JEE Chemistry Electrochemistry MCQs Set 16

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

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Multiple Choice Questions

Question 1: The standard cell potential of the Zn | Zn\( ^{2+} \) || Cu\( ^{2+} \) | Cu cell is 1.10 V. The standard Gibbs energy change of the cell reaction (the maximum electrical work obtained from this cell is its magnitude) is
(a) 106.15 kJ
(b) -212.30 kJ
(c) -318.45 kJ
(d) -424.60 kJ
Show Answer & Explanation

Answer: (b) -212.30 kJ

Explanation:
1. n = 2 electrons are transferred.
2. \( \Delta G^\circ=-nFE^\circ=-2\times96500\times1.10 \) = -212.3 kJ.

Teacher's Note:
1. \( \Delta G^\circ=-nFE^\circ \) and \( \log K=\frac{nE^\circ}{0.0591} \) at 298 K.
2. Take F = 96500 C/mol.

Question 2: The molar conductivities of NaCl, HCl and CH3COONa at infinite dilution are 126.45, 426.16 and 91 \( ohm^{-1}cm^2mol^{-1} \) respectively. The molar conductivity of CH3COOH at infinite dilution is
(a) 201.28
(b) 390.71
(c) 698.28
(d) 540.48
Show Answer & Explanation

Answer: (b) 390.71

Explanation:
1. Kohlrausch's law: \( \Lambda^\infty(CH_3COOH)=\Lambda^\infty(CH_3COONa)+\Lambda^\infty(HCl)-\Lambda^\infty(NaCl) \).
2. = 91 + 426.16 - 126.45 = 390.71.

Teacher's Note:
1. Conductivity: \( \kappa=G\times\frac{l}{a} \) and the cell constant is \( l/a \) in cm\(^{-1}\).
2. Kohlrausch's law: add and subtract the limiting molar conductivities of the ions.

Question 3: The electrodes of a conductivity cell are 3 cm apart and have a cross-sectional area of 4 cm\( ^2 \). The cell constant (in cm\( ^{-1} \)) is
(a) 4 x 3
(b) 4/3
(c) 3/4
(d) 9/4
Show Answer & Explanation

Answer: (c) 3/4

Explanation:
1. Cell constant = l/a = 3/4 cm\( ^{-1} \).

Teacher's Note:
1. Conductivity: \( \kappa=G\times\frac{l}{a} \) and the cell constant is \( l/a \) in cm\(^{-1}\).
2. Kohlrausch's law: add and subtract the limiting molar conductivities of the ions.

Question 4: The specific conductance of a solution is 0.2 \( ohm^{-1}cm^{-1} \) and its conductance is 0.04 \( ohm^{-1} \). The cell constant is
(a) 1 cm\( ^{-1} \)
(b) 0 cm\( ^{-1} \)
(c) 5 cm\( ^{-1} \)
(d) 0.2 cm\( ^{-1} \)
Show Answer & Explanation

Answer: (c) 5 cm\( ^{-1} \)

Explanation:
1. Cell constant = specific conductance / conductance = 0.2/0.04 = 5 cm\( ^{-1} \).

Teacher's Note:
1. Conductivity: \( \kappa=G\times\frac{l}{a} \) and the cell constant is \( l/a \) in cm\(^{-1}\).
2. Kohlrausch's law: add and subtract the limiting molar conductivities of the ions.

Question 5: In balancing the half reaction \( S_2O_3^{2-}\rightarrow S(s) \), the number of electrons that must be added is
(a) 4 on the left
(b) 3 on the right
(c) 2 on the left
(d) 2 on the right
Show Answer & Explanation

Answer: (a) 4 on the left

Explanation:
1. Sulphur in \( S_2O_3^{2-} \) has oxidation state +2 and becomes 0 in S(s).
2. For 2 S atoms: \( S_2O_3^{2-}+6H^++4e^-\rightarrow2S+3H_2O \), so 4 electrons are added on the left.

Teacher's Note:
1. Cell potential: \( E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode} \) using reduction potentials.
2. A positive cell potential means the cell reaction is spontaneous.

Question 6: The anode half-reaction occurring during the discharge of a lead storage battery is
(a) \( Pb(s)+SO_2+O_2\rightarrow PbSO_4(s) \)
(b) \( Pb(s)+SO_4^{2-}(aq)\rightarrow PbSO_4(s)+2e^- \)
(c) \( PbO_2+4H^++2e^-+SO_4^{2-}\rightarrow PbSO_4+2H_2O \)
(d) \( Pb^{2+}(aq)+SO_4^{2-}(aq)\rightarrow PbSO_4(s) \)
Show Answer & Explanation

Answer: (b) \( Pb(s)+SO_4^{2-}(aq)\rightarrow PbSO_4(s)+2e^- \)

Explanation:
1. Oxidation occurs at the anode: Pb is oxidised to Pb(II) and precipitates as PbSO4.
2. Option (c) is the cathode reaction.

Teacher's Note:
1. Cell potential: \( E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode} \) using reduction potentials.
2. A positive cell potential means the cell reaction is spontaneous.

Question 7: If the Zn\( ^{2+} \)/Zn electrode is diluted 100 times, the change in emf is
(a) increase of 59 mV
(b) decrease of 59 mV
(c) increase of 29.5 mV
(d) decrease of 29.5 mV
Show Answer & Explanation

Answer: (b) decrease of 59 mV

Explanation:
1. For the electrode reduction: \( E=E^\circ+\frac{0.0591}{2}\log[Zn^{2+}] \).
2. A 100-fold dilution changes log[Zn\( ^{2+} \)] by -2, so \( \Delta E=\frac{0.0591}{2}\times(-2)=-0.059 \) V: a decrease of 59 mV.

Teacher's Note:
1. Nernst equation at 298 K: \( E=E^\circ-\frac{0.0591}{n}\log Q \).
2. Dilute the ions on the reactant side and the emf falls; dilute the product side and it rises.

Question 8: A hydrogen electrode dipped in solution of pH = 3 and another of pH = 6 are connected by a salt bridge. The emf of the resulting cell is
(a) 0.177 V
(b) 0.3 V
(c) 0.052 V
(d) 0.104 V
Show Answer & Explanation

Answer: (a) 0.177 V

Explanation:
1. Each hydrogen electrode has \( E=-0.0591\,pH \).
2. emf = 0.0591 x (6 - 3) = 0.177 V.

Teacher's Note:
1. Nernst equation at 298 K: \( E=E^\circ-\frac{0.0591}{n}\log Q \).
2. Dilute the ions on the reactant side and the emf falls; dilute the product side and it rises.

Question 9: The potential of a hydrogen electrode dipped in a solution of pH = 3 at 25 °C (2.303RT/F = 0.059 V) is
(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. \( E=-0.059\,pH=-0.059\times3 \) = -0.177 V.

Teacher's Note:
1. Nernst equation at 298 K: \( E=E^\circ-\frac{0.0591}{n}\log Q \).
2. Dilute the ions on the reactant side and the emf falls; dilute the product side and it rises.

Question 10: The standard electrode potentials of Zn\( ^{2+} \)/Zn and Ag\( ^+ \)/Ag are -0.763 V and +0.799 V respectively. The standard potential of the cell is
(a) 1.56 V
(b) 0.036 V
(c) -1.562 V
(d) 0.799 V
Show Answer & Explanation

Answer: (a) 1.56 V

Explanation:
1. E(cell) = 0.799 - (-0.763) = 1.562 V, i.e. about 1.56 V.

Teacher's Note:
1. Cell potential: \( E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode} \) using reduction potentials.
2. A positive cell potential means the cell reaction is spontaneous.

Question 11: The standard reduction potentials at 298 K are Zn\( ^{2+} \)/Zn = -0.762 V, Cr\( ^{3+} \)/Cr = -0.740 V, \( H^+/H_2 \) = 0.00 V and Fe\( ^{3+} \)/Fe\( ^{2+} \) = 0.770 V. The strongest reducing agent is
(a) Zn(s)
(b) Cr(s)
(c) \( H_2(g) \)
(d) Fe\( ^{2+} \)(aq)
Show Answer & Explanation

Answer: (a) Zn(s)

Explanation:
1. The most negative reduction potential belongs to the best reducing agent.
2. Zn (-0.762 V) is the lowest, so Zn(s) is the strongest reducing agent.

Teacher's Note:
1. The lower (more negative) the reduction potential, the stronger the reducing agent.
2. A metal displaces another from solution only if its reduction potential is lower.

Question 12: A solution contains one mole per litre each of Cu(NO3)2, AgNO3, Hg2(NO3)2 and Mg(NO3)2 and is electrolysed with inert electrodes. The standard reduction potentials are Ag/Ag\( ^+ \) = +0.80 V, 2Hg/Hg\( _2^{2+} \) = +0.79 V, Cu/Cu\( ^{2+} \) = +0.34 V, Mg/Mg\( ^{2+} \) = -2.37 V. With increasing voltage the sequence of deposition of metals on the cathode will be
(a) Ag, Hg, Cu, Mg
(b) Mg, Cu, Hg, Ag
(c) Ag, Hg, Cu
(d) Cu, Hg, Ag
Show Answer & Explanation

Answer: (c) Ag, Hg, Cu

Explanation:
1. The ion with the highest reduction potential is reduced first: Ag, then Hg, then Cu.
2. Mg\( ^{2+} \) is not reduced from aqueous solution (water is reduced first), so Mg is not deposited.

Teacher's Note:
1. The lower (more negative) the reduction potential, the stronger the reducing agent.
2. A metal displaces another from solution only if its reduction potential is lower.

Question 13: The standard oxidation potentials of zinc and silver in water at 298 K are Zn to Zn\( ^{2+} \) + 2e: 0.76 V and Ag to Ag\( ^+ \) + e: -0.80 V. Which of the following reactions actually takes place?
(a) \( \ce{Zn(s)+2Ag^+(aq)->Zn^{2+}(aq)+2Ag(s)} \)
(b) \( \ce{Zn^{2+}(aq)+2Ag(s)->2Ag^+(aq)+Zn(s)} \)
(c) \( \ce{Zn(s)+Ag(s)->Zn^{2+}(aq)+Ag^+(aq)} \)
(d) \( \ce{Zn^{2+}(aq)+Ag^+(aq)->Zn(s)+Ag(s)} \)
Show Answer & Explanation

Answer: (a) \( \ce{Zn(s)+2Ag^+(aq)->Zn^{2+}(aq)+2Ag(s)} \)

Explanation:
1. For Zn + 2Ag\( ^+ \) the cell emf = 0.76 + 0.80 = +1.56 V, which is positive.
2. So zinc displaces silver from solution; the reverse reaction (b) is non-spontaneous.

Teacher's Note:
1. Cell potential: \( E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode} \) using reduction potentials.
2. A positive cell potential means the cell reaction is spontaneous.

Question 14: The standard reduction potentials for Fe\( ^{2+} \)/Fe and Sn\( ^{2+} \)/Sn are -0.44 V and -0.14 V respectively. For the reaction \( \ce{Fe^{2+}+Sn->Fe+Sn^{2+}} \), the standard emf is
(a) +0.30 V
(b) -0.58 V
(c) +0.58 V
(d) -0.30 V
Show Answer & Explanation

Answer: (d) -0.30 V

Explanation:
1. Cathode: Fe\( ^{2+} \)/Fe (-0.44 V); anode: Sn/Sn\( ^{2+} \) (-0.14 V).
2. E = -0.44 - (-0.14) = -0.30 V, so the reaction is not spontaneous.

Teacher's Note:
1. Cell potential: \( E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode} \) using reduction potentials.
2. A positive cell potential means the cell reaction is spontaneous.

Question 15: Powdered lead and iron are added to a solution that is 1.0 M in both Pb\( ^{2+} \) and Fe\( ^{2+} \) ions. Given E(Fe\( ^{2+} \)/Fe) = -0.44 V and E(Pb\( ^{2+} \)/Pb) = -0.13 V, the reaction that occurs results in
(a) more iron and Pb\( ^{2+} \) ions being formed
(b) more lead and Fe\( ^{2+} \) ions being formed
(c) an increase in concentration of both Pb\( ^{2+} \) and Fe\( ^{2+} \) ions
(d) no net change
Show Answer & Explanation

Answer: (b) more lead and Fe\( ^{2+} \) ions being formed

Explanation:
1. Fe is the stronger reducing agent: \( \ce{Fe+Pb^{2+}->Fe^{2+}+Pb} \), E = -0.13 + 0.44 = +0.31 V.
2. So lead metal is deposited and Fe\( ^{2+} \) ions are formed.

Teacher's Note:
1. The lower (more negative) the reduction potential, the stronger the reducing agent.
2. A metal displaces another from solution only if its reduction potential is lower.

Question 16: Given Fe\( ^{2+} \) + 2e to Fe, E = -0.440 V and Fe\( ^{3+} \) + 3e to Fe, E = -0.036 V, the standard electrode potential for Fe\( ^{3+} \) + e to Fe\( ^{2+} \) is
(a) -0.476 V
(b) -0.404 V
(c) +0.404 V
(d) +0.772 V
Show Answer & Explanation

Answer: (d) +0.772 V

Explanation:
1. \( \Delta G^\circ \) is additive, not E: (Fe\( ^{3+} \) to Fe) - (Fe\( ^{2+} \) to Fe).
2. 1 x E = 3(-0.036) - 2(-0.440) = -0.108 + 0.880 = +0.772 V.

Teacher's Note:
1. \( \Delta G^\circ=-nFE^\circ \) and \( \log K=\frac{nE^\circ}{0.0591} \) at 298 K.
2. Take F = 96500 C/mol.

Question 17: The reduction potentials (in volts) of four elements P, Q, R, S are -2.90, +0.34, +1.20 and -0.76 respectively. The order of decreasing reactivity (as metals, i.e. reducing power) is
(a) P > Q > R > S
(b) Q > P > R > S
(c) R > Q > S > P
(d) P > S > Q > R
Show Answer & Explanation

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

Explanation:
1. The more negative the reduction potential, the stronger the reducing agent and the more reactive the metal.
2. Order: P (-2.90) > S (-0.76) > Q (+0.34) > R (+1.20).

Teacher's Note:
1. The lower (more negative) the reduction potential, the stronger the reducing agent.
2. A metal displaces another from solution only if its reduction potential is lower.

Question 18: The standard electrode potentials of Mg\( ^{2+} \)/Mg and Cu\( ^{2+} \)/Cu are -2.37 V and +0.33 V. The emf of a cell made of these two half cells is
(a) -2.03 V
(b) 1.36 V
(c) 2.70 V
(d) 2.03 V
Show Answer & Explanation

Answer: (c) 2.70 V

Explanation:
1. E = 0.33 - (-2.37) = 2.70 V.

Teacher's Note:
1. Cell potential: \( E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode} \) using reduction potentials.
2. A positive cell potential means the cell reaction is spontaneous.

Question 19: A cell constructed by coupling a standard copper electrode and a standard magnesium electrode has an emf of 2.7 V. If the standard reduction potential of copper is +0.34 V, that of magnesium electrode is
(a) +3.04 V
(b) -3.04 V
(c) +2.36 V
(d) -2.36 V
Show Answer & Explanation

Answer: (d) -2.36 V

Explanation:
1. 2.7 = 0.34 - E(Mg), so E(Mg) = 0.34 - 2.7 = -2.36 V.

Teacher's Note:
1. Cell potential: \( E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode} \) using reduction potentials.
2. A positive cell potential means the cell reaction is spontaneous.

Question 20: The emf of the cell in which the reaction \( \ce{Zn(s)+Ni^{2+}(a=1.0)<=>Zn^{2+}(a=10)+Ni(s)} \) occurs is found to be 0.5105 V at 298 K. The standard emf of the cell is
(a) 0.5400 V
(b) 0.4810 V
(c) 0.5696 V
(d) -0.5105 V
Show Answer & Explanation

Answer: (a) 0.5400 V

Explanation:
1. Q = 10/1 = 10, n = 2.
2. E = E° - (0.0591/2) log 10, so E° = 0.5105 + 0.0295 = 0.5400 V.

Teacher's Note:
1. Nernst equation at 298 K: \( E=E^\circ-\frac{0.0591}{n}\log Q \).
2. Dilute the ions on the reactant side and the emf falls; dilute the product side and it rises.

Question 21: For the redox reaction \( \ce{Zn(s)+Cu^{2+}(0.1 M)->Zn^{2+}(1 M)+Cu(s)} \) taking place in a cell, E° of the cell is 1.10 V. E(cell) will be (2.303RT/F = 0.0591)
(a) 2.14 V
(b) 1.80 V
(c) 1.07 V
(d) 0.82 V
Show Answer & Explanation

Answer: (c) 1.07 V

Explanation:
1. Q = 1/0.1 = 10.
2. E = 1.10 - (0.0591/2) log 10 = 1.10 - 0.0295 = 1.07 V.

Teacher's Note:
1. Nernst equation at 298 K: \( E=E^\circ-\frac{0.0591}{n}\log Q \).
2. Dilute the ions on the reactant side and the emf falls; dilute the product side and it rises.

Question 22: The emf of a Daniell cell at 298 K is E1: Zn | ZnSO4 (0.01 M) || CuSO4 (1.0 M) | Cu. When the concentration of ZnSO4 is 1.0 M and that of CuSO4 is 0.01 M, the emf changes to E2. The relationship between E1 and E2 is
(a) E2 = 0 and not equal to E1
(b) E1 > E2
(c) E1 < E2
(d) E1 = E2
Show Answer & Explanation

Answer: (b) E1 > E2

Explanation:
1. E1: Q = 0.01/1 = 0.01, so E = E° + 0.0591.
2. E2: Q = 1/0.01 = 100, so E = E° - 0.0591.
3. Hence E1 > E2.

Teacher's Note:
1. Nernst equation at 298 K: \( E=E^\circ-\frac{0.0591}{n}\log Q \).
2. Dilute the ions on the reactant side and the emf falls; dilute the product side and it rises.

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Electrochemistry Objective Questions & Solutions for JEE Chemistry

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