Class 12 Chemistry Ch 2 Electrochemistry Exemplar - Download PDF

Correct option: (iii).

Concept used. A standard electrode potential E^∘ is the potential of an electrode measured against the Standard Hydrogen Electrode (SHE) under standard conditions: all gases at 1 bar, all solutes at 1 M activity, and temperature 298 K. Any departure (gas at 0.1 bar, H+ at 0.1 M, Cu²⁺ at 2 M) makes the measured cell emf non-standard and the Nernst correction must be applied. So the ``standard E^∘_Cu²⁺/Cu'' is recorded only when every ion/gas in the cell is at unit activity.

1. Recall the cell notation rule: anode (oxidation) on the left, cathode (reduction) on the right, salt bridge (||) in the middle. For measuring E^∘_Cu²⁺/Cu, SHE is the reference (left) and the copper half-cell is on the right.
2. Standard conditions demand: p_H2=1 bar (rules out (i)), [H+]=1 M (rules out (iv)), and [Cu²⁺]=1 M (rules out (ii) which has 2 M).
3. Only option (iii) keeps every species at standard activity. With SHE giving E^∘=0 V by definition, the voltmeter reads E^∘_cell = E^∘_cathode - E^∘_anode = E^∘_Cu²⁺/Cu - 0 = +0.34 V.

Option (iii): Pt(s) | H2(1 bar) | H+(1 M) || Cu²⁺(1 M) | Cu.

Correct option: (ii).

Concept used. The Nernst equation for a metal electrode Mⁿ⁺|M relates the electrode potential to the ion concentration: E = E^∘ - 0.059nlog1[Mⁿ⁺] = E^∘ + 0.059nlog[Mⁿ⁺]. This is a straight-line equation in log[Mⁿ⁺] of the form y = mx + c, where the slope m = +0.059/n is positive and the y-intercept c = E^∘_Mⁿ⁺|M.

1. Start from the given form: E = E^∘ - 0.0592log1[Mg²⁺]. Use the log identity log(1/x) = -log x: E = E^∘ + 0.0592log[Mg²⁺].
2. Identify slope and intercept: slope = +0.0592 = +0.0295 > 0, intercept = E^∘_Mg²⁺|Mg = -2.37 V (negative).
3. The graph is a straight line with positive slope (rising left to right) and a negative y-intercept (below the log[Mg²⁺] axis). Among the four sketches, only (ii) shows a straight line cutting the y-axis below the origin and rising.

Option (ii): straight line with positive slope and negative y-intercept.

Correct option: (iii).

Concept used. An intensive property does not depend on the amount of substance (examples: temperature, density, electrode potential, molar volume). An extensive property scales with the amount of substance (examples: mass, volume, energy, Gibbs free energy, enthalpy). The cell emf E_cell is a potential difference per unit charge, so it stays the same whether the cell delivers 1 mole or 100 moles of electrons. In contrast, _r G = -nFE_cell scales linearly with n, the moles of electrons transferred.

1. Test E_cell. Define it as the potential per unit charge: E = W_electrical/Q. Doubling the cell size doubles both W and Q, so E stays the same. E_cell is intensive.
2. Test _r G. By definition, _r G = -nFE_cell. Doubling n (i.e. doubling the moles of electrons in the balanced equation) doubles _r G. Hence _r G is extensive.
3. Combining: E_cell intensive, _r G extensive. Only option (iii) matches.

Option (iii): E_cell is intensive, _r G is extensive.

Correct option: (ii).

Concept used. The electromotive force (emf) of a cell is precisely the potential difference between the two terminals measured when no current is drawn (i.e. on open circuit, with an ideal high-resistance voltmeter or a potentiometer). The cell potential (or terminal voltage) is the potential difference under operating conditions and is less than the emf by the IR drop across the internal resistance: V_cell = _cell - Ir.

1. Define each term carefully. The emf ε is the maximum potential the cell can deliver, achieved only when I → 0 (no current). The cell potential V is what a real voltmeter reads while current flows, and includes the drop Ir across the internal resistance.
2. The phrase ``when no current is drawn'' is the defining feature of emf. So the answer is (ii) Cell emf.
3. ``Cell potential'', ``Potential difference'', and ``Cell voltage'' are loose synonyms for the operating terminal voltage and don't carry the ``zero-current'' restriction.

Option (ii): Cell emf.

Correct option: (iv).

Concept used. An inert electrode (typically platinum, graphite, or gold) is one that does not take part in the chemical reaction. Its job is purely to: (a) provide a conducting surface where electrons can be exchanged with the solution, and (b) host either an oxidation half-reaction or a reduction half-reaction, but not both on the same surface. A single electrode is always associated with one half-reaction (oxidation or reduction); a redox (full) reaction needs two electrodes.

1. Read each statement carefully.
   (i) ``Does not participate'' –- correct, this is the definition of inert.
   (ii) ``Provides surface for oxidation OR reduction'' –- correct, it hosts exactly one half-cell reaction.
   (iii) ``Provides surface for conduction of electrons'' –- correct, inert electrodes are good conductors.
   (iv) ``Provides surface for redox reaction'' –- incorrect, because a redox (both halves) reaction occurs in the whole cell across two electrodes, not at one inert electrode.
2. Hence statement (iv) is not correct.
3. Example: in Pt(s)|H2(g)|H+(aq), the Pt only hosts the H2 <=> 2H+ + 2e- half-reaction; the matching reduction (or oxidation) happens at the OTHER electrode.

Option (iv): it provides surface for redox reaction –- INCORRECT (an electrode hosts only one half-cell reaction).

Correct option: (iii).

Concept used. A galvanic cell converts chemical energy into electrical energy spontaneously; an electrolytic cell consumes external electrical energy to drive a non-spontaneous reaction. A galvanic cell can be forced to behave electrolytically by applying an external EMF E_ext greater than its own cell emf E_cell and in the opposite direction. The applied voltage then dominates, reverses the current, and pushes the cell reaction backwards (non-spontaneous direction).

1. Test each condition.
   (i) E_cell = 0: cell is at equilibrium; no current flows; not electrolytic, but ``dead''.
   (ii) E_cell > E_ext: the cell still drives current outward, behaves as a galvanic cell.
   (iii) E_ext > E_cell: external voltage dominates; the cell reaction is reversed; electrolytic behaviour.
   (iv) E_cell = E_ext: exact balance; no current flows in either direction (this is how a potentiometer measures emf).
2. Hence (iii) is the only condition causing electrolytic behaviour.

Option (iii): E_ext > E_cell.

Correct option: (iii).

Concept used. The conductivity κ of an electrolyte solution measures how easily ions move through it under an applied field. Three things slow the ions down: their size (bigger ions experience more drag), the solvent's viscosity, and the solvation sheath (a tightly bound shell of solvent molecules around each ion, which effectively enlarges the moving species). Temperature increases ionic mobility (less viscosity, more thermal energy), so κ rises with T.

1. Check (i). Larger ions move slower (more frictional drag), so κ depends on ionic size. Statement correct.
2. Check (ii). κ ∝ 1/η approximately (Walden's rule: Λ^∘η = const). So κ depends on viscosity. Statement correct.
3. Check (iii). Solvation increases the effective radius of an ion. For example, Li+ has a smaller crystal radius than Cs+ but larger hydrated radius (because Li+ is more polarising and binds more water molecules). Result: Cs+ moves faster through water. So conductivity does depend on solvation. Statement (iii) is INCORRECT.
4. Check (iv). At higher T, ion mobility rises and viscosity drops, so κ rises by ∼ 2% per ^∘C. Statement correct.

Option (iii) is incorrect –- conductivity does depend on ion solvation.

Correct option: (ii) Cr.

Concept used. The reducing strength of a species is its tendency to lose electrons (i.e. get oxidised). A species is a strong reducing agent when its oxidised form has a very negative reduction potential, because the species is then eager to give up electrons. Equivalently, the oxidation potential -E^∘_red must be large and positive. The most negative E^∘_red in the list belongs to the strongest reducing agent.

1. Pair each candidate with the reduction half-reaction in the table:
   Cl- comes from Cl2 + 2e- -> 2Cl-, E^∘ = +1.36 V.
   Cr comes from Cr³⁺ + 3e- -> Cr, E^∘ = -0.74 V.
   Cr³⁺ comes from Cr2O7^2-/Cr³⁺, E^∘ = +1.33 V.
   Mn²⁺ comes from MnO4-/Mn²⁺, E^∘ = +1.51 V.
2. For each, write the oxidation potential E^∘_ox = -E^∘_red:
   Cl-: -1.36 V;   Cr: +0.74 V;   Cr³⁺: -1.33 V;   Mn²⁺: -1.51 V.
3. The largest positive oxidation potential is for Cr (+0.74 V), so Cr is most ready to be oxidised Cr -> Cr³⁺ + 3e-, and is therefore the strongest reducing agent in the set.

Option (ii) Cr is the strongest reducing agent.

Correct option: (iii) MnO4-.

Concept used. The strongest oxidising agent has the most positive standard reduction potential, because a high positive E^∘ means the oxidised form is very eager to accept electrons (get reduced). So we identify the candidate with the largest E^∘_red.

1. Pair each candidate with its couple from the Q.8 data:
   Cl- is the reduced form of Cl2, so for oxidising power look at Cl2/Cl-: E^∘ = +1.36 V.
   Mn²⁺ is the reduced form of MnO4-, but Mn²⁺ itself as oxidising agent corresponds to Mn²⁺ + 2e- -> Mn, which is not given.
   MnO4- is the oxidising form, couple MnO4-/Mn²⁺ with E^∘ = +1.51 V.
   Cr³⁺ as oxidising form belongs to Cr2O7^2-/Cr³⁺ reduced form; as oxidiser look at Cr³⁺/Cr with E^∘ = -0.74 V (negative, poor oxidiser).
2. The four oxidising couples give +1.36, –- (not given), +1.51, and -0.74 V. The largest positive value is +1.51 V for MnO4-/Mn²⁺.
3. Therefore MnO4- is the strongest oxidising agent among the four.

Option (iii) MnO4-, largest E^∘_red = +1.51 V.

Correct option: (ii).

Concept used. To rank reducing power, list the oxidation potentials E^∘_ox = -E^∘_red of each species. The species with the LARGEST positive E^∘_ox is the strongest reducing agent; the SMALLEST (most negative) is the weakest. Order from weakest to strongest by listing E^∘_ox in increasing order.

1. Compute E^∘_ox for each species (using its relevant couple):
   Mn²⁺ → MnO4-: E^∘_ox = -(+1.51) = -1.51 V.
   Cl- → Cl2: E^∘_ox = -(+1.36) = -1.36 V.
   Cr³⁺ → Cr2O7^2-: E^∘_ox = -(+1.33) = -1.33 V.
   Cr → Cr³⁺: E^∘_ox = -(-0.74) = +0.74 V.
2. Order from least to most reducing power (least to greatest E^∘_ox): Mn²⁺ (-1.51) < Cl- (-1.36) < Cr³⁺ (-1.33) < Cr (+0.74).
3. This matches option (ii) exactly.

Option (ii): Mn²⁺ < Cl- < Cr³⁺ < Cr.

Correct option: (iv) Mn²⁺.

Concept used. The ``most stable reduced form'' is the species that is hardest to oxidise back. Equivalently, its oxidised partner has the highest reduction potential , because if the forward (reduction) reaction is highly favourable (E^∘_red very positive), then the reverse (oxidation) reaction is very unfavourable, leaving the reduced form locked in.

1. Identify the couple each candidate sits on the reduced side of:
   Cl-: Cl2/Cl-, E^∘_red = +1.36 V.
   Cr³⁺: as reduced form, Cr2O7^2-/Cr³⁺, E^∘_red = +1.33 V.
   Cr: Cr³⁺/Cr, E^∘_red = -0.74 V.
   Mn²⁺: MnO4-/Mn²⁺, E^∘_red = +1.51 V.
2. The HIGHEST E^∘_red is +1.51 V for Mn²⁺. This means MnO4- very strongly wants to become Mn²⁺, hence Mn²⁺ is the hardest to oxidise back, most stable in its reduced form.

Option (iv) Mn²⁺, most stable reduced form.

Correct option: (i) Cr³⁺.

Concept used. The ``most stable oxidised species'' is the oxidised form that is hardest to reduce further. Its E^∘_red as the oxidised partner of a deeper reduction must be the LOWEST (least positive or most negative), because a low E^∘_red means there is little driving force to take it further down.

1. Identify each candidate's role:
   Cr³⁺, sits as oxidised form of the couple Cr³⁺/Cr, E^∘_red = -0.74 V.
   MnO4-, oxidised form of MnO4-/Mn²⁺, E^∘_red = +1.51 V.
   Cr2O7^2-, oxidised form of Cr2O7^2-/Cr³⁺, E^∘_red = +1.33 V.
   Mn²⁺, sits as oxidised form of Mn²⁺/Mn (data NOT given; the Mn²⁺ here only appears as reduced form of MnO4-/Mn²⁺).
2. The LOWEST E^∘_red as oxidised partner is -0.74 V for Cr³⁺/Cr, with Cr³⁺ on the oxidised side. So Cr³⁺ has the smallest tendency to accept further electrons.
3. Therefore Cr³⁺ is the most stable oxidised species.

Option (i) Cr³⁺, lowest E^∘_red as oxidised partner = most stable.

Correct option: (iii) 3F.

Concept used. Faraday's first law of electrolysis: the mass deposited at an electrode is proportional to the charge passed, w = (M/nF) Q. Equivalently, to deposit 1 mole of a metal Mⁿ⁺, you need n moles of electrons = nF coulombs, where F = 96,500 C/mol is the Faraday constant (the charge on 1 mole of electrons).

1. Identify the oxidation state of aluminium in Al2O3. Oxygen is -2, the compound is neutral, so two Al atoms contribute +6 between them; each Al is Al³⁺.
2. Write the cathode half-reaction during electrolysis: Al³⁺ + 3e- -> Al. To produce 1 mole of Al requires 3 moles of electrons.
3. Convert to charge: 3 moles of electrons = 3 F = 3 × 96,500 = 289,500 C.

Option (iii): 3F = 289,500 C of charge is needed per mole of Al.

Correct option: (iv).

Concept used. The cell constant of a conductivity cell is a purely geometric quantity: G^* = lA, where l is the distance between the two parallel electrodes and A is the cross-sectional area of each electrode. Because l and A are fixed by the construction of the cell, G^* is independent of the electrolyte, its concentration, and the temperature. The only way to change G^* is to physically modify the cell (move the plates or change their size).

1. Recall the relation κ = G^*R, where κ is conductivity (S cm^-1), R is measured resistance (Ω), and G^* is cell constant (cm^-1).
2. Examine each option.
   (i) Change electrolyte: κ changes, R changes, but G^* = l/A stays the same, geometry unchanged. False.
   (ii) Change concentration: κ and R change, but l, A don't change. False.
   (iii) Temperature: κ and R change, l and A change only by tiny thermal expansion (negligible). False to good approximation.
   (iv) Remains constant, TRUE.
3. Hence (iv) is correct.

Option (iv): cell constant is a geometric property and stays constant.

Correct option: (i).

Concept used. A lead storage battery consists of a Pb electrode (anode) and a PbO2 electrode (cathode) dipped in ∼ 38% H2SO4. During discharge both electrodes are coated with PbSO4, and the overall reaction is Pb + PbO2 + 2H2SO4 -> 2PbSO4 + 2H2O. During charging, an external EMF reverses every step:

• At what was the anode during discharge (Pb side), PbSO4 is reduced back to Pb: PbSO4 + 2e- -> Pb + SO4^2-.
• At what was the cathode during discharge (PbO2 side), PbSO4 is oxidised back to PbO2: PbSO4 + 2H2O -> PbO2 + SO4^2- + 4H+ + 2e-.

1. During discharge, Pb plate (anode) becomes PbSO4 by oxidation, and PbO2 plate (cathode) becomes PbSO4 by reduction.
2. During charging, the directions reverse. On the Pb-plate side (still called ``anode'' by the question), PbSO4 is REDUCED back to Pb. This matches statement (i).
3. Statements (ii), (iii) refer to the cathode side and claim it forms Pb. Wrong, the cathode side reforms PbO2, not Pb.
4. Statement (iv) gets the cathode side reaction right (PbSO4 → PbO2) but mislabels it as ``anode''. Wrong.

Option (i): during charging, PbSO4 at the anode (Pb plate) is reduced back to Pb.

Correct option: (ii).

Concept used. Kohlrausch's law of independent migration of ions: at infinite dilution, the limiting molar conductivity Λ⁰_m of any electrolyte equals the sum of the limiting ionic conductivities of its constituent ions: Λ⁰_m(AB) = λ⁰₊(A) + λ⁰_-(B). This lets us compute Λ⁰_m of a weak electrolyte (like NH4OH) by combining Λ⁰_m values of strong electrolytes that share the same ions.

1. Write the ionic contributions for NH4OH: Λ⁰_m(NH4OH) = λ⁰_NH4+ + λ⁰_OH-.
2. Build λ⁰_NH4+ + λ⁰_OH- from strong electrolytes:
   Λ⁰_m(NH4Cl) = λ⁰_NH4+ + λ⁰_Cl-    (gives NH4+).
   Λ⁰_m(NaOH) = λ⁰_Na+ + λ⁰_OH-    (gives OH-).
   Λ⁰_m(NaCl) = λ⁰_Na+ + λ⁰_Cl-    (extra Na+ + Cl- to subtract).
3. Combine: Λ⁰_m(NH4Cl) + Λ⁰_m(NaOH) - Λ⁰_m(NaCl) = (λ⁰_NH4+ + λ⁰_Cl-) + (λ⁰_Na+ + λ⁰_OH-) - (λ⁰_Na+ + λ⁰_Cl-) = λ⁰_NH4+ + λ⁰_OH- = Λ⁰_m(NH4OH). Matches option (ii).

Option (ii): Λ⁰_m(NH4OH) = Λ⁰_m(NH4Cl) + Λ⁰_m(NaOH) - Λ⁰_m(NaCl).

Correct option: (ii).

Concept used. At the anode of an electrolytic cell, oxidation occurs. For aqueous NaCl, the two species competing to be oxidised are Cl- (giving Cl2, E^∘_ox = -1.36 V) and water (giving O2, E^∘_ox = -1.23 V). On E^∘ alone, water should be oxidised first because its oxidation potential (-1.23) is less negative than that of Cl- (-1.36). However, the kinetic phenomenon of overpotential for O2 evolution at electrodes (a substantial activation barrier) reduces the rate of water oxidation, so Cl- ends up being oxidised preferentially in practice. The question asks which half-reaction occurs (in the thermodynamic sense the NCERT answer expects), so the textbook answer keys to the water-oxidation half-reaction (ii) as the half-reaction that would occur at the anode purely on the basis of E^∘.

(The textbook answer is (ii). The reason Cl- wins in practice is the overpotential issue addressed in Q34.)

1. At the anode, oxidation happens. Reduction half-reactions (i) and (iii) immediately rule themselves out as anode candidates.
2. Between water oxidation (ii) and Cl- oxidation (iv): E^∘_ox(H2O) = -1.23 V; E^∘_ox(Cl-) = -1.36 V. Water oxidation is thermodynamically easier (less negative oxidation potential).
3. Hence on a purely thermodynamic basis, the half-reaction that ``should'' occur at the anode is (ii) 2H2O -> O2 + 4H+ + 4e-.

Option (ii) (textbook answer based on E^∘, ignoring kinetic overpotential).

Correct options: (ii), (iv).

Concept used. A positive standard reduction potential E^∘_Cu²⁺/Cu = +0.34 V means Cu²⁺ has a greater tendency to accept electrons than H+ does. So Cu²⁺ is a stronger oxidising agent than H+, which is equivalent to saying that Cu metal is a weaker reducing agent than H2. Consequently, Cu cannot reduce H+ to H2, i.e. Cu cannot displace hydrogen from acid.

1. Statement (i): ``Cu/Cu²⁺ stronger reducing agent than H2''. Wrong. Cu lies below H in the electrochemical series, so it's a weaker reducing agent.
2. Statement (ii): ``Cu²⁺ stronger oxidising agent than H+''. Correct, E^∘_Cu²⁺/Cu > E^∘_H+/H2 (i.e. 0.34 > 0).
3. Statement (iii): ``Cu can displace H2 from acid''. Wrong, since Cu cannot reduce H+.
4. Statement (iv): ``Cu cannot displace H2 from acid''. Correct, the experimental observation matches.

Correct options: (ii) and (iv).

Correct options: (i), (iii).

Concept used. In aqueous electrolysis, the species oxidised at the anode and reduced at the cathode are determined by the standard potentials (subject to overvoltage). For H2SO4(aq): at the cathode, H+ reduces to H2 (highest E^∘_red available among reducible species); at the anode, in DILUTE solution water is oxidised to O2 (water oxidation requires only 1.23 V vs 1.96 V for SO4^2- → S2O8^2-). Only in CONCENTRATED H2SO4 does SO4^2- become the favoured anode reactant.

1. Cathode in dilute H2SO4: 2H+ + 2e- -> H2 (E^∘ = 0). Statement (i) correct.
2. Anode in dilute H2SO4: water is oxidised (E^∘_cell = 1.23 V is less than the 1.96 V needed for SO4^2- oxidation). Statement (iii) correct.
3. Statement (ii): in CONCENTRATED H2SO4, the situation reverses; SO4^2- is oxidised to S2O8^2- (peroxydisulfate), so water is NOT oxidised at the anode. Statement (ii) wrong.
4. Statement (iv): tetrathionate is S4O6^2-, not the product here (S2O8^2- is peroxydisulfate). Even ignoring the chemistry, (iv) is wrong because in dilute solution water is oxidised first.

Correct options: (i) and (iii).

Correct options: (ii), (iii).

Concept used. At equilibrium, the cell emf E_cell = 0, so the Nernst equation gives 0 = E^∘_cell - RTnFln K_c E^∘_cell = RTnFln K_c. At T = 298 K, RTF = 0.02569 V, so 2.303 RTF = 0.0591 V ≈ 0.059 V. For the Daniel cell, n = 2 (two electrons transferred per Zn + Cu²⁺ -> Zn²⁺ + Cu).

1. Apply the Nernst-equilibrium relation: E^∘_cell = 2.303 RTnFlog K_c. Substitute n = 2, E^∘_cell = 1.1 V: 1.1 = 2.303 RT2Flog K_c. That is statement (ii). Correct.
2. Rearrange using the 298 K shortcut 2.303 RT/F = 0.059: log K_c = n · E^∘_cell0.059 = 2 × 1.10.059 = 2.20.059. That is statement (iii). Correct.
3. Statement (i) ``1.1 = K_c'' is nonsense, K_c has no voltage units. Wrong.
4. Statement (iv) ``log K_c = 1.1'' ignores the factor n/0.059. Wrong.

Correct options: (ii) and (iii).

Correct options: (i), (ii).

Concept used. Conductivity κ is an intrinsic property of an electrolyte solution: it depends on what is dissolved (the ions present), how concentrated they are, the temperature, the solvent, and the degree of dissociation, but NOT on the geometry of the measuring cell or the power of the AC source used to measure it. κ is defined as κ = R^-1 · (l/A), where any change in l or A is cancelled by the corresponding change in measured R.

1. Statement (i): nature of electrolyte. Strong vs weak, univalent vs divalent, these change ionic mobilities and therefore κ. Correct.
2. Statement (ii): concentration. As c rises, more ions per unit volume, κ rises (up to a maximum, then falls due to ion pairing in very concentrated solutions). Correct.
3. Statement (iii): power of AC source. AC just measures the resistance; the magnitude of the applied current/voltage doesn't change κ (within the linear range). Wrong.
4. Statement (iv): distance between electrodes. Affects R and G^* = l/A together, but κ = G^*/R is unchanged. Wrong.

Correct options: (i) and (ii).

Correct options: (i), (iv).

Concept used. Kohlrausch's law. For H2O (treated as the very weak electrolyte H+ + OH-), Λ⁰_m(H2O) = λ⁰_H+ + λ⁰_OH-. Build this from any pair of strong electrolytes that together contain H+ and OH-, then subtract a third whose ions are the spectators left over.

1. Option (i): Λ⁰_m(HCl) + Λ⁰_m(NaOH) - Λ⁰_m(NaCl) = (H+ + Cl-) + (Na+ + OH-) - (Na+ + Cl-) = H+ + OH- = Λ⁰_m(H2O). CORRECT.
2. Option (iv): Λ⁰_m(NH4OH) + Λ⁰_m(HCl) - Λ⁰_m(NH4Cl) = (NH4+ + OH-) + (H+ + Cl-) - (NH4+ + Cl-) = H+ + OH- = Λ⁰_m(H2O). CORRECT.
3. Option (ii): Λ⁰_m(HNO3) + Λ⁰_m(NaNO3) - Λ⁰_m(NaOH) = (H+ + NO3-) + (Na+ + NO3-) - (Na+ + OH-) = H+ + 2NO3- - OH-. Wrong.
4. Option (iii) algebraically reduces to H+ + OH- but NCERT key omits it.

Per official NCERT Exemplar key: options (i) and (iv).

Correct options: (i), (iii).

Concept used. With inert (platinum) electrodes, the electrolysis of CuSO4(aq) has two competitions. At the cathode: Cu²⁺ (E^∘_red = +0.34 V) vs H+ from water (0.00 V). Cu²⁺ wins; Cu deposits. At the anode: water oxidation (H2O → O2, E^∘_ox = -1.23 V) vs SO4^2- oxidation (very negative, impractical). Water wins; O2 is released. Platinum is inert, so it does not dissolve.

1. Cathode: Cu²⁺(aq) + 2e- -> Cu(s). Copper plates out. Statement (i) correct.
2. Anode: 2H2O(l) -> O2(g) + 4H+(aq) + 4e-. Oxygen evolves. Statement (iii) correct.
3. Statement (ii) ``Cu deposits at anode'', wrong; reduction happens at the cathode, not anode.
4. Statement (iv) ``Cu dissolves at anode'', wrong; the anode is Pt (inert), and there's no Cu metal there to dissolve.

Correct options: (i) and (iii).

Correct options: (i), (ii).

Concept used. With copper electrodes (i.e. a reactive anode), the easiest oxidation at the anode is now the dissolution of Cu metal itself: Cu(s) → Cu²⁺(aq) + 2e- with E^∘_ox = -0.34 V, which is much less negative than water oxidation (-1.23 V) or SO4^2- oxidation (-1.96 V). So the anode dissolves; at the cathode, Cu²⁺ deposits as before. Net effect: copper is transferred from anode to cathode, an essential industrial step in copper electrorefining.

1. Cathode: Cu²⁺(aq) + 2e- -> Cu(s). Statement (i) correct.
2. Anode: Cu(s) -> Cu²⁺(aq) + 2e-. Statement (ii) correct.
3. Statement (iii) ``O2 at anode'', wrong, because anode-Cu dissolution is much easier than water oxidation.
4. Statement (iv) ``Cu deposits at anode'', wrong; cathode is where Cu deposits.

Correct options: (i) and (ii).

Correct options: (i), (ii).

Concept used. Conductivity κ is the reciprocal of resistivity ρ: κ = 1ρ = 1R·lA = G^*R, where G^* = l/A is the cell constant of the conductivity cell. Units of κ: S cm^-1 (or S m^-1). _m is the molar conductivity, related to κ by _m = κ × 1000 / c, so _m ≠ κ. l/A alone is the cell constant, not the conductivity.

1. Start from the resistance of a uniform conductor: R = ρ lA, so ρ = R Al.
2. Conductivity is κ = 1/ρ = 1R lA, which matches statement (i).
3. Identify the cell constant G^* = l/A. Substituting: κ = G^*/R, which matches statement (ii).
4. Statement (iii) κ = _m is wrong (different quantities, different units).
5. Statement (iv) κ = l/A is wrong (l/A alone is the cell constant, not the conductivity).

Correct options: (i) and (ii).

Correct options: (i), (iii).

Concept used. Molar conductivity _m = κ × 1000 / c is the conductivity per mole of electrolyte. Like κ, it depends on intrinsic solution properties, temperature (ion mobility rises with T) and concentration (more ions per mole at higher dilution, because the strong-electrolyte dissociation is complete and ion-ion interactions weaken). It does NOT depend on the cell geometry (electrode distance, electrode area), because κ is geometry-independent and c is intensive.

1. (i) Temperature. κ increases ∼ 2% per K, and c is roughly constant, so _m rises with T. Correct.
2. (ii) Distance between electrodes. Affects R and G^* equally; κ unchanged; _m unchanged. Wrong.
3. (iii) Concentration. At low c, _m approaches Λ⁰_m. At higher c, _m falls (ion-ion drag for strong electrolytes; reduced dissociation for weak ones). So _m depends on c. Correct.
4. (iv) Surface area of electrodes. Same reasoning as (ii), κ unaffected. Wrong.

Correct options: (i) and (iii).

Correct options: (ii), (iii).

Concept used. In standard cell notation, the left half-cell is the anode (oxidation) and the right half-cell is the cathode (reduction). So for Mg|Mg²⁺||Cu²⁺|Cu: Mg is the anode (Mg -> Mg²⁺ + 2e-), Cu is the cathode (Cu²⁺ + 2e- -> Cu). The overall (net) cell reaction is the sum of the two half-reactions: Mg + Cu²⁺ -> Mg²⁺ + Cu. Cu²⁺ (not Cu) is the oxidising agent, because it accepts electrons.

1. (i) ``Mg is cathode''. Wrong, Mg is on the left, it's the anode.
2. (ii) ``Cu is cathode''. Correct, right-hand half-cell.
3. (iii) Cell reaction Mg + Cu²⁺ -> Mg²⁺ + Cu. Correct, sum of the two half-reactions.
4. (iv) ``Cu is the oxidising agent''. Wrong, Cu is the product of reduction; Cu²⁺ is the oxidising agent.

Correct options: (ii) and (iii).

Concept used. An electrode potential is the potential difference between a metal (or other electrode material) and the surrounding electrolyte solution. Measuring the absolute potential of a single electrode would require inserting a probe (a second electrode) into the same solution, because a voltmeter always measures a difference between two terminals. So you can never read a half-cell's potential directly; you can only measure it relative to another half-cell.

1. Connect a voltmeter across one half-cell. To complete the circuit you must introduce a second conductor (a wire and another electrode), which itself sits at some potential. The reading is the difference, never the absolute value.
2. By international convention the Standard Hydrogen Electrode (SHE) is assigned E^∘ = 0 V exactly. All ``standard electrode potentials'' tabulated for other electrodes are measured against SHE, so they too are relative numbers.
3. Therefore the absolute electrode potential of any single electrode cannot be measured experimentally; only the difference relative to a reference electrode can.

No. Only the relative electrode potential (with respect to a reference such as SHE) is measurable.

Concept used. E^∘_cell is the standard emf of the cell (all species at unit activity). _r G^∘ = -nFE^∘_cell is the corresponding standard Gibbs energy change. Both relate to the overall reaction's spontaneity in the standard state. A spontaneous reaction has E^∘_cell > 0 and _r G^∘ < 0; a non-spontaneous reaction in the opposite sense. If E^∘_cell = 0, then _r G^∘ = 0 too, meaning the standard reaction has no driving force, no useful work can be obtained, and the cell would not function as a galvanic cell.

1. E^∘_cell = 0 would mean E^∘_cathode = E^∘_anode. Physically, the two half-cells have identical thirst for electrons; no net electron flow occurs in the standard state.
2. Substituting into _r G^∘ = -nFE^∘_cell = 0. The reaction has K = 1 (from _r G^∘ = -RTln K).
3. A cell with E^∘_cell = 0 does not act as a useful galvanic cell. The textbook answer is: no, E^∘_cell (and _r G^∘) cannot be zero for a working cell.

No. E^∘_cell = 0 implies _r G^∘ = 0, meaning the cell has no driving force and would not function.

Concept used. E_cell is the actual (non-standard) cell potential at the moment of measurement; _r G is the Gibbs energy change at the same instant. Both depend on the reaction quotient Q via the Nernst equation E_cell = E^∘_cell - RTnFln Q and _r G = -nFE_cell. When the cell discharges, ion concentrations change continuously until Q reaches the equilibrium constant K. At that point, the Nernst-equation expression collapses to E_cell = E^∘_cell - RTnFln K = 0, which means the cell has reached equilibrium.

1. Start with Nernst: E_cell = E^∘_cell - RTnFln Q.
2. Setting E_cell = 0 gives ln Q = nF E^∘_cellRT. But this is also the expression for ln K (from E^∘_cell = RTnFln K). Therefore Q = K — the cell is at equilibrium.
3. At equilibrium, _r G = -nFE_cell = 0 simultaneously.

E_cell = 0 and _r G = 0 when the cell reaction has reached equilibrium (Q = K).

Concept used. A standard reduction potential is the electrode's potential measured against SHE. By convention all half-reactions are written as reductions. A negative E^∘ means the reduction half-reaction has LESS tendency to proceed than the SHE reaction does; equivalently, the oxidation of the metal is favoured over the oxidation of H2. Concretely: when a Zn electrode is dipped in 1 M Zn²⁺ and coupled to SHE, the Zn electrode becomes the anode (Zn oxidises to Zn²⁺), while H+ gets reduced at the SHE side.

1. E^∘_H+/H2 = 0 (by convention) and E^∘_Zn²⁺/Zn = -0.76 V < 0.
2. Compare driving forces: H+ has a STRONGER tendency to be reduced than Zn²⁺, so in the coupled cell, H+ gains electrons (cathode) and Zn loses them (anode).
3. Conclusion: Zn is MORE reactive than H2, i.e. Zn can displace H2 from dilute acid: Zn + 2HCl -> ZnCl2 + H2.

Zn is more reactive than H2. When connected to SHE, Zn is oxidised (Zn -> Zn²⁺ + 2e-) and H+ is reduced (2H+ + 2e- -> H2).

Concept used. Faraday's first law: the mass deposited at the cathode is w = M· QnF, where M is the atomic mass, Q = It is the charge passed, n is the number of electrons per atom in the cathode reaction, and F is Faraday's constant. Since Q is the same in both cells (same I, same t), but M/n differs (Cu has M = 63.5, n = 2; Ag has M = 108, n = 1), the masses deposited will be different.

1. Compute the common charge: Q = I · t = 1 A × 10 × 60 s = 600 C.
2. Mass of Cu deposited (cathode: Cu²⁺ + 2e- -> Cu, n = 2): w_Cu = 63.52 × 96500 × 600 = 63.5 × 600193000 = 0.197 g.
3. Mass of Ag deposited (cathode: Ag+ + e- -> Ag, n = 1): w_Ag = 1081 × 96500 × 600 = 108 × 60096500 = 0.671 g.
4. 0.197 ≠ 0.671, so the deposited masses differ. Silver deposits more because Ag has higher atomic mass AND only needs 1 e^- per atom.

Different. w_Cu ≈ 0.197 g, w_Ag ≈ 0.671 g (Ag is heavier per electron because n=1 vs Cu's n=2).

Concept used. A galvanic cell is depicted in the standard cell notation: anode | anode electrolyte || cathode electrolyte | cathode, where ``|'' marks a phase boundary, ``||'' is the salt bridge, anode is on the left (oxidation), cathode on the right (reduction). Identify which species is oxidised and which is reduced from the cell reaction.

1. Read the cell reaction Cu + 2Ag+ -> 2Ag + Cu²⁺. Cu goes from 0 to +2 oxidation state — oxidation. Ag goes from +1 to 0 — reduction.
2. Oxidation at the anode: Cu(s) -> Cu²⁺(aq) + 2e-. Reduction at the cathode: Ag+(aq) + e- -> Ag(s) (multiply by 2 to balance electrons).
3. Write the cell notation: Cu(s) | Cu²⁺(aq) || Ag+(aq) | Ag(s).

Cu(s) | Cu²⁺(aq) || Ag+(aq) | Ag(s).

Concept used. Overpotential (or overvoltage) is the extra voltage beyond the thermodynamic E^∘ needed to make a half-reaction proceed at a useful rate. It arises from kinetic factors: high activation energy for gas-evolution reactions, sluggish electron transfer, slow nucleation of gas bubbles. Oxygen evolution at most electrode materials (Pt, graphite) has a substantial overpotential of ∼ 0.4 V, so even though water oxidation is thermodynamically easier than Cl- oxidation, in practice Cl- wins because Cl2 has a much lower overpotential.

1. Thermodynamic comparison (oxidation potentials): H2O -> O2: E^∘_ox = -1.23 V; Cl- -> Cl2: E^∘_ox = -1.36 V. Naively, water oxidises more easily.
2. Add overpotential. For oxygen evolution, the kinetic barrier on Pt or graphite is ∼ +0.4 to +0.6 V. Effective oxidation potential of water becomes ∼ -1.23 - 0.5 = -1.73 V (worse than Cl- at -1.36 V).
3. Hence Cl- is actually oxidised first at the anode, producing Cl2 gas. This is the industrial chlor-alkali process.

Because oxygen evolution requires high overpotential, while Cl2 evolution does not. So Cl- is oxidised at anode despite the less favourable E^∘.

Concept used. An electrode potential is the potential difference that develops at the interface between an electrode (metal, graphite, gas+inert electrode, etc.) and its electrolyte solution. It arises because, when a metal is dipped in a solution of its own ions, two opposing processes occur at the surface: oxidation (M -> Mⁿ⁺ + ne-, sending ions into solution) and reduction (Mⁿ⁺ + ne- -> M, pulling ions out). At equilibrium, the metal acquires a net charge relative to the solution: more positive if reduction dominates, more negative if oxidation dominates. This charge separation is the electrode potential.

1. Define operationally: ``the tendency of an electrode to lose or gain electrons when in contact with a solution of its own ions, expressed as a potential relative to a standard reference (SHE).''
2. If the metal tends to oxidise (lose electrons) MORE than SHE: electrode potential is NEGATIVE (e.g. Zn: -0.76 V).
3. If the metal tends to reduce (gain electrons) MORE than SHE: electrode potential is POSITIVE (e.g. Cu: +0.34 V).
4. At standard conditions (all activities = 1, T = 298 K) the electrode potential is called the standard electrode potential, E^∘, and is tabulated for reduction half-reactions.

Electrode potential: potential difference at the metal-solution interface, measured vs SHE; reflects the metal's tendency to lose or gain electrons.

Concept used. When an electrochemical (galvanic) cell is connected to an electrolytic cell, the galvanic cell acts as the DC power source. Its positive terminal pushes current outward into the external circuit; the negative terminal receives current from the circuit. In the electrolytic cell, the electrode connected to the galvanic cell's POSITIVE terminal becomes the ANODE (positive), and the electrode connected to the negative terminal becomes the CATHODE (negative). In a Zn|Cu galvanic cell, Cu (right side) is the positive terminal (cathode) and Zn (left side) is the negative terminal (anode).

1. In the upper cell (galvanic, Zn|ZnSO4||CuSO4|Cu), Zn is the anode (negative) and Cu is the cathode (positive).
2. The galvanic cell's Cu plate (positive) connects to electrode A of the lower electrolytic cell; its Zn plate (negative) connects to electrode B. (The exact connection is shown by which wire goes where in the figure; following the NCERT figure, the Zn side feeds electrode A.)
3. Polarity in the electrolytic cell (following NCERT answer key): A becomes negative (cathode of the electrolytic cell), B becomes positive (anode).

Electrode A: NEGATIVE polarity (cathode of electrolytic cell). Electrode B: POSITIVE polarity (anode of electrolytic cell).

Concept used. When direct current (DC) is passed through an electrolyte, ions actually migrate to the electrodes and undergo electrolysis (oxidation/reduction). This changes the concentration of ions near the electrodes and at the bulk, changing the resistance over time and altering the very quantity being measured. Alternating current (AC), with its periodic reversal of polarity, prevents net ion accumulation and net electrolysis at the electrodes, so the ionic composition stays constant and the measured resistance is reproducible.

1. DC through an electrolyte causes continuous electrolysis: Cu²⁺ deposits on the cathode, water oxidises at the anode, etc. The ion concentration changes, and so does κ, in real time.
2. AC (typically ∼ 1 kHz, low voltage) reverses every half-cycle, so any ion that drifts toward an electrode is pulled back the next half-cycle. No net accumulation.
3. No electrolysis, no concentration change, so the measured resistance is intrinsic to the solution at the moment of measurement.

AC prevents electrolysis and ion accumulation, keeping the solution composition (and hence the measured resistance) constant during measurement.

Concept used. A galvanic cell with emf E_cell drives current I through an external circuit. If an external opposing potential E_ext is applied in the opposite direction, the net driving force is E_net = E_cell - E_ext. When E_ext = E_cell (exact balance), the net driving force is zero, no current flows, and the cell reaction stops (neither forward nor backward). This is the principle of the potentiometer, used to measure emf without drawing current.

1. Daniel cell: E_cell = 1.1 V. Without any external opposition, it drives current through the external circuit.
2. Apply E_ext = 1.1 V in opposition: E_net = 1.1 - 1.1 = 0 V.
3. Current I = E_net/(R_ext + r) = 0. No current flows.
4. Cell reaction Zn + Cu²⁺ -> Zn²⁺ + Cu stops (no net forward or backward progress).

No current flows; the cell reaction stops entirely. This is the basis of potentiometric emf measurement.

Concept used. Electrolysis of brine (chlor-alkali process) produces Cl2 at the anode and H2 at the cathode: aligned Anode (oxidation): &2Cl^-(aq) -> Cl2(g) + 2e^-.
Cathode (reduction): &2H2O(l) + 2e^- -> H2(g) + 2OH^-(aq). aligned The cathode reaction produces OH- ions in the bulk solution, while Na+ stays as spectator (less easily reduced than water). The net result: the solution becomes strongly basic, and the pH RISES from neutral 7 to typically 13–14 in industrial cells.

1. Anode reaction removes Cl- as Cl2 gas. Doesn't affect pH directly.
2. Cathode reaction generates OH-. Each mole of H2 produced accompanies 2 moles of OH- in solution.
3. Bulk solution now contains Na+ and OH- — effectively NaOH. The pH rises.

pH rises (becomes basic) because NaOH is formed at the cathode region as OH^- is generated.

Concept used. The cell potential is determined by the Nernst equation: E_cell = E^∘_cell - RTnFln Q, where Q depends on the activities of the species involved. If the overall cell reaction involves only solids and insoluble products (no aqueous ions), then Q stays essentially constant as the cell discharges, and so does E_cell. The mercury cell has exactly such a reaction: Zn(Hg) + HgO(s) -> ZnO(s) + Hg(l). All four species are solids or pure liquids; no ionic activities appear in Q. The cell potential is constant at ∼ 1.35 V throughout its life. In a dry cell, however, the reaction involves Zn²⁺, NH4+ and NH3, whose concentrations change as the cell discharges, so the emf drops progressively.

1. Identify the mercury-cell reaction (cathode + anode + overall):
   Anode: Zn(Hg) + 2OH- -> ZnO(s) + H2O + 2e-.
   Cathode: HgO(s) + H2O + 2e- -> Hg(l) + 2OH-.
   Overall: Zn(Hg) + HgO(s) -> ZnO(s) + Hg(l).
2. The OH- ions consumed at the anode are regenerated at the cathode (note matching coefficients). Net change in [OH-]: zero. No ion concentration in the overall reaction.
3. Q depends only on activities of solid/liquid pure phases, all of which are constant at 1. So log Q = 0 and E_cell = E^∘_cell = 1.35 V for the entire discharge.

Mercury-cell reaction involves only solids and pure liquid, so Q doesn't change with discharge, and E stays at ∼ 1.35 V.

Concept used. On dilution, the molar conductivity _m rises differently for strong vs weak electrolytes. For a STRONG electrolyte (fully dissociated, e.g. NaCl), the number of ions per mole is fixed; dilution only reduces ion-ion interactions, so _m rises slowly toward Λ⁰_m (perhaps 1.2–2 times over a 100-fold dilution). For a WEAK electrolyte (partially dissociated, e.g. CH3COOH), dilution dramatically increases the degree of dissociation α, so the number of ions per mole rises strongly; _m can rise by a factor of 10–100 between concentrated and dilute solutions.

1. Electrolyte A: _m rises 25 times on dilution. This is a huge increase, characteristic of a weak electrolyte (degree of dissociation α increases substantially with dilution).
2. Electrolyte B: _m rises only 1.5 times. A modest rise, characteristic of a strong electrolyte (already fully dissociated; only ion-ion drag relaxes on dilution).
3. Conclusion: B is the STRONG electrolyte; A is weak.

B is the strong electrolyte. Its _m rises only modestly (1.5x) on dilution; strong electrolytes are already fully dissociated, so dilution only relaxes ion-ion interactions.

Concept used. In the electrolysis of dilute H2SO4, the half-reactions are: aligned Anode: &2H2O(l) -> O2(g) + 4H⁺(aq) + 4e^-.
Cathode: &4H⁺(aq) + 4e^- -> 2H2(g). aligned The overall reaction is 2H2O -> 2H2 + O2, i.e. the net electrolysis of water. The H+ produced at the anode exactly equals the H+ consumed at the cathode (4 each per 4e^-). So [H+] in the bulk solution stays constant, and the pH is unchanged.

1. Anode produces 4 H+ per 4 e^-.
2. Cathode consumes 4 H+ per 4 e^-.
3. Net change in [H+]: zero. pH unaffected.
4. Equivalent statement: the overall electrolysis is just 2H2O -> 2H2 + O2. Water is consumed; ions are not produced or consumed.

No, pH is NOT affected. H+ is produced at anode and consumed at cathode in equal amounts; the net process is just water splitting.

Concept used. The specific conductivity (conductivity) κ is the conductivity per unit volume of the solution. It depends on (a) the number of charge carriers per unit volume (which scales with concentration c), and (b) ionic mobility. On dilution by adding water, the concentration c falls; therefore the number of ions per unit volume falls; therefore κ DECREASES.

1. κ has the form κ = c × _m / 1000 (with c in mol/L, _m in S cm² mol^-1).
2. On dilution, c drops faster than _m rises (for strong electrolytes, _m rises only slightly; for weak, _m rises but c drops by more).
3. Net effect: κ decreases with dilution. This is specific conductivity, in contrast to molar conductivity which behaves oppositely.

κ decreases on dilution because the number of ions per unit volume falls (concentration drops).

Concept used. A reference electrode is one whose potential is fixed and known, against which other electrodes can be measured. The internationally accepted reference electrode for standard electrode potentials is the Standard Hydrogen Electrode (SHE), defined as:

• Pt black (catalytic platinum) electrode,
• H2 gas at 1 bar pressure,
• H+ ions at 1 M activity (effectively 1 M HCl),
• Temperature 298 K.

1. SHE is the primary reference. Construction: Pt foil coated with finely-divided Pt (Pt black) immersed in 1 M HCl, with H2 gas bubbled at 1 bar.
2. Half-reaction: 2H+(aq, 1 M) + 2e- <=> H2(g, 1 bar); E^∘ = 0 V by definition.
3. To measure another electrode's E^∘, build the cell SHE || unknown electrode; the voltmeter reading (open circuit) gives E^∘_unknown directly.

Standard Hydrogen Electrode (SHE), E^∘ = 0 V by definition; all other electrode potentials are measured vs SHE.

Concept used. Cell notation convention: anode on the left, cathode on the right. The left half-cell undergoes oxidation, the right undergoes reduction.

1. Identify the left half-cell: Cu | Cu²⁺. Half-reaction (oxidation): Cu(s) -> Cu²⁺(aq) + 2e-. Cu is the anode, getting oxidised.
2. Identify the right half-cell: Cl-|Cl2,Pt. Pt is the inert electrode. Half-reaction (reduction): Cl2(g) + 2e- -> 2Cl-(aq). Pt is the cathode surface (the actual electron acceptor is Cl2, getting reduced).
3. Sum: Cu + Cl2 -> Cu²⁺ + 2Cl- (overall cell reaction).

Anode (oxidation): Cu -> Cu²⁺ + 2e-. Cathode (reduction): Cl2 + 2e- -> 2Cl-.

Concept used. The Daniel cell is Zn|Zn²⁺||Cu²⁺|Cu, with the net reaction Zn + Cu²⁺ -> Zn²⁺ + Cu and n = 2 electrons transferred. The Nernst equation for this cell at 298 K is E_cell = E^∘_cell - 0.059nlog Q = E^∘_cell - 0.0592log[Zn²⁺][Cu²⁺], where Q is the reaction quotient. Solids (Zn, Cu) have activity 1 and don't enter Q. Increasing [Zn²⁺] INCREASES Q, which INCREASES log Q, which DECREASES E_cell.

1. Write the Nernst equation: E_cell = 1.10 - 0.0592log[Zn²⁺][Cu²⁺].
2. Test: if [Zn²⁺] = [Cu²⁺] = 1 M (standard): log Q = 0, E_cell = 1.10 V.
3. Increase [Zn²⁺] to 10 M (with [Cu²⁺] fixed at 1 M): Q = 10, log Q = 1, Nernst correction = -0.059/2 = -0.0295, E_cell = 1.10 - 0.0295 = 1.0705 V.
4. So increasing [Zn²⁺] decreases E_cell (the cell becomes less spontaneous).

E_cell = E^∘_cell - 0.0592log[Zn²⁺][Cu²⁺].
[0.4em] Increasing [Zn²⁺] DECREASES E_cell.

Concept used. A primary battery (e.g. dry cell) contains a fixed amount of reactants and cannot be recharged; when depleted, it is discarded. A secondary battery (e.g. lead storage) can be recharged but takes time and degrades over many cycles. A fuel cell (e.g. H2/O₂) runs continuously as long as reactants are supplied externally, with no down-time and no internal degradation from cycling.

1. Primary batteries: limited reactant; once exhausted, discarded. Examples: Leclanche dry cell, mercury cell.
2. Secondary batteries: rechargeable but require interruption of service while charging; storage degrades with cycle number. Examples: lead-acid, Ni-Cd, Li-ion.
3. Fuel cells: reactants (H2 + O2, or methanol + O2) are supplied externally and continuously; products (H2O) are removed continuously. No internal exhaustion. Runs as long as fuel is available.

Fuel cells run continuously as long as fuel is supplied (no exhaustion, no recharge needed); they don't store reactants internally.

Concept used. The lead storage battery consists of Pb (anode) and PbO2 (cathode) in ∼ 38% H2SO4 (d ≈ 1.28 g/mL). During DISCHARGE, the half-reactions are: aligned Anode: &Pb(s) + SO4^2-(aq) -> PbSO4(s) + 2e^-,
Cathode: &PbO2(s) + 4H+(aq) + SO4^2-(aq) + 2e^- -> PbSO4(s) + 2H2O(l). aligned Overall: Pb + PbO2 + 2H2SO4 -> 2PbSO4 + 2H2O. Note that H2SO4 is CONSUMED (2 moles per cell reaction) and H2O is PRODUCED (2 moles per cell reaction). So the electrolyte becomes more dilute, and its density DECREASES.

1. Anode (oxidation): Pb -> Pb²⁺, with Pb²⁺ precipitating as PbSO4 on the anode.
2. Cathode (reduction): Pb⁴⁺ in PbO2 goes to Pb²⁺ as PbSO4.
3. Overall: Pb(s) + PbO2(s) + 2H2SO4(aq) -> 2PbSO4(s) + 2H2O(l).
4. As discharge proceeds, H2SO4 is consumed and H2O is added, so the electrolyte dilutes and its density falls (from ∼ 1.28 g/mL fully charged to ∼ 1.10 g/mL fully discharged).

Overall: Pb + PbO2 + 2H2SO4 -> 2PbSO4 + 2H2O. Density of electrolyte DECREASES (dilution by H2O, consumption of H2SO4).

Concept used. CH3COOH (acetic acid) is a weak electrolyte: at finite concentration it dissociates only partially, CH3COOH + H2O <=> CH3COO- + H3O+, with degree of dissociation α ≪ 1. On dilution, the equilibrium shifts right (Le Chatelier: more solvent favours more ions), so α rises substantially, dramatically increasing the number of ions per mole. Thus _m rises by a large factor.

CH3COONa (sodium acetate) is a strong electrolyte: in solution it is fully dissociated into Na+ and CH3COO- at all reasonable concentrations. The number of ions per mole is already at its maximum. On dilution, only the ion-ion attractive interactions weaken (reduced interionic drag), so _m rises only slightly (gradually) toward Λ⁰_m.

1. For CH3COOH (weak): α = moles dissociated / moles dissolved. At 0.1 M, α ∼ 0.013, so only 1.3% of molecules are ionised. At 0.001 M, α ∼ 0.13, ionisation is 10× higher, _m is 10× higher.
2. For CH3COONa (strong): α = 1 at all concentrations. Dilution from 0.1 M to 0.001 M increases _m by ∼ 10–15%, not by a factor.
3. Hence _m(CH3COOH) rises drastically; _m(CH3COONa) rises gradually.

CH3COOH is a weak electrolyte: dilution raises α sharply, so _m rises drastically. CH3COONa is strong: α = 1 already, so dilution only reduces interionic drag, giving a gradual rise.

Correct matches: (i)→(c), (ii)→(d), (iii)→(a), (iv)→(b).

Concept used. Dimensional check for each electrochemical quantity, derived from its defining equation. Each symbol carries its own distinctive units; matching is straightforward once the formulas are recalled.

1. _m = κ × 1000 / c. Units: (S cm^-1)(cm³/mol) = S cm² mol^-1. So (i)→(c).
2. E_cell is a potential difference, measured in volts (V). (ii)→(d).
3. κ (conductivity) is reciprocal resistivity: (S cm^-1). (iii)→(a).
4. G^* = l/A has units of cm^-1 (or m^-1). (iv)→(b).

(i)→(c), (ii)→(d), (iii)→(a), (iv)→(b).

Correct matches: (i)→(d), (ii)→(a), (iii)→(b), (iv)→(c).

Concept used. Classify each quantity as intensive (does not depend on amount), extensive (depends on amount), or related to specific physical behaviours (dilution, ions per volume).

1. _m: rises with dilution (sharply for weak, gradually for strong). Matches (d).
2. E^∘_cell: per unit charge ⇒ intensive. Matches (a).
3. κ (conductivity): depends on number of ions per unit volume (i.e. concentration). Matches (b).
4. _r G_cell: total Gibbs free energy change, scales with n, so extensive. Matches (c).

(i)→(d), (ii)→(a), (iii)→(b), (iv)→(c).

Correct matches: (i)→(d), (ii)→(c), (iii)→(a), (iv)→(b).

Concept used. Each cell or process has a distinctive feature that uniquely identifies it.

1. Lead storage: Pb is anode, PbO2 is cathode in H2SO4. Matches (d).
2. Mercury cell: constant potential (1.35 V) throughout life because reaction involves only solid/liquid species. Matches (c).
3. Fuel cell: thermodynamic efficiency Δ G / Δ H ≈ 83% for H2 + O2 cell. Maximum efficiency among electrochemical devices. Matches (a).
4. Rusting: Fe corrodes in moist air. Prevented by coating with Zn (galvanisation), which is more reactive and acts as sacrificial anode. Matches (b).

(i)→(d), (ii)→(c), (iii)→(a), (iv)→(b).

Correct matches: (i)→(d), (ii)→(c), (iii)→(b), (iv)→(a).

Concept used. Match each quantity to its defining expression.

1. κ = G^*/R (conductivity = cell constant / resistance). Matches (d).
2. _m = κ × 1000 / c, often written _m = κ / c in appropriate units. Matches (c).
3. α = _m / Λ⁰_m (Arrhenius degree of dissociation). Matches (b).
4. Q = I × t (Faraday's first law). Matches (a).

(i)→(d), (ii)→(c), (iii)→(b), (iv)→(a).

Correct matches: (i)→(d), (ii)→(c), (iii)→(a) and (e), (iv)→(b).

Concept used. Match each named cell to its identifying feature.

1. Lechlanche (dry) cell: Zn -> Zn²⁺ + 2e- at the Zn anode. Matches (d).
2. Ni-Cd cell: rechargeable; used in cordless tools. Matches (c).
3. Fuel cell (H2/O2): net reaction 2H2 + O2 -> 2H2O, and converts combustion energy into electricity. Matches both (a) and (e).
4. Mercury cell: no ions involved in net reaction (Q40); used in hearing aids and pacemakers. Matches (b).

(i)→(d), (ii)→(c), (iii)→(a) and (e), (iv)→(b).

Correct matches: (i)→(c), (ii)→(a), (iii)→(g), (iv)→(e), (v)→(d), (vi)→(b), (vii)→(f).

Concept used. Use the four given E^∘ values to identify the strongest oxidiser, strongest reducer, etc. Rules: strongest oxidiser = highest E^∘_red on the oxidised side; strongest reducer = most negative E^∘_red on the reduced side.

1. (i) F2: highest E^∘ = +2.87 V, the best oxidising agent, non-metal. Matches (c).
2. (ii) Li: most negative E^∘ = -3.05 V, strongest reducing metal. Matches (a).
3. (iii) Au³⁺: E^∘ = +1.4 V, oxidising metal ion. Matches (g).
4. (iv) Br-: anion that Au³⁺ can oxidise (1.4 > 1.09). Matches (e).
5. (v) Au: metal at E^∘ = +1.4 V, very unreactive. Matches (d).
6. (vi) Li+: corresponds to most negative E^∘, weakest oxidising metal ion. Matches (b).
7. (vii) F-: anion of strongest oxidiser, weakest reducing anion. Matches (f).

(i)→(c), (ii)→(a), (iii)→(g), (iv)→(e), (v)→(d), (vi)→(b), (vii)→(f).

Correct option: (iii) Assertion true, reason false.

Concept used. Reactivity of a metal is judged by its position in the electrochemical series; a metal with NEGATIVE E^∘_red is more reactive than H2 (displaces H2 from acid), and a metal with POSITIVE E^∘_red is less reactive than H2.

1. Assertion check: E^∘_Cu²⁺/Cu = +0.34 V > 0. Cu cannot displace H2 from acid; it is less reactive than H2. Assertion TRUE.
2. Reason check: E^∘_Cu²⁺/Cu is given as ``negative''. But +0.34 V is POSITIVE, not negative. Reason FALSE.
3. Assertion true, reason false. Option (iii).

Option (iii): assertion true (Cu is below H in activity series); reason false (E^∘_Cu²⁺/Cu = +0.34 is positive, not negative).

Correct option: (iii) Assertion true, reason false.

Concept used. For a galvanic cell, E^∘_cell > 0, which requires E^∘_cathode > E^∘_anode (by the definition E_cell = E_cathode - E_anode). The reason's inequality is reversed.

1. Assertion: E_cell positive for spontaneous cell operation. TRUE.
2. Reason: E_cathode < E_anode. This is the OPPOSITE of what's required. A galvanic cell needs E_cathode > E_anode. Reason FALSE.
3. Option (iii).

Option (iii): assertion true, reason is the reversed inequality (should be E_cathode > E_anode).