Electrochemistry – JEE Advanced Previous Year Questions with Solutions

JEE Advanced Electrochemistry Questions with Solutions are available here. Going through the questions and solutions and at the same time practising solving the question from this chapter will help JEE aspirants in realizing the question types, patterns and difficulty levels.

We have compiled some of the most important questions on Electrochemistry from the past JEE Advanced exams. These questions and solutions will further help JEE aspirants to develop better problem-solving skills as well as come up with the most accurate solutions. Additionally, the JEE Advanced Electrochemistry question with solutions given here is available in the form of a PDF which can be downloaded instantly from below.

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JEE Advanced Previous Year Questions on Electrochemistry

Question 1. Molar conductivity (∧_m) of an aqueous solution of sodium stearate, which behaves as a strong electrolyte, is recorded at varying concentrations (c) of sodium stearate. Which one of the following plots provides the correct representation of micelle formation in the solution? (critical micelle concentration (CMC) is marked with an arrow in the figures).

Solution: (B)

Sodium stearate → CH₃(CH₂)16COO^–Na⁺

At normal or low concentration, it behaves as a strong electrolyte and for strong electrolytes, molar conductance (∧_m) decreases with an increase in concentration. But for the above particular concentration, sodium stearate forms aggregates and that concentration is called CMC.

However, beyond CMC, the ions get associated and form micelles. So the number of charge carriers decreases even further. So further beyond CMC, the molar conductivity decreases more abruptly. Since the number of ions decreases, hence ∧_m decreases.

Question 2. The conductance of a 0.0015M aqueous solution of a weak monobasic acid was determined by using a conductivity cell consisting of platinized Pt electrodes. The distance between the electrodes is 120 cm with an area of cross-section of 1cm². The conductance of this solution was found to be 5 × 10^-7S. The pH of the solution is 4. The value of limiting molar conductivity (Λom) of this weak monobasic acid in an aqueous solution is Z × 10² S cm^-1 mol^-1. The value of Z is.

Solution: (6)

Let’s assume a weak monobasic acid HA. It dissociates as per the equation:

HA(aq) ⇌ H⁺ (aq) + A^– (aq)

From this equation:

And as pH is given to be equal to 4

∴ [H⁺] = 10^-4 M = = c × α

From the above α can be calculated as:

α = [H⁺] / [HA] = 10^-4 / 0.0015

α = 1/15

Also, the conductivity k = G × ℓ / A

Putting the values we get,

K = 5 × 10^-7 × 120 / 1

= 6 × 10^-5 S cm^-1

And,

λ_m = k × 1000 / M

Putting Values:

= 5 × 10^-5 × 10³ / 15 × 10^-4

= 40 S cm^-1 mol^-1

Now from the expression;

α = λ_m / λ_m^o

Putting the values:

1 / 15 = 40 / Z × 10²

Therefore, Z = 40 × 15 / 100 = 6

Question 3. Consider an electrochemical cell: A(s) | An+ (aq, 2 M) || B2n+ (aq, 1 M) | B(s). The value of ΔHθ for the cell reaction is twice that of ΔGθ at 300 K. If the emf of the cell is zero, the ΔSθ (in J K−1 mol−1) of the cell reaction per mole of B formed at 300 K is ____. (Given: ln(2) = 0.7, R (universal gas constant) = 8.3 J K−1 mol−1. H, S and G are enthalpies, entropy and Gibbs energy, respectively.)

Solution: (-11.62)

A(s) | An⁺ (aq, 2 M) || B²ⁿ⁺ (aq, 1 M) | B(s)

2A(s) → Anⁿ⁺ + 2ne^–

B²ⁿ⁺ + 2ne^– → B(s)

2A(s)⁺ + B²ⁿ⁺→ Anⁿ⁺ + B(s)

Q = [Aⁿ⁺]² / B²ⁿ⁺ = 2 × 2/ 1 = 4

ΔG^o = ΔH^o − TΔS^o

ΔH^o = 2ΔG^o

E^o_cell = RT / 2nF ln 4

ΔG^o = -2n × F × RT / 2nF ln 4

ΔG^o = -RT ln 4

ΔS^o = ΔH^o – ΔG^o / T = ΔG^o / T = RT ln 4 / T

ΔS^o = – 8314 × 1.4

ΔS^o = -11.62 J mol ^-1 K^-1

Question 4. For the electrochemical cell, Mg(s)|Mg2+(aq, 1M)||Cu2+ (aq, 1M) | Cu(s) the standard emf of the cell is 2.70 V at 300 K. When the concentration of Mg2+ is changed to x M, the cell potential changes to 2.67 V at 300 K. The value of x is____.

(given, F/R = 11500 KV–1, where F is the Faraday constant and R is the gas constant, ln(10) = 2.30).

Solution: (10)

Mg(s) + Cu²⁺(aq) → Mg²⁺(aq) + Cu(s)

E^o_cell = 2.70; E_cell = 2.67; Mg²⁺ = x M; Cu²⁺ = 1 M

E_cell = E^o_cell − RT / nF lnx

2.67 = 2.70 − RT / 2Flnx

−0.03=−2FR×300×lnx

lnx = 0.03 × 2 / 300 × F / R

lnx = 0.03 × 2 × 11500 / 300×1

lnx = 2.30 = ln(10)

∴ x = 10

Question 5. The molar conductivity of a solution of a weak acid HX(0.01 M) is 10 times smaller than the molar conductivity of a solution of a weak acid HY(0.1 M). If λ^o_x ≈ λ^o_y, the difference in their pK_a values, pK_a(HX) − pK_a(HY), is (consider the degree of ionization of both acids to be <<1).

Solution: (3)

Given ∧_m(HX) / ∧_m(HY) = 1 / 10

α₁ / α₂ = ∧_m(HX) / ∧^o_m (HX) / ∧_m(HY) / ∧^o_m (HY) = 1 / 10

K_a(HX) / K_a(HY) = C₁α²₁ / C₂α²₂

= 0.01 / 0.1 × (1 / 10)² =1 / 1000

log K_a(HX) − logK_a(HY) = −3

∴ pK_a(HX) − pK_a(HY) = 3

Question 6. All the energy released from the reaction X → Y, ∆_r G^o = -193J mol^-1 is used for oxidizing M⁺ as M⁺ → M3⁺ + 2e^–, E^o = -0.25V. Under standard conditions, the number of moles of M⁺ oxidized when one mole of X is converted to Y is [F = 96500 C mol^-1].

Solution: (4)

M⁺ → M3⁺ + 2e^–

Given X → Y, ∆_r G^o = -193J mol^-1

△G^o for above reaction is:

△G^o = −nF E^o

= −2 × 96500 × (−0.25)

= 48250 J/mol

= 48.25k J/mol

∴ The number of moles of M⁺ oxidised from (X→Y).

= 193 / 48.25 = 4

Question 7. In a galvanic cell, the salt bridge:

A. Does not participate chemically in the cell reaction

B. Stops the diffusion of ions from one electrode to another

C. Is necessary for the occurrence of the cell reaction

D. Ensures mixing of the two electrolytic solutions

Solution: (A, B)

A salt bridge, in electrochemistry, is a laboratory device used to connect the oxidation and reduction half cells of a galvanic cell(voltaic cell), a type of electrochemical cell. It maintains electrical neutrality within the internal circuit, preventing the cell from rapidly running its reaction to equilibrium. If no salt bridge were present, the solution in the one-half cell would accumulate a negative charge and the solution in the other half cell would accumulate a positive charge as the reaction proceeded, quickly preventing further reaction, and hence the production of electricity.

Salt bridge keeps the solutions in two half-cells electrically neutral. It prevents transference or diffusion of the ions from one half-cell to the other.

Question 8. Consider a 70% efficient hydrogen-oxygen fuel cell working under standard conditions at 1 bar and 298 K. Its cell reaction is;

H₂(g) + ½ O₂(g) → H₂O(l)

The work derived from the cell on the consumption of 1.0 × 10^-3 mol of H₂(g) is used to compress 1.00 mol of a monoatomic ideal gas in a thermally insulated container. What is the change in the temperature (in K) of the ideal gas?

The standard reduction potentials for the two half-cells are given below.

O₂(g) + 4H+(aq) + 4e^– → 2 H₂O (l), E^o = 1.23 V

2H+(aq) + 2^– → H₂(g), E^o = 0.00 V

Use F = 96500 C mol^-1, R = 8.314 J mol^-1 K^-1

Solution: (13.32)

For given reaction: H₂(g) + ½ O₂(g) → H₂O(l)

E^o = 1.23 V

∆G^o = –nF E^o_cell {-2 x 96500 x 1.23} J

Since, Work derived from this fuel cell = 70 / 100 x (-∆G^o_cell) x 10^-3 = x J

Since insulated vessel, hence q = 0

From equation, for monoatomic gas,

w = ∆U ⇒ x = nC_v, _m∆T; {C_V,m = 3R / 2}

Or

70 / 100 x (2 × 96500 × 1.23) × 10^-3 = 1 × 3 / 2 × 8.314 × ∆T

∴ ∆T = 13.32

Question 9. For the following cell, Zn(s) | ZnSO₄(aq) || CuSO₄(aq) | Cu(s) when the concentration of Zn²⁺ is 10 times the concentration of Cu²⁺, the expression for ΔG (in J mol–1) is [F is Faraday constant; R is gas constant; T is temperature; E° (cell) = 1.1V].

A. 1.1F

B. 2.303RT – 2.2F

C. 2.303RT + 1.1F

D. – 2.2F

Solution: (B)

Zn + Cu²⁺(aq) → Zn²⁺(aq) + Cu

We know that,

E = E^o − RT / nF ln [Zn²⁺] / [Cu²⁺]

∴ E = 1.1 − RT / 2F ln10

Now, ΔG = −nFE

Here n = 2

∴ △G = −2F[1.1 − RT / 2F ln10]

= −2.2F + RT log₁₀ 10 × 2.303

∴ △G = 2.303RT − 2.2F

Question 10. AgNO₃(aq.) was added to an aqueous KCl solution gradually and the conductivity of the solution was measured. The plot of conductance (Λ) versus the volume of AgNO₃ is.

A. (P)

B. (Q)

C. (R)

D. (S)

Solution: (D)

Initially, to the solution of KCl, solution of AgNO₃ added, AgCl(s) is formed.

Hence, the conductivity of the solution is almost compensated (or slightly increased) by the formation of KNO₃. After the endpoint, conductivity increases more rapidly because of the addition of excess AgNO₃ solution.

The plot of conductance (Λ) versus the volume of AgNO₃ is as given by (S). This is as explained below.

Initial conductance (Λ) of the solution was due to K(aq.)+ and Cl(aq.)−.

On addition of AgNO₃, the reaction occurs as,

AgNO₃(aq.) + KCl(aq.) → AgCl↓ + KNO₃

AgNO₃ acts as a limiting reagent up to complete precipitation. The conductance up to precipitation is a straight line parallel to the x-axis due to the replacement of Cl^– ions with NO₃^– ions (ionic mobility of NO₃^– and Cl^– are almost the same). After complete precipitation, with further addition of AgNO₃, an increase in the conductance is observed due to added Ag⁺, NO₃^– ions.

Hence, option D is correct.

Electrochemistry – Important Questions