JEE Chemistry Electrochemistry MCQs Set 25

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

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: If E°A2+/A = −0.30 V and E°A3+/A2+ = 0.40 V, the standard EMF of the reaction A + 2A3+ → 3A2+ will be:
(a) 0.30 V
(b) 0.40 V
(c) 0.70 V
(d) 0.10 V
Show Answer & Explanation

Answer: (c) 0.70 V

Explanation:
1. Anode: A → A2+ + 2e−, oxidation potential +0.30 V. Cathode: A3+ + e− → A2+, 0.40 V.
2. E° = 0.40 + 0.30 = 0.70 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 2: On the basis of the following E° values, the strongest oxidising agent is: K+ + e− → K, E° = −2.923 V; Mg2+ + 2e− → Mg, E° = −2.337 V.
(a) K+
(b) Mg2+
(c) K
(d) Mg
Show Answer & Explanation

Answer: (b) Mg2+

Explanation:
1. An oxidising agent is the species on the reactant side, with the higher reduction potential.
2. Between K+ and Mg2+, Mg2+ (−2.337 V) is the stronger oxidant.
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: The equivalent conductance of a M/20 solution of a weak monobasic acid is 10 S cm2/eq. and at infinite dilution is 200 S cm2/eq. The dissociation constant of this acid is:
(a) 1.25 × 10−4
(b) 1.25 × 10−5
(c) 1.25 × 10−6
(d) 6.26 × 10−4
Show Answer & Explanation

Answer: (a) 1.25 × 10−4

Explanation:
1. α = Λ/Λ° = 10/200 = 0.05.
2. Ka = Cα2 = 0.05 × (0.05)2 = 1.25 × 10−4.
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 4: Given Fe2+ + 2e− → Fe(s), E° = −0.447 V and Fe3+ + e− → Fe2+, E° = +0.771 V. Find E° for the reaction Fe3+ + 3e− → Fe(s).
(a) 0.34 V
(b) −0.041 V
(c) +0.39 V
(d) −0.47 V
Show Answer & Explanation

Answer: (b) −0.041 V

Explanation:
1. Add free energies (not potentials): ΔG° = −1(0.771)F − 2(−0.447)F = +0.123F.
2. E° = −ΔG°/3F = −0.041 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 5: An increase in molar conductance of a strong electrolyte with dilution is mainly due to:
(a) increase in the number of ions
(b) increase in ionic mobility of ions
(c) 100% ionisation of the electrolyte at normal dilution
(d) increase in both the number of ions and ionic mobility
Show Answer & Explanation

Answer: (b) increase in ionic mobility of ions

Explanation:
1. A strong electrolyte is already completely ionised, so the number of ions per mole does not change.
2. On dilution interionic attraction falls, so ionic mobility increases.
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 6: Which relation of EMF of an electrochemical cell is correct?
(a) EMF of cell = oxidation potential of anode − reduction potential of cathode
(b) EMF of cell = oxidation potential of anode + reduction potential of cathode
(c) EMF of cell = reduction potential of anode + reduction potential of cathode
(d) all of these
Show Answer & Explanation

Answer: (b) EMF of cell = oxidation potential of anode + reduction potential of cathode

Explanation:
1. Cell EMF = Ecathode(reduction) − Eanode(reduction).
2. Since oxidation potential = −reduction potential, this equals oxidation potential of anode + reduction potential of cathode.
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: The standard electrode potential of the Au3+/Au couple is 1.428 V and that for the Na+/Na couple is −2.714 V. These two couples in their standard state are connected to make a cell. The cell potential will be:
(a) 4.142 V
(b) 1.286 V
(c) −1.286 V
(d) 3.376 V
Show Answer & Explanation

Answer: (a) 4.142 V

Explanation:
1. Au3+/Au is the cathode and Na+/Na the anode.
2. E°cell = 1.428 − (−2.714) = 4.142 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 8: The limiting molar conductivity of CH3COOH (i.e. Λ°m(CH3COOH)) is equal to:
(a) Λ°(CH3COOH) + Λ°(CH3COONa) − Λ°(NaOH)
(b) Λ°(CH3COONa) + Λ°(HCl) − Λ°(NaCl)
(c) Λ°(CH3COONa) + Λ°(NaCl) − Λ°(NaOH)
(d) Λ°(NaOH) + Λ°(NaCl) − Λ°(CH3COONa)
Show Answer & Explanation

Answer: (b) Λ°(CH3COONa) + Λ°(HCl) − Λ°(NaCl)

Explanation:
1. By Kohlrausch's law: CH3COO− + H+.
2. Λ°(CH3COONa) + Λ°(HCl) − Λ°(NaCl) = λ(CH3COO−) + λ(Na+) + λ(H+) + λ(Cl−) − λ(Na+) − λ(Cl−).
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 9: (a) H2O2 + O3 → H2O + 2O2; (b) H2O2 + Ag2O → 2Ag + H2O + O2. The role of hydrogen peroxide in the above reactions is respectively:
(a) oxidising agent in (a) and reducing agent in (b)
(b) reducing agent in (a) and oxidising agent in (b)
(c) reducing agent in (a) and (b)
(d) oxidising agent in (a) and (b)
Show Answer & Explanation

Answer: (c) reducing agent in (a) and (b)

Explanation:
1. In both reactions O of H2O2 (−1) goes to O2 (0), i.e. it is oxidised.
2. So H2O2 acts as a reducing agent in both.
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 10: The weight of copper (atomic weight 63.5) displaced by a quantity of electricity which displaces 5600 mL of O2 at STP will be:
(a) 63.5 g
(b) 31.75 g
(c) 15.875 g
(d) 127 g
Show Answer & Explanation

Answer: (b) 31.75 g

Explanation:
1. 5600 mL O2 = 0.25 mol = 1 equivalent (4 F per mol).
2. 1 equivalent of Cu = 63.5/2 = 31.75 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 11: When 0.1 mole of Cr2O72− is reduced to Cr3+ completely, the quantity of electricity required is:
(a) 9650 C
(b) 96500 C
(c) 57900 C
(d) 54900 C
Show Answer & Explanation

Answer: (c) 57900 C

Explanation:
1. Cr2O72− + 14H+ + 6e− → 2Cr3+ + 7H2O.
2. Q = 0.1 × 6 × 96500 = 57900 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 12: An aqueous solution of which of the following compounds is the best conductor of electric current?
(a) CH3COOH
(b) C12H22O11
(c) H2C2O4
(d) NaOH
Show Answer & Explanation

Answer: (d) NaOH

Explanation:
1. NaOH is a strong electrolyte and ionises completely.
2. Acetic acid and oxalic acid are weak, and sucrose is a non-electrolyte.
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: The number of electrons delivered at the cathode during electrolysis by a current of 2 ampere in 120 seconds is (charge on electron = 1.6 × 10−19 C):
(a) 3.75 × 1020
(b) 7.48 × 1020
(c) 1.5 × 1021
(d) 3 × 1020
Show Answer & Explanation

Answer: (c) 1.5 × 1021

Explanation:
1. Q = It = 2 × 120 = 240 C.
2. Number of electrons = 240/(1.6 × 10−19) = 1.5 × 1021.
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 14: The molar conductivity of a 0.1 mol/dm3 solution of KNO3 with electrolytic conductivity 4 × 10−3 S cm−1 at 298 K is:
(a) 11.52 S cm2 mol−1
(b) 20 S cm2 mol−1
(c) 40 S cm2 mol−1
(d) 13.48 S cm2 mol−1
Show Answer & Explanation

Answer: (c) 40 S cm2 mol−1

Explanation:
1. Λm = κ × 1000/C with C in mol/L.
2. = 4 × 10−3 × 1000/0.1 = 40 S cm2 mol−1.
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 15: The resistance of a conductivity cell containing 0.001 M KCl solution at 298 K is 1500 ohm. What is the cell constant if the conductivity of 0.001 M KCl solution at 298 K is 0.146 × 10−3 S cm−1?
(a) 0.419 cm−1
(b) 0.219 cm−1
(c) 0.45 cm−1
(d) 0.75 cm−1
Show Answer & Explanation

Answer: (b) 0.219 cm−1

Explanation:
1. Cell constant = κ × R.
2. = 0.146 × 10−3 × 1500 = 0.219 cm−1.
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 16: Calculate the ΔG° of the reaction Fe2+(aq) + Ag+(aq) → Fe3+(aq) + Ag(s). E°Ag+/Ag = 0.8 V and E°Fe3+/Fe2+ = 0.77 V.
(a) −2895 J mol−1
(b) −3845 J mol−1
(c) −1874 J mol−1
(d) −375 J mol−1
Show Answer & Explanation

Answer: (a) −2895 J mol−1

Explanation:
1. E°cell = 0.80 − 0.77 = 0.03 V with n = 1.
2. ΔG° = −96500 × 0.03 = −2895 J mol−1.
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 17: Calculate the EMF of the cell Fe(s) + 2H+(1 M) → Fe2+(0.001 M) + H2(g, 1 atm). (Given E°Fe2+/Fe = −0.44 V)
(a) 0.215 V
(b) 0.38 V
(c) −0.48 V
(d) 0.5285 V
Show Answer & Explanation

Answer: (d) 0.5285 V

Explanation:
1. E° = 0 − (−0.44) = 0.44 V and n = 2.
2. E = 0.44 − (0.0591/2) log(0.001/1) = 0.44 + 0.0887 = 0.5285 V.
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 18: A solution of CuSO4 is electrolysed for 10 minutes with a current of 1.5 amperes. What is the mass of copper deposited at the cathode? (Molar mass of Cu = 63.5 g/mol)
(a) 0.492 g
(b) 0.325 g
(c) 0.873 g
(d) 0.296 g
Show Answer & Explanation

Answer: (d) 0.296 g

Explanation:
1. Q = 1.5 × 600 = 900 C.
2. Mass = (63.5/2 × 900)/96500 = 0.296 g.
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: The conductance is the:
(a) reciprocal of specific resistance
(b) reciprocal of resistance
(c) reciprocal of current
(d) reciprocal of concentration
Show Answer & Explanation

Answer: (b) reciprocal of resistance

Explanation:
1. Conductance G = 1/R, measured in siemens.
2. The reciprocal of specific resistance is conductivity.
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 20: The ion which has the lowest ionic mobility in aqueous solution is:
(a) Li+
(b) Na+
(c) K+
(d) Rb+
Show Answer & Explanation

Answer: (a) Li+

Explanation:
1. Li+ is the smallest and the most heavily hydrated, so the hydrated ion is the largest.
2. It therefore moves most slowly.
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 21: An electric current of 100 ampere is passed through a molten liquid of sodium chloride for 5 hours. Calculate the volume of chlorine gas liberated at the electrode at NTP.
(a) 208.91 L
(b) 310 L
(c) 40.8 L
(d) 150.2 L
Show Answer & Explanation

Answer: (a) 208.91 L

Explanation:
1. Q = 100 × 5 × 3600 = 1.8 × 106 C.
2. Moles of Cl2 = Q/(2F) = 9.33, so volume = 9.33 × 22.4 = 208.9 L.
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 22: Zn can displace:
(a) Mg2+ from its aqueous solution
(b) Cu2+ from its aqueous solution
(c) Na+ from its aqueous solution
(d) Al3+ from its aqueous solution
Show Answer & Explanation

Answer: (b) Cu2+ from its aqueous solution

Explanation:
1. A metal displaces ions of metals lower in the electrochemical series.
2. Zn is above Cu but below Mg, Na and Al.
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: Which of the following is incorrect regarding the salt bridge solution?
(a) the solution must be of a strong electrolyte
(b) the solution should be inert towards both electrodes
(c) the size of the cation and anion of the salt should be much different
(d) the salt bridge solution is prepared in gelatin or agar-agar to make it semi-solid
Show Answer & Explanation

Answer: (c) the size of the cation and anion of the salt should be much different

Explanation:
1. The ions must have nearly equal mobilities so that the liquid junction potential is minimal.
2. So their sizes should be similar, not much different (KCl is the usual choice).
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 24: The cell reaction during discharging of a lead storage battery at the cathode is:
(a) Pb + SO42− − 2e− → PbSO4
(b) PbSO4 + 2e− → Pb + SO42−
(c) PbO2 + SO42− + 4H+ + 2e− → PbSO4 + 2H2O
(d) PbSO4 + 2H2O → PbO2 + SO42− + 4H+ + 2e−
Show Answer & Explanation

Answer: (c) PbO2 + SO42− + 4H+ + 2e− → PbSO4 + 2H2O

Explanation:
1. During discharge the cathode (PbO2) is reduced.
2. PbO2 + SO42− + 4H+ + 2e− → PbSO4 + 2H2O.
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 25: In the electrolysis of NaCl, when a Pt electrode is taken, H2 is liberated at the cathode, while with a Hg cathode it forms sodium amalgam. This is because:
(a) Hg is more inert than Pt
(b) Na is dissolved in Hg while it does not dissolve in Pt
(c) more voltage is required to reduce H+ at Hg than at Pt
(d) the concentration of H+ ions is larger when Pt electrode is taken
Show Answer & Explanation

Answer: (c) more voltage is required to reduce H+ at Hg than at Pt

Explanation:
1. Hydrogen has a large overvoltage at a mercury surface, so H+ is not discharged there.
2. Na+ is then discharged and dissolves in Hg as amalgam.
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.

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