NEET Electrochemistry Previous Year Questions with Solutions

Because standard reduction potential is an intensive property, potentials cannot be directly added or subtracted. We must convert each half-reaction to its extensive standard Gibbs free energy change: ΔG° = -nFE°

• • Reaction 1: Fe³⁺ + 3e⁻ → Fe, n₁ = 3, E₁° = -0.04 V.
• • Reaction 2: Fe²⁺ + 2e⁻ → Fe, n₂ = 2, E₂° = -0.44 V.
• • Target Reaction 3: Fe³⁺ + e⁻ → Fe²⁺, n₃ = 1, E₃° = ?

Subtracting Reaction 2 from Reaction 1 yields Target Reaction 3: [Fe³⁺ + 3e⁻ → Fe] - [Fe²⁺ + 2e⁻ → Fe] = Fe³⁺ + e⁻ → Fe²⁺ Therefore: ΔG₃° = ΔG₁° - ΔG₂°

• • Faraday's constant F cancels from all terms.

Substitute n₁ = 3, E₁° = -0.04 V, n₂ = 2, E₂° = -0.44 V, and n₃ = 1 into the equation.

• • E°(Fe³⁺/Fe²⁺) = +0.76 V.
• • Matches Option (2).

For strong electrolytes at low concentrations, molar conductivity varies linearly with the square root of concentration according to the Debye-Hückel-Onsager equation: Λ_m = Λ_m° - A√c

• • Plot of Λ_m vs √c is a straight line with slope = -A and y-intercept = Λ_m°.
• • Given slope = -90.0 S cm² mol⁻³/² L¹/² ⇒ A = 90.0 S cm² mol⁻³/² L¹/².

Substitute c = 0.01 M and Λ_m = 145.0 S cm² mol⁻¹ into the Onsager equation.

• • Λ_m° = 145.0 + 9.0 = 154.0 S cm² mol⁻¹.

According to Kohlrausch's Law, the limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation: Λ_m°(XY) = λ°(X⁺) + λ°(Y⁻)

• • Limiting molar conductivity of Y⁻ = 80.0 S cm² mol⁻¹.
• • Matches Option (1).

For a monobasic weak acid HA dissociating into H⁺ and A⁻, the limiting molar conductivity at infinite dilution is the sum of the limiting molar conductivities of its constituent ions:

• • Λ_m° represents the theoretical conductivity at 100% complete dissociation.

The degree of dissociation α of a weak electrolyte at concentration C is the ratio of its molar conductivity Λ_m to its limiting molar conductivity Λ_m°:

• • Expressed as a percentage, α = 22.5%.

The calculated degree of dissociation is 0.225:

• • Matches Option (3) exactly.

One Faraday (1 F ≈ 96487 C) is the electrical charge carried by 1 mole of electrons. The number of Faradays required to oxidize or reduce a substance is given by:

In water, oxygen has oxidation state -2. In O2, oxygen has oxidation state 0. Writing the half-reaction for 1 mole of H2O:

In permanganate ion MnO4⁻, manganese is in the +7 oxidation state. In acidic reduction to Mn²⁺:

Calcium in molten CaCl2 is Ca²⁺. Deposition of 1 mole of Ca requires 2 moles of electrons:

In FeO, iron is in the +2 oxidation state. In Fe2O3, iron is in the +3 oxidation state. For 1 mole of FeO:

In the electrolysis of copper sulphate (CuSO4) solution, cupric ions (Cu²⁺) are reduced at the cathode by accepting two electrons per atom:

Using Q = I × t where current I = 9.6487 A and time t = 100 s:

According to Faraday’s First Law of Electrolysis, the mass m of substance deposited is proportional to charge Q:

Substituting M = 63 g/mol, z = 2, F = 96487 C, and Q = 964.87 C:

The maximum non-expansion electrical work (w_elec) obtainable from a galvanic cell operating reversibly at constant temperature and pressure is equal to the decrease in Gibbs free energy of the cell reaction:

Consider the Daniell cell reaction. For the single-electron representation vs two-electron representation, the electrical potential E_cell remains identical (+1.10 V), but the moles of electrons transferred n changes, directly doubling Δ_r G:

An intensive property is one whose value is independent of the size or quantity of matter in the system (e.g., electrical potential E_cell, temperature, pressure). An extensive property is one whose value depends on the mass or quantity of matter (e.g., Gibbs energy G, enthalpy H, volume V). Therefore, E_cell is intensive, and Δ_r G is extensive.

Because Δ_r G is extensive while E_cell is intensive, Δ_r G must scale with the stoichiometric extent of reaction, which is mathematically embodied by the stoichiometric factor n. Thus, Reason R correctly explains Assertion A.

The cell constant G* represents the geometric ratio of electrode separation (l) to electrode surface area (A):

Conductivity κ (specific conductance) is defined as the reciprocal of resistivity ρ:

Substitute the given conductivity κ = 0.0210 Ω⁻¹ cm⁻¹ and resistance R = 60 Ω:

Kohlrausch's law states that the limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte: $$\Lambda_m^\circ = \nu_+ \lambda_+^\circ + \nu_- \lambda_-^\circ$$

To calculate $\Lambda_m^\circ(CH_3COOH)$, we combine the limiting molar conductivities of the three strong electrolytes ($CH_3COONa$, $HCl$, and $NaCl$): $$\Lambda_m^\circ(CH_3COOH) = \lambda^\circ(CH_3COO^-) + \lambda^\circ(H^+)$$ Expressing in terms of the given salts:

Substitute the given numerical values: - $\Lambda_m^\circ(CH_3COONa) = 91.0\text{ S cm}^2\text{ mol}^{-1}$ - $\Lambda_m^\circ(HCl) = 426.16\text{ S cm}^2\text{ mol}^{-1}$ - $\Lambda_m^\circ(NaCl) = 126.45\text{ S cm}^2\text{ mol}^{-1}$

According to Kohlrausch's law of independent migration of ions, the limiting molar conductivity of acetic acid is the sum of the individual ionic limiting molar conductivities:

The degree of dissociation $\alpha$ of a weak electrolyte at concentration $c$ is the ratio of its molar conductivity $\Lambda_m$ to its limiting molar conductivity $\Lambda_m^\circ$:

Applying Ostwald's dilution law for weak monobasic acids: $$K_a = \frac{c \alpha^2}{1 - \alpha}$$ Because $\alpha = 0.05$, $1 - \alpha = 0.95 \approx 1$. Using the standard approximation:

Dilute sulphuric acid dissociates into hydrogen ions and sulphate ions, while water also self-ionizes: $$\text{H}_2\text{SO}_4 \text{(aq)} \to 2\text{H}^+ + \text{SO}_4^{2-}$$ $$\text{H}_2\text{O (l)} \rightleftharpoons \text{H}^+ + \text{OH}^-$$

Hydrogen ions are the only reducible cations present. They migrate to the inert platinum cathode and undergo reduction: $$2\text{H}^+ \text{(aq)} + 2e^- \to \text{H}_2\text{ (g)}$$ Hydrogen gas is evolved at the cathode.

At the inert platinum anode, two competing oxidation half-reactions are possible: 1. Oxidation of water: $$2\text{H}_2\text{O (l)} \to \text{O}_2\text{ (g)} + 4\text{H}^+ + 4e^- \quad (E^\circ_{\text{cell}} = +1.23\text{ V})$$ 2. Oxidation of sulphate ions: $$2\text{SO}_4^{2-} \text{(aq)} \to \text{S}_2\text{O}_8^{2-} \text{(aq)} + 2e^- \quad (E^\circ_{\text{cell}} = +1.96\text{ V})$$

The reaction with the lower standard potential ($$E^\circ = +1.23\text{ V}$$) is thermodynamically favored. In dilute sulphuric acid, oxidation of water takes preference over sulphate oxidation, so oxygen gas ($$\text{O}_2$$) is evolved at the anode.

In molten calcium chloride ($$\text{CaCl}_2$$), calcium exists as divalent cations ($$\text{Ca}^{2+}$$). At the cathode, each calcium ion requires two electrons to deposit as metallic calcium: $$\text{Ca}^{2+} + 2e^- \to \text{Ca (s)}$$

By Faraday's second law of electrolysis, deposition of 1 mole of an element requires $$n$$ moles of electrons, where $$n$$ is the valency of the ion. Since 1 mole of electrons carries a charge of 1 Faraday ($$1\text{ F} = 96500\text{ C}$$): $$1\text{ mol of Ca (40 g)} \iff 2\text{ moles of electrons} = 2\text{ F}$$

$$\text{Moles of Ca} = \frac{\text{Given Mass}}{\text{Molar Mass}} = \frac{20\text{ g}}{40\text{ g mol}^{-1}} = 0.5\text{ mol}$$

$$\text{Number of Faradays} = \text{Moles} \times n = 0.5 \times 2 = 1\text{ F}$$ Thus, exactly 1 Faraday of charge is required.