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The Vault
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The Vault
CBSE Class 12 Chemistry · 11 questions · 27 marks
Every battery in your pocket is a controlled redox reaction with the electrons routed through a wire instead of jumping directly between chemicals. This chapter connects cell potentials to the Nernst equation, conductance to molar conductivity, and electricity to the mass deposited at an electrode through Faraday's laws.
For the cell Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s), which change will increase Ecell?
Answer
Increasing [Cu²⁺] at constant [Zn²⁺]. The Nernst equation gives Ecell = E°cell − (0.0591/2) log([Zn²⁺]/[Cu²⁺]). Raising [Cu²⁺] makes the ratio smaller, so log Q becomes more negative and the subtracted term adds to E°cell. Electrode area and salt-bridge thickness affect current and internal resistance but not the cell potential, which is an intensive property.
The same quantity of electricity is passed through solutions of AgNO₃ and CuSO₄ in series. If 2.16 g of silver (atomic mass 108) is deposited, the mass of copper (atomic mass 63.5) deposited is:
Answer
0.635 g. Moles of Ag = 2.16/108 = 0.02 mol, and Ag⁺ + e⁻ → Ag needs one electron per atom, so 0.02 mol of electrons flowed. Cu²⁺ + 2e⁻ → Cu needs two electrons per atom, so moles of Cu = 0.02/2 = 0.01 mol, giving a mass of 0.01 × 63.5 = 0.635 g. This illustrates Faraday's second law: equal charge deposits masses in the ratio of equivalent masses.
Molar conductivity of a weak electrolyte increases sharply on dilution mainly because:
Answer
The degree of dissociation increases, producing more ions per mole. Λm is defined per mole of electrolyte, so what matters is how many ions each mole actually releases. Dilution shifts the ionisation equilibrium of a weak electrolyte to the right (Ostwald's dilution law), so α and therefore Λm rise steeply and approach Λ°m only at infinite dilution. Note that conductivity κ, which counts ions per unit volume, actually falls on dilution.
For a cell reaction with n = 2 and E°cell = +0.34 V, the value of ΔG° is (F = 96500 C mol⁻¹):
Answer
−65.6 kJ mol⁻¹. Using ΔG° = −nFE°cell = −(2)(96500)(0.34) = −65 620 J mol⁻¹ ≈ −65.6 kJ mol⁻¹. The negative sign confirms that a cell with a positive standard potential drives a spontaneous reaction.
Assertion (A): The value of Λ°m for acetic acid cannot be found by extrapolating a plot of Λm against √C. Reason (R): Acetic acid is a weak electrolyte whose degree of dissociation changes steeply near infinite dilution, so the plot does not approach a straight line.
Answer
Both A and R are true and R is the correct explanation of A. For a strong electrolyte Λm = Λ°m − A√C is linear, so the intercept gives Λ°m directly. For acetic acid, α itself rises steeply as C → 0 and the curve shoots up almost vertically, giving no reliable intercept. Λ°m must instead be assembled from strong-electrolyte data using Kohlrausch's law: Λ°m(CH₃COOH) = Λ°m(CH₃COONa) + Λ°m(HCl) − Λ°m(NaCl).
Calculate the emf at 298 K of the cell Zn(s) | Zn²⁺(0.010 M) || Cu²⁺(0.100 M) | Cu(s), given E°(Zn²⁺/Zn) = −0.76 V and E°(Cu²⁺/Cu) = +0.34 V.
Answer
Step 1 — cell reaction and n: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), with n = 2. Step 2 — standard potential: E°cell = E°cathode − E°anode = 0.34 − (−0.76) = 1.10 V. Step 3 — Nernst equation: Ecell = E°cell − (0.0591/n) log([Zn²⁺]/[Cu²⁺]). Step 4 — substitute: [Zn²⁺]/[Cu²⁺] = 0.010/0.100 = 0.10, and log 0.10 = −1. Ecell = 1.10 − (0.0591/2)(−1) = 1.10 + 0.0296 = 1.13 V. The emf is slightly above E°cell because the product ion Zn²⁺ is more dilute than the reactant ion Cu²⁺, which pushes the reaction forward.
The conductivity of a 0.20 M solution of a strong electrolyte is 2.48 ×10⁻² Ω⁻¹ cm⁻¹. Calculate its molar conductivity. If a 0.050 M solution of a weak acid has Λm = 19.5 Ω⁻¹ cm² mol⁻¹ and Λ°m = 390 Ω⁻¹ cm² mol⁻¹, calculate its degree of dissociation and dissociation constant.
Answer
Part 1 — Λm = 1000κ/C = (1000 × 2.48 ×10⁻²)/0.20 = 24.8/0.20 = 124 Ω⁻¹ cm² mol⁻¹. Part 2 — degree of dissociation: α = Λm/Λ°m = 19.5/390 = 0.050, i.e. 5.0% dissociated. Part 3 — dissociation constant by Ostwald's dilution law: Ka = Cα²/(1 − α) = (0.050)(0.050)²/(1 − 0.050) = (0.050)(2.5 ×10⁻³)/0.950 = 1.25 ×10⁻⁴/0.950 = 1.32 ×10⁻⁴. The small Ka confirms that the acid is weak.
State the products obtained at each electrode when an aqueous solution of sodium sulphate is electrolysed using inert platinum electrodes, and explain why sodium metal is not deposited.
Answer
At the cathode, water is reduced: 2H₂O + 2e⁻ → H₂(g) + 2OH⁻, so hydrogen gas is released. At the anode, water is oxidised: 2H₂O → O₂(g) + 4H⁺ + 4e⁻, so oxygen gas is released; the sulphate ion is not oxidised because sulphur is already in its highest oxidation state. Sodium is not deposited because E°(Na⁺/Na) = −2.71 V is far more negative than the reduction potential of water, so water is reduced preferentially at the cathode — the species with the higher (less negative) reduction potential is discharged first.
(i) Write the Nernst equation for the electrode reaction M^n+(aq) + ne⁻ → M(s) and explain each term. (ii) For the cell Ni(s) | Ni²⁺(0.10 M) || Ag⁺(0.0010 M) | Ag(s), given E°(Ni²⁺/Ni) = −0.25 V and E°(Ag⁺/Ag) = +0.80 V, write the cell reaction, calculate E°cell and Ecell at 298 K, and state whether the reaction is spontaneous.
Answer
(i) For M^n+ + ne⁻ → M(s), the Nernst equation at 298 K is E(M^n+/M) = E°(M^n+/M) − (0.0591/n) log(1/[M^n+]). Here E° is the standard electrode potential when [M^n+] = 1 M, n is the number of electrons gained per ion, and the activity of the pure solid metal is taken as 1, which is why it does not appear in the log term. The factor 0.0591 comes from 2.303RT/F evaluated at 298 K. Raising [M^n+] makes log(1/[M^n+]) more negative, so the electrode potential increases — a more concentrated ion solution is easier to reduce. (ii) Step 1 — identify electrodes. Silver has the higher reduction potential, so it is the cathode and nickel is the anode. Anode: Ni(s) → Ni²⁺ + 2e⁻. Cathode: 2Ag⁺ + 2e⁻ → 2Ag(s). Overall: Ni(s) + 2Ag⁺(aq) → Ni²⁺(aq) + 2Ag(s), n = 2. Step 2 — E°cell = E°cathode − E°anode = 0.80 − (−0.25) = 1.05 V. Step 3 — reaction quotient Q = [Ni²⁺]/[Ag⁺]² = 0.10/(1.0 ×10⁻³)² = 0.10/1.0 ×10⁻⁶ = 1.0 ×10⁵. Step 4 — Nernst: Ecell = 1.05 − (0.0591/2) log(1.0 ×10⁵) = 1.05 − (0.02955)(5) = 1.05 − 0.148 = 0.90 V. Step 5 — conclusion: Ecell = +0.90 V is positive, so ΔG = −nFEcell = −(2)(96500)(0.90) = −1.74 ×10⁵ J mol⁻¹ is negative and the reaction is spontaneous as written. The emf is lower than E°cell because the very dilute Ag⁺ makes the forward reaction less favourable than under standard conditions.
(i) State Faraday's two laws of electrolysis. (ii) A current of 2.50 A is passed through a silver nitrate solution for 30.0 minutes using silver electrodes. Calculate the charge passed, the moles of electrons, and the mass of silver deposited (atomic mass of Ag = 108, F = 96500 C mol⁻¹). (iii) Explain why a lead storage battery is called a secondary cell and write the reaction at its anode during discharge.
Answer
(i) First law: the mass of a substance deposited or liberated at an electrode is directly proportional to the quantity of electricity passed, w ∝ Q, i.e. w = Z I t where Z is the electrochemical equivalent. Second law: when the same quantity of electricity is passed through different electrolytes in series, the masses deposited are proportional to their chemical equivalent masses. (ii) Step 1 — charge: Q = I × t = 2.50 A × (30.0 × 60) s = 2.50 × 1800 = 4500 C. Step 2 — moles of electrons: n(e⁻) = Q/F = 4500/96500 = 0.0466 mol. Step 3 — silver deposition needs one electron per atom, Ag⁺ + e⁻ → Ag, so moles of Ag = 0.0466 mol. Step 4 — mass: w = 0.0466 × 108 = 5.04 g of silver. (iii) A lead storage battery is a secondary cell because the electrode reactions can be reversed by passing an external current, so it can be recharged and reused many times. During discharge the anode is spongy lead: Pb(s) + SO₄²⁻(aq) → PbSO₄(s) + 2e⁻. The cathode is lead dioxide, PbO₂ + SO₄²⁻ + 4H⁺ + 2e⁻ → PbSO₄ + 2H₂O. On charging, the PbSO₄ on both plates is converted back to Pb and PbO₂ and the sulphuric acid is regenerated.
A student assembles a cell using a magnesium strip in 1.0 M Mg²⁺ solution and a copper strip in 1.0 M Cu²⁺ solution, joined by a KCl salt bridge and a voltmeter. Given E°(Mg²⁺/Mg) = −2.37 V and E°(Cu²⁺/Cu) = +0.34 V, F = 96500 C mol⁻¹. (a) Write the cell notation and identify the anode. (b) Calculate E°cell. (c) Calculate ΔG° for the cell reaction. (d) The student removes the salt bridge and the voltmeter reading drops to zero. Explain why.
Answer
(a) Cell notation: Mg(s) | Mg²⁺(1.0 M) || Cu²⁺(1.0 M) | Cu(s). Magnesium has the more negative reduction potential, so it is oxidised and acts as the anode (negative terminal); copper is the cathode. (b) E°cell = E°cathode − E°anode = 0.34 − (−2.37) = 2.71 V. (c) The cell reaction Mg(s) + Cu²⁺(aq) → Mg²⁺(aq) + Cu(s) transfers n = 2 electrons. ΔG° = −nFE°cell = −(2)(96500)(2.71) = −523 030 J mol⁻¹ ≈ −523 kJ mol⁻¹. The large negative value shows a strongly spontaneous reaction. (d) Without the salt bridge the circuit is broken and charge cannot balance. As Mg²⁺ accumulates in the anode compartment it builds a positive charge, while loss of Cu²⁺ leaves excess sulphate and a negative charge in the cathode compartment. This charge separation immediately opposes further electron flow, so the current and the measured potential fall to zero. The salt bridge supplies K⁺ to the cathode side and Cl⁻ to the anode side, keeping both solutions electrically neutral and completing the internal circuit.
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