Electrochemistry

What to remember

• In a galvanic cell oxidation occurs at the anode (negative) and reduction at the cathode (positive); E°cell = E°cathode − E°anode, and a positive E°cell means a spontaneous cell reaction.
• Nernst equation at 298 K: E = E° − (0.059/n) log Q; at equilibrium E = 0, so log K = nE°/0.059 and ΔG° = −nFE°cell.
• Conductivity (κ) falls and molar conductivity (Λm) rises on dilution; Kohlrausch's law, Λ°m = λ°(cation) + λ°(anion), is used to find Λ°m of weak electrolytes.
• Faraday's law: m = ZIt = (M/nF) × I × t; 1 faraday = 96500 C = charge of 1 mol of electrons. Products of electrolysis depend on the electrolyte, the electrodes and the concentration.

Electrochemical cells and electrode potential

A galvanic (voltaic) cell converts the chemical energy of a spontaneous redox reaction into electrical energy. The Daniell cell is the standard example. A zinc rod dips in ZnSO₄ solution and a copper rod dips in CuSO₄ solution. The two solutions are joined by a salt bridge.

• Anode (oxidation, negative electrode): Zn(s) → Zn²⁺(aq) + 2e⁻
• Cathode (reduction, positive electrode): Cu²⁺(aq) + 2e⁻ → Cu(s)
• Overall: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)
• Electrons flow in the external wire from anode to cathode. Conventional current flows from cathode to anode.

The salt bridge is a U-tube holding a gel with an inert electrolyte such as KCl, KNO₃ or NH₄NO₃. Its ions have nearly equal speeds. It completes the circuit by ion flow and keeps each half-cell electrically neutral. Electrons do not travel through it.

Cell representation: the anode is written on the left, the cathode on the right. A single vertical line shows a phase boundary. A double vertical line shows the salt bridge. Example: Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s).

Electrode potential is the potential developed at the metal–solution interface. The standard electrode potential E° is measured at 1 M ion concentration, 1 bar pressure for gases and 298 K. By convention, tables list reduction potentials.

The standard hydrogen electrode (SHE) is the reference. A platinum foil coated with platinum black is dipped in 1 M H⁺ solution. Pure H₂ gas at 1 bar is bubbled over it. Its potential is fixed as 0.00 V. Half-reaction: H⁺(aq) + e⁻ ⇌ ½ H₂(g). To find an unknown E°, the electrode is joined to SHE and the cell voltage is measured.

E°cell = E°(cathode) − E°(anode), both as reduction potentials. E° is an intensive property. It does not change when the half-reaction is multiplied by a number.

Electrochemical series. Species are arranged in order of their standard reduction potentials.

Reading the series: a more positive E° means a stronger oxidising agent (F₂ is the strongest). A more negative E° means a stronger reducing agent (Li is the strongest among these). A metal with lower E° displaces a metal with higher E° from its salt solution. Metals with negative E° (above hydrogen) liberate H₂ from dilute acids. Copper, silver and gold do not.

Nernst equation, EMF, equilibrium constant and Gibbs energy

For the half-reaction Mⁿ⁺ + ne⁻ → M:

E(Mⁿ⁺/M) = E° − (2.303RT/nF) log (1/[Mⁿ⁺]) = E° + (0.059/n) log [Mⁿ⁺] at 298 K.

For the cell reaction aA + bB → cC + dD, the reaction quotient is Q = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ. Pure solids and pure liquids are not included.

Ecell = E°cell − (2.303RT/nF) log Q = E°cell − (0.059/n) log Q (at 298 K)

Here n is the number of electrons transferred in the balanced cell reaction. As the cell works, Q rises and Ecell falls. When Ecell = 0 the cell is "dead" and the reaction is at equilibrium. Then Q = K.

At equilibrium: E°cell = (2.303RT/nF) log K, so log K = nE°cell / 0.059 at 298 K.

Gibbs energy: ΔG = −nFEcell and ΔG° = −nFE°cell. Also ΔG° = −2.303RT log K. A positive E°cell gives a negative ΔG° and a spontaneous reaction.

Constants: R = 8.314 J K⁻¹ mol⁻¹, F = 96500 C mol⁻¹.

Worked example 1 (Nernst). Find Ecell for Zn | Zn²⁺(0.1 M) || Cu²⁺(0.01 M) | Cu. Given E°Zn²⁺/Zn = −0.76 V and E°Cu²⁺/Cu = +0.34 V.

E°cell = 0.34 − (−0.76) = 1.10 V. n = 2. Q = [Zn²⁺]/[Cu²⁺] = 0.1/0.01 = 10, so log Q = 1.

Ecell = 1.10 − (0.059/2) × 1 = 1.10 − 0.0295 = 1.0705 V ≈ 1.07 V.

Worked example 2 (K and ΔG°). Cell: Ni | Ni²⁺ || Ag⁺ | Ag. E°Ni²⁺/Ni = −0.25 V, E°Ag⁺/Ag = +0.80 V.

E°cell = 0.80 − (−0.25) = 1.05 V. Cell reaction: Ni + 2Ag⁺ → Ni²⁺ + 2Ag, so n = 2.

log K = (2 × 1.05)/0.059 = 35.59, so K ≈ 10³⁵·⁶.

ΔG° = −2 × 96500 × 1.05 = −202650 J mol⁻¹ ≈ −202.7 kJ mol⁻¹.

Conductance of electrolytic solutions

Ions carry current in an electrolytic solution. Quantities and units:

Relations: κ = G × G* = G × (l/A). The cell constant is found by filling the cell with a standard solution, such as KCl of known κ, and measuring R.

Molar conductivity is the conductivity of all the ions produced by 1 mole of electrolyte in a solution of volume V, kept between two electrodes 1 cm (or 1 m) apart. If concentration c is in mol L⁻¹ and κ is in S cm⁻¹, then Λm (S cm² mol⁻¹) = κ × 1000 / c. Also 1 S m² mol⁻¹ = 10⁴ S cm² mol⁻¹.

Variation with concentration.

• Conductivity κ decreases on dilution for all electrolytes, because the number of ions per unit volume falls.
• Molar conductivity Λm increases on dilution for all electrolytes.
• Strong electrolytes are fully ionised. Λm rises slowly on dilution, because inter-ionic attraction falls. The relation is Λm = Λ°m − A√c, so Λm against √c is a straight line. The intercept at c → 0 gives Λ°m.
• Weak electrolytes have low Λm at high concentration, because few ions are present. Λm rises steeply on dilution, because the degree of dissociation α increases. Λm against √c is not linear. Λ°m cannot be found by extrapolation.

Kohlrausch's law

Law of independent migration of ions: the limiting molar conductivity of an electrolyte is the sum of the limiting molar conductivities of its cation and its anion, each multiplied by the number of ions in the formula unit.

Λ°m = ν₊λ°₊ + ν₋λ°₋

Applications.

• 1. Λ°m of a weak electrolyte: Λ°m(CH₃COOH) = Λ°m(CH₃COONa) + Λ°m(HCl) − Λ°m(NaCl).
• 2. Degree of dissociation: α = Λm / Λ°m.
• 3. Dissociation constant of a weak acid HA: Ka = cα² / (1 − α), where c is the concentration.

Worked example 3. For a weak acid, Λ°m = 390 S cm² mol⁻¹. In 0.01 M solution, Λm = 39 S cm² mol⁻¹. Then α = 39/390 = 0.1. Ka = cα²/(1 − α) = 0.01 × 0.01 / 0.9 = 1.11 × 10⁻⁴.

Worked example 4 (cell constant). A cell filled with a solution of κ = 0.0125 S cm⁻¹ has R = 80 Ω. Cell constant = κ × R = 0.0125 × 80 = 1.0 cm⁻¹. The same cell with another solution gives R = 400 Ω. Then κ = 1.0/400 = 0.0025 S cm⁻¹.

Electrolytic cells and Faraday's laws

An electrolytic cell uses electrical energy from an external source to drive a non-spontaneous reaction. The anode is positive and the cathode is negative. Oxidation still occurs at the anode and reduction at the cathode.

Faraday's first law: the mass of a substance deposited at an electrode is directly proportional to the quantity of charge passed. m = ZQ = ZIt, where Q = It (charge in coulomb = current in ampere × time in second).

Faraday's second law: for the same charge, masses deposited are proportional to their equivalent masses (M/n). Z = M/(nF), the electrochemical equivalent.

One faraday = 96500 C = charge on 1 mol of electrons. Cu²⁺ needs 2 F per mole of Cu. Al³⁺ needs 3 F per mole of Al. Ag⁺ needs 1 F per mole of Ag.

Worked example 5. A current of 9.65 A passes for 50 minutes through molten Al₂O₃ (Al³⁺ ions). Q = 9.65 × 3000 = 28950 C = 0.3 F. Moles of Al = 0.3/3 = 0.1 mol. Mass of Al = 0.1 × 27 = 2.7 g.

Products of electrolysis. At the cathode, the species with the higher reduction potential is reduced. At the anode, the species with the lower reduction potential is oxidised. Water can also be reduced or oxidised.

In aqueous NaCl, Na⁺ is not reduced, because water is reduced more easily than Na⁺ (E° of Na⁺/Na is −2.71 V). Cl₂ forms at the anode rather than O₂, because oxidation of water needs an extra potential (overpotential). With copper electrodes in CuSO₄ the net effect is transfer of copper from anode to cathode. This is the basis of purification of copper.

Batteries and fuel cells

A primary cell cannot be recharged. A secondary cell can be recharged by passing current in the reverse direction.

Dry cell: anode Zn → Zn²⁺ + 2e⁻; cathode MnO₂ + NH₄⁺ + e⁻ → MnO(OH) + NH₃.

Mercury cell: anode Zn(Hg) + 2OH⁻ → ZnO + H₂O + 2e⁻; cathode HgO + H₂O + 2e⁻ → Hg + 2OH⁻. Overall: Zn(Hg) + HgO → ZnO + Hg. Ion concentration in the electrolyte does not change, so the voltage stays constant.

Lead storage battery.

• Discharge, anode: Pb(s) + SO₄²⁻(aq) → PbSO₄(s) + 2e⁻
• Discharge, cathode: PbO₂(s) + SO₄²⁻(aq) + 4H⁺(aq) + 2e⁻ → PbSO₄(s) + 2H₂O(l)
• Overall discharge: Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O
• On charging, the reaction is reversed: 2PbSO₄ + 2H₂O → Pb + PbO₂ + 2H₂SO₄.
• During discharge H₂SO₄ is used up and water forms, so the density of the acid falls.

Nickel–cadmium cell: Cd + 2Ni(OH)₃ → CdO + 2Ni(OH)₂ + H₂O during discharge.

Fuel cell (H₂–O₂). Hydrogen and oxygen are supplied continuously to porous carbon electrodes containing a catalyst such as finely divided platinum or palladium. The electrolyte is an aqueous alkali (KOH or NaOH).

• Anode: 2H₂(g) + 4OH⁻(aq) → 4H₂O(l) + 4e⁻
• Cathode: O₂(g) + 2H₂O(l) + 4e⁻ → 4OH⁻(aq)
• Overall: 2H₂ + O₂ → 2H₂O

The only product is water, and the cell does not need recharging while fuel is supplied. Its efficiency is about 70%, much higher than that of a thermal power plant. It has been used in space vehicles.

Corrosion

Corrosion is the slow eating away of a metal by air and moisture. Rusting of iron is an electrochemical process. A drop of water on iron forms a tiny galvanic cell.

• Anode (oxidation, at one spot): Fe(s) → Fe²⁺(aq) + 2e⁻ (E° of Fe²⁺/Fe = −0.44 V)
• Cathode (reduction, at another spot): O₂(g) + 4H⁺(aq) + 4e⁻ → 2H₂O(l) (E° = +1.23 V)
• Overall: 2Fe(s) + O₂(g) + 4H⁺(aq) → 2Fe²⁺(aq) + 2H₂O(l)
• Fe²⁺ is further oxidised to Fe³⁺ and forms rust, Fe₂O₃·xH₂O.

Both oxygen and moisture are needed. Dissolved CO₂ (carbonic acid) and salts speed up rusting.

Prevention.

• 1. Barrier protection: paint, grease, plastic or a metal coating, such as tin or chromium plating.
• 2. Galvanisation: coating iron with zinc. Even if the coating is scratched, zinc protects iron, because zinc has a lower reduction potential and is oxidised first.
• 3. Sacrificial protection: connecting iron to a more reactive metal such as magnesium or zinc.
• 4. Using anti-rust solutions and alloys such as stainless steel.

NEET traps

• 1. Anode is negative in a galvanic cell but positive in an electrolytic cell. Oxidation is at the anode in both.
• 2. Electrons never flow through the salt bridge or the solution. Only ions do.
• 3. E° does not change when the half-reaction is multiplied by a number. ΔG° does change.
• 4. Both κ and Λm are different quantities. κ decreases on dilution and Λm increases.
• 5. The Λm against √c line can be extrapolated to Λ°m for a strong electrolyte but not for a weak electrolyte.
• 6. Aqueous NaCl gives H₂ at the cathode, not Na. Molten NaCl gives Na.
• 7. Pt electrodes in CuSO₄ give O₂ at the anode. Cu electrodes give dissolution of the anode.
• 8. The mercury cell has a constant voltage. The lead storage battery's acid density falls on discharge.
• 9. In Q for the Nernst equation, solids and pure liquids are left out.
• 10. The factor 0.059 is the value of 2.303RT/F at 298 K. Do not use it at other temperatures.

One-liners

• 1. SHE has E° = 0.00 V by convention.
• 2. E°cell = E°cathode − E°anode.
• 3. Salt bridge keeps electrical neutrality; it does not carry electrons.
• 4. Ecell = E°cell − (0.059/n) log Q at 298 K.
• 5. log K = nE°cell/0.059 at 298 K.
• 6. ΔG° = −nFE°cell.
• 7. 1 F = 96500 C mol⁻¹.
• 8. SI unit of κ is S m⁻¹ and of Λm is S m² mol⁻¹.
• 9. Cell constant = l/A; κ = G × cell constant.
• 10. Λm = κ × 1000/c in S cm² mol⁻¹ when c is in mol L⁻¹.
• 11. Λm of a strong electrolyte = Λ°m − A√c.
• 12. α = Λm/Λ°m for a weak electrolyte.
• 13. Dry cell cathode is a graphite rod with MnO₂; anode is the zinc container.
• 14. Lead storage battery uses about 38% H₂SO₄ and forms PbSO₄ on both plates in discharge.
• 15. Rust is Fe₂O₃·xH₂O; zinc coating gives sacrificial protection.

Practice questions

1. Which electrode is assigned a standard electrode potential of exactly zero by convention?

   1. Standard hydrogen electrode
   2. Zinc electrode in 1 M Zn²⁺ solution
   3. Copper electrode in 1 M Cu²⁺ solution
   4. Silver electrode in 1 M Ag⁺ solution
   Answer

   A. Standard hydrogen electrode

   The standard hydrogen electrode (Pt, H₂ at 1 bar, 1 M H⁺) is the reference, with E° = 0.00 V.

2. In the Daniell cell Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s), the electrons flow

   1. from the copper electrode to the zinc electrode through the external wire
   2. from the zinc electrode to the copper electrode through the external wire
   3. from the copper electrode to the zinc electrode through the salt bridge
   4. from the zinc electrode to the copper electrode through the salt bridge
   Answer

   B. from the zinc electrode to the copper electrode through the external wire

   Zn is the anode (oxidation) so electrons leave it and travel through the external wire to the cathode. The salt bridge carries ions only.

3. The SI unit of molar conductivity is

   1. S m⁻² mol
   2. S m⁻¹
   3. Ω m mol⁻¹
   4. S m² mol⁻¹
   Answer

   D. S m² mol⁻¹

   Λm = κ/c = (S m⁻¹)/(mol m⁻³) = S m² mol⁻¹.

4. The standard reduction potentials are E°(Cu²⁺/Cu) = +0.34 V and E°(Ag⁺/Ag) = +0.80 V. The E°cell of the cell Cu(s) | Cu²⁺(aq) || Ag⁺(aq) | Ag(s) is

   1. 1.14 V
   2. −0.46 V
   3. 0.46 V
   4. 0.92 V
   Answer

   C. 0.46 V

   E°cell = E°cathode − E°anode = 0.80 − 0.34 = 0.46 V.

5. For the cell Zn | Zn²⁺(0.01 M) || Cu²⁺(1 M) | Cu, E°cell = 1.10 V at 298 K. Taking 2.303RT/F = 0.059 V, the EMF of the cell is

   1. 1.041 V
   2. 1.159 V
   3. 1.218 V
   4. 1.10 V
   Answer

   B. 1.159 V

   E = 1.10 − (0.059/2) log(0.01/1) = 1.10 − 0.0295 × (−2) = 1.10 + 0.059 = 1.159 V.

6. A cell reaction involves 2 electrons and has E°cell = 0.59 V at 298 K. Using log K = nE°/0.059, the equilibrium constant K is

   1. 10²⁰
   2. 10⁴⁰
   3. 10⁵
   4. 10¹⁰
   Answer

   A. 10²⁰

   log K = (2 × 0.59)/0.059 = 20, so K = 10²⁰.

7. For the reaction Zn + Cu²⁺ → Zn²⁺ + Cu, E°cell = 1.10 V. Taking F = 96500 C mol⁻¹, ΔG° is

   1. +212.3 kJ mol⁻¹
   2. −106.15 kJ mol⁻¹
   3. −424.6 kJ mol⁻¹
   4. −212.3 kJ mol⁻¹
   Answer

   D. −212.3 kJ mol⁻¹

   ΔG° = −nFE° = −2 × 96500 × 1.10 = −212300 J = −212.3 kJ mol⁻¹.

8. A current of 2 A is passed for 4825 s through aqueous CuSO₄ using inert electrodes (Cu = 63.5, F = 96500 C mol⁻¹). The mass of copper deposited is

   1. 1.5875 g
   2. 31.75 g
   3. 3.175 g
   4. 6.35 g
   Answer

   C. 3.175 g

   Q = 2 × 4825 = 9650 C = 0.1 F. Cu²⁺ + 2e⁻ → Cu needs 2 F per mol, so moles of Cu = 0.05 and mass = 0.05 × 63.5 = 3.175 g.

9. A conductivity cell has a cell constant of 0.5 cm⁻¹. When filled with a solution it shows a resistance of 250 Ω. The conductivity of the solution is

   1. 500 S cm⁻¹
   2. 0.002 S cm⁻¹
   3. 125 S cm⁻¹
   4. 0.02 S cm⁻¹
   Answer

   B. 0.002 S cm⁻¹

   κ = cell constant/R = 0.5/250 = 0.002 S cm⁻¹.

10. A 0.1 M solution has a conductivity of 0.02 S cm⁻¹. Its molar conductivity is

    1. 20 S cm² mol⁻¹
    2. 2 S cm² mol⁻¹
    3. 2000 S cm² mol⁻¹
    4. 200 S cm² mol⁻¹
    Answer

    D. 200 S cm² mol⁻¹

    Λm = κ × 1000/c = 0.02 × 1000/0.1 = 200 S cm² mol⁻¹.

11. The limiting molar conductivities (S cm² mol⁻¹) are CH₃COONa 91, HCl 426 and NaCl 126. The limiting molar conductivity of CH₃COOH is

    1. 391
    2. 643
    3. 552
    4. 209
    Answer

    A. 391

    Λ°(CH₃COOH) = Λ°(CH₃COONa) + Λ°(HCl) − Λ°(NaCl) = 91 + 426 − 126 = 391 S cm² mol⁻¹.

12. In the electrolysis of molten NaCl using inert electrodes, the product at the cathode is

    1. hydrogen gas
    2. sodium oxide
    3. sodium metal
    4. chlorine gas
    Answer

    C. sodium metal

    Na⁺ + e⁻ → Na at the cathode. Cl₂ forms at the anode.

13. In the electrolysis of concentrated aqueous NaCl using inert electrodes, the gas liberated at the anode is

    1. oxygen
    2. chlorine
    3. carbon dioxide
    4. hydrogen
    Answer

    B. chlorine

    Cl⁻ is oxidised to Cl₂ at the anode because oxidation of water to O₂ needs an extra potential (overpotential).

14. During the discharge of a lead storage battery, the solid that forms on both the anode and the cathode is

    1. PbO
    2. PbO₂
    3. Pb
    4. PbSO₄
    Answer

    D. PbSO₄

    Anode: Pb + SO₄²⁻ → PbSO₄ + 2e⁻. Cathode: PbO₂ + SO₄²⁻ + 4H⁺ + 2e⁻ → PbSO₄ + 2H₂O.

15. During the electrolysis of aqueous CuSO₄ using copper electrodes, what happens at the anode?

    1. Copper dissolves to give Cu²⁺ ions
    2. Oxygen gas is liberated
    3. Hydrogen gas is liberated
    4. Copper metal is deposited
    Answer

    A. Copper dissolves to give Cu²⁺ ions

    Copper is oxidised more easily than water or sulphate, so Cu → Cu²⁺ + 2e⁻ at the anode.

16. Given E°(Mg²⁺/Mg) = −2.36 V and E°(Zn²⁺/Zn) = −0.76 V, the E°cell of the cell Mg | Mg²⁺ || Zn²⁺ | Zn is

    1. 3.12 V
    2. −1.60 V
    3. 1.60 V
    4. 0.80 V
    Answer

    C. 1.60 V

    E°cell = −0.76 − (−2.36) = 1.60 V.

17. Given below are two statements. Statement I: On dilution, the conductivity (κ) of an electrolytic solution decreases. Statement II: On dilution, the molar conductivity (Λm) of an electrolytic solution increases.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    A. Both Statement I and Statement II are correct

    κ falls because the number of ions per unit volume falls; Λm rises because the volume containing 1 mol increases and inter-ionic attraction falls or dissociation increases.

18. Given below are two statements. Statement I: The limiting molar conductivity of a strong electrolyte can be obtained by extrapolating the plot of Λm against √c to zero concentration. Statement II: The limiting molar conductivity of a weak electrolyte can be obtained in the same way because its plot is also a straight line.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    C. Statement I is correct but Statement II is incorrect

    For a weak electrolyte Λm rises steeply at low concentration and the plot is not linear, so Kohlrausch's law is used instead.

19. Given below are two statements. Statement I: A mercury cell gives a constant potential of about 1.35 V throughout its life. Statement II: Mercury cells are used in small devices such as watches and hearing aids.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    A. Both Statement I and Statement II are correct

    The overall reaction Zn(Hg) + HgO → ZnO + Hg involves no change in ion concentration, so the voltage is constant; this makes it suitable for small devices.

20. Which of the following statements about the dry cell (Leclanché cell) are correct? 1. The zinc container acts as the anode. 2. The cathode is a graphite rod surrounded by MnO₂ and carbon. 3. It can be recharged by reversing the current.

    1. 1 and 3 only
    2. 2 and 3 only
    3. 1, 2 and 3
    4. 1 and 2 only
    Answer

    D. 1 and 2 only

    The dry cell is a primary cell: Zn container is the anode and the graphite rod with MnO₂ is the cathode. It cannot be recharged.

21. Given below are two statements. Statement I: In an electrolytic cell the cathode is the positive electrode. Statement II: In a galvanic cell oxidation takes place at the cathode.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    B. Both Statement I and Statement II are incorrect

    Electrolytic cell: cathode is negative. Galvanic cell: oxidation is always at the anode, reduction at the cathode.

22. Given below are two statements. Statement I: Electrons flow from one half-cell to the other through the salt bridge. Statement II: The salt bridge keeps the two half-cells electrically neutral by the movement of ions.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    D. Statement I is incorrect but Statement II is correct

    Electrons flow only through the external wire. Ions move through the salt bridge and maintain neutrality.

23. Given below are two statements. Statement I: The E° of the Zn²⁺/Zn half-cell remains −0.76 V when its half-reaction is multiplied by 2. Statement II: The ΔG° of the half-reaction also remains unchanged when it is multiplied by 2.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    C. Statement I is correct but Statement II is incorrect

    E° is intensive and does not change; ΔG° = −nFE° doubles when n doubles.

24. Given below are two statements. Statement I: In the H₂–O₂ fuel cell the reactants are stored inside the cell and the cell must be recharged when they are used up. Statement II: The product of the H₂–O₂ fuel cell is hydrogen peroxide.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    B. Both Statement I and Statement II are incorrect

    Reactants are supplied continuously from outside and no recharging is needed; the product is water.

25. Given below are two statements. Statement I: The mass of a substance deposited at an electrode is directly proportional to the quantity of charge passed. Statement II: One faraday is the charge carried by 1 mol of electrons, about 96500 C.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    A. Both Statement I and Statement II are correct

    Both are Faraday's first law and the definition of the faraday.

26. Given below are two statements. Statement I: Metals that liberate hydrogen from dilute acids have positive standard reduction potentials. Statement II: Metals with negative standard reduction potentials liberate hydrogen from dilute acids.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    D. Statement I is incorrect but Statement II is correct

    Metals above hydrogen in the series have negative E° (lower than 0.00 V) and are oxidised by H⁺; so Statement I is wrong.

27. Assertion (A): The molar conductivity of acetic acid increases sharply on dilution. Reason (R): The degree of dissociation of acetic acid increases on dilution.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    A. Both A and R are true and R is the correct explanation of A

    More ions are produced per mole as α rises, so Λm rises steeply.

28. Assertion (A): The lead storage battery is a secondary cell. Reason (R): The electrolyte of the lead storage battery is sulphuric acid.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    B. Both A and R are true but R is not the correct explanation of A

    It is secondary because the discharge reaction can be reversed by charging; the electrolyte being H₂SO₄ does not explain that.

29. Assertion (A): The H₂–O₂ fuel cell is a pollution-free source of energy. Reason (R): A fuel cell must be recharged at regular intervals to keep working.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    C. A is true but R is false

    A is true because the product is water. R is false: reactants are supplied continuously and no recharging is needed.

30. Assertion (A): The molar conductivity of a strong electrolyte increases on dilution because the number of ions per mole of the electrolyte increases. Reason (R): Inter-ionic attractions decrease on dilution.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    D. A is false but R is true

    A strong electrolyte is already fully ionised, so ions per mole do not increase; Λm rises because inter-ionic attraction falls. A is false, R is true.

31. Assertion (A): A scratched zinc-coated iron article does not rust easily. Reason (R): Zinc has a lower reduction potential than iron and is oxidised in preference to iron.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    A. Both A and R are true and R is the correct explanation of A

    Zinc (−0.76 V) is more easily oxidised than iron (−0.44 V) and acts as the sacrificial anode, protecting iron.

32. Assertion (A): Electrolysis of molten NaCl gives sodium metal at the cathode. Reason (R): Chlorine gas is liberated at the anode in this electrolysis.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    B. Both A and R are true but R is not the correct explanation of A

    Both are true, but the formation of Cl₂ at the anode does not explain why Na⁺ is reduced at the cathode.

33. Match List I with List II. List I: a. Dry cell b. Mercury cell c. Lead storage battery d. H₂–O₂ fuel cell. List II: i. Pb anode, PbO₂ cathode, about 38% H₂SO₄ ii. Zn–Hg amalgam anode and HgO cathode iii. Porous carbon electrodes with catalyst and continuous supply of gases iv. Zinc container with a paste of NH₄Cl and ZnCl₂

    1. a-i, b-ii, c-iv, d-iii
    2. a-ii, b-iv, c-i, d-iii
    3. a-iv, b-ii, c-i, d-iii
    4. a-iv, b-ii, c-iii, d-i
    Answer

    C. a-iv, b-ii, c-i, d-iii

    Each cell is matched by its construction.

34. Match List I with List II. List I: a. Conductivity b. Molar conductivity c. Cell constant d. Resistivity. List II: i. Ω m ii. S m⁻¹ iii. m⁻¹ iv. S m² mol⁻¹

    1. a-iv, b-ii, c-i, d-iii
    2. a-ii, b-iv, c-iii, d-i
    3. a-i, b-iv, c-iii, d-ii
    4. a-ii, b-iii, c-iv, d-i
    Answer

    B. a-ii, b-iv, c-iii, d-i

    κ: S m⁻¹; Λm: S m² mol⁻¹; cell constant l/A: m⁻¹; ρ: Ω m.

35. Match List I with List II. List I: a. Anode product from molten NaCl b. Anode product from aqueous CuSO₄ with Pt electrodes c. Anode change in aqueous CuSO₄ with Cu electrodes d. Cathode product from aqueous NaCl. List II: i. H₂ ii. Cu²⁺ ions in solution iii. O₂ iv. Cl₂

    1. a-iv, b-ii, c-iii, d-i
    2. a-iii, b-iv, c-ii, d-i
    3. a-i, b-iii, c-ii, d-iv
    4. a-iv, b-iii, c-ii, d-i
    Answer

    D. a-iv, b-iii, c-ii, d-i

    Molten NaCl: Cl₂; Pt/CuSO₄: water oxidised to O₂; Cu anode dissolves; aqueous NaCl cathode: water reduced to H₂.

36. Match List I with List II. List I: a. Li⁺/Li b. Zn²⁺/Zn c. Cu²⁺/Cu d. Ag⁺/Ag. List II (standard reduction potential): i. +0.80 V ii. −0.76 V iii. −3.05 V iv. +0.34 V

    1. a-iii, b-iv, c-ii, d-i
    2. a-ii, b-iii, c-iv, d-i
    3. a-iii, b-ii, c-iv, d-i
    4. a-iii, b-ii, c-i, d-iv
    Answer

    C. a-iii, b-ii, c-iv, d-i

    Li −3.05, Zn −0.76, Cu +0.34, Ag +0.80 V.

37. Match List I with List II. List I: a. Anode reaction of rusting b. Cathode reaction of rusting c. Rust d. Sacrificial protection of iron. List II: i. Fe₂O₃·xH₂O ii. Iron connected to magnesium iii. O₂ + 4H⁺ + 4e⁻ → 2H₂O iv. Fe → Fe²⁺ + 2e⁻

    1. a-iv, b-iii, c-ii, d-i
    2. a-iv, b-iii, c-i, d-ii
    3. a-i, b-iii, c-iv, d-ii
    4. a-iii, b-iv, c-i, d-ii
    Answer

    B. a-iv, b-iii, c-i, d-ii

    Anode: Fe oxidised; cathode: O₂ reduced; rust is hydrated Fe₂O₃; a more reactive metal protects iron sacrificially.

38. Match List I with List II. List I (ion reduced at the cathode to the metal): a. Ag⁺ b. Cu²⁺ c. Al³⁺ d. Na⁺. List II (mass deposited by 1 faraday): i. 9 g ii. 23 g iii. 31.75 g iv. 108 g. (Ag = 108, Cu = 63.5, Al = 27, Na = 23)

    1. a-iii, b-iv, c-i, d-ii
    2. a-iv, b-i, c-iii, d-ii
    3. a-iv, b-iii, c-i, d-ii
    4. a-iv, b-iii, c-ii, d-i
    Answer

    C. a-iv, b-iii, c-i, d-ii

    Mass per faraday = M/n: Ag 108/1, Cu 63.5/2 = 31.75, Al 27/3 = 9, Na 23/1.

39. At 298 K a 0.01 M solution of a weak monobasic acid has Λm = 39.1 S cm² mol⁻¹, and its Λ°m is 391 S cm² mol⁻¹. The dissociation constant Ka of the acid is about

    1. 1.00 × 10⁻⁴
    2. 1.00 × 10⁻³
    3. 1.11 × 10⁻³
    4. 1.11 × 10⁻⁴
    Answer

    D. 1.11 × 10⁻⁴

    α = 39.1/391 = 0.1. Ka = cα²/(1 − α) = 0.01 × 0.01/0.9 = 1.11 × 10⁻⁴.

40. A copper electrode dips in 0.01 M Cu²⁺ solution at 298 K. Given E°(Cu²⁺/Cu) = +0.34 V and 2.303RT/F = 0.059 V, the electrode potential is

    1. 0.281 V
    2. 0.399 V
    3. 0.222 V
    4. 0.340 V
    Answer

    A. 0.281 V

    E = 0.34 + (0.059/2) log 0.01 = 0.34 − 0.059 = 0.281 V.

41. Using the electrochemical series, which of the following metals will liberate hydrogen gas from dilute hydrochloric acid?

    1. Copper
    2. Silver
    3. Gold
    4. Zinc
    Answer

    D. Zinc

    Zn (−0.76 V) is above hydrogen and is oxidised by H⁺. Cu, Ag and Au have positive E° and do not react.

42. A current of 9.65 A is passed for 1000 s through molten MgCl₂ with inert electrodes (Mg = 24, F = 96500 C mol⁻¹). The mass of magnesium deposited is

    1. 4.8 g
    2. 2.4 g
    3. 1.2 g
    4. 0.6 g
    Answer

    C. 1.2 g

    Q = 9.65 × 1000 = 9650 C = 0.1 F. Mg²⁺ + 2e⁻ → Mg needs 2 F per mol, so 0.05 mol × 24 = 1.2 g.

43. A concentration cell is made as Cu | Cu²⁺(0.01 M) || Cu²⁺(1.0 M) | Cu at 298 K. Taking 2.303RT/F = 0.059 V, its EMF is

    1. 0.00 V
    2. 0.059 V
    3. 0.0295 V
    4. 0.118 V
    Answer

    B. 0.059 V

    E°cell = 0. E = 0 − (0.059/2) log(0.01/1.0) = −0.0295 × (−2) = 0.059 V.

44. A cell filled with 0.1 M KCl (κ = 0.0100 S cm⁻¹) has a resistance of 150 Ω. The same cell filled with a sulphuric acid solution has a resistance of 50 Ω. The conductivity of the acid solution is

    1. 0.03 S cm⁻¹
    2. 0.30 S cm⁻¹
    3. 3.0 S cm⁻¹
    4. 0.0033 S cm⁻¹
    Answer

    A. 0.03 S cm⁻¹

    Cell constant = κR = 0.0100 × 150 = 1.5 cm⁻¹. κ(acid) = 1.5/50 = 0.03 S cm⁻¹.

45. During the discharge of a lead storage battery, the density of the sulphuric acid electrolyte

    1. increases, because water is used up
    2. remains constant, because the electrolyte is not consumed
    3. increases, because PbSO₄ dissolves in the acid
    4. decreases, because H₂SO₄ is used up and water is formed
    Answer

    D. decreases, because H₂SO₄ is used up and water is formed

    Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O consumes acid and makes water, so the density falls.