NCERT Solutions Class 12 Chemistry Ch 2 Electrochemistry PDF

Strategic angle. A metal A displaces a metal B from B's salt solution if and only if A is the stronger reducing agent –- equivalently, if E^∘(Aⁿ⁺/A) < E^∘(B^m+/B). The more negative E^∘ sits to the left of the electrochemical series.

Alternative approach: thermodynamic check. For a displacement A + B^m+ → Aⁿ⁺ + B, compute E^∘_cell = E^∘(B^m+/B) - E^∘(Aⁿ⁺/A). For Mg + Cu²⁺ -> Mg²⁺ + Cu: E^∘_cell = 0.34 - (-2.37) = +2.71 V, hugely positive so _r G^∘ = -nFE^∘_cell ≪ 0. The reverse, Cu + Mg²⁺, gives -2.71 V and is impossible.

1. Rank by E^∘. Mg^2+/Mg (-2.37) is the most negative, so Mg is the strongest reducer; Cu^2+/Cu (+0.34) the most positive, so Cu the weakest reducer.
2. Apply pairwise. 3 Mg(s) + 2 AlCl3(aq) -> 3 MgCl2(aq) + 2 Al(s), 2 Al(s) + 3 ZnSO4(aq) -> Al2(SO4)3(aq) + 3 Zn(s), Zn(s) + FeSO4(aq) -> ZnSO4(aq) + Fe(s), Fe(s) + CuSO4(aq) -> FeSO4(aq) + Cu(s).
3. No reverse reactions. Cu cannot displace any of the other four; their E^∘ values are all more negative than +0.34 V.
4. Concept linkage. The same ranking decides which Daniell-type galvanic cells are spontaneous: anode = more negative-E^∘ metal; cathode = more positive-E^∘ metal. A Mg|Mg^2+|Cu^2+|Cu cell delivers E^∘_cell = +2.71 V –- nearly twice the Daniell cell.

JEE/NEET relevance. Activity-series ordering recurs every year: ``which metals displace copper from CuSO4?'' The reliable answer: every metal above Cu in the series. The same series predicts which metals corrode in acidic rainwater (those above hydrogen) and which can be extracted by carbon reduction (those below carbon).

Mg > Al > Zn > Fe > Cu in decreasing reducing power; Mg displaces all four, Cu displaces none.

Quick reading. Reducing power runs opposite to standard reduction potential. Sort E^∘ in decreasing order and the metals appear in increasing order of reducing power: lower E^∘(Mⁿ⁺/M) stronger reducer.

Alternative approach: oxidation-potential view. Flip each E^∘_red sign: Ag: -0.80, Hg: -0.79, Cr: +0.74, Mg: +2.37, K: +2.93 V. The metal with the largest E^∘_ox is the strongest reducer. K (+2.93) sits far above; Ag (-0.80) is weakest.

1. Pair the metal with its E^∘. tabularll Ag+/Ag & +0.80 V
   Hg^2+/Hg & +0.79 V
   Cr^3+/Cr & -0.74 V
   Mg^2+/Mg & -2.37 V
   K+/K & -2.93 V
   tabular
2. Sort decreasing E^∘: Ag, Hg, Cr, Mg, K.
3. Invert for increasing reducing power. Ag < Hg < Cr < Mg < K.
4. Endpoints from chemistry. K reacts violently with cold water; Mg with hot water. Cr passivates. Ag resists dilute non-oxidising acids. The chemistry matches the ranking.

Numerical cross-check. Cell K K+ |  Ag+ Ag: E^∘_cell = 0.80 - (-2.93) = +3.73 V (very feasible). Cell Hg Hg^2+ |  Ag+ Ag: E^∘_cell = 0.80 - 0.79 = +0.01 V (barely feasible). The huge potassium-mercury gap vs the tiny mercury-silver gap confirms that potassium is in a different league.

JEE/NEET relevance. Reduction-potential ranking shows up every year, often disguised as ``which metal will not displace hydrogen from dilute acid?'' (answer: E<sup>∘</sup> > 0, here Ag, Hg) or ``arrange in increasing order of oxidation potential'' (reverse of reduction ranking).

Increasing reducing power: Ag < Hg < Cr < Mg < K.

Picture-first. Two beakers, two electrodes, two solutions, one salt bridge, one external wire with a voltmeter. The metal whose ion is being reduced is the cathode; the other is the anode. In Zn(s) + 2 Ag+(aq) -> Zn²⁺(aq) + 2 Ag(s), Ag+ is reduced, so Ag is the cathode and Zn is the anode. Once you fix the sides, every other answer follows from charge balance.

Alternative approach: build the cell from E^∘ values. E^∘(Zn²⁺/Zn) = -0.76 V and E^∘(Ag+/Ag) = +0.80 V. The more negative reduction potential always goes to the anode (better reducer); the more positive to the cathode (better oxidising agent). So Zn is the anode (negative terminal) and Ag the cathode (positive terminal), without even reading the reaction direction.

1. Split into half-cells. aligned Oxidation (anode): &Zn(s) -> Zn²⁺(aq) + 2 e^-
   &E^∘_red(Zn²⁺/Zn) = -0.76 V
   Reduction (cathode): &2 Ag+(aq) + 2 e^- -> 2 Ag(s)
   &E^∘_red(Ag+/Ag) = +0.80 V aligned Electrons balance: 2 released by oxidation = 2 absorbed by reduction.
2. Cell EMF (sanity check). E^∘_cell = E^∘_cathode - E^∘_anode = 0.80 - (-0.76) = +1.56 V. Positive: reaction as written is spontaneous, exactly as a galvanic cell should be.
3. Cell notation (IUPAC). Zn(s) Zn²⁺(aq) Ag+(aq) Ag(s). Anode metal anodic solution | cathodic solution cathode metal. Single = phase boundary; | = salt bridge.
4. Direction of charge –- external circuit. Electrons leave Zn (becoming electron-rich, the negative terminal). They flow through the wire to Ag, where Ag+ strips them off (so Ag is the positive terminal). Conventional current is opposite to electron flow.
5. Current inside the cell. Charge must close the loop through the solution and salt bridge, carried by ions: cations (Zn^2+, K+ from the bridge) drift toward the cathode compartment; anions (NO3^-) drift toward the anode compartment. The bridge keeps each half-cell electrically neutral, preventing the charge build-up that would otherwise stop the cell.

Numerical cross-check. _r G^∘ = -nFE^∘_cell = -(2)(96500)(1.56) = -301,080 J/mol = -301.08 kJ/mol. Strongly negative; Zn spontaneously reduces Ag+. The equilibrium constant log K = 2 × 1.56 / 0.0591 = 52.79, so K ∼ 10⁵³ –- reaction goes essentially to completion.

JEE/NEET relevance. Cell-diagram-from-reaction is a very common short-answer question. The marking scheme is mechanical: anode on the left (-), cathode on the right (+), | for salt bridge, phase labels (s, aq, l, g) included. Missing any of these costs marks.

Generalisation. Replace the metals and electrolytes and the same template works for any galvanic cell. The electrode with the more negative E^∘ is always the anode (negative); the more positive E^∘ is always the cathode (positive). Cations move toward the cathode, anions toward the anode –- through the solution and the bridge.

Zn(s) Zn²⁺(aq) Ag+(aq) Ag(s); Zn is the negative terminal; ions carry current inside the cell, electrons outside; Zn -> Zn²⁺ + 2e^- at the anode, 2 Ag+ + 2 e^- -> 2 Ag at the cathode.

Strategic angle. Three quantities, one chain: E^∘_cell from the difference of tabulated reduction potentials; _r G^∘ scales linearly with n and E^∘; K is exponential in nE^∘/0.0591. Determine n from the balanced cell reaction before substituting –- mis-counting n is the most common error.

Alternative approach: per-electron view. Write everything per mole of electrons transferred: E^∘_cell unchanged, but Δ G^∘ per electron is -FE^∘ (-33 kJ for part (i), -2.9 kJ for part (ii)). Multiplying by n gives the per-reaction value. This view makes clear why n controls magnitude.

1. (i) Confirm n = 6. Cr: Cr -> Cr³⁺ + 3 e^- × 2 = 6 electrons out. Cd: Cd²⁺ + 2 e^- -> Cd × 3 = 6 electrons in. Charges balance with n = 6.
2. Cell EMF. E^∘_cell = (-0.40) - (-0.74) = +0.34 V. Positive, spontaneous.
3. Recompute _r G^∘ (i). -n F E^∘ = -6 × 96500 × 0.34. First 6 × 96500 = 579,000; then 579000 × 0.34 = 196,860 J. So _r G^∘ = -196.86 kJ/mol.
4. Equilibrium constant (i). log K = (6 × 0.34)/0.0591 = 2.04/0.0591 ≈ 34.52, hence K ≈ 3.3 × 10³⁴.
5. (ii) n = 1. Fe²⁺ -> Fe³⁺ + e^- and Ag+ + e^- -> Ag. E^∘_cell = 0.80 - 0.77 = +0.03 V. _r G^∘ = -96500 × 0.03 = -2895 J = -2.895 kJ. log K = 0.03/0.0591 = 0.508; K = 10^0.508 ≈ 3.22.

Cross-check (i). Each electron costs 96500 × 0.34 ≈ 32.8 kJ. With 6 electrons, 6 × 32.8 = 196.8 kJ, matching |_r G^∘| to three figures.

Reading the two answers together. Part (i) is highly spontaneous: huge |_r G^∘|, astronomical K. Part (ii) is marginally spontaneous: small _r G^∘, K of order unity, so reactants and products coexist in similar amounts at equilibrium. The contrast illustrates how log K = nE^∘/0.0591 amplifies even tiny EMF differences when n is large.

JEE/NEET relevance. Three-in-one problems (E^∘ → _r G^∘ → K) are JEE Mains favourites: they test definition recall, balancing-electron skills, and arithmetic. Watch for the trap of using n = electrons in one half-reaction.

(i) E^∘_cell = 0.34 V; _r G^∘ = -196.86 kJ/mol; K ≈ 3.3 × 10³⁴. (ii) E^∘_cell = 0.03 V; _r G^∘ = -2.895 kJ/mol; K ≈ 3.22.

Strategic angle. The Nernst equation E = E^∘ - (0.0591/n)log Q at 298 K has one tricky part: writing the correct Q. For each cell, identify anode (oxidation) on the left and cathode (reduction) on the right, then build Q as products over reactants of the net cell reaction. Pure solids/liquids and 1-bar gases give unity, not zero.

Alternative approach: half-cell Nernst. Apply Nernst to each half-reaction separately, then subtract to get E_cell. The algebra collapses to the same form, but the half-cell view helps when gas pressures or aqueous redox couples are involved.

1. (i) Recap. Anode Mg, cathode Cu. E^∘ = 0.34 - (-2.37) = +2.71 V, n = 2, Q = [Mg²⁺]/[Cu²⁺] = 0.001/0.0001 = 10. log 10 = 1. E = 2.71 - 0.05912(1) = 2.71 - 0.02955 ≈ 2.68 V.
2. (ii) Recap. Anode Fe, cathode SHE with [H+] = 1 M. E^∘ = 0 - (-0.44) = +0.44 V, n = 2. Q = [Fe²⁺] p_H2/[H+]² = 10^-3. log(10^-3) = -3. E = 0.44 - 0.05912(-3) = 0.44 + 0.0887 ≈ 0.53 V.
3. (iii) Recap. Anode Sn, cathode SHE at 0.02 M. E^∘ = 0 - (-0.14) = +0.14 V, n = 2. Q = 0.05/(0.02)² = 125. log 125 = 2.0969. E = 0.14 - 0.02955 × 2.0969 ≈ 0.078 V.
4. (iv) Recap. Anode (on the left): 2 Br^- -> Br2(l) + 2 e^-; cathode SHE at 0.03 M. Net: 2 Br^- + 2 H+ -> Br2(l) + H2(g) with Br2 liquid (activity 1). E^∘ = 0 - 1.08 = -1.08 V, n = 2. Q = 1/[(0.01)²(0.03)²] = 1/(9 × 10^-8) ≈ 1.11 × 10⁷. log Q = 7.046. E = -1.08 - 0.02955 × 7.046 ≈ -1.29 V.

Numerical cross-check (i). If concentrations were both 1 M (Q = 1), E = E^∘ = 2.71 V. We sit ∼ 30 mV below standard, consistent with the 10:1 ratio and n = 2.

Sanity check (iv). A negative cell EMF means the reaction as written is non-spontaneous: bromide cannot reduce hydrogen ions. The spontaneous direction is the reverse (H2 reduces Br2 to Br^-), and that cell would deliver +1.29 V.

Concept linkage –- dilution and EMF. In cell (ii) diluting the anode compartment from 1 M to 10^-3 M raises EMF from 0.44 to 0.53 V. The cell wants more Fe^2+: making the product scarce drives the reaction forward (Le Chatelier), equivalent to log Q < 0.

JEE/NEET relevance. The Nernst equation generates 1-mark questions on whether E increases on diluting a compartment, on finding Q from E and E^∘, and on concentration cells. Memorise the 0.0591/n form for 298 K to save 30 seconds per question.

(i) 2.68 V; (ii) 0.53 V; (iii) 0.078 V; (iv) -1.29 V.

Strategic angle. Two ingredients: cell EMF and electron count. EMF comes from the difference of standard reduction potentials of the cathode and anode. Electron count follows from the balanced half-reactions; both halves here are written with 2 e^-, so n = 2.

Alternative approach via half-reaction E^∘. If you recall only E^∘(Ag+/Ag) = +0.80 V, the Ag2O/Ag couple can be reconstructed from the formation free energy of Ag2O; the result is ≈ +0.344 V. The ∼ 0.46 V drop from +0.80 comes from the energy needed to break the Ag-O-Ag bond and the strongly alkaline product.

1. Half-cells. aligned Cathode (reduction): &Ag2O(s) + H2O(l) + 2 e^- -> 2 Ag(s) + 2 OH^-(aq)
   Anode (oxidation): &Zn(s) -> Zn²⁺(aq) + 2 e^- aligned Adding (with Zn^2+ combining with 2 OH^- to give Zn(OH)2/ZnO) yields the net cell reaction in the question. n = 2.
2. Cell EMF. E^∘_cell = (+0.344) - (-0.76) = +1.104 V.
3. Gibbs energy. _r G^∘ = -nFE^∘_cell = -(2)(96500)(1.104) = -213,072 J/mol = -213.07 kJ/mol.
4. Equilibrium constant (extra). log K = nE^∘/0.0591 = (2 × 1.104)/0.0591 = 2.208/0.0591 ≈ 37.36, so K ≈ 2.3 × 10³⁷ –- the reaction proceeds essentially to completion.

Engineering note –- why voltage is so stable. None of the species in the net reaction is aqueous in large amounts (Zn^2+ stays complexed/precipitated as ZnO/Zn(OH)2 in alkaline electrolyte), so Q stays near 1 and operating voltage drifts very little as the cell discharges. This is exactly why button cells run watches: a clock crystal needs a stable supply to keep time.

Numerical cross-check. Energy per gram of Zn consumed: |Δ G|/M(Zn) = 213,072/65.4 ≈ 3260 J/g. Times typical 0.1 g of Zn in a button cell: ∼ 326 J = 0.09 Wh –- in the right ballpark for a watch battery.

JEE/NEET relevance. Battery questions cluster around ``calculate E^∘ and _r G^∘'' for button cell, lead-acid cell, NiCd cell. The recipe is the same: identify half-reactions, look up E^∘, substitute into -nFE^∘.

E^∘_cell ≈ +1.104 V; _r G^∘ ≈ -213.07 kJ/mol.

Structural observation. Conductivity is a bulk property of the electrolyte solution at a given concentration; molar conductivity normalises by the amount of electrolyte so we can compare strong vs weak, or two different salts, on the same axis. The two quantities differ in physical meaning and in how they respond to dilution.

Alternative approach: per-ion view. Think of _m as a sum of ionic conductivities. Kohlrausch's law, Λ^∘_m = ₊λ^∘₊ + _-λ^∘_-, says that at infinite dilution each ion contributes independently: λ^∘₊ for the cation, λ^∘_- for the anion. Ionic conductivities are characteristic of the ion (H+ biggest, OH^- next), not the parent salt. This is why Λ^∘_m(HCl) > Λ^∘_m(NaCl) > Λ^∘_m(KNO3).

• κ has units of S/cm (or S/m). It rises with concentration in dilute solutions, peaks, and may even fall at very high concentrations (ion-ion attraction and ion-pairing impede motion).
• _m has units of S cm²/mol. It always rises on dilution because each mole of ions is spread over a larger volume with fewer competing ions, and (for weak electrolytes) α grows.

1. Strong electrolytes (NaCl, HCl, KNO3, KCl): fully dissociated. _m decrease with √c is small and linear –- the Debye-H"uckel-Onsager equation, _m = Λ^∘_m - A√c. Extrapolate the plot to √c = 0 to get Λ^∘_m.
2. Weak electrolytes (CH3COOH, NH4OH): partially dissociated; α ≪ 1 at finite c but α → 1 as c → 0. Curves sharply upward near √c = 0. Cannot extrapolate linearly –- use Kohlrausch's law.
3. Degree of dissociation. α = _m/Λ^∘_m, from which K_a = cα²/(1-α) (Ostwald's dilution law).
4. Why κ falls but _m rises. On dilution, ions per cm³ drop (so κ falls), but per mole of electrolyte the same number of ions spread over a larger volume with fewer competing neighbours, so the ratio κ/c grows.

Practical use. κ is the measured quantity (cell constant × conductance). The downstream chemistry (degree of dissociation, dissociation constant, ionic mobilities) flows from converting κ to _m and applying the appropriate dilution law.

Numerical anchor. At 298 K, Λ^∘_m(KCl) = 149.86, Λ^∘_m(HCl) = 425.9, Λ^∘_m(CH3COOH) = 390.5 S cm²/mol; yet at 0.1 M, _m(CH3COOH) falls below 10 because α is only ∼ 0.013. Strong vs weak is unmistakable from a single concentration measurement.

JEE/NEET relevance. _m-vs-√c plots are a favourite NEET question: identify the strong vs the weak electrolyte from two curves. Strong = nearly horizontal straight line; weak = steep curve rising near origin.

κ = G · l/A in S/cm; _m = κ × 1000/c in S cm²/mol. κ falls on dilution; _m rises –- linearly for strong (Debye-H"uckel-Onsager), sharply for weak (Kohlrausch + Ostwald).

Quick reading. A single substitution: divide κ by c, multiply by 1000 to handle L → cm³.

Alternative approach: SI route. κ = 0.0248 S/cm = 2.48 S/m; c = 0.20 mol/L = 200 mol/m³. _m(SI) = 2.48/200 = 1.24 × 10^-2 S m²/mol; × 10⁴ = 124 S cm²/mol.

1. Working formula: _m = κ × 1000/c, κ in S/cm, c in mol/L, _m in S cm²/mol.
2. Substitute: _m = 0.0248 × 10000.20 = 24.80.20 = 124 S cm²mol^-1.
3. Order-of-magnitude check. Λ^∘_m(KCl) = 149.86 S cm²/mol. Our 124 sits below this, expected at non-zero concentration where ion-ion attraction reduces mobility.

Cross-check via Debye-H"uckel-Onsager. Gap 149.86 - 124 = 25.86 = A√c, so A = 25.86/√0.20 = 57.8 S cm² mol^-1 M^-1/2, close to the textbook ≈ 60 for 1:1 strong electrolytes in water at 298 K.

JEE/NEET relevance. The × 1000 unit trick is exactly the kind of detail JEE Mains tests. Forgetting it produces 0.124 instead of 124 –- a 1000× error.

_m(KCl, 0.20 M) = 124 S cm² mol^-1.

Quick reading. The cell constant converts conductance (instrument reading) into conductivity (a solution property). Algebraically G^* = κ R.

Alternative approach: dimensional check. [κ] = S/cm; [R] = Ω = 1/S. So [κ R] = cm^-1 –- exactly [l/A], length over area. Dimensional self-consistency confirms the formula.

1. Start from κ = G · (l/A) = G · G^*.
2. Solve: G^* = κ/G = κ · R since G = 1/R.
3. Substitute κ = 1.46 × 10^-4 S/cm, R = 1500 Ω: 1.46 × 1500 = 2190; 2190 × 10^-4 = 0.219. G^* = 0.219 cm^-1.
4. Sanity check. Typical Pt cell: l ∼ 1 cm, A ∼ 4 cm², so l/A ≈ 0.25 cm^-1. Our 0.219 matches.

Numerical cross-check. For an unknown solution showing R = 600 Ω in the same cell: κ = G^*/R = 0.219/600 = 3.65 × 10^-4 S/cm. One calibration unlocks every later measurement.

Why 0.001 M KCl. Dilute KCl is a primary standard: (i) its κ is tabulated with high accuracy over a wide T range, (ii) it is easy to prepare from solid KCl, (iii) at this concentration ion-pairing is negligible.

G^* = 0.219 cm^-1.

Picture-first. Convert every conductivity, divide by the matching molar concentration, plot _m vs √c, and extrapolate to √c = 0. Working in SI keeps the arithmetic clean; convert at the end. NaCl is a strong electrolyte, so the plot is essentially a straight line and a two-point extrapolation gives a good Λ^∘_m.

Alternative approach: cm-based per-row. Multiply listed 10²κ by 10^-2 to get S/m, again by 10^-2 to get S/cm, then _m = κ · 1000/c.

1. SI conversion. Listed 10² κ in S/m ⇒ κ = value/100 in S/m. Molarity × 1000 gives mol/m³. Divide for _m in S m²/mol, then × 10⁴ for S cm²/mol.
2. Row-by-row. _m(0.001) = 1.237 × 10^-2/1 = 1.237 × 10^-2 S m²/mol = 123.7 S cm²/mol. Similarly _m(0.010) = 118.5; _m(0.020) = 115.8; _m(0.050) = 111.1; _m(0.100) = 106.74.
3. Tabulate √c. tabularlrrrrr c / M & 0.001 & 0.010 & 0.020 & 0.050 & 0.100
   √c & 0.0316 & 0.1000 & 0.1414 & 0.2236 & 0.3162
   _m & 123.7 & 118.5 & 115.8 & 111.1 & 106.74
   tabular
4. Extrapolate. First and last points: slope = (123.7 - 106.74)/(0.0316 - 0.3162) = 16.96/(-0.2846) = -59.6 S cm² mol^-1 M^-1/2. Intercept (using first point): Λ^∘_m = 123.7 - (-59.6)× 0.0316 = 123.7 + 1.88 = 125.6 ≈ 126 S cm²/mol.
5. Linearity check. Predicted vs actual at middle points: √c = 0.1 predicts 125.6 - 5.96 = 119.6 (actual 118.5, -1%); √c = 0.1414 predicts 117.2 (actual 115.8, -1%). Excellent fit –- strong-electrolyte signature.

Comparison with literature. Λ^∘_m(NaCl) = 126.45 S cm²/mol at 298 K (NCERT Table 2.4). Our extrapolation ≈ 126 matches within 0.5%.

Cross-check via Kohlrausch. Λ^∘(Na+) = 50.1 and Λ^∘(Cl^-) = 76.3 at 298 K; sum = 126.4 S cm²/mol –- excellent agreement. The per-ion view rebuilds the salt's limiting conductivity.

Concept linkage. The DHO slope |A| = 59.6 matches the universal value ≈ 60 for 1:1 strong electrolytes in water at 298 K. The slope is theory-determined by long-range ion-ion interaction, so its closeness to the textbook value validates the data.

JEE/NEET relevance. ``Plot, then extrapolate'' is a stock NEET diagram task. Never try to extrapolate a weak-electrolyte curve linearly –- the curvature near √c = 0 gives garbage; use Kohlrausch's law instead.

_m values: 123.7, 118.5, 115.8, 111.1, 106.74 S cm²/mol; Λ^∘_m ≈ 126 S cm²/mol.

Strategic angle. A single chain: κ → _m → α → K_a. Each arrow uses one definition; keep significant figures through the chain so K_a does not lose precision.

Alternative approach: small-α shortcut. Since α ≈ 0.08 ≪ 1, K_a ≈ cα² = 1.70 × 10^-5, within 8% of the exact value. Useful when α < 0.05.

1. _m. _m = κ · 1000/c = 7.896 × 10^-5 × 1000/(2.41 × 10^-3) = 0.07896/0.00241 = 32.76 S cm²/mol.
2. α. α = _m/Λ^∘_m = 32.76/390.5 = 0.0839. Only ∼ 8.4% is dissociated.
3. K_a. K_a = cα²/(1 - α). (0.0839)² = 0.007039. cα² = 2.41 × 10^-3 × 0.007039 = 1.697 × 10^-5. Divide by 0.9161: K_a = 1.853 × 10^-5 ≈ 1.85 × 10^-5.

Cross-check via pH. [H+] = cα = 2.02 × 10^-4; pH = 3.69. Substitute into K_a = [H+]²/([HA] - [H+]): (2.02 × 10^-4)²/(2.21 × 10^-3) = 1.85 × 10^-5. Match.

Reading the result. K_a ≈ 1.8 × 10^-5 corresponds to pK_a ≈ 4.74, the textbook value. Λ^∘_m = 390.5 is built from ionic conductivities λ^∘(H+) = 349.6 and λ^∘(CH3COO^-) = 40.9; almost 90% of Λ^∘_m comes from H+ hopping via the Grotthuss mechanism.

JEE/NEET relevance. Computing K_a from _m is JEE Mains classic. Often only κ and c are given –- remember to fetch Λ^∘_m from the data table or compute it via Kohlrausch.

_m = 32.76 S cm²/mol; α ≈ 0.084; K_a ≈ 1.85 × 10^-5.

Quick reading. For each reduction, read the electron count from the balanced half-reaction and multiply by Faraday's constant. The whole problem reduces to ``what is n?''.

Alternative approach: oxidation-state arithmetic. (i) Al: +3 → 0, drop 3, n = 3. (ii) Cu: +2 → 0, drop 2, n = 2. (iii) Mn in MnO4^-: +7 → +2, drop 5, n = 5.

1. (i) Al. Al³⁺ + 3 e^- -> Al: Q = 3 × 96500 = 2.895 × 10⁵ C.
2. (ii) Cu. Cu²⁺ + 2 e^- -> Cu: Q = 2 × 96500 = 1.93 × 10⁵ C.
3. (iii) Permanganate. MnO4^- + 8 H+ + 5 e^- -> Mn²⁺ + 4 H2O: n = 5 (Mn: +7 → +2). Q = 5 × 96500 = 4.825 × 10⁵ C.

Reading the numbers. Each electron costs ∼ 9.65 × 10⁴ C per mole. The result scales linearly with n: Al needs 50% more charge than Cu for the same mole count; permanganate needs 2.5× as much as Cu.

Cross-check (iii). 4.825 × 10⁵ C = 134 Ah. A 1 A current must run for 134 hours to reduce one mole (∼ 100 g) of MnO4^- –- a real industrial number.

JEE/NEET relevance. ``Charge to deposit x moles of M'' is a classroom staple. The trap: mis-counting electrons for couples like Cr2O7^2-/Cr^3+ (6, not 3 or 7).

(i) 2.895 × 10⁵ C; (ii) 1.93 × 10⁵ C; (iii) 4.825 × 10⁵ C.

Strategic angle. Moles of product → moles of electrons. Convert at the end if coulombs are needed. ``1 Faraday'' = 1 mole of electrons = 96,500 C.

Alternative approach: equivalents. Mass per Faraday = M_w/z. For Ca: 40/2 = 20 g/F; Al: 27/3 = 9 g/F. So 20 g Ca → 1 F; 40 g Al → 40/9 = 4.44 F. The equivalent route is the fastest mental computation.

1. (i) Ca. Ca²⁺ + 2 e^- -> Ca, z = 2. Moles of Ca = 20/40 = 0.5. Moles of e^- = 1. ⇒ 1 Faraday.
2. (ii) Al. Al³⁺ + 3 e^- -> Al, z = 3. Moles of Al = 40/27 = 1.481. Moles of e^- = 4.444. ⇒ 4.44 Faradays.
3. Sanity check (i). One Faraday gives half a mole of any divalent metal; half a mole of Ca = 20 g.
4. Sanity check (ii). One Faraday gives one-third mole of any trivalent metal; 4.44/3 = 1.48 mol Al = 40 g.

Numerical cross-check (industrial). 1 tonne of Al (3.70 × 10⁴ mol) needs 3 × 3.70 × 10⁴ = 1.11 × 10⁵ F ≈ 1.07 × 10¹⁰ C. At 4.5 V cell voltage, energy = 4.83 × 10¹⁰ J = 13.4 MWh per tonne. This matches the ``∼ 15,000 kWh/tonne'' industrial figure. The 3-electron reduction is exactly why aluminium is so energy-intensive vs sodium (z = 1) or magnesium (z = 2).

(i) 1.0 Faraday = 96,500 C; (ii) 4.44 Faradays ≈ 4.29 × 10⁵ C.

Strategic angle. Balance the half-reaction, count electrons per mole of starting material, multiply by F. The trap is the basis: the question says ``per mole of H2O'', not ``per mole of O2''.

Alternative approach: oxidation-state arithmetic. (i) O: -2 → 0 per atom, 2 per H2O; n = 2. (ii) Fe: +2 → +3, n = 1 per FeO.

1. (i) Water oxidation. 2 H2O -> O2 + 4 H+ + 4 e^-: 4 electrons for 2 mol H2O, i.e. 2 per mole of H2O. Q = 2 × 96500 = 1.93 × 10⁵ C.
2. (ii) FeO to Fe2O3. Fe goes +2 → +3: 1 e^- per Fe, 1 per mole of FeO. Q = 1 × 96500 = 9.65 × 10⁴ C.

Connection to corrosion. Both reactions appear together in rusting: Fe/FeO is oxidised by atmospheric O2 in moist air. Each turnover transfers exactly the electrons we counted –- 4 from the O2-reduction cathode (matching 4 Fe^2+-to-Fe^3+ oxidations).

Numerical cross-check. Industrial water electrolysis at ∼ 2.0 V producing 1 mol/s of H2 draws 4 × 96500/(2) = 193,000 A per cell (using n = 4 per net reaction 2 H2O -> 2 H2 + O2). The factor-of-2 between H2O and O2 specifications is exactly why ``per mole of'' phrasing matters.

JEE/NEET relevance. ``Per mole of O2'' would give Q = 4 × 96500 = 3.86 × 10⁵ C, exactly double. A 1-mark difference between right and wrong.

(i) 1.93 × 10⁵ C per mole of H2O; (ii) 9.65 × 10⁴ C per mole of FeO.

Strategic angle. Charge → moles of electrons → moles of metal → mass. Each arrow is one division. The composite formula w = M_w I t / (n F) collapses the chain, but knowing the chain helps debug arithmetic errors.

Alternative approach: equivalent weight. Ni's chemical equivalent: 58.7/2 = 29.35 g/F. Coulombs = 5 × 1200 = 6000; Faradays = 6000/96500 = 0.0622. Mass = 0.0622 × 29.35 = 1.83 g. Fastest mental computation.

1. Charge passed. t = 20 min × 60 = 1200 s. Q = I t = 5 × 1200 = 6000 C.
2. Moles of e^-. n_e = Q/F = 6000/96500 = 0.0622 mol.
3. Moles of Ni. Ni²⁺ + 2 e^- -> Ni, so n_Ni = 0.0622/2 = 0.0311 mol.
4. Mass. w = 0.0311 × 58.7 = 1.83 g.

Cross-check via master formula. w = M_w I t / (n F) = (58.7 × 5 × 1200)/(2 × 96500) = 352,200/193,000 = 1.825 g; three sig. fig. = 1.83 g.

Reading the result. 5 A for 20 min deposits under 2 g of Ni. For a watch-case coat of ∼ 5 g, you'd need either 25 min at 10 A or 50 min at 5 A. Industry uses the linear scaling to size plants.

Concept linkage –- inert electrodes. The Pt electrodes are inert; they pass current without dissolving. If the anode were Ni metal instead, the anode reaction Ni -> Ni²⁺ + 2 e^- would replenish bulk Ni^2+, an electrorefining cell.

JEE/NEET relevance. End-to-end Faraday-first-law problems appear in nearly every JEE Mains paper. The most common trap: not converting minutes to seconds. Q = I t needs t in seconds.

Mass of Ni deposited ≈ 1.83 g.

Strategic angle. Three cells in series receive the same charge in the same time. Compute charge from one cell (here cell B, silver), then use it to find masses in the other two.

Alternative approach: equivalents. eq deposited = Q/F, same for all three cells. eq = 1.45/108 = 0.01343. w(Cu) = 0.01343 × 31.75 = 0.426 g. w(Zn) = 0.01343 × 32.7 = 0.439 g. Faraday's second law in three lines.

1. Charge from Ag. Moles Ag = 1.45/108 = 0.01343; moles e^- = 0.01343 (since Ag+ + e^- -> Ag); Q = 0.01343 × 96500 = 1295.6 C.
2. Time. t = Q/I = 1295.6/1.5 = 863.7 s = 14 min 24 s.
3. Mass of Cu. 2 e^- per Cu, so moles Cu = 0.01343/2 = 0.006716. w(Cu) = 0.006716 × 63.5 = 0.426 g.
4. Mass of Zn. 2 e^- per Zn, so moles Zn = 0.006716. w(Zn) = 0.006716 × 65.4 = 0.439 g.

Cross-check. Sum of moles of electrons across the three cells must equal Q/F = 0.01343. Independent calculation: Ag 0.01343 × 1 = 0.01343; Cu 0.006716 × 2 = 0.01343; Zn 0.006716 × 2 = 0.01343. All identical, as required by series connection.

Generalisation. The series circuit is the classic demonstration of Faraday's second law: same charge through different electrolytes deposits masses in the ratio of chemical equivalents M_w/z. Zn is heavier than Cu, but only slightly more Zn is deposited because M_w/z values (32.7 vs 31.75) are close.

JEE/NEET relevance. Series-cell problems are favourite multi-part JEE Mains questions: they bundle time, charge, mass, and the second law. A common variant: given two deposit masses, find the ratio of atomic weights or charges.

t ≈ 864 s (≈ 14 min 24 s); w(Cu) ≈ 0.426 g; w(Zn) ≈ 0.439 g.

Strategic angle. The species written first in the question is the oxidising agent (cathode side); the second is the reducing agent (anode side). Compute E^∘_cell; positive means feasible: E^∘_cell = E^∘_cathode - E^∘_anode > 0 ⇔ _r G^∘ < 0 ⇔ spontaneous.

Alternative approach: ``higher reduces lower''. The couple with the higher E^∘_red will oxidise the couple with the lower one. Use this to predict the spontaneous direction first, then check if the postulated reaction matches.

1. (i) Fe^3+ (0.77) vs I2/I^- (0.54): higher is Fe^3+. E^∘_cell = 0.77 - 0.54 = +0.23 V. Feasible.
2. (ii) Ag+ (0.80) vs Cu^2+/Cu (0.34): higher is Ag+. E^∘_cell = 0.80 - 0.34 = +0.46 V. Feasible.
3. (iii) Fe^3+ (0.77) vs Br2/Br^- (1.08): higher is Br2, so Br2 would oxidise Fe^2+, but the question asks the reverse. E^∘_cell = 0.77 - 1.08 = -0.31 V. Not feasible.
4. (iv) Fe^3+ (0.77) vs Ag+/Ag (0.80): higher is Ag+ (just barely). E^∘_cell = 0.77 - 0.80 = -0.03 V. Not feasible (barely).
5. (v) Br2 (1.08) vs Fe^3+/Fe^2+ (0.77): higher is Br2, so Br2 oxidises Fe^2+. E^∘_cell = 1.08 - 0.77 = +0.31 V. Feasible.

Numerical sanity (iv). E^∘_cell = -0.03 V gives _r G^∘ = -F × (-0.03) = +2.9 kJ/mol. With RT = 2.5 kJ/mol at 298 K, K = e^-2.9/2.5 = 0.31 –- only ∼ 31% of equilibrium completion in the forward direction. Not feasible in the standard state, but at non-standard concentrations (high Fe^3+, very low Ag+) the Nernst equation could flip the sign.

Concept linkage –- the ladder. On the standard reduction- potential ladder: Br2/Br^- > Ag+/Ag > Fe³⁺/Fe²⁺ > I2/I^- > Cu²⁺/Cu. Higher couples oxidise lower; the converse is non-spontaneous.

JEE/NEET relevance. ``Predict feasibility from E^∘'' is a 1-mark MCQ workhorse. The trick: identify which species is being oxidised in the postulated reaction –- it is the reducing agent, not the oxidising one.

Feasible: (i), (ii), (v). Not feasible: (iii), (iv).

Strategic angle. For aqueous electrolysis, list every species that could be oxidised (or reduced) at each electrode and pick by the rule: highest reduction potential wins at the cathode; lowest reduction potential wins at the anode, with overpotential as a real-world tiebreaker. Reactive electrodes (Ag, Cu) can also dissolve in preference to water.

Alternative approach. For each cell, list (a) cathode candidates as reductions and (b) anode candidates as oxidations, each with its E^∘_red. Cathode picks the highest E^∘_red; anode picks the lowest (equivalently, highest E^∘_ox). Overpotential reorders winners when two candidates are within ∼ 0.5 V.

1. (i) AgNO3 / Ag electrodes. Cathode: Ag+ (+0.80) vs H2O (-0.83 at pH 7). Ag+ wins. Anode: dissolution of Ag electrode (E^∘_ox = -0.80) vs H2O (-1.23) vs NO3^- (very inert). Ag dissolves (least-negative oxidation potential). Effect: Ag moves from anode to cathode; bulk AgNO3 concentration unchanged. This is the basis of silver electroplating and electrorefining.
2. (ii) AgNO3 / Pt electrodes. Cathode: same as (i); Ag deposits on Pt. Anode: Pt is inert; H2O oxidation wins over inert NO3^-: 2 H2O -> O2 + 4 H+ + 4 e^-. Products: Ag at cathode, O2 at anode. Solution slowly becomes acidic and depleted in silver.
3. (iii) Dilute H2SO4 / Pt. Cathode: only H+ and H2O compete; H+ wins. 2 H+ + 2 e^- -> H2. Anode: SO4^2- oxidation to peroxodisulphate needs E^∘_red = +2.05 V (very hard); H2O wins: 2 H2O -> O2 + 4 H+ + 4 e^-. Products: H2/O2. Net: 2 H2O -> 2 H2 + O2. H2SO4 merely carries current –- electrolysis of water.
4. (iv) Aqueous CuCl2 / Pt. Cathode: Cu^2+ (+0.34) vs H2O (-0.83); Cu^2+ wins. Copper deposits. Anode: thermodynamic comparison gives 2 Cl^- -> Cl2 (E^∘_ox = -1.36) vs H2O (-1.23). H2O would win on thermodynamics –- but the overpotential for O2 at Pt is large (∼ 0.5 V), so practical anode product is Cl2. Products: Cu / Cl2.

Numerical anchor for overpotential. O2 evolution at Pt needs E_anode ≥ 1.23 + η(O2) ≈ 1.73 V. Cl2 evolution needs ≥ 1.36 + η(Cl2) ≈ 1.41 V –- lower. Cl2 wins on kinetics. This is exactly why the chlor-alkali industry electrolyses brine for chlorine.

Reading the four cases together. Cases (i) and (ii) differ only at the anode: the choice of electrode material flips whether the salt is consumed (Pt) or merely shuttled (Ag). Cases (iii) and (iv) emphasise that the dissolved anion matters: stable SO4^2- forces H2O oxidation; kinetically active Cl^- wins over H2O thanks to overpotential, even though thermodynamics favours H2O.

Concept linkage. The same overpotential effect explains why brine (not pure water) is electrolysed for chlorine, why Hg-cathode chlor-alkali cells produce NaOH without Na escape, and why electrolytic vs catalytic synthesis depends on the electrode metal.

JEE/NEET relevance. ``Predict products of electrolysis'' appears every year and rewards systematic candidate listing. Two classic surprises tested: (a) reactive metal electrodes (Ag, Cu) dissolving instead of water oxidising, and (b) Cl2 winning over O2 at Pt anodes due to overpotential.

(i) Ag/Ag (plating); (ii) Ag/O2; (iii) H2/O2 (water electrolysis); (iv) Cu/Cl2 (overpotential).