NCERT Solutions for Class 12 Chemistry Chapter 2: Electrochemistry

Principle. The potential of a single electrode cannot be measured on its own; only the potential difference between two electrodes (the emf of a cell) can be measured. So the electrode whose potential is wanted is combined with a reference electrode, the standard hydrogen electrode (SHE), whose potential is arbitrarily fixed at 0.00 V at all temperatures: Pt(s)|H₂(g, 1 bar)|H⁺(aq, 1 M).

Procedure.

1. Prepare the Mg²⁺|Mg half-cell: a clean magnesium rod dipped in a solution of Mg²⁺ ions of 1 M concentration (e.g. 1 M MgSO₄) at 298 K.
2. Prepare the SHE: a platinum foil coated with platinum black, immersed in 1 M H⁺ (e.g. 1 M HCl) with pure H₂ gas at 1 bar bubbled over it.
3. Join the two solutions by a salt bridge and connect the electrodes through a high-resistance voltmeter, giving the cell Pt(s)|H₂(g, 1 bar)|H⁺(aq, 1 M)||Mg²⁺(aq, 1 M)|Mg(s).
4. Read the emf. It is found that the Mg electrode is the negative terminal (electrons flow from Mg to the SHE), so Mg is oxidised and the SHE is the cathode.

Calculation. E°cell = E°(right) − E°(left). The measured magnitude is 2.36 V, and because Mg is the negative electrode (it is the anode: Mg → Mg²⁺ + 2e⁻), the standard reduction potential of the Mg²⁺|Mg couple is negative:

E°(Mg²⁺/Mg) = −2.36 V (NCERT Table 2.1)

The large negative value shows that Mg²⁺ is very difficult to reduce and that magnesium metal is a very strong reducing agent, much stronger than hydrogen.

Extra depth. The SHE is a primary reference but is awkward to use (platinised Pt, pure H₂ at exactly 1 bar, acid of unit activity), so in routine work secondary references such as the calomel electrode or Ag/AgCl electrode are calibrated against it. Whatever reference is used, the reduction potential of the test half-cell is its emf relative to the SHE with the SHE written on the left, so a negative value like −2.36 V for Mg²⁺/Mg tells you that Mg is a much stronger reductant than H₂.

Conclusion: E°(Mg²⁺/Mg) is determined by measuring the emf of the cell formed with the SHE under standard conditions; the result is −2.36 V.

Answer: No. Zinc is a stronger reducing agent than copper, so it will displace Cu²⁺ ions from the solution.

Reasoning with standard electrode potentials (Table 2.1). E°(Zn²⁺/Zn) = −0.76 V and E°(Cu²⁺/Cu) = +0.34 V. The metal with the lower (more negative) reduction potential is oxidised more readily.

Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

E°cell = E°cathode − E°anode = 0.34 V − (−0.76 V) = +1.10 V

A positive E°cell means the reaction is spontaneous (Δ_rG° = −nFE°cell = −2 × 96487 × 1.10 = −212.3 kJ mol⁻¹). The zinc pot would therefore be eaten away (Zn dissolves as Zn²⁺), the blue colour of CuSO₄ would fade and a reddish-brown deposit of copper would form on the inside of the pot — the Daniell-cell reaction occurring by direct contact.

Extra depth. This is the reason galvanised (zinc-coated) iron is used: zinc, being more electropositive than iron (−0.76 V against −0.44 V), is oxidised in preference to iron. The same displacement reaction, Zn + Cu²⁺ → Zn²⁺ + Cu, is exactly the one that runs in the Daniell cell, so the spontaneity (E°cell = +1.10 V; Δ_rG° = −212 kJ mol⁻¹) can be read directly from Table 2.1.

Conclusion: Copper sulphate solution cannot be stored in a zinc pot because Zn (E° = −0.76 V) displaces Cu²⁺ (E° = +0.34 V), with E°cell = +1.10 V.

Concept. Ferrous ion is oxidised to ferric ion: Fe²⁺ → Fe³⁺ + e⁻. The standard potential of the Fe³⁺/Fe²⁺ couple is +0.77 V (Table 2.1). A substance can oxidise Fe²⁺ if its own reduction potential is more positive than +0.77 V, because then E°cell = E°(oxidising agent) − 0.77 V is positive.

Illustration. With acidified dichromate: 6Fe²⁺ + Cr₂O₇²⁻ + 14H⁺ → 6Fe³⁺ + 2Cr³⁺ + 7H₂O. With permanganate (used in redox titration of iron): 5Fe²⁺ + MnO₄⁻ + 8H⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O.

Extra depth. Fe³⁺/Fe²⁺ has E° = +0.77 V, so only couples above it in Table 2.1 can oxidise Fe²⁺: F₂ (2.87), Co³⁺ (1.81), H₂O₂ (1.78), MnO₄⁻ (1.51), Au³⁺ (1.40), Cl₂ (1.36), Cr₂O₇²⁻ (1.33) and O₂ (1.23) are all suitable. The conditions matter: MnO₄⁻ and Cr₂O₇²⁻ need an acidic medium (H⁺ appears on the left of their half-reactions), which is why permanganate titrations of Fe²⁺ are carried out in dilute H₂SO₄.

Conclusion: Any three of Cl₂, Br₂, Cr₂O₇²⁻ (acidic), MnO₄⁻ (acidic), O₂ (acidic), H₂O₂ (acidic) or F₂ can oxidise Fe²⁺, since each has E° greater than +0.77 V.

Half-reaction. H⁺(aq) + e⁻ → ½H₂(g), with E°(H⁺/H₂) = 0.00 V, n = 1, p(H₂) = 1 bar.

Nernst equation. E(H⁺/H₂) = E° − (0.059/n) log {p(H₂)^½ ÷ [H⁺]} = 0 − 0.059 log (1/[H⁺]) = 0.059 log [H⁺].

Since pH = −log [H⁺]: E = −0.059 × pH.

pH = 10 ⇒ [H⁺] = 10⁻¹⁰ M; E = 0.059 × log 10⁻¹⁰ = −0.059 × 10 = −0.59 V

The more alkaline the solution, the more negative the hydrogen electrode potential (the electrode becomes a better reducing agent).

Extra depth. The result E = −0.059 pH makes the hydrogen electrode a pH sensor: each unit rise in pH lowers the potential by 59 mV at 298 K. This is the principle behind the glass electrode of the pH meter and shows how Nernst’s equation links electrochemistry with acid–base chemistry.

Conclusion: The potential of the hydrogen electrode at pH 10 is −0.59 V.

Cell. Ni(s)|Ni²⁺(0.160 M)||Ag⁺(0.002 M)|Ag(s); n = 2 electrons are transferred (Ni → Ni²⁺ + 2e⁻; 2Ag⁺ + 2e⁻ → 2Ag).

Nernst equation (298 K). E(cell) = E°cell − (0.059/n) log Q, where Q = [Ni²⁺]/[Ag⁺]² (solids omitted).

E(cell) = 1.05 − (0.059/2) log (0.160 ÷ (0.002)²) = 1.05 − 0.0295 × log (4 × 10⁴)

log (4 × 10⁴) = 4.602 ⇒ E(cell) = 1.05 − 0.0295 × 4.602 = 1.05 − 0.1358 = 0.914 V ≈ 0.91 V

The emf is lower than E° because the product concentration is high and the reactant Ag⁺ concentration is low (Q > 1).

Extra depth. The cell potential falls from 1.05 V to 0.91 V because Q = [Ni²⁺]/[Ag⁺]² is very large (4 × 10⁴); the cell is “partly discharged” by concentration effects alone. When Q becomes equal to K the potential becomes zero — the cell is dead — which links the Nernst equation to the equilibrium constant.

Conclusion: The emf of the cell is 0.91 V.

Step 1 — n. Two electrons are transferred: 2I⁻ → I₂ + 2e⁻ and 2Fe³⁺ + 2e⁻ → 2Fe²⁺; so n = 2.

Step 2 — standard Gibbs energy. Δ_rG° = −nFE°cell

Δ_rG° = −2 × 96487 C mol⁻¹ × 0.236 V = −45541 J mol⁻¹ = −45.54 kJ mol⁻¹

Step 3 — equilibrium constant. Δ_rG° = −2.303 RT log K_c

log K_c = 45541 ÷ (2.303 × 8.314 × 298) = 45541 ÷ 5705.8 = 7.98 ⇒ K_c ≈ 9.62 × 10⁷

(Using E°cell = (0.059/n) log K_c: log K_c = 2 × 0.236 ÷ 0.059 = 8.0, i.e. K_c ≈ 10⁸ — the same within rounding.) A large K_c means the reaction goes almost to completion.

Extra depth. A positive E°cell (0.236 V) gives a negative Δ_rG° and a large K, but the numbers show that the equilibrium lies on the product side only moderately (K ≈ 10⁸); Fe³⁺ oxidising I⁻ is the basis of the well-known iodometric analysis of iron(III). Note that Δ_rG° depends on the number of electrons written in the equation, whereas E° does not.

Conclusion: Δ_rG° = −45.54 kJ mol⁻¹; K_c = 9.62 × 10⁷.

Reason. Conductivity κ is the conductance of a unit volume of solution (1 cm³ or 1 m³) held between two electrodes of unit area at unit distance; it depends on the number of ions present per unit volume that carry the current.

When the solution is diluted, the concentration falls and the number of ions per unit volume decreases (the same quantity of solute is spread over a larger volume). Fewer ions carry the current across the unit-volume column, so κ decreases. This is true for both strong and weak electrolytes.

Contrast. Molar conductivity Λ_m = κ/c, however, increases on dilution, because the volume V containing one mole of electrolyte increases and this more than compensates for the decrease in κ (for weak electrolytes the degree of dissociation also increases).

Extra depth. Conductivity is an intensive property per unit volume, so it must decrease as the same number of ions are spread through a larger volume; this is true even for strong electrolytes that are fully dissociated. The more useful quantity for comparing electrolytes is the molar conductivity Λ_m = κ/c, which accounts for dilution and increases with dilution.

Conclusion: Conductivity decreases with dilution because the number of current-carrying ions per unit volume of the solution decreases.

Concept. Water is a very weak electrolyte (H₂O ⇌ H⁺ + OH⁻), so its limiting molar conductivity cannot be found by extrapolating Λ_m to zero concentration; Kohlrausch’s law of independent migration of ions must be used.

Λ°m(H₂O) = λ°(H⁺) + λ°(OH⁻)

Equivalently, the individual ion values can be obtained from strong electrolytes: Λ°m(H₂O) = Λ°m(HCl) + Λ°m(NaOH) − Λ°m(NaCl), because HCl + NaOH − NaCl = (H⁺ + Cl⁻) + (Na⁺ + OH⁻) − (Na⁺ + Cl⁻) = H⁺ + OH⁻. Each of these strong electrolytes can be measured accurately at several concentrations and extrapolated to c → 0 using Λ_m = Λ°m − Ac^½.

Λ°m(H₂O) = 349.6 + 199.1 = 548.7 S cm² mol⁻¹ (Table 2.4 values)

Check with strong electrolytes: Λ°m(HCl) = 425.9; Λ°m(NaOH) = 50.1 + 199.1 = 249.2; Λ°m(NaCl) = 126.4; 425.9 + 249.2 − 126.4 = 548.7 S cm² mol⁻¹ ✔.

Extra depth. Pure water has a very low conductivity (3.5 × 10⁻⁵ S m⁻¹ in Table 2.2) because only about 10⁻⁷ M H⁺ and OH⁻ are present, so a direct measurement is dominated by trace impurities; this is why Kohlrausch’s indirect method is used. The value Λ°m = 548.7 S cm² mol⁻¹ shows how highly mobile H⁺ (349.6) and OH⁻ (199.1) are, thanks to the Grotthuss (proton-hopping) mechanism.

Conclusion: Λ°m(H₂O) = Λ°m(HCl) + Λ°m(NaOH) − Λ°m(NaCl) = 548.7 S cm² mol⁻¹.

Step 1 — limiting molar conductivity (Kohlrausch).

Λ°m(HCOOH) = λ°(H⁺) + λ°(HCOO⁻) = 349.6 + 54.6 = 404.2 S cm² mol⁻¹

Step 2 — degree of dissociation. α = Λ_m/Λ°m

α = 46.1 ÷ 404.2 = 0.114

Step 3 — dissociation constant for HCOOH ⇌ H⁺ + HCOO⁻: K_a = cα²/(1 − α)

K_a = 0.025 × (0.114)² ÷ (1 − 0.114) = 3.249 × 10⁻⁴ ÷ 0.886 = 3.67 × 10⁻⁴ mol L⁻¹

Extra depth. Methanoic (formic) acid is a weak electrolyte, so its Λ_m (46.1) is far below Λ° (404.2); the degree of dissociation α = 0.114 and K_a = 3.67 × 10⁻⁴ mol L⁻¹ are of the same order as the accepted thermodynamic K_a of HCOOH (about 1.8 × 10⁻⁴ at 298 K); the NCERT data give a value roughly twice as large because the figures supplied are approximate. Ostwald’s dilution law, K_a = cα²/(1 − α), is derived from the same idea.

Conclusion: α = 0.114 (11.4 % dissociated); K_a = 3.67 × 10⁻⁴ mol L⁻¹.

Step 1 — charge. Q = I × t = 0.5 A × (2 × 3600 s) = 0.5 × 7200 = 3600 C.

Step 2 — number of electrons. Charge of one electron = 1.6021 × 10⁻¹⁹ C (equivalently 1 mol of electrons = 96487 C).

N = Q ÷ e = 3600 ÷ 1.6021 × 10⁻¹⁹ = 2.25 × 10²² electrons

Check via the faraday: 3600 ÷ 96487 = 0.0373 mol of electrons; × 6.022 × 10²³ = 2.25 × 10²².

Extra depth. The charge Q = It = 3600 C corresponds to 0.0373 mol of electrons, which is a tiny fraction of the Faraday (96487 C); the large number of electrons (2.25 × 10²²) shows how enormous the Avogadro constant is. The calculation is used for coulometers and for determining the Avogadro number experimentally from electrolysis.

Conclusion: About 2.25 × 10²² electrons pass through the wire.

Concept. Highly electropositive metals (large negative E°) cannot be obtained from their compounds by chemical reducing agents, so they are extracted by electrolytic reduction of their fused (molten) salts or oxides.

Electrolysis is also used for the refining of copper (impure Cu as anode, pure Cu deposited on cathode, in CuSO₄ solution).

Extra depth. Metals that are more electropositive than hydrogen and cannot be reduced chemically are made electrolytically — K, Na, Ca, Mg, Al — and so are copper (refining), zinc and chlorine (from brine). Aluminium needs a lot of electrical energy (about 15 kWh per kg) because Al₂O₃ must be dissolved in molten cryolite at about 1230 K to lower its melting point and increase conductivity.

Conclusion: Na, K, Li, Mg, Ca and Al are extracted electrolytically (and Cu is purified electrolytically).

Concept. From the balanced half-reaction, reduction of 1 mol of Cr₂O₇²⁻ needs 6 mol of electrons, i.e. 6 faradays.

Q = n F = 6 × 96487 C = 578922 C ≈ 5.79 × 10⁵ C

(Using F ≈ 96500 C mol⁻¹ gives 5.79 × 10⁵ C as well.) Here Cr changes from +6 to +3 in each of two Cr atoms: 2 × 3 = 6 electrons.

Extra depth. The ratio Q/F = 6 shows that the reduction of dichromate is a six-electron process; for titrations this means that 1 mol of Cr₂O₇²⁻ is equivalent to 6 equivalents of reductant (e.g. 6 mol Fe²⁺). The equivalent mass of K₂Cr₂O₇ is therefore M/6 = 294/6 = 49 g.

Conclusion: 5.79 × 10⁵ C (6 F) is needed.

Discharge (battery in use). Anode: Pb(s) + SO₄²⁻(aq) → PbSO₄(s) + 2e⁻. Cathode: PbO₂(s) + SO₄²⁻(aq) + 4H⁺(aq) + 2e⁻ → PbSO₄(s) + 2H₂O(l). Overall: Pb(s) + PbO₂(s) + 2H₂SO₄(aq) → 2PbSO₄(s) + 2H₂O(l). After discharge both plates are coated with insoluble PbSO₄ and the H₂SO₄ concentration (initially 38 %) has fallen.

Recharging. A direct current source with a voltage greater than the battery’s emf (≈ 2 V per cell) is connected in the opposite direction: the positive terminal of the charger to the PbO₂ (cathode during discharge) plate and the negative terminal to the Pb plate. The cell now acts as an electrolytic cell and the discharge reaction is reversed:

Materials involved: PbSO₄ (reactant on both plates), water (consumed), Pb (regenerated at the negative plate), PbO₂ (regenerated at the positive plate), and H₂SO₄ (regenerated, so the density of the acid rises back to the full-charge value ≈ 1.28 g mL⁻¹ and acts as a measure of the state of charge), driven by the external DC source.

Extra depth. The lead–acid battery runs at about 2.0 V per cell and the six cells of a car battery give 12 V; as the battery discharges, the sulphuric acid is consumed and its density falls (from about 1.28 to 1.10 g mL⁻¹), which is why a hydrometer indicates the state of charge. Overcharging causes electrolysis of water and loss of electrolyte.

Conclusion: Recharging reverses the discharge reaction: 2PbSO₄ + 2H₂O → Pb + PbO₂ + 2H₂SO₄.

Concept. A fuel cell converts the energy of combustion of a fuel directly into electricity; any fuel that can be oxidised at the anode with the supply of an oxidant (O₂) at the cathode can be used.

• Methane (CH₄): CH₄ + 2O₂ → CO₂ + 2H₂O, anode reaction (acidic medium) CH₄ + 2H₂O → CO₂ + 8H⁺ + 8e⁻.
• Methanol (CH₃OH): CH₃OH + 3/2 O₂ → CO₂ + 2H₂O (direct methanol fuel cell), anode CH₃OH + H₂O → CO₂ + 6H⁺ + 6e⁻.

Other possibilities are ethanol, natural gas, hydrazine (N₂H₄) and carbon monoxide. Such cells have efficiencies of around 70 % compared with ≈ 40 % for thermal power plants and give pollution-free power (the product is mainly water and CO₂).

Extra depth. Fuel cells are efficient because they avoid the Carnot limit of heat engines; the efficiency of H₂–O₂ cells is about 70 % compared with about 40 % in thermal power. Methanol is easier to store and transport than hydrogen, which is why direct-methanol fuel cells are being developed for portable electronics.

Conclusion: Methane and methanol are two fuels (besides hydrogen) used in fuel cells.

Concept. Rusting needs water and air (O₂, with dissolved CO₂). On the surface of an iron object, impurities and differences in stress/oxygen concentration create tiny anodic and cathodic regions, and the water film acts as the electrolyte — a small galvanic cell is set up (NCERT Fig. 2.13).

The positive E°cell (1.67 V) shows the process is spontaneous. H⁺ is believed to come from H₂CO₃ formed when CO₂ from air dissolves in water (and from other acidic oxides). The Fe²⁺ ions are further oxidised by atmospheric oxygen to Fe³⁺, which separate as hydrated ferric oxide, rust: 2Fe²⁺(aq) + 2H₂O(l) + ½O₂(g) → Fe₂O₃(s) + 4H⁺(aq), Fe₂O₃·xH₂O. The H⁺ ions regenerated keep the process going.

Extra depth. Rusting is faster in saline or acidic water because ions improve the conductivity of the electrolyte (coastal corrosion), and it can be prevented by galvanising, tin plating, painting, using sacrificial Mg or Zn electrodes (cathodic protection of pipelines and ship hulls) or forming alloys such as stainless steel. In the overall reaction, the Fe²⁺ formed is oxidised by O₂ to Fe³⁺, and hydrated ferric oxide Fe₂O₃·xH₂O (rust) precipitates.

Conclusion: Rusting is an electrochemical process in which iron acts as the anode (Fe → Fe²⁺ + 2e⁻) and O₂ is reduced at the cathode; E°cell = 1.67 V.

Concept. A metal displaces another metal from the solution of its salt if it lies above it in the electrochemical series, i.e. it has a more negative standard reduction potential and so is a stronger reducing agent. Standard electrode potentials (Table 2.1): Mg²⁺/Mg = −2.36 V; Al³⁺/Al = −1.66 V; Zn²⁺/Zn = −0.76 V; Fe²⁺/Fe = −0.44 V; Cu²⁺/Cu = +0.34 V.

Mg > Al > Zn > Fe > Cu

Meaning of the order. Each metal displaces all the metals to its right from their salt solutions: Mg displaces Al, Zn, Fe and Cu; Al displaces Zn, Fe and Cu; Zn displaces Fe and Cu; Fe displaces only Cu; Cu displaces none of these. Example: Fe(s) + Cu²⁺(aq) → Fe²⁺(aq) + Cu(s), E°cell = 0.34 − (−0.44) = +0.78 V (spontaneous), whereas Cu(s) + Fe²⁺(aq) → no reaction (E°cell = −0.78 V).

Extra depth. This series is the electrochemical (activity) series of metals. The practical use of the order is for predicting which metals can be used to displace others in metallurgy (thermite-type reductions: Al reduces Fe₂O₃) and for choosing sacrificial metals for corrosion protection.

Conclusion: Order in which the metals displace each other (strongest first): Mg, Al, Zn, Fe, Cu.

Concept. The lower (more negative) the standard reduction potential, the greater the tendency of the metal to lose electrons, i.e. the stronger it is as a reducing agent.

Ag < Hg < Cr < Mg < K

Silver and mercury have positive potentials, so their ions are easily reduced and the metals hardly lose electrons (they are noble). Potassium has the most negative potential and is the most powerful reducing agent among the five.

Extra depth. Reducing power runs opposite to E°; K, with E° = −2.93 V, is the strongest reducing agent and silver the weakest, so silver and mercury occur native or are easily obtained from their compounds, while potassium requires electrolysis. The values given in the question are rounded (e.g. Mg²⁺/Mg is −2.37 here against −2.36 V in Table 2.1) but the order is unaffected.

Conclusion: Increasing order of reducing power: Ag < Hg < Cr < Mg < K.

Cell diagram (anode on the left, cathode on the right):

Zn(s) | Zn²⁺(aq) || Ag⁺(aq) | Ag(s)

(i) Negatively charged electrode. The zinc electrode is the anode where oxidation occurs and electrons are released, so it is negative with respect to the solution; the silver electrode (cathode) is positive.

(ii) Carriers of current. In the external circuit, electrons flow from Zn to Ag (and conventional current flows from Ag to Zn). Inside the cell, current is carried by the ions in the solutions and in the salt bridge: Zn²⁺, Ag⁺ and NO₃⁻/K⁺ ions (cations move towards the cathode, anions towards the anode).

(iii) Electrode reactions.

E°cell = E°(Ag⁺/Ag) − E°(Zn²⁺/Zn) = 0.80 − (−0.76) = +1.56 V. (The silver half-reaction is multiplied by 2 only to balance electrons; E° is an intensive property and does not change.)

Extra depth. The salt bridge (KCl or KNO₃ in agar) completes the circuit and prevents liquid-junction potential and the build-up of charge; its ions migrate, so the current is carried by ions inside the cell and by electrons in the external wire. The double line in the cell notation represents the salt bridge and a single line represents a phase boundary.

Conclusion: Cell: Zn|Zn²⁺||Ag⁺|Ag; Zn is negative; electrons flow in the wire and ions in the solution; Zn → Zn²⁺ + 2e⁻ at the anode and Ag⁺ + e⁻ → Ag at the cathode.

Formulae. E°cell = E°cathode − E°anode; Δ_rG° = −nFE°cell; log K_c = nE°cell/0.059 (298 K).

(i) Anode (oxidation): Cr → Cr³⁺ + 3e⁻ (E° = −0.74 V); cathode: Cd²⁺ + 2e⁻ → Cd (E° = −0.40 V; this value is from the standard tables — it is not listed in Table 2.1 of the textbook). Electrons balanced: 2Cr + 3Cd²⁺ → 2Cr³⁺ + 3Cd, n = 6.

E°cell = −0.40 − (−0.74) = +0.34 V

Δ_rG° = −6 × 96487 × 0.34 = −196833 J mol⁻¹ ≈ −196.8 kJ mol⁻¹

log K = 6 × 0.34 ÷ 0.059 = 34.58 ⇒ K ≈ 3.8 × 10³⁴

(ii) Anode: Fe²⁺ → Fe³⁺ + e⁻ (E° of Fe³⁺/Fe²⁺ = +0.77 V); cathode: Ag⁺ + e⁻ → Ag (E° = +0.80 V); n = 1.

E°cell = 0.80 − 0.77 = +0.03 V

Δ_rG° = −1 × 96487 × 0.03 = −2894.6 J mol⁻¹ ≈ −2.89 kJ mol⁻¹

log K = 0.03 ÷ 0.059 = 0.508 ⇒ K ≈ 3.2

Extra depth. Part (i) uses E° of Cd²⁺/Cd = −0.40 V, a value from standard tables that is not printed in this edition’s Table 2.1; this is stated clearly so that no value is assumed silently. Notice that Δ_rG° of (i) (−196.8 kJ mol⁻¹) is huge compared with that of (ii) (−2.89 kJ mol⁻¹) because n = 6 and E° is 0.34 V against 0.03 V; K of (ii) is only 3.2, so this reaction does not go to completion.

Conclusion: (i) 0.34 V, −196.8 kJ mol⁻¹, K ≈ 3.8 × 10³⁴; (ii) 0.03 V, −2.89 kJ mol⁻¹, K ≈ 3.2.

General form. E(cell) = E°cell − (0.059/n) log Q at 298 K, where Q is the reaction quotient (solids, liquids and gases at 1 bar are taken as 1).

(i) Reaction: Mg + Cu²⁺ → Mg²⁺ + Cu, n = 2; E°cell = 0.34 − (−2.36) = 2.70 V.

E = 2.70 − (0.059/2) log ([Mg²⁺]/[Cu²⁺]) = 2.70 − 0.0295 log (0.001/0.0001) = 2.70 − 0.0295 = 2.67 V

(ii) Reaction: Fe + 2H⁺ → Fe²⁺ + H₂, n = 2; E°cell = 0 − (−0.44) = 0.44 V.

E = 0.44 − 0.0295 log ([Fe²⁺] ÷ [H⁺]²) = 0.44 − 0.0295 log (0.001/1) = 0.44 + 0.0885 = 0.53 V

(iii) Reaction: Sn + 2H⁺ → Sn²⁺ + H₂, n = 2; E°cell = 0 − (−0.14) = 0.14 V.

E = 0.14 − 0.0295 log (0.050 ÷ (0.020)²) = 0.14 − 0.0295 log 125 = 0.14 − 0.0295 × 2.097 = 0.14 − 0.062 = 0.078 V ≈ 0.08 V

(iv) Anode: 2Br⁻ → Br₂(l) + 2e⁻; cathode: 2H⁺ + 2e⁻ → H₂. Reaction: 2Br⁻ + 2H⁺ → Br₂(l) + H₂(g), n = 2; E°cell = 0 − 1.09 = −1.09 V (a non-spontaneous cell as written).

E = −1.09 − 0.0295 log (1 ÷ ([Br⁻]²[H⁺]²)) = −1.09 − 0.0295 log (1 ÷ (10⁻⁴ × 9 × 10⁻⁴)) = −1.09 − 0.0295 × 7.046 = −1.298 V ≈ −1.30 V

Extra depth. The Nernst equation applies to each cell as written; a negative E (cell iv) means that the reaction as written is non-spontaneous and the reverse would occur. The number of electrons n must be the number in the balanced equation (2 here for every cell), and pure solids, liquids (Br₂) and gases at 1 bar do not appear in Q.

Conclusion: (i) 2.67 V; (ii) 0.53 V; (iii) 0.08 V; (iv) −1.30 V.

Half-reactions. Anode: Zn → Zn²⁺ + 2e⁻ (E° = −0.76 V). Cathode: Ag₂O + H₂O + 2e⁻ → 2Ag + 2OH⁻ (E° = +0.344 V — a standard tabulated value, not listed in Table 2.1 of the textbook). n = 2.

E° = 0.344 − (−0.76) = 1.104 V

Δ_rG° = −nFE° = −2 × 96487 × 1.104 = −213046 J mol⁻¹ ≈ −213 kJ mol⁻¹

The large negative Δ_rG° shows the cell reaction is strongly spontaneous. Zn is the reducing agent and Ag₂O the oxidising agent; the cell has a nearly constant potential because the net reaction involves no species whose concentration changes appreciably in the electrolyte (KOH).

Extra depth. The silver oxide (button) cell gives a constant voltage of about 1.5 V and is used in watches and hearing aids because the overall reaction does not change any ion concentration appreciably. The use of E°(Ag₂O/Ag) = 0.344 V is flagged because it is not given in Table 2.1.

Conclusion: E° = 1.104 V; Δ_rG° = −213 kJ mol⁻¹.

Conductivity (κ). The reciprocal of resistivity (ρ): κ = 1/ρ. It is the conductance of a solution contained in a cube of 1 m (or 1 cm) edge length, i.e. between two electrodes of unit area of cross-section kept at unit distance. Since R = ρl/A, G = 1/R = κA/l. SI unit: S m⁻¹ (1 S cm⁻¹ = 100 S m⁻¹); κ = G × (l/A) = G × cell constant.

Molar conductivity (Λ_m). The conductivity of the solution divided by the molar concentration: Λ_m = κ/c. It is the conductance of the volume V of solution containing one mole of electrolyte, placed between two electrodes unit distance apart and with area of cross-section large enough to hold that volume: Λ_m = κV. Units: S m² mol⁻¹ (1 S m² mol⁻¹ = 10⁴ S cm² mol⁻¹); Λ_m (S cm² mol⁻¹) = κ (S cm⁻¹) × 1000 ÷ molarity (mol L⁻¹).

Variation with concentration.

• Conductivity decreases with decrease in concentration (dilution) for both strong and weak electrolytes, because the number of ions per unit volume that carry the current decreases.
• Molar conductivity increases with decrease in concentration, because the volume V containing one mole of electrolyte increases and this more than compensates for the fall in κ (Λ_m = κV).
• Strong electrolytes (e.g. KCl, HCl): Λ_m increases slowly on dilution and obeys Λ_m = Λ°m − A c^½ — a plot of Λ_m against c^½ is a straight line with intercept Λ°m (limiting molar conductivity, the value when c → 0) and slope −A. The increase arises from the decrease in inter-ionic attractions as ions get farther apart, so ions move faster.
• Weak electrolytes (e.g. CH₃COOH): Λ_m is low at high concentration (small degree of dissociation) and rises very steeply at low concentration because the degree of dissociation α increases towards 1 and the number of ions in the total volume containing 1 mol rises sharply. Λ°m cannot be got by extrapolation; it is found from Kohlrausch’s law.

Extra depth. The unit of κ is S m⁻¹ (or S cm⁻¹) whereas Λ_m has S m² mol⁻¹ (or S cm² mol⁻¹), the latter dimension showing that molar conductivity is “conductance per mole”. For strong electrolytes the Debye–Hückel–Onsager equation Λ = Λ° − A√c is a limiting law valid at low concentration, with A depending on the charges on the ions (1-1, 2-1, 2-2 electrolytes).

Conclusion: κ decreases and Λ_m increases on dilution; Λ_m = Λ°m − A√c for strong electrolytes while weak electrolytes show a steep rise at high dilution.

Formula. Λ_m = κ × 1000 cm³ L⁻¹ ÷ molarity (mol L⁻¹) (κ in S cm⁻¹ gives Λ_m in S cm² mol⁻¹).

Λ_m = (0.0248 S cm⁻¹ × 1000 cm³ L⁻¹) ÷ 0.20 mol L⁻¹ = 124 S cm² mol⁻¹

In SI units: κ = 2.48 S m⁻¹, c = 200 mol m⁻³ ⇒ Λ_m = 2.48 ÷ 200 = 1.24 × 10⁻² S m² mol⁻¹ = 124 × 10⁻⁴ S m² mol⁻¹ (same value).

Extra depth. The molar conductivity of KCl at 0.20 M is 124 S cm² mol⁻¹, slightly less than the limiting value 150.0 (Example 2.6), consistent with the trend Λ_m = Λ° − A√c. Using SI units (S m² mol⁻¹) and non-SI units (S cm² mol⁻¹) is permitted, the conversion factor being 10⁴.

Conclusion: Λ_m = 124 S cm² mol⁻¹.

Concept. Cell constant G* = l/A = R × κ (the conductivity of the standard solution is known, so the cell constant is obtained from the measured resistance).

G* = 1500 Ω × 0.146 × 10⁻³ S cm⁻¹ = 0.219 cm⁻¹

In SI units G* = 21.9 m⁻¹. Once G* is known, the conductivity of any other solution in the same cell is κ = G*/R.

Extra depth. The cell constant is a property of the cell (distance between the electrodes ÷ their area) that is calibrated with a standard KCl solution whose conductivity is known accurately; once calibrated the same cell gives κ = G*/R for any other solution. AC (about 550–5000 Hz) is used to avoid electrolysis and polarisation of the electrodes.

Conclusion: Cell constant = 0.219 cm⁻¹.

Method. Convert c to mol m⁻³ (c × 1000) and use Λ_m = κ/c (S m² mol⁻¹); multiply by 10⁴ to express in S cm² mol⁻¹. Also compute √c in (mol L⁻¹)^½.

Plot. Λ_m (y-axis) against c^½ (x-axis) gives a nearly straight line (strong electrolyte, Λ_m = Λ°m − A c^½) with negative slope; the points and the best-fit line are shown below.

Λ°m. The least-squares line has slope ≈ −59 S cm² mol⁻¹ per (mol L⁻¹)^½ (so A ≈ 59) and intercept at c^½ = 0:

Λ°m(NaCl) ≈ 124.7 ≈ 125 S cm² mol⁻¹

Cross-check from Kohlrausch’s law with Table 2.4: Λ°m(NaCl) = λ°(Na⁺) + λ°(Cl⁻) = 50.1 + 76.3 = 126.4 S cm² mol⁻¹, in good agreement (graphical extrapolation from five points has an uncertainty of 1–2 units).

Extra depth. The linear plot shows the Debye–Hückel–Onsager behaviour of a strong electrolyte; extrapolating to zero concentration is acceptable only for strong electrolytes — for weak electrolytes the curve is too steep. The intercept (≈ 125) is close to the sum of the ionic conductivities (50.1 + 76.3 = 126.4), a direct verification of Kohlrausch’s law.

Conclusion: Λ_m falls from 123.7 to 106.7 S cm² mol⁻¹ as c increases; extrapolation gives Λ°m ≈ 125 S cm² mol⁻¹ (accepted value 126.4).

Step 1 — molar conductivity.

Λ_m = κ × 1000 ÷ c = (7.896 × 10⁻⁵ × 1000) ÷ 0.00241 = 32.76 S cm² mol⁻¹

Step 2 — degree of dissociation. α = Λ_m/Λ°m = 32.76 ÷ 390.5 = 0.0839.

Step 3 — dissociation constant. CH₃COOH ⇌ CH₃COO⁻ + H⁺; K_a = cα²/(1 − α).

K_a = 0.00241 × (0.0839)² ÷ (1 − 0.0839) = 1.696 × 10⁻⁵ ÷ 0.9161 = 1.85 × 10⁻⁵ mol L⁻¹

This agrees with the accepted K_a of acetic acid (1.8 × 10⁻⁵) and with NCERT Example 2.9 (1.78 × 10⁻⁵ at slightly different concentration).

Extra depth. Compare the dissociation constant of acetic acid (1.85 × 10⁻⁵) with that for formic acid (3.67 × 10⁻⁴ in intext 2.9): formic acid is about 20 times stronger. The method — measuring conductivity to find α — is Ostwald’s method, and it works because only the ions carry current.

Conclusion: Λ_m = 32.76 S cm² mol⁻¹; α = 0.0839; K_a = 1.85 × 10⁻⁵ mol L⁻¹.

Concept. Charge = (number of moles of electrons) × F, with F = 96487 C mol⁻¹. Write the half-reaction to find the electrons per mole.

In (iii) manganese changes oxidation state from +7 to +2 (gain of 5 electrons).

Extra depth. These calculations are Faraday’s second law in disguise: the charge needed is proportional to the number of electrons transferred per mole of substance. With 1 F = 96487 C mol⁻¹ the three values (3F, 2F, 5F) show that charge needed is proportional to the change in oxidation number: Al +3 → 0, Cu +2 → 0, Mn +7 → +2.

Conclusion: (i) 3F = 2.89 × 10⁵ C; (ii) 2F = 1.93 × 10⁵ C; (iii) 5F = 4.82 × 10⁵ C.

(i) Ca²⁺ + 2e⁻ → Ca. Moles of Ca = 20.0 ÷ 40 g mol⁻¹ = 0.5 mol. Electricity required = 0.5 mol × 2 F/mol = 1 F.

(ii) Al³⁺ + 3e⁻ → Al. Moles of Al = 40.0 ÷ 27 g mol⁻¹ = 1.481 mol. Electricity = 1.481 × 3 F = 4.44 F (≈ 4.28 × 10⁵ C).

(i) 1 F; (ii) 4.44 F

Extra depth. The same stoichiometric approach is used in industry: 1 kg of Al is 37 mol and needs 3 F per mole, i.e. about 111 F ≈ 1.07 × 10⁷ C (≈ 3 kWh of electrical work at 1 V; commercial cells operate at about 4–5 V, which is why roughly 13–15 kWh is consumed per kg). Faraday’s laws (1833–34) connect the mass deposited with the quantity of electricity.

Conclusion: 1 faraday is needed for 20.0 g Ca and about 4.44 faradays for 40.0 g Al.

(i) Oxidation of water: H₂O → ½O₂ + 2H⁺ + 2e⁻. One mole of H₂O releases 2 mol of electrons.

Q = 2F = 2 × 96487 = 192974 C ≈ 1.93 × 10⁵ C

(ii) Balanced per mole of FeO: FeO + ½O²⁻ → ½Fe₂O₃ + e⁻, i.e. the only change is Fe²⁺ → Fe³⁺ + e⁻ (overall 2FeO + O²⁻ → Fe₂O₃ + 2e⁻). One mole of FeO (one Fe²⁺) releases 1 mol of electrons.

Q = 1F = 96487 C ≈ 9.65 × 10⁴ C

Extra depth. The electron count is determined by the change in oxidation state: O⁻² → O⁰ (2e⁻ per O atom) in water, and Fe²⁺ → Fe³⁺ (1e⁻) in FeO → Fe₂O₃. Writing the half-reaction for 1 mole of the named species avoids mistakes with balancing.

Conclusion: (i) 1.93 × 10⁵ C; (ii) 9.65 × 10⁴ C.

Step 1 — charge. Q = I t = 5 A × (20 × 60 s) = 6000 C.

Step 2 — electrode reaction. Ni²⁺ + 2e⁻ → Ni(s): 2F (2 × 96487 C) deposit 1 mol Ni (58.7 g).

m = (58.7 g mol⁻¹ × 6000 C) ÷ (2 × 96487 C mol⁻¹) = 352200 ÷ 192974 = 1.825 g

The platinum electrodes are inert and only supply or accept electrons. The question assumes that the whole current (100 % current efficiency) is used to reduce Ni²⁺ (E° = −0.25 V) at the cathode.

Extra depth. The electrolysis of Ni(NO₃)₂ at platinum electrodes in practice deposits nickel only at a suitable potential and pH; here the problem is a stoichiometry exercise (current efficiency 100 %). Electroplating of nickel uses similar conditions, and the mass deposited is directly proportional to Q.

Conclusion: 1.825 g of nickel is deposited.

Concept. In a series circuit the same charge passes through every cell, so the amounts deposited are in the ratio of their equivalent weights (Faraday’s second law).

Step 1 — charge from silver. Ag⁺ + e⁻ → Ag; 96487 C deposit 108 g (1 mol). Q = (1.45 ÷ 108) × 96487 = 0.013426 × 96487 = 1295.4 C.

Step 2 — time.

t = Q ÷ I = 1295.4 C ÷ 1.5 A = 863.6 s ≈ 864 s (≈ 14.4 min)

Step 3 — copper. Cu²⁺ + 2e⁻ → Cu; 2F deposit 63.5 g, so m(Cu) = (63.5 × 1295.4) ÷ (2 × 96487) = 0.426 g.

Step 4 — zinc. Zn²⁺ + 2e⁻ → Zn; 2F deposit 65.4 g, so m(Zn) = (65.4 × 1295.4) ÷ (2 × 96487) = 0.439 g.

Extra depth. The result shows Faraday’s second law: equal charge deposits masses proportional to equivalent weights — Ag (108/1), Cu (63.5/2) and Zn (65.4/2) — in the ratio 108 : 31.75 : 32.7. Using the exact ratio with 1.45 g Ag gives 0.426 g Cu and 0.439 g Zn.

Conclusion: Time = ≈ 864 s; mass of Cu = 0.426 g and of Zn = 0.439 g.

Criterion. A redox reaction is feasible (spontaneous) if E°cell = E°cathode − E°anode is positive (Δ_rG° negative). Write the oxidising agent as the cathode (reduction) and the reducing agent as the anode (oxidation).

Extra depth. Feasibility from E° tells only the thermodynamic spontaneity; the reaction may still be slow (kinetics), as with the oxidation of Fe²⁺ by dilute acid H⁺. E°cell > 0.2 V is usually a reliable indication of a practical reaction. The reverse of each non-feasible reaction is feasible — e.g. Fe³⁺ will not oxidise Br⁻, but Br₂ will oxidise Fe²⁺.

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

Rules. At the cathode the species with the higher reduction potential is reduced; at the anode the species whose couple has the lower reduction potential (the one that is easiest to oxidise) is oxidised, subject to overpotential effects; a reactive (active) electrode itself may be oxidised. Relevant values: Ag⁺/Ag +0.80 V; Cu²⁺/Cu +0.34 V; H⁺/H₂ 0.00 V (H₂O/H₂, OH⁻ −0.83 V); O₂, H⁺/H₂O +1.23 V; Cl₂/Cl⁻ +1.36 V.

(i) AgNO₃(aq), Ag electrodes. Species: Ag⁺, NO₃⁻, H₂O. Cathode: Ag⁺ + e⁻ → Ag (+0.80 V, higher than water) — silver is deposited. Anode: the silver electrode itself is oxidised (Ag → Ag⁺ + e⁻) in preference to water or NO₃⁻. Result: silver dissolves from the anode and deposits on the cathode (silver plating / refining); the concentration of the solution is unchanged.

(ii) AgNO₃(aq), Pt electrodes. Cathode: Ag⁺ + e⁻ → Ag — silver is deposited. Anode: Pt is inert, so water is oxidised: 2H₂O → O₂ + 4H⁺ + 4e⁻ (NO₃⁻ is not oxidised). Result: Ag at the cathode, O₂ gas at the anode, and the solution becomes acidic (HNO₃ forms).

(iii) Dilute H₂SO₄, Pt electrodes. Cathode: 2H⁺ + 2e⁻ → H₂ (the only reducible species except water). Anode: 2H₂O → O₂ + 4H⁺ + 4e⁻ (1.23 V) is preferred over SO₄²⁻ → S₂O₈²⁻ + 2e⁻ (1.96 V) in dilute solution. Result: H₂ at the cathode and O₂ at the anode (volume ratio 2 : 1) — effectively electrolysis of water; the acid becomes more concentrated.

(iv) CuCl₂(aq), Pt electrodes. Cathode: Cu²⁺ + 2e⁻ → Cu (+0.34 V, more positive than H⁺/water) — copper is deposited. Anode: 2Cl⁻ → Cl₂ + 2e⁻ — chlorine gas is evolved (because of the overpotential of oxygen at the anode, Cl⁻ is oxidised in preference to water). Result: Cu at the cathode, Cl₂ at the anode.

Extra depth. The products of electrolysis depend on the electrode material and on the competition between species at each electrode, including overpotential: for aqueous NaCl the chlorine is obtained at the anode (not O₂) because of the overpotential of oxygen. In electrolysis of dilute H₂SO₄ the gases are in the ratio 2 : 1 (H₂ : O₂), confirming the composition of water.

Conclusion: (i) Ag deposits / Ag dissolves; (ii) Ag and O₂; (iii) H₂ and O₂; (iv) Cu and Cl₂.