JEE Chemistry Electrochemistry MCQs Set 19

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

Navigate directly to the 50 objective questions for Electrochemistry using the digital viewer below. Each practice set includes verified answer keys, allowing students to instantly cross-check their work and identify areas requiring further revision.

Multiple Choice Questions

Question 1: The emf of the cell Ni | Ni2+ (1.0 M) || Ag+ (1.0 M) | Ag, given E(Ni2+/Ni) = -0.25 V and E(Ag+/Ag) = +0.80 V, is:
(a) -0.25 - 0.80 = -1.05 V
(b) 0.80 - (-0.25) = +1.05 V
(c) -0.80 - (-0.25) = -0.55 V
(d) -0.25 + 0.80 = 0.55 V
Show Answer & Explanation

Answer: (b) 0.80 - (-0.25) = +1.05 V

Explanation:
1. E(cell) = E(cathode) - E(anode), both as reduction potentials
2. Ag+ is reduced at the cathode and Ni is oxidised at the anode
3. 0.80 - (-0.25) = 1.05 V

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

Question 2: For F2 + 2e- giving 2F-, E = 2.8 V. The value of E for (1/2)F2 + e- giving F- is:
(a) -1.4 V
(b) 1.4 V
(c) 2.8 V
(d) -2.8 V
Show Answer & Explanation

Answer: (c) 2.8 V

Explanation:
1. Electrode potential is an intensive property
2. Multiplying or dividing a half reaction does not change E

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

Question 3: Given E(Mg2+/Mg) = -2.37 V and E(Fe3+/Fe) = -0.04 V, the best reducing agent among Mg2+, Fe3+, Mg and Fe is:
(a) Fe3+
(b) Mg2+
(c) Fe
(d) Mg
Show Answer & Explanation

Answer: (d) Mg

Explanation:
1. Reducing agents are the reduced forms of couples, so the choice is between Mg and Fe
2. The more negative reduction potential means the stronger reducing agent, so Mg

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

Question 4: A copper spoon is placed in a solution of ferrous sulphate. What happens?
(a) copper precipitates out
(b) iron precipitates
(c) copper and iron both precipitate
(d) no reaction takes place
Show Answer & Explanation

Answer: (d) no reaction takes place

Explanation:
1. E(Cu2+/Cu) = +0.34 V is higher than E(Fe2+/Fe) = -0.44 V
2. Cu cannot reduce Fe2+, so there is no reaction

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

Question 5: The metals Mg > Al > Zn > Cu > Ag are in decreasing electropositive character. What happens if a copper spoon is used to stir a solution of aluminium nitrate?
(a) an alloy of aluminium and copper forms
(b) there is no reaction
(c) the solution becomes blue
(d) the spoon gets coated with aluminium
Show Answer & Explanation

Answer: (b) there is no reaction

Explanation:
1. Copper is less electropositive than aluminium
2. Cu cannot displace Al3+, so nothing happens

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

Question 6: For Zn2+/Zn, E = -0.76 V and for Ag+/Ag, E = +0.799 V. The correct statement is:
(a) Zn is reduced and Ag is oxidised spontaneously
(b) Zn undergoes reduction and Ag is oxidised
(c) Zn undergoes oxidation and Ag+ gets reduced
(d) no suitable answer
Show Answer & Explanation

Answer: (c) Zn undergoes oxidation and Ag+ gets reduced

Explanation:
1. The couple with the lower reduction potential (Zn) is oxidised
2. Ag+ is reduced at the cathode
3. E(cell) = 0.799 + 0.76 = 1.56 V

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

Question 7: Fe3+ + e- giving Fe2+ has E = +0.77 V, Al3+ + 3e- giving Al has E = -1.66 V and Br2 + 2e- giving 2Br- has E = +1.08 V. The reducing power of Fe2+, Al and Br- increases in the order:
(a) Br- < Fe2+ < Al
(b) Fe2+ < Al < Br-
(c) Al < Fe2+ < Br-
(d) Al < Br- < Fe2+
Show Answer & Explanation

Answer: (a) Br- < Fe2+ < Al

Explanation:
1. A lower reduction potential of the couple means a stronger reducing agent
2. The order of reduction potential is Br2/Br- (1.08) > Fe3+/Fe2+ (0.77) > Al3+/Al (-1.66)
3. So reducing power increases Br- < Fe2+ < Al

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

Question 8: KCl can be used as the salt bridge electrolyte in which of the following cells?
(a) Pb | Pb(NO3)2 || Cu(NO3)2 | Cu
(b) Fe | FeSO4 || Pb(NO3)2 | Pb
(c) Zn | ZnCl2 || AgNO3 | Ag
(d) Cu | CuSO4 || AuCl3 | Au
Show Answer & Explanation

Answer: (d) Cu | CuSO4 || AuCl3 | Au

Explanation:
1. Cl- precipitates Ag+ as AgCl and Pb2+ as PbCl2
2. The Cu-Au cell contains neither Ag+ nor Pb2+, so KCl does not interfere

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

Question 9: Given E(Fe3+/Fe2+) = +0.77 V and E(Sn2+/Sn) = -0.14 V, the standard potential for Sn + 2Fe3+ giving 2Fe2+ + Sn2+ is:
(a) 1.68 V
(b) 0.63 V
(c) 1.40 V
(d) 0.91 V
Show Answer & Explanation

Answer: (d) 0.91 V

Explanation:
1. E = 0.77 - (-0.14) = 0.91 V

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

Question 10: Given Fe3+ + 3e- giving Fe, E = -0.036 V, and Fe2+ + 2e- giving Fe, E = -0.440 V, the E for Fe3+ + e- giving Fe2+ is:
(a) 0.440 V
(b) -0.476 V
(c) -0.404 V
(d) 0.772 V
Show Answer & Explanation

Answer: (d) 0.772 V

Explanation:
1. Add free energies: n E = 3(-0.036) - 2(-0.440)
2. E = (-0.108 + 0.880)/1 = +0.772 V

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

Question 11: Cu+ + e- giving Cu has E = x1 V and Cu2+ + 2e- giving Cu has E = x2 V. The E for Cu2+ + e- giving Cu+ is:
(a) 2x2 - x1
(b) x1 - x2
(c) x1 - 2x2
(d) x1 + 2x2
Show Answer & Explanation

Answer: (a) 2x2 - x1

Explanation:
1. Cu2+ + 2e- gives Cu (2 x2) minus Cu+ + e- gives Cu (x1)
2. n E = 2x2 - x1 for n = 1, so E = 2x2 - x1

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

Question 12: For Zn + Cu2+ giving Zn2+ + Cu, E(cell) = 1.10 V at 25 C. When 0.1 M Cu2+ and 0.1 M Zn2+ are used, the emf is:
(a) -1.10 V
(b) 0.110 V
(c) -0.110 V
(d) 1.10 V
Show Answer & Explanation

Answer: (d) 1.10 V

Explanation:
1. Q = [Zn2+]/[Cu2+] = 1
2. log Q = 0, so E = E(cell) = 1.10 V

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

Question 13: For the cell Pt | H2 (1 atm) | pH = 1 || pH = 2 | H2 (1 atm) | Pt, the cell reaction is:
(a) spontaneous
(b) non-spontaneous
(c) at equilibrium
(d) none of these
Show Answer & Explanation

Answer: (b) non-spontaneous

Explanation:
1. E = 0.059 log([H+]cathode/[H+]anode) = 0.059 log(0.01/0.1) = -0.059 V
2. A negative emf means the reaction is non-spontaneous as written

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

Question 14: The emf of a concentration cell with two zinc electrodes dipping in M/4 and M/16 zinc sulphate solutions at 25 C is:
(a) 0.0178 V
(b) 0.0356 V
(c) 0.0125 V
(d) 0.0250 V
Show Answer & Explanation

Answer: (a) 0.0178 V

Explanation:
1. E = (0.059/2) log(0.25/0.0625) = 0.0295 x log 4
2. = 0.0295 x 0.602 = 0.0178 V

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

Question 15: In an electrolytic cell Ag | AgNO3 | Ag, when current is passed, the concentration of AgNO3:
(a) increases
(b) decreases
(c) remains the same
(d) none of these
Show Answer & Explanation

Answer: (c) remains the same

Explanation:
1. Ag dissolves at the anode and the same amount of Ag deposits at the cathode
2. The electrolyte concentration does not change

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

Question 16: If 0.224 L of H2 gas is formed at the cathode (water electrolysis), the volume of O2 gas formed at the anode under identical conditions is:
(a) 0.224 L
(b) 0.112 L
(c) 1.12 L
(d) 0.448 L
Show Answer & Explanation

Answer: (b) 0.112 L

Explanation:
1. 2H2O gives 2H2 + O2, so O2 volume is half that of H2
2. 0.224/2 = 0.112 L

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

Question 17: Aqueous solutions of A (AgNO3) and B (LiCl) are electrolysed with Pt electrodes. The pH of the resulting solutions will:
(a) increase in A and decrease in B
(b) decrease in both
(c) increase in both
(d) decrease in A and increase in B
Show Answer & Explanation

Answer: (d) decrease in A and increase in B

Explanation:
1. In A, water is oxidised at the anode, giving H+ (HNO3 forms), so pH falls
2. In B, water is reduced at the cathode, giving OH- (LiOH forms), so pH rises

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

Question 18: During electrolysis of CuSO4 solution using Pt electrodes, the pH of the solution:
(a) increases
(b) decreases
(c) cannot be predicted
(d) remains unchanged
Show Answer & Explanation

Answer: (b) decreases

Explanation:
1. Cathode: Cu2+ + 2e- gives Cu
2. Anode: 2H2O gives O2 + 4H+ + 4e-
3. H2SO4 accumulates, so pH decreases

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

Question 19: How many faradays are required to reduce one mole of MnO4- to Mn2+?
(a) 5
(b) 1
(c) 2
(d) 3
Show Answer & Explanation

Answer: (a) 5

Explanation:
1. Mn changes from +7 to +2, a gain of 5 electrons
2. So 5 faradays

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

Question 20: Three faradays of electricity are passed through an aqueous solution of iron(II) bromide. The mass of iron deposited at the cathode (at. mass 56) is:
(a) 56 g
(b) 84 g
(c) 112 g
(d) 168 g
Show Answer & Explanation

Answer: (b) 84 g

Explanation:
1. Fe2+ + 2e- gives Fe, so 2 F deposits 1 mol
2. 3 F deposits 1.5 mol = 1.5 x 56 = 84 g

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

Question 21: A current of 2 A is passed for 1 h through a solution of CuSO4 and 0.237 g of Cu is deposited at the cathode. The current efficiency is:
(a) 42.2 percent
(b) 10 percent
(c) 40.01 percent
(d) 26.1 percent
Show Answer & Explanation

Answer: (b) 10 percent

Explanation:
1. Theoretical mass = (63.5/2) x (2 x 3600)/96500 = 2.369 g
2. Efficiency = 0.237/2.369 x 100 = 10 percent

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

Question 22: A current of 9.65 A is passed through an aqueous NaCl solution for 1000 s. The amount of NaOH formed during electrolysis is:
(a) 2.0 g
(b) 4.0 g
(c) 6.0 g
(d) 8.0 g
Show Answer & Explanation

Answer: (b) 4.0 g

Explanation:
1. Charge = 9.65 x 1000 = 9650 C = 0.1 F
2. Each F produces 1 mol OH-, so 0.1 mol NaOH = 4.0 g

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

Question 23: Salts of A (atomic mass 15), B (27) and C (48) are electrolysed with the same charge. When 4.5 g of A is deposited, 2.7 g of B and 9.6 g of C are deposited. The valencies of A, B and C are respectively:
(a) 1, 3 and 2
(b) 2, 6 and 3
(c) 3, 1 and 3
(d) 3, 1 and 2
Show Answer & Explanation

Answer: (b) 2, 6 and 3

Explanation:
1. Equivalents deposited are equal: 4.5 vA/15 = 2.7 vB/27 = 9.6 vC/48
2. 0.3 vA = 0.1 vB = 0.2 vC
3. vA = 2 gives vB = 6 and vC = 3

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

Question 24: During discharge of a lead storage cell, the density of sulphuric acid in the cell:
(a) remains unchanged
(b) increases
(c) decreases
(d) first increases and then decreases
Show Answer & Explanation

Answer: (c) decreases

Explanation:
1. Pb + PbO2 + 2H2SO4 gives 2PbSO4 + 2H2O
2. Acid is consumed and water is formed, so the density falls

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

Question 25: Which is not a correct method for preventing iron from rusting?
(a) connecting to a sacrificial magnesium electrode
(b) galvanisation
(c) making the medium acidic
(d) making the medium alkaline
Show Answer & Explanation

Answer: (c) making the medium acidic

Explanation:
1. Acid medium speeds up rusting (more H+ for the cathodic reaction)
2. The other three methods protect iron

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

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Practice MCQs for JEE Chemistry Electrochemistry

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