JEE Chemistry Electrochemistry MCQs Set 26

Download JEE MCQs for JEE Chemistry: Electrochemistry

Explore reliable objective questions for Electrochemistry tailored for JEE learners. Utilizing these Chemistry multiple-choice formats ensures thorough preparation and strengthens problem-solving speed for upcoming school assessments.

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 minimum equivalent conductance in the fused state is shown by:
(a) MgCl2
(b) BeCl2
(c) CaCl2
(d) SrCl2
Show Answer & Explanation

Answer: (b) BeCl2

Explanation:
1. BeCl2 is largely covalent, so it hardly ionises in the molten state.
2. MgCl2, CaCl2 and SrCl2 are ionic and conduct well.
3. Answer: (b)

Teacher's Note:
1. Write the standard relation first (Nernst, Faraday or conductivity) and substitute carefully. Check the sign and units of the result.

Question 2: On electrolysis of dilute sulphuric acid using a platinum electrode, the product obtained at the anode will be:
(a) hydrogen
(b) oxygen
(c) sulphur dioxide
(d) hydrogen sulphide
Show Answer & Explanation

Answer: (b) oxygen

Explanation:
1. Water is oxidised at the anode in preference to sulphate.
2. 2H2O → O2 + 4H+ + 4e−, so oxygen is evolved.
3. Answer: (b)

Teacher's Note:
1. Write the standard relation first (Nernst, Faraday or conductivity) and substitute carefully. Check the sign and units of the result.

Question 3: Rust is a mixture of:
(a) FeO, Fe(OH)2
(b) FeO, Fe(OH)3
(c) Fe2O3, Fe(OH)3
(d) Fe3O4, Fe(OH)3
Show Answer & Explanation

Answer: (c) Fe2O3, Fe(OH)3

Explanation:
1. Rust is hydrated ferric oxide, Fe2O3·xH2O.
2. It is a mixture of Fe2O3 and Fe(OH)3.
3. Answer: (c)

Teacher's Note:
1. Write the standard relation first (Nernst, Faraday or conductivity) and substitute carefully. Check the sign and units of the result.

Question 4: Which metal will dissolve if the cell works: Cu|Cu2+ ‖ Ag+|Ag?
(a) Cu
(b) Ag
(c) both Cu and Ag
(d) neither of these
Show Answer & Explanation

Answer: (a) Cu

Explanation:
1. The left electrode (Cu) is the anode, where oxidation occurs: Cu → Cu2+ + 2e−.
2. Ag+ is reduced and deposits on the Ag electrode.
3. Answer: (a)

Teacher's Note:
1. Write the standard relation first (Nernst, Faraday or conductivity) and substitute carefully. Check the sign and units of the result.

Question 5: On the basis of the electrochemical theory of aqueous corrosion, the reaction occurring at the cathode is:
(a) O2 + 4H+ + 4e− → 2H2O(l)
(b) Fe(s) → Fe2+(aq) + 2e−
(c) Fe2+ → Fe3+ + e−
(d) H2(g) + 2OH−(aq) → 2H2O(l) + 2e−
Show Answer & Explanation

Answer: (a) O2 + 4H+ + 4e− → 2H2O(l)

Explanation:
1. At the cathode, dissolved oxygen is reduced in acidic water.
2. The oxidation of iron (Fe → Fe2+ + 2e−) occurs at the anode.
3. Answer: (a)

Teacher's Note:
1. Write the standard relation first (Nernst, Faraday or conductivity) and substitute carefully. Check the sign and units of the result.

Question 6: When a lead storage battery is charged, it acts as a:
(a) fuel cell
(b) electrolytic cell
(c) galvanic cell
(d) concentration cell
Show Answer & Explanation

Answer: (b) electrolytic cell

Explanation:
1. On charging, external electrical energy drives the reverse reaction.
2. The battery then behaves as an electrolytic cell.
3. Answer: (b)

Teacher's Note:
1. Write the standard relation first (Nernst, Faraday or conductivity) and substitute carefully. Check the sign and units of the result.

Question 7: Zinc acts as a sacrificial or cathodic protection to prevent rusting of iron because:
(a) E°OP of Zn < E°OP of Fe
(b) E°OP of Zn > E°OP of Fe
(c) E°OP of Zn = E°OP of Fe
(d) Zn is cheaper than Fe
Show Answer & Explanation

Answer: (b) E°OP of Zn > E°OP of Fe

Explanation:
1. Zn has the higher oxidation potential (it is more easily oxidised).
2. So Zn corrodes in place of iron.
3. Answer: (b)

Teacher's Note:
1. Write the standard relation first (Nernst, Faraday or conductivity) and substitute carefully. Check the sign and units of the result.

Question 8: The molar conductivity of a 0.01 M solution of a weak acid is 16.6 Ω−1 cm2 mol−1. Its molar conductivity at infinite dilution is 390.7 Ω−1 cm2 mol−1. The degree of dissociation of the weak acid is:
(a) 0.24
(b) 0.42
(c) 0.042
(d) 0.024
Show Answer & Explanation

Answer: (c) 0.042

Explanation:
1. α = Λm/Λ°m = 16.6/390.7.
2. = 0.0425 ≈ 0.042.
3. Answer: (c)

Teacher's Note:
1. Write the standard relation first (Nernst, Faraday or conductivity) and substitute carefully. Check the sign and units of the result.

Question 9: The strongest oxidising agent among the species having SRP values In3+ (−1.34 V), Au3+ (1.4 V), Hg2+ (0.86 V) and Cr3+ (−0.74 V) is:
(a) Cr3+
(b) Au3+
(c) Hg2+
(d) In3+
Show Answer & Explanation

Answer: (b) Au3+

Explanation:
1. A higher standard reduction potential means a stronger oxidising agent.
2. Au3+ has the highest SRP (1.4 V).
3. Answer: (b)

Teacher's Note:
1. Write the standard relation first (Nernst, Faraday or conductivity) and substitute carefully. Check the sign and units of the result.

Question 10: Given the standard electrode potentials: (a) K+/K = −3.02 V, (b) Cu2+/Cu = +0.34 V, (c) Hg2+/Hg = 0.92 V, (d) Cr3+/Cr = −0.74 V. The decreasing order of reducing power of these elements is:
(a) a > b > c > d
(b) a > d > c > b
(c) a > d > b > c
(d) c > b > d > a
Show Answer & Explanation

Answer: (c) a > d > b > c

Explanation:
1. The lower the reduction potential, the stronger the reducing agent.
2. Order of potentials: K (−3.02) < Cr (−0.74) < Cu (+0.34) < Hg (+0.92), giving a > d > b > c.
3. Answer: (c)

Teacher's Note:
1. Write the standard relation first (Nernst, Faraday or conductivity) and substitute carefully. Check the sign and units of the result.

Question 11: Calculate Ecell for Mg(s) + 2Ag+(0.0001 M) → Mg2+(0.100 M) + 2Ag(s), given E°cell = 3.17 V.
(a) −2.96 V
(b) +2.96 V
(c) 3.38 V
(d) −3.38 V
Show Answer & Explanation

Answer: (b) +2.96 V

Explanation:
1. E = E° − (0.0591/2) log([Mg2+]/[Ag+]2).
2. = 3.17 − 0.02955 × log(0.1/10−8) = 3.17 − 0.02955 × 7 = 2.96 V.
3. Answer: (b)

Teacher's Note:
1. Write the standard relation first (Nernst, Faraday or conductivity) and substitute carefully. Check the sign and units of the result.

Question 12: Which of the following statements is correct for an electrolytic cell?
(a) electrons flow from cathode to anode through the external battery
(b) electrons flow from cathode to anode within the electrolyte solution
(c) migration of ions occurs along with oxidation at the cathode and reduction at the anode
(d) migration of ions occurs along with reduction at the cathode and oxidation at the anode
Show Answer & Explanation

Answer: (d) migration of ions occurs along with reduction at the cathode and oxidation at the anode

Explanation:
1. In any electrochemical cell reduction occurs at the cathode and oxidation at the anode.
2. Cations migrate to the cathode and anions to the anode, and electrons do not flow through the solution.
3. Answer: (d)

Teacher's Note:
1. Write the standard relation first (Nernst, Faraday or conductivity) and substitute carefully. Check the sign and units of the result.

Question 13: In the electrolysis of acidulated water, 4.48 L of hydrogen at STP was produced by passing a current of 2.14 A. For how many hours was the current passed?
(a) 4
(b) 3
(c) 6
(d) 5
Show Answer & Explanation

Answer: (d) 5

Explanation:
1. 4.48 L = 0.2 mol H2, needing 0.4 F = 38600 C.
2. t = 38600/2.14 = 18037 s ≈ 5 hours.
3. Answer: (d)

Teacher's Note:
1. Write the standard relation first (Nernst, Faraday or conductivity) and substitute carefully. Check the sign and units of the result.

Question 14: When a dilute aqueous solution of Li2SO4 is electrolysed, the products formed at the anode and cathode are respectively:
(a) H2 and O2
(b) O2 and Li
(c) SO2 and H2
(d) O2 and H2
Show Answer & Explanation

Answer: (d) O2 and H2

Explanation:
1. Water is oxidised in preference to sulphate at the anode, giving O2.
2. Water is reduced in preference to Li+ at the cathode, giving H2.
3. Answer: (d)

Teacher's Note:
1. Write the standard relation first (Nernst, Faraday or conductivity) and substitute carefully. Check the sign and units of the result.

Question 15: How much time is required for complete decomposition of 4 mol of water using a current of 4 ampere?
(a) 96500 s
(b) 3.85 × 104 s
(c) 1.93 × 105 s
(d) 2.92 × 105 s
Show Answer & Explanation

Answer: (c) 1.93 × 105 s

Explanation:
1. 2H2O → 2H2 + O2 needs 4 F per 2 mol of water, so 4 mol of water needs 8 F = 8 × 96500 = 772000 C.
2. t = 772000/4 = 1.93 × 105 s.
3. Answer: (c)

Teacher's Note:
1. Write the standard relation first (Nernst, Faraday or conductivity) and substitute carefully. Check the sign and units of the result.

Question 16: The amount of metal deposited at the cathode on passing an electric current of 0.75 ampere in aqueous ferric sulphate solution for 30 minutes will be (atomic weight of Fe = 56):
(a) 0.00435 g
(b) 0.261 g
(c) 0.783 g
(d) 0.522 g
Show Answer & Explanation

Answer: (b) 0.261 g

Explanation:
1. Q = 0.75 × 1800 = 1350 C and Fe3+ needs 3 electrons.
2. Mass = (56/3 × 1350)/96500 = 0.261 g.
3. Answer: (b)

Teacher's Note:
1. Write the standard relation first (Nernst, Faraday or conductivity) and substitute carefully. Check the sign and units of the result.

Question 17: For the cell Pt|H2(1 atm)|H+(0.001 M)‖H+(0.1 M)|H2(1 atm)|Pt, what will be the value of Ecell?
(a) 0.1182 V
(b) −0.1182 V
(c) 0.0591 V
(d) −0.0591 V
Show Answer & Explanation

Answer: (a) 0.1182 V

Explanation:
1. For a hydrogen concentration cell E = 0.0591 log([H+]cathode/[H+]anode).
2. = 0.0591 log(0.1/0.001) = 0.0591 × 2 = 0.1182 V.
3. Answer: (a)

Teacher's Note:
1. Write the standard relation first (Nernst, Faraday or conductivity) and substitute carefully. Check the sign and units of the result.

Question 18: Consider the electrode potentials: Mg2+ + 2e− → Mg, E° = −2.37 V; V2+ + 2e− → V, E° = −1.18 V; Cu2+ + e− → Cu+, E° = 0.15 V. Which of the following reactions will proceed spontaneously from left to right?
(a) Mg2+ + V → Mg + V2+
(b) Mg2+ + 2Cu+ → Mg + 2Cu2+
(c) V2+ + 2Cu+ → V + 2Cu2+
(d) V + 2Cu2+ → V2+ + 2Cu+
Show Answer & Explanation

Answer: (d) V + 2Cu2+ → V2+ + 2Cu+

Explanation:
1. Spontaneous when E°cell = E°(cathode) − E°(anode) > 0.
2. V + 2Cu2+: 0.15 − (−1.18) = +1.33 V. The other three give negative values.
3. Answer: (d)

Teacher's Note:
1. Write the standard relation first (Nernst, Faraday or conductivity) and substitute carefully. Check the sign and units of the result.

Question 19: A metal bucket is to be electroplated using ZnCl2 as the electrolyte. How many moles of zinc are deposited in 20 min by a constant current of 10 A?
(a) 0.01
(b) 0.03
(c) 0.06
(d) 0.10
Show Answer & Explanation

Answer: (c) 0.06

Explanation:
1. Q = 10 × 1200 = 12000 C.
2. Moles of Zn = Q/(2F) = 12000/193000 = 0.062 ≈ 0.06.
3. Answer: (c)

Teacher's Note:
1. Write the standard relation first (Nernst, Faraday or conductivity) and substitute carefully. Check the sign and units of the result.

Question 20: The quantity of electricity required to release 112 cm3 of hydrogen at STP from acidified water is:
(a) 0.1 F
(b) 1 F
(c) 965 C
(d) 96500 C
Show Answer & Explanation

Answer: (c) 965 C

Explanation:
1. 112 cm3 = 0.005 mol H2 and each mole needs 2 F.
2. Q = 0.01 F = 0.01 × 96500 = 965 C.
3. Answer: (c)

Teacher's Note:
1. Write the standard relation first (Nernst, Faraday or conductivity) and substitute carefully. Check the sign and units of the result.

Question 21: If E°Au+/Au = 1.69 V and E°Au3+/Au = 1.40 V, then the value of E°Au3+/Au+ will be:
(a) 0.19 V
(b) 2.945 V
(c) 1.255 V
(d) 1.19 V
Show Answer & Explanation

Answer: (c) 1.255 V

Explanation:
1. Au3+ + 3e− → Au (3 × 1.40 = 4.20) minus Au+ + e− → Au (1.69).
2. For Au3+ + 2e− → Au+, nE° = 4.20 − 1.69 = 2.51, so E° = 2.51/2 = 1.255 V.
3. Answer: (c)

Teacher's Note:
1. Write the standard relation first (Nernst, Faraday or conductivity) and substitute carefully. Check the sign and units of the result.

Question 22: Given standard electrode potentials K+/K = −2.93 V, Ag+/Ag = 0.80 V, Hg2+/Hg = 0.79 V, Mg2+/Mg = −2.37 V, Cr3+/Cr = −0.74 V. The correct order of their reducing power is:
(a) Ag > Hg > Cr > Mg > K
(b) K > Mg > Cr > Hg > Ag
(c) K > Cr > Mg > Ag > Hg
(d) Cr > Mg > K > Hg > Ag
Show Answer & Explanation

Answer: (b) K > Mg > Cr > Hg > Ag

Explanation:
1. The lower the E°, the stronger the reducing agent.
2. Order of E°: K (−2.93) < Mg (−2.37) < Cr (−0.74) < Hg (0.79) < Ag (0.80).
3. Answer: (b)

Teacher's Note:
1. Write the standard relation first (Nernst, Faraday or conductivity) and substitute carefully. Check the sign and units of the result.

Question 23: The conductivity of electrolytic solutions depends on:
(a) size of the ions produced and their solvation
(b) the nature of the solvent and its viscosity
(c) temperature
(d) all of the above
Show Answer & Explanation

Answer: (d) all of the above

Explanation:
1. Ionic mobility depends on ion size, solvation, solvent viscosity and temperature.
2. Conductivity depends on all of these.
3. Answer: (d)

Teacher's Note:
1. Write the standard relation first (Nernst, Faraday or conductivity) and substitute carefully. Check the sign and units of the result.

Question 24: The volume of hydrogen gas liberated at STP when a current of 5.36 ampere is passed through dilute H2SO4 for 5 hours will be:
(a) 5.6 L
(b) 11.2 L
(c) 16.8 L
(d) 22.4 L
Show Answer & Explanation

Answer: (b) 11.2 L

Explanation:
1. Q = 5.36 × 5 × 3600 = 96480 C ≈ 1 F.
2. 1 F liberates 0.5 mol of H2 = 11.2 L at STP.
3. Answer: (b)

Teacher's Note:
1. Write the standard relation first (Nernst, Faraday or conductivity) and substitute carefully. Check the sign and units of the result.

Question 25: The electrolysis of silver nitrate solution is carried out using silver electrodes. Which of the following reactions occurs at the anode?
(a) Ag → Ag+ + e−
(b) Ag+ + e− → Ag
(c) 2H2O → 4H+ + O2 + 4e−
(d) 4OH− → 2H2 + O2 + 4e−
Show Answer & Explanation

Answer: (a) Ag → Ag+ + e−

Explanation:
1. A reactive (non-inert) silver anode is oxidised in preference to water.
2. Ag → Ag+ + e− occurs at the anode.
3. Answer: (a)

Teacher's Note:
1. Write the standard relation first (Nernst, Faraday or conductivity) and substitute carefully. Check the sign and units of the result.

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FAQs

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