JEE Chemistry Electrochemistry MCQs Set 24

Practice MCQs for JEE Chemistry Electrochemistry

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

Access Electrochemistry Questions and Solutions

Access the complete set of multiple-choice questions for Electrochemistry below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official JEE textbooks.

Multiple Choice Questions

Question 1: E° for the reaction Fe + Zn2+ → Zn + Fe2+ is −0.35 V. The given cell reaction is:
(a) Feasible
(b) Not feasible
(c) In equilibrium
(d) Can't say anything
Show Answer & Explanation

Answer: (b) Not feasible

Explanation:
1. A negative E° means ΔG° > 0, so the reaction is not spontaneous under standard conditions.
2. Answer: (b)

Teacher's Note:
1. Write the relevant formula first.
2. Check signs and units before choosing.

Question 2: For a reaction A(s) + 2B+ → A2+ + 2B(s), Kc has been found to be 1012. The E°cell is:
(a) 0.354 V
(b) 0.708 V
(c) 0.0098 V
(d) 1.36 V
Show Answer & Explanation

Answer: (a) 0.354 V

Explanation:
1. E° = (0.0591/n) log K = (0.0591/2) × 12 = 0.3546 V.
2. Answer: (a)

Teacher's Note:
1. Write the relevant formula first.
2. Check signs and units before choosing.

Question 3: The standard reduction potentials of Cu2+/Cu and Cu2+/Cu+ are 0.337 and 0.153 V respectively. The standard electrode potential of the Cu+/Cu half cell is:
(a) 0.184 V
(b) 0.827 V
(c) 0.521 V
(d) 0.490 V
Show Answer & Explanation

Answer: (c) 0.521 V

Explanation:
1. Cu2+ + 2e− → Cu (ΔG = −2 × 0.337F); Cu2+ + e− → Cu+ (−0.153F).
2. Subtract: Cu+ + e− → Cu has nE° = 0.674 − 0.153 = 0.521, n = 1.
3. E° = 0.521 V.
4. Answer: (c)

Teacher's Note:
1. Write the relevant formula first.
2. Check signs and units before choosing.

Question 4: A hydrogen electrode (PH2 = 1 atm) is dipped in a solution of pH = 3 at 25°C. The potential of the electrode would be:
(a) 0.177 V
(b) −0.177 V
(c) 0.087 V
(d) 0.059 V
Show Answer & Explanation

Answer: (b) −0.177 V

Explanation:
1. E = −0.0591 × pH = −0.0591 × 3 = −0.177 V.
2. Answer: (b)

Teacher's Note:
1. Write the relevant formula first.
2. Check signs and units before choosing.

Question 5: What is the potential of the cell containing two hydrogen electrodes represented by Pt, H2(g) | H+(10−8 M) || H+(0.001 M) | H2(g), Pt?
(a) −0.295 V
(b) −0.0591 V
(c) 0.295 V
(d) 0.0591 V
Show Answer & Explanation

Answer: (c) 0.295 V

Explanation:
1. This is a concentration cell: E = 0.0591 log([H+]cathode/[H+]anode) = 0.0591 log(10−3/10−8).
2. E = 0.0591 × 5 = 0.2955 V.
3. Answer: (c)

Teacher's Note:
1. Write the relevant formula first.
2. Check signs and units before choosing.

Question 6: For the cell Cu | Cu2+ || Ag+ | Ag, if the concentrations of Cu2+ and Ag+ ions both become ten times, the emf of the cell:
(a) Becomes 10 times
(b) Remains the same
(c) Increases by 0.0295 V
(d) Decreases by 0.0295 V
Show Answer & Explanation

Answer: (c) Increases by 0.0295 V

Explanation:
1. E = E° − (0.0591/2) log([Cu2+]/[Ag+]2).
2. The ratio changes by 10/102 = 1/10, so log falls by 1.
3. E increases by 0.0591/2 = 0.0295 V.
4. Answer: (c)

Teacher's Note:
1. Write the relevant formula first.
2. Check signs and units before choosing.

Question 7: When an electric current is passed through acidified water, 112 mL of hydrogen gas at STP collects at the cathode in 965 seconds. The current passed, in ampere, is:
(a) 1.0
(b) 0.5
(c) 0.1
(d) 2.0
Show Answer & Explanation

Answer: (a) 1.0

Explanation:
1. 112 mL H2 = 0.005 mol, needing 0.01 F = 965 C.
2. I = 965 C/965 s = 1.0 A.
3. Answer: (a)

Teacher's Note:
1. Write the relevant formula first.
2. Check signs and units before choosing.

Question 8: Two 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 the cells will be:
(a) 3 : 1
(b) 2 : 1
(c) 1 : 1
(d) 3 : 2
Show Answer & Explanation

Answer: (d) 3 : 2

Explanation:
1. Moles of Fe deposited ∝ 1/n: Fe2+ needs 2F and Fe3+ needs 3F per mole.
2. Ratio = 1/2 : 1/3 = 3 : 2.
3. Answer: (d)

Teacher's Note:
1. Write the relevant formula first.
2. Check signs and units before choosing.

Question 9: A current of 9.65 A flowing for 10 minutes deposits 3.0 g of a metal. The equivalent weight of the metal is:
(a) 10
(b) 30
(c) 50
(d) 96.5
Show Answer & Explanation

Answer: (c) 50

Explanation:
1. Q = 9.65 × 600 = 5790 C = 0.06 F.
2. Equivalent weight = 3.0/0.06 = 50.
3. Answer: (c)

Teacher's Note:
1. Write the relevant formula first.
2. Check signs and units before choosing.

Question 10: How many coulombs of electricity are required for the oxidation of 1 mole of H2O to O2?
(a) 9.65 × 104 C
(b) 4.825 × 105 C
(c) 1.93 × 105 C
(d) 1.93 × 104 C
Show Answer & Explanation

Answer: (c) 1.93 × 105 C

Explanation:
1. 2H2O → O2 + 4H+ + 4e−: 2 mol H2O release 4 mol e−.
2. 1 mol H2O needs 2 F = 2 × 96500 = 1.93 × 105 C.
3. Answer: (c)

Teacher's Note:
1. Write the relevant formula first.
2. Check signs and units before choosing.

Question 11: 10800 C of electricity deposited 2.977 g of a metal of atomic mass 106.4 g mol−1. The charge on the metal cation is:
(a) +4
(b) +3
(c) +2
(d) +1
Show Answer & Explanation

Answer: (a) +4

Explanation:
1. Moles of metal = 2.977/106.4 = 0.02798.
2. Faradays = 10800/96500 = 0.1119.
3. n = 0.1119/0.02798 = 4, so the charge is +4.
4. Answer: (a)

Teacher's Note:
1. Write the relevant formula first.
2. Check signs and units before choosing.

Question 12: On passing electricity through dilute H2SO4 solution, the amounts of substance liberated at the cathode and anode are in the ratio (by mass):
(a) 1 : 8
(b) 8 : 1
(c) 16 : 1
(d) 1 : 16
Show Answer & Explanation

Answer: (a) 1 : 8

Explanation:
1. Cathode: H2 (equivalent weight 1); anode: O2 (equivalent weight 8).
2. Mass ratio 1 : 8.
3. Answer: (a)

Teacher's Note:
1. Write the relevant formula first.
2. Check signs and units before choosing.

Question 13: A silver cup is plated with silver by passing 965 A current for one second. The mass of Ag deposited is (at. wt. Ag = 107.87):
(a) 9.89 g
(b) 107.87 g
(c) 1.0787 g
(d) 100.2 g
Show Answer & Explanation

Answer: (c) 1.0787 g

Explanation:
1. Q = 965 C = 0.01 F.
2. Mass = 0.01 × 107.87 = 1.0787 g.
3. Answer: (c)

Teacher's Note:
1. Write the relevant formula first.
2. Check signs and units before choosing.

Question 14: When electricity is passed through a solution of AlCl3, 13.5 g of Al are deposited. The number of faradays that must have been passed is:
(a) 5.0
(b) 1.0
(c) 1.5
(d) 3.0
Show Answer & Explanation

Answer: (c) 1.5

Explanation:
1. Moles of Al = 13.5/27 = 0.5.
2. Al3+ needs 3F per mole: 0.5 × 3 = 1.5 F.
3. Answer: (c)

Teacher's Note:
1. Write the relevant formula first.
2. Check signs and units before choosing.

Question 15: A solution of sodium sulphate in water is electrolysed using inert electrodes. The products at the cathode and anode are respectively:
(a) H2, SO2
(b) O2, H2
(c) O2, Na
(d) H2, O2
Show Answer & Explanation

Answer: (d) H2, O2

Explanation:
1. Water is reduced more easily than Na+, so H2 forms at the cathode.
2. Water is oxidised more easily than SO42−, so O2 forms at the anode.
3. Answer: (d)

Teacher's Note:
1. Write the relevant formula first.
2. Check signs and units before choosing.

Question 16: When 96500 coulombs of electricity are passed through molten barium chloride, the amount of barium deposited will be:
(a) 0.5 mol
(b) 1.0 mol
(c) 1.5 mol
(d) 2.0 mol
Show Answer & Explanation

Answer: (a) 0.5 mol

Explanation:
1. Ba2+ + 2e− → Ba needs 2F per mole.
2. 1 F gives 0.5 mol of Ba.
3. Answer: (a)

Teacher's Note:
1. Write the relevant formula first.
2. Check signs and units before choosing.

Question 17: A factory produces 40 kg of calcium in two hours by electrolysis. How much aluminium can be produced by the same current in two hours? (At. wt. Ca = 40, Al = 27)
(a) 22 kg
(b) 18 kg
(c) 9 kg
(d) 27 kg
Show Answer & Explanation

Answer: (b) 18 kg

Explanation:
1. Equivalent weights: Ca 20, Al 9.
2. Mass of Al = 40 × 9/20 = 18 kg.
3. Answer: (b)

Teacher's Note:
1. Write the relevant formula first.
2. Check signs and units before choosing.

Question 18: What would be the ratio of moles of Ag+, Cu2+ and Fe3+ deposited by the passage of the same quantity of electricity through solutions of their salts?
(a) 1 : 1 : 1
(b) 1 : 1/2 : 1/3
(c) 1/3 : 1/2 : 1
(d) 1 : 2 : 3
Show Answer & Explanation

Answer: (b) 1 : 1/2 : 1/3

Explanation:
1. Moles deposited = Q/(nF): Ag+ n = 1, Cu2+ n = 2, Fe3+ n = 3.
2. Ratio = 1 : 1/2 : 1/3.
3. Answer: (b)

Teacher's Note:
1. Write the relevant formula first.
2. Check signs and units before choosing.

Question 19: Electrolysis of aqueous CuSO4 (inert electrodes) causes:
(a) An increase in pH
(b) A decrease in pH
(c) Either decrease or increase
(d) None of these
Show Answer & Explanation

Answer: (b) A decrease in pH

Explanation:
1. Cathode: Cu2+ + 2e− → Cu. Anode: 2H2O → O2 + 4H+ + 4e−.
2. H+ accumulates, so pH decreases (H2SO4 forms).
3. Answer: (b)

Teacher's Note:
1. Write the relevant formula first.
2. Check signs and units before choosing.

Question 20: The passage of current liberates H2 at the cathode and Cl2 at the anode. The solution is:
(a) CuSO4(aq)
(b) CuCl2(aq)
(c) NaCl(aq)
(d) Water
Show Answer & Explanation

Answer: (c) NaCl(aq)

Explanation:
1. Aqueous NaCl: H2 (from water) at the cathode and Cl2 at the anode.
2. CuCl2 would deposit Cu at the cathode.
3. Answer: (c)

Teacher's Note:
1. Write the relevant formula first.
2. Check signs and units before choosing.

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Practice MCQs for JEE Chemistry Electrochemistry

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FAQs

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