JEE Chemistry Electrochemistry MCQs Set 23

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

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: How many faraday are required to obtain 1000 g of sodium perchlorate from the electrolysis of hot aqueous NaCl: \( \ce{NaCl + 4H2O -> NaClO4 + 4H2} \)?
(a) 40.3
(b) 18.3
(c) 66
(d) 31.6
Show Answer & Explanation

Answer: (c) 66

Explanation:
1. Chlorine changes from -1 in NaCl to +7 in \( \ce{NaClO4} \): 8 electrons per formula unit.
2. Moles of \( \ce{NaClO4} \) (122.5 g per mol) in 1000 g = 8.16.
3. Faradays = \( 8.16\times8\approx65.3\approx66 \).

Teacher's Note:
1. Faradays = moles x electrons per formula unit.

Question 2: For which of these oxidation/reduction pairs will the reduction potential vary with pH? I. \( \ce{AmO2^{2+}/AmO2+} \); II. \( \ce{AmO2^{2+}/Am^{4+}} \); III. \( \ce{Am^{4+}/Am^{2+}} \)
(a) II only
(b) I only
(c) I and II only
(d) I, II and III
Show Answer & Explanation

Answer: (a) II only

Explanation:
1. The potential depends on pH only if \( \ce{H+} \) (or \( \ce{OH-} \)) appears in the half-reaction.
2. \( \ce{AmO2^{2+} + 4H+ + 2e- -> Am^{4+} + 2H2O} \) contains H+, whereas the other two half-reactions are simple electron transfers without H+.

Teacher's Note:
1. Only half-cell reactions that involve H+ or OH- depend on pH.

Question 3: In the concentration cell \( \text{Pt(H}_2)\,|\,\text{HA (0.1 M), NaA (1 M)}\,\|\,\text{HA (1 M), NaA (1 M)}\,|\,\text{(H}_2)\text{Pt} \) (\( pK_a \) of HA = 4), the cell potential will be:
(a) 0.03 V
(b) -0.06 V
(c) -0.03 V
(d) 0.06 V
Show Answer & Explanation

Answer: (d) 0.06 V

Explanation:
1. Left: \( [H^+]=K_a\frac{0.1}{1}=10^{-5} \) M. Right: \( [H^+]=K_a\frac11=10^{-4} \) M.
2. \( E=0.0591\log\frac{[H^+]_{right}}{[H^+]_{left}}=0.0591\log10=0.0591\approx0.06 \) V.

Teacher's Note:
1. Each side is a buffer; find [H+] from Henderson's equation.

Question 4: The pressure of hydrogen gas is increased from 1 atm to 100 atm, keeping \( \ce{H+} \) (1 M) constant. The voltage of the hydrogen half-cell at 25 \(^\circ\)C will be:
(a) 0.059 V
(b) -0.0591 V
(c) 0.59 V
(d) 0.0259 V
Show Answer & Explanation

Answer: (b) -0.0591 V

Explanation:
1. For \( \ce{2H+ + 2e- -> H2} \): \( E=0-\frac{0.0591}{2}\log\frac{P_{H_2}}{[H^+]^2} \).
2. \( E=-0.02955\times\log100=-0.0591 \) V.

Teacher's Note:
1. Higher product pressure makes the reduction potential more negative.

Question 5: For the cell reaction \( \ce{Mg(s) + 2Ag+(aq) <=> Mg^{2+}(aq) + 2Ag(s)} \), \( E^\circ_{cell}=3.17 \) V at 298 K. The values of \( E_{cell} \), \( \Delta G^\circ \) and Q at \( \ce{Ag+} \) and \( \ce{Mg^{2+}} \) concentrations of 0.001 M and 0.02 M respectively are:
(a) 3.04 V, -605.8 kJ per mol, 20000
(b) 3.04 V, -611.8 kJ per mol, 20000
(c) 3.04 V, +611.8 kJ per mol, 20000
(d) 3.13 V, -604 kJ per mol, 20
Show Answer & Explanation

Answer: (b) 3.04 V, -611.8 kJ per mol, 20000

Explanation:
1. \( Q=\frac{[Mg^{2+}]}{[Ag^+]^2}=\frac{0.02}{(0.001)^2}=2\times10^4 \).
2. \( E=3.17-\frac{0.0591}{2}\log(2\times10^4)=3.17-0.127=3.04 \) V.
3. \( \Delta G^\circ=-nFE^\circ=-2\times96500\times3.17=-611.8 \) kJ per mol.

Teacher's Note:
1. dG-standard uses E-standard, not the actual E.

Question 6: Consider the following four electrodes: P = \( \ce{Cu^{2+}} \) (0.0001 M)/Cu(s); Q = \( \ce{Cu^{2+}} \) (0.1 M)/Cu(s); R = \( \ce{Cu^{2+}} \) (0.01 M)/Cu(s); S = \( \ce{Cu^{2+}} \) (0.001 M)/Cu(s). If the standard reduction potential of \( \ce{Cu^{2+}/Cu} \) is 0.34 V, the reduction potential in volt of the above electrodes follows the order:
(a) P > S > R > Q
(b) R > S > Q > P
(c) P > Q > R > S
(d) Q > R > S > P
Show Answer & Explanation

Answer: (d) Q > R > S > P

Explanation:
1. \( E=0.34+\frac{0.0591}{2}\log[Cu^{2+}] \).
2. Greater \( [Cu^{2+}] \) gives a higher reduction potential: 0.1 M > 0.01 M > 0.001 M > 0.0001 M.

Teacher's Note:
1. Higher ion concentration favours reduction.

Question 7: Which is correct about fuel cells?
(a) All of these
(b) The cell runs continuously as long as the fuels are supplied
(c) They are more efficient and free from pollution
(d) They are used to provide power and drinking water to astronauts in space programmes
Show Answer & Explanation

Answer: (a) All of these

Explanation:
1. Fuel cells run continuously while reactants are fed in.
2. They convert chemical energy directly with high efficiency and, in the hydrogen-oxygen cell, the only product is water, which is used as drinking water in space missions.
3. So all three statements are correct.

Teacher's Note:
1. H2-O2 fuel cells were used on spacecraft for power and water.

Question 8: Salts of A (atomic weight 7), B (atomic weight 27) and C (atomic weight 48) were electrolysed under identical conditions using the same quantity of electricity. When 2.1 g of A was deposited, the masses of B and C deposited were 2.7 g and 7.2 g. The valencies of A, B and C are respectively:
(a) 3, 1 and 2
(b) 3, 1 and 3
(c) 1, 3 and 2
(d) 2, 3 and 2
Show Answer & Explanation

Answer: (c) 1, 3 and 2

Explanation:
1. Equal charge deposits equal numbers of equivalents, so the masses are proportional to the equivalent weights: 2.1 : 2.7 : 7.2 = 7 : 9 : 24.
2. Equivalent weight = atomic weight / valency: A = 7/1, B = 27/3 = 9, C = 48/2 = 24.
3. Valencies are 1, 3 and 2.

Teacher's Note:
1. Faraday's second law: mass is proportional to equivalent weight.

Question 9: An electrochemical cell is set up as follows: \( \text{Pt(H}_2\text{, 1 atm)}\,/\,0.1\text{ M HCl}\,\|\,0.1\text{ M acetic acid}\,/\,\text{(H}_2\text{, 1 atm) Pt} \). The EMF of the cell will not be zero because:
(a) Different acids are used in the two compartments
(b) The EMF of a cell depends on the molarities of the acids used
(c) The pH of 0.1 M HCl and 0.1 M acetic acid is not the same
(d) The temperature is constant
Show Answer & Explanation

Answer: (c) The pH of 0.1 M HCl and 0.1 M acetic acid is not the same

Explanation:
1. Both electrodes are hydrogen electrodes at 1 atm, so the EMF depends only on the difference in \( [H^+] \).
2. HCl is strong (\( [H^+]=0.1 \) M) while acetic acid is weak (\( [H^+]\approx1.3\times10^{-3} \) M), so a potential develops.

Teacher's Note:
1. A concentration cell has zero EMF only if the two sides have equal [H+].

Question 10: Standard electrode potentials of some half cells are: (A) \( \ce{Sn^{4+} + 2e- -> Sn^{2+}} \), +0.15 V; (B) \( \ce{2Hg^{2+} + 2e- -> Hg2^{2+}} \), +0.92 V; (C) \( \ce{PbO2 + 4H+ + 2e- -> Pb^{2+} + 2H2O} \), +1.45 V. For these, which statement is correct?
(a) \( \ce{Sn^{2+}} \) is a more powerful reducing agent than \( \ce{Hg2^{2+}} \)
(b) \( \ce{Pb^{2+}} \) is a more powerful reducing agent than \( \ce{Sn^{2+}} \)
(c) \( \ce{Sn^{4+}} \) is a more powerful oxidising agent than \( \ce{Pb^{4+}} \)
(d) \( \ce{Hg^{2+}} \) is a more powerful oxidising agent than \( \ce{Pb^{4+}} \)
Show Answer & Explanation

Answer: (a) \( \ce{Sn^{2+}} \) is a more powerful reducing agent than \( \ce{Hg2^{2+}} \)

Explanation:
1. The lower the reduction potential of a couple, the stronger the reducing agent of its reduced form.
2. \( \ce{Sn^{2+}} \) (0.15 V) is a stronger reductant than \( \ce{Hg2^{2+}} \) (0.92 V), and \( \ce{Pb^{2+}} \) (1.45 V) is a weaker reductant than Sn2+.
3. Pb(IV) (1.45 V) is a stronger oxidant than both Hg(II) (0.92 V) and Sn(IV) (0.15 V).

Teacher's Note:
1. High E means strong oxidant (oxidised form) and weak reductant (reduced form).

Question 11: If the same quantity of electricity is passed through CuCl and \( \ce{CuSO4} \), the ratio of the weights of Cu deposited from \( \ce{CuSO4} \) and CuCl is:
(a) 2 : 1
(b) 1 : 1
(c) 4 : 1
(d) 1 : 2
Show Answer & Explanation

Answer: (d) 1 : 2

Explanation:
1. CuCl contains \( \ce{Cu+} \): 1 F deposits 63.5 g.
2. \( \ce{CuSO4} \) contains \( \ce{Cu^{2+}} \): 1 F deposits 31.75 g.
3. Ratio \( \ce{CuSO4} \) : CuCl = 1 : 2.

Teacher's Note:
1. Mass deposited is inversely proportional to the charge on the ion for the same charge passed.

Question 12: The specific conductance k is \( 4.95\times10^{-5} \) S cm\(^{-1}\) for a 0.00099 M acetic acid solution. The degree of dissociation of acetic acid, if \( \Lambda_M^\infty \) is 400 S cm\(^2\) mol\(^{-1}\), is:
(a) 0.25
(b) 0.125
(c) 0.0125
(d) 1.25
Show Answer & Explanation

Answer: (b) 0.125

Explanation:
1. \( \Lambda_M=\frac{1000k}{C}=\frac{1000\times4.95\times10^{-5}}{0.00099}=50 \) S cm\(^2\) mol\(^{-1}\).
2. \( \alpha=\frac{\Lambda_M}{\Lambda_M^\infty}=\frac{50}{400}=0.125 \).

Teacher's Note:
1. Degree of dissociation = molar conductance / limiting molar conductance.

Question 13: The value of \( \Lambda_m^\infty \) for \( \ce{SrCl2} \) in water at 25 \(^\circ\)C from the following data: concentration 0.25 mol per L gives \( \Lambda_m=260 \) and concentration 1 mol per L gives \( \Lambda_m=250 \) (ohm\(^{-1}\) cm\(^2\) mol\(^{-1}\)) is:
(a) 260
(b) 270
(c) 250
(d) 255
Show Answer & Explanation

Answer: (b) 270

Explanation:
1. Kohlrausch's law for strong electrolytes: \( \Lambda_m=\Lambda_m^\infty-A\sqrt C \).
2. \( 260=\Lambda^\infty-0.5A \) and \( 250=\Lambda^\infty-A \): subtracting gives \( A=20 \).
3. \( \Lambda^\infty=260+10=270 \).

Teacher's Note:
1. Plot or solve two equations in the square root of concentration.

Question 14: The standard reduction potentials of the following reactions are: (i) \( \ce{MnO4- + 8H+ + 5e- -> Mn^{2+} + 4H2O} \), 1.51 V; (ii) \( \ce{Sn^{4+} + 2e- -> Sn^{2+}} \), 0.15 V; (iii) \( \ce{Cr2O7^{2-} + 14H+ + 6e- -> 2Cr^{3+} + 7H2O} \), 1.33 V; (iv) \( \ce{Ce^{4+} + e- -> Ce^{3+}} \), 1.61 V. The oxidising power of the various species decreases in the order:
(a) \( \ce{Ce^{4+}} > \ce{Cr2O7^{2-}} > \ce{Sn^{4+}} > \ce{MnO4-} \)
(b) \( \ce{Cr2O7^{2-}} > \ce{Sn^{4+}} > \ce{Ce^{4+}} > \ce{MnO4-} \)
(c) \( \ce{MnO4-} > \ce{Ce^{4+}} > \ce{Sn^{4+}} > \ce{Cr2O7^{2-}} \)
(d) \( \ce{Ce^{4+}} > \ce{MnO4-} > \ce{Cr2O7^{2-}} > \ce{Sn^{4+}} \)
Show Answer & Explanation

Answer: (d) \( \ce{Ce^{4+}} > \ce{MnO4-} > \ce{Cr2O7^{2-}} > \ce{Sn^{4+}} \)

Explanation:
1. A higher reduction potential means a stronger oxidising agent.
2. Order of E: 1.61 (Ce4+) > 1.51 (MnO4-) > 1.33 (Cr2O7 2-) > 0.15 (Sn4+).

Teacher's Note:
1. Rank by standard reduction potential.

Question 15: The number of faradays required to produce one mole of water from a hydrogen-oxygen fuel cell containing aqueous alkali as electrolyte is:
(a) 2
(b) 1
(c) 3
(d) 4
Show Answer & Explanation

Answer: (a) 2

Explanation:
1. \( \ce{2H2 + O2 -> 2H2O} \) involves 4 electrons for 2 mol of water.
2. So 1 mol of water needs 2 mol of electrons, that is 2 faraday.

Teacher's Note:
1. Water has oxygen going from 0 to -2: two electrons per mole of water.

Question 16: 4 mol of electrons were transferred from anode to cathode in an experiment on electrolysis of water. The total volume of the two dry gases (at STP) produced is approximately (in L):
(a) 22.4
(b) 44.8
(c) 67.2
(d) 89.4
Show Answer & Explanation

Answer: (c) 67.2

Explanation:
1. \( \ce{2H2O -> 2H2 + O2} \) requires 4 mol of electrons for 2 mol \( \ce{H2} \) and 1 mol \( \ce{O2} \).
2. Total gas = 3 mol = \( 3\times22.4=67.2 \) L.

Teacher's Note:
1. 4 mol of e- gives 2 mol H2 at the cathode and 1 mol O2 at the anode.

Question 17: The ionisation constant of a weak electrolyte is \( 25\times10^{-6} \) while the equivalent conductance of its 0.01 M solution is 19.6 S cm\(^2\) eq\(^{-1}\). The equivalent conductance of the electrolyte at infinite dilution (in S cm\(^2\) eq\(^{-1}\)) is approximately:
(a) 250
(b) 196
(c) 392
(d) 384
Show Answer & Explanation

Answer: (c) 392

Explanation:
1. For a weak electrolyte, \( \alpha=\sqrt{K/C}=\sqrt{\frac{25\times10^{-6}}{0.01}}=0.05 \).
2. \( \Lambda^\infty=\frac{\Lambda}{\alpha}=\frac{19.6}{0.05}=392 \).

Teacher's Note:
1. Use alpha = root of K over C for a weak electrolyte.

Question 18: Three moles of electrons are passed through three solutions in succession containing \( \ce{AgNO3} \), \( \ce{CuSO4} \) and \( \ce{AuCl3} \) respectively. The molar ratio of the amounts of cations reduced at the cathode will be:
(a) 6 : 3 : 2
(b) 1 : 2 : 3
(c) 2 : 1 : 3
(d) 3 : 2 : 1
Show Answer & Explanation

Answer: (a) 6 : 3 : 2

Explanation:
1. Ag+ needs 1 e-: 3 mol Ag. Cu2+ needs 2 e-: 1.5 mol Cu. Au3+ needs 3 e-: 1 mol Au.
2. Ratio = 3 : 1.5 : 1 = 6 : 3 : 2.

Teacher's Note:
1. Moles deposited = moles of electrons / charge on the ion.

Question 19: Given the limiting equivalent conductances of \( \ce{BaCl2} \) = \( x_1 \), \( \ce{H2SO4} \) = \( x_2 \) and HCl = \( x_3 \), and the specific conductance of a saturated \( \ce{BaSO4} \) solution = y S cm\(^{-1}\), the \( K_{sp} \) of \( \ce{BaSO4} \) is:
(a) \( \frac{10^3y}{2(x_1+x_2-2x_3)} \)
(b) \( \frac{10^6y^2}{(x_1+x_2-2x_3)^2} \)
(c) \( \frac{x_1+x_2-2x_3}{10^6y} \)
(d) \( \frac{10^6y^2}{4(x_1+x_2-x_3)^2} \)
Show Answer & Explanation

Answer: (d) \( \frac{10^6y^2}{4(x_1+x_2-x_3)^2} \)

Explanation:
1. \( \Lambda^\infty_{eq}(\ce{BaSO4})=x_1+x_2-x_3 \) (Kohlrausch's law with equivalent conductances).
2. Normality of the saturated solution \( N=\frac{1000y}{x_1+x_2-x_3} \); molar solubility \( S=\frac N2 \) because 1 mol gives 2 equivalents.
3. \( K_{sp}=S^2=\frac{10^6y^2}{4(x_1+x_2-x_3)^2} \).

Teacher's Note:
1. For a sparingly soluble salt, Lambda at the saturation concentration is about Lambda-infinity.

Question 20: Select the correct statement based on the half-cell reactions: \( \ce{Fe^{2+} -> Fe^{3+} + e-} \), \( E^\circ=-0.77 \) V; \( \ce{O2 + 4H+ + 4e- -> 2H2O} \), \( E^\circ=1.23 \) V; \( \ce{Fe -> Fe^{2+} + 2e-} \), \( E^\circ=0.44 \) V.
(a) Only: Fe\(^{2+}\) solution is not stable in air
(b) All of these: Fe\(^{2+}\) solution is not stable in air, Fe nails can prevent oxidation of Fe\(^{2+}\) to Fe\(^{3+}\) by air, and corrosion of iron takes place by air
(c) Only: Fe nails can prevent oxidation of Fe\(^{2+}\) to Fe\(^{3+}\) by air
(d) Only: corrosion of iron takes place by air
Show Answer & Explanation

Answer: (b) All of these: Fe\(^{2+}\) solution is not stable in air, Fe nails can prevent oxidation of Fe\(^{2+}\) to Fe\(^{3+}\) by air, and corrosion of iron takes place by air

Explanation:
1. \( \ce{O2} \) oxidising \( \ce{Fe^{2+}} \): \( E=1.23-0.77=+0.46 \) V, spontaneous, so Fe2+ is not stable in air.
2. Fe reducing \( \ce{Fe^{3+}} \): \( E=0.77+0.44=+1.21 \) V, spontaneous, so iron nails keep iron in the +2 state.
3. Fe oxidised by \( \ce{O2} \): \( E=1.23+0.44=+1.67 \) V, so iron corrodes in air. All three are correct.

Teacher's Note:
1. Check the sign of the cell EMF for each pairing.

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