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Class 12 Chemistry — questions with answers and worked explanations.
These questions are drawn from the Pariksha Sutra question bank for Electrochemistry, part of the Class 12 Chemistry syllabus. Each one shows the correct answer and, where a method helps, the working behind it.
Read the question, decide your answer before looking, then check the explanation — that is what turns practice into marks. If a question catches you out, the explanation is the part worth re-reading.
Oxidation takes place at the electrode that loses electrons, which is the anode in a galvanic cell.
Electrons flow from the anode through the external circuit to the cathode, making the anode negative.
Therefore the electrode where oxidation occurs is the anode, and it is the negative terminal. Anode, and it is the negative terminal
Write the Nernst equation as E = E° – (RT/nF) ln Q.
Convert natural log to base‑10 log: ln Q = 2.303 log Q, giving the factor (2.303 RT/nF).
At 298 K, 2.303 RT/F ≈ 0.0591 V, so the term becomes (0.0591/n) log (1/[Mn⁺]).
Thus the 0.0591 factor comes from 2.303 RT/F at 298 K.
ΔG° = –n F Ecell, so –RT ln K = –n F Ecell ⇒ ln K = n F Ecell / (R T).
Substituting n=2, F=96485 C mol⁻¹, Ecell=0.295 V, R=8.314 J mol⁻¹ K⁻¹, T=298 K gives ln K ≈ 22.96.
Convert to base‑10 log: log K = ln K / 2.303 ≈ 22.96 / 2.303 ≈ 9.97 ≈ 10.
Hence log K ≈ 10.
Λm = Λ° – A√c (Debye‑Hückel‑Onsager) shows molar conductivity decreases linearly with √c.
For a strong electrolyte Λ° is constant, A is positive, so the term A√c subtracts from Λ°.
Thus a plot of Λm against √c gives a straight line whose slope is –A (negative).
Hence the correct choice is a straight line with negative slope.
Al³⁺ gains 3 electrons to become Al metal, so 3 F of charge are needed per mole of Al formed.
One Faraday equals one mole of electrons (1 F = 96 485 C).
Therefore depositing 1 mol Al requires 3 mol e⁻ = 3 F.
Answer: 3
In aqueous solution the anode is the positive electrode where oxidation occurs.
Water is oxidized preferentially over sulfate ions: 2H₂O → O₂ + 4H⁺ + 4e⁻.
Thus oxygen gas is liberated at the anode during electrolysis of dilute H₂SO₄ with Pt electrodes.
Electrode potential is defined for a half‑cell under standard conditions; it depends on the activity (concentration) of the reacting ions via the Nernst equation.
Temperature appears in the Nernst term (RT/nF) and changes the potential.
The metal’s nature determines its standard reduction potential, a fundamental property.
Changing the electrode’s surface area does not alter the thermodynamic potential, only the current magnitude; therefore the size of the electrode (surface area) does not affect the single electrode potential. The size of the electrode (surface area)
ΔG° = –n F Ecell, and ΔG° = –RT ln K.
Equate: n F Ecell = RT ln K → ln K = (n F Ecell)/(RT).
At 298 K, (F/RT)= (96485)/(8.314×298)=38.9, so ln K = 38.9×0.0591≈2.30.
e^{2.30}=10, therefore K = 10.
Dilution reduces the number of ions per unit volume, so the ion concentration (c) decreases.
Specific conductance κ = λ c, where λ (molar conductivity) is nearly constant for a given electrolyte at a fixed temperature.
Since κ is directly proportional to c, lowering c lowers κ.
Thus on dilution the conductivity of the electrolyte solution decreases.
Faraday’s second law relates the mass of substance deposited to the charge passed and its equivalent weight.
Mass deposited ∝ (charge × equivalent mass)/F, where F is Faraday’s constant.
Since the same charge is used for different electrolytes, the ratio of masses equals the ratio of their equivalent masses.
Thus the masses are proportional to equivalent masses.
Cu²⁺ ions from CuSO₄ are reduced at the cathode: Cu²⁺ + 2e⁻ → Cu(s).
Water reduction to H₂ requires a higher overpotential than Cu²⁺ reduction, so Cu deposits preferentially.
Thus the cathodic product is copper metal.
In a hydrogen‑oxygen fuel cell, H₂ is oxidized at the anode to 2H⁺ + 2e⁻ and O₂ is reduced at the cathode to 4e⁻ + 4H⁺ → 2H₂O.
Balancing the electrons gives the overall cell reaction: 2H₂ + O₂ → 2H₂O.
Thus the only product formed is water, not hydrogen peroxide, ozone or carbon dioxide. Water.
In a galvanic cell the oxidation and reduction half‑reactions generate excess positive charge at the anode and excess negative charge at the cathode.
If these charges are not balanced, the cell quickly stops producing current.
A salt bridge contains inert electrolyte ions that can move into each half‑cell, neutralising the charge buildup.
Thus its primary function is to maintain electrical neutrality by allowing ion migration.
E = E° – (0.0592 / n) log [Cu²⁺] (Nernst equation, 298 K)
n = 2 for Cu²⁺ + 2e⁻ → Cu, so subtract (0.0592/2) log [Cu²⁺] from 0.34 V
[Cu²⁺] = 0.01 ⇒ log 0.01 = –2, giving –(0.0592/2)(–2) = +0.0592 V
E = 0.34 V – 0.0592 V = 0.2808 V ≈ 0.281 V, which matches the listed answer.
Maximum work = change in Gibbs free energy for the cell reaction.
ΔG = –n F E_cell where n = moles of electrons transferred, F = Faraday constant and E_cell = cell emf.
Electrical work obtainable = –ΔG, so W_max = –(–n F E_cell) = –n F E_cell.
Thus the correct choice is –nFE_cell.
Limiting molar conductivity (Λ₀) is measured at infinite dilution where ions do not interact.
At this condition each ion moves independently, so its contribution to conductivity is additive.
Thus Λ₀ equals the ionic conductivity of the cation plus that of the anion.
Hence the correct answer is the sum of individual ionic contributions of cation and anion.
1 F corresponds to 1 mol of electrons (96 485 C).
Cu²⁺ gains 2 e⁻ to form Cu, so 1 F deposits ½ mol of Cu (31.75 g).
2 F provide twice the electrons, depositing 1 mol of Cu.
Thus 2 F give 63.5 g of copper.
At the anode oxidation occurs; in concentrated NaCl solution the chloride ion concentration is high, so Cl⁻ is oxidised preferentially over water.
Oxidation half‑reaction: 2Cl⁻ → Cl₂(g) + 2e⁻ (E° = –1.36 V) is easier than water oxidation to O₂ (E° = –1.23 V) because Cl⁻ is more abundant.
Thus chlorine gas is liberated at the anode.
In an electrolytic cell external electricity is applied to force a reaction that would not occur on its own. The supplied electrical energy supplies the Gibbs free‑energy needed for the non‑spontaneous process. Hence the cell’s purpose is that electrical energy drives a non‑spontaneous reaction. Electrical energy drives a non-spontaneous reaction.
Use Nernst equation: Ecell = E°cell – (RT/nF) ln Q.
At 298 K, (RT/F) = 0.0257 V, so (RT/nF) = 0.0257/2 = 0.01285 V.
If Q = 1, ln Q = 0, thus the second term vanishes and Ecell = E°cell.
Therefore the cell potential equals the standard cell potential.
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