanation

Answer: (a) 7 x 10^-3

Explanation:
1. In SI units, 0.2 M = 200 mol/m3 AgNO3 and 100 mol/m3 Ba(NO3)2
2. NO3- = 400 mol/m3
3. 5.3 = 200(6 x 10^-3) + 100(13 x 10^-3) + 400 lambda
4. lambda = 2.8/400 = 7 x 10^-3

Teacher's Note:
1. Check the sign of the potential and the number of electrons before substituting.

Question 17: The molar conductivity of a weak monobasic acid HA at infinite dilution is 390.7 S cm2/mol, and at 0.01 M it is 3.907 S cm2/mol. The pH of 0.01 M HA is:
(a) 2
(b) 4
(c) 5
(d) 3
Show Answer & Explanation

Answer: (b) 4

Explanation:
1. Degree of dissociation = 3.907/390.7 = 0.01
2. [H+] = 0.01 x 0.01 = 10^-4 M
3. pH = 4

Teacher's Note:
1. Check the sign of the potential and the number of electrons before substituting.

Question 18: For a saturated solution of AgCl at 25 C the conductivity is 3.4 x 10^-6 ohm-1 cm-1 and that of the water used is 2.02 x 10^-6. Molar conductivity of AgCl is 138 ohm-1 cm2 mol-1. The solubility of AgCl in mmol per m3 is:
(a) 1
(b) 10
(c) 100
(d) 1000
Show Answer & Explanation

Answer: (b) 10

Explanation:
1. kappa(AgCl) = 3.4 - 2.02 = 1.38 x 10^-6 S/cm
2. S = kappa x 1000/Lambda = 1.38 x 10^-6 x 1000/138 = 10^-5 mol/L
3. = 10^-2 mol/m3 = 10 mmol/m3

Teacher's Note:
1. Check the sign of the potential and the number of electrons before substituting.

Question 19: The standard reduction potentials of Pt/Fe3+, Fe2+ and Pt/Sn4+, Sn2+ are +0.77 V and +0.15 V respectively at 25 C. The standard emf of the reaction Sn4+ + 2Fe2+ giving Sn2+ + 2Fe3+ is:
(a) -0.62 V
(b) +0.31 V
(c) -0.92 V
(d) +0.85 V
Show Answer & Explanation

Answer: (a) -0.62 V

Explanation:
1. Sn4+ is reduced (E = 0.15) and Fe2+ is oxidised (E of Fe3+/Fe2+ = 0.77)
2. E = 0.15 - 0.77 = -0.62 V (non-spontaneous)

Teacher's Note:
1. Check the sign of the potential and the number of electrons before substituting.

Question 20: By how much does the reduction potential of a hydrogen electrode change when its solution at pH 0 is neutralised to pH 7 at 25 C?
(a) increases by 0.41 V
(b) increases by 0.059 V
(c) decreases by 0.059 V
(d) decreases by 0.41 V
Show Answer & Explanation

Answer: (d) decreases by 0.41 V

Explanation:
1. E = -0.059 pH, so going from pH 0 to 7 gives E = -0.413 V
2. It decreases by 0.41 V

Teacher's Note:
1. Check the sign of the potential and the number of electrons before substituting.

Question 21: The cost of electricity for producing X litres of H2 at NTP at the cathode is Rs. X. The cost for producing X litres of O2 at NTP at the anode would be (1 mol of electrons = 1 unit):
(a) 4X
(b) 2X
(c) 32X
(d) 16X
Show Answer & Explanation

Answer: (b) 2X

Explanation:
1. H2: 2 mol e- per 22.4 L, that is 1 mol e- per 11.2 L
2. O2: 4 mol e- per 22.4 L, that is 1 mol e- per 5.6 L
3. For the same volume O2 needs twice as many electrons, so the cost is 2X

Teacher's Note:
1. Check the sign of the potential and the number of electrons before substituting.

Question 22: A current of 0.1 A is passed for 965 s through a solution of Cu+ and 0.03175 g of copper is deposited. The current efficiency (Cu = 63.5) is:
(a) 79 percent
(b) 50 percent
(c) 39.5 percent
(d) 63.25 percent
Show Answer & Explanation

Answer: (b) 50 percent

Explanation:
1. Charge = 96.5 C = 0.001 F
2. Theoretical mass = 0.001 x 63.5 = 0.0635 g (one electron per Cu+)
3. Efficiency = 0.03175/0.0635 = 50 percent

Teacher's Note:
1. Check the sign of the potential and the number of electrons before substituting.

Question 23: A current of 9.95 A flowing for 10 minutes deposits 3 g of a metal. The equivalent weight of the metal is:
(a) 12.5
(b) 18.5
(c) 48.5
(d) 21.5
Show Answer & Explanation

Answer: (c) 48.5

Explanation:
1. Charge = 9.95 x 600 = 5970 C
2. E = 3 x 96500/5970 = 48.5

Teacher's Note:
1. Check the sign of the potential and the number of electrons before substituting.

Question 24: The specific conductance of N/10 KCl at 25 C is 0.0112 ohm-1 cm-1. The resistance of the cell containing this solution is 55 ohm. The cell constant is:
(a) 0.616 cm-1
(b) 6.16 cm-1
(c) 616 cm-1
(d) 0.0616 cm-1
Show Answer & Explanation

Answer: (a) 0.616 cm-1

Explanation:
1. Cell constant = kappa x R = 0.0112 x 55 = 0.616 cm-1

Teacher's Note:
1. Check the sign of the potential and the number of electrons before substituting.

Question 25: The equivalent conductance of N/10 NaCl at 25 C is 10^-2 S m2/eq. The resistance of the solution in the cell is 50 ohm. The cell constant is:
(a) 50 m-1
(b) 50 x 10^-6 m-1
(c) 50 x 10^-3 m-1
(d) 50 x 10^3 m-1
Show Answer & Explanation

Answer: (a) 50 m-1

Explanation:
1. Concentration = 0.1 eq/L = 100 eq/m3
2. kappa = Lambda x C = 10^-2 x 100 = 1 S/m
3. Cell constant = kappa x R = 1 x 50 = 50 m-1

Teacher's Note:
1. Check the sign of the potential and the number of electrons before substituting.

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Multiple Choice Questions (MCQs) for JEE Chemistry Electrochemistry

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