Electrochemistry: Conductance, Kohlrausch's Law, Cell Potentials, the Nernst Equation and Cell Thermodynamics

This chapter answers Section XL0.6, “Electrochemistry”, of XL0 Chemistry, the part of the Life Sciences (XL) paper that every candidate answers: “Conductance, Kohlrausch law, cell potentials, EMF, Nernst equation, thermodynamic aspects and their applications.” It teaches conductance, conductivity and molar conductivity; Kohlrausch's laws and their uses (limiting conductivities of weak electrolytes, degree of dissociation, solubility, conductometric titrations); galvanic cells and standard electrode potentials; the Nernst equation with concentration cells, pH electrodes and membrane potentials; and the thermodynamics of cells, ΔG = −nFE, equilibrium constants from E° and ΔH and ΔS from the temperature coefficient, ending with biological redox potentials. The numericals are molar conductivity and Ka from conductance data, cell EMFs, Nernst potentials, ΔG°, K and Ksp from E°, and potentials of combined half-reactions.

1. Conductance, conductivity and molar conductivity

An electrolyte solution conducts by the movement of ions. Its resistance R (ohm) gives the conductance G = 1/R (siemens, S). Because R depends on the cell, the property of the solution itself is the conductivity κ = G × (l/A), where l/A, the cell constant (cm⁻¹), is set by the separation and area of the electrodes. In practice the cell constant is found with a standard KCl solution of known κ: cell constant = κ(KCl) × R(KCl). Conductance is measured with alternating current so that no electrolysis or polarisation occurs at the electrodes.

To compare electrolytes, conductivity is divided by concentration to give the molar conductivity, Λm = κ/c. With κ in S cm⁻¹ and c in mol L⁻¹, Λm = 1000κ/c in S cm² mol⁻¹ (the 1000 converts litres to cm³). On dilution κ falls, because each cm³ holds fewer ions, but Λm rises, because the ions of a strong electrolyte drag less on one another and a weak electrolyte ionises more. Worked example: a cell filled with 0.0100 M KCl (κ = 1.413 × 10⁻³ S cm⁻¹ at 25 °C) reads 250 Ω, so its cell constant is 1.413 × 10⁻³ × 250 = 0.3533 cm⁻¹. Filled with 0.0200 M NaCl it reads 150 Ω, so κ = 0.3533/150 = 2.355 × 10⁻³ S cm⁻¹ and Λm = 1000 × 2.355 × 10⁻³/0.0200 = 117.8 S cm² mol⁻¹.

2. Kohlrausch's laws and their applications

Plotting Λm against √c separates two kinds of electrolyte. For a strong electrolyte (KCl, HCl, CH₃COONa) the plot is a straight line at low concentration, Λm = Λ°m − A√c (Kohlrausch's square-root law, explained by interionic attraction), and extrapolating to c = 0 gives the limiting molar conductivity Λ°m. For a weak electrolyte (CH₃COOH, NH₃) Λm stays low and then shoots up as c approaches zero because ionisation increases, so Λ°m cannot be found by extrapolation.

Law of independent migration of ions. At infinite dilution each ion moves independently of its partner, so Λ°m = ν₊λ°₊ + ν₋λ°₋, where λ° are the limiting ionic molar conductivities and ν the number of each ion in the formula. The Λ°m of a weak electrolyte is then built from strong ones: Λ°(CH₃COOH) = Λ°(CH₃COONa) + Λ°(HCl) − Λ°(NaCl) = 91.0 + 426.1 − 126.4 = 390.7 S cm² mol⁻¹.

Limiting ionic molar conductivities in water at 25 °C (S cm² mol⁻¹, approximate)
Cationλ°Anionλ°
H⁺349.8OH⁻198.6
K⁺73.5Cl⁻76.3
Na⁺50.1CH₃COO⁻40.9
🎯 Why H⁺ and OH⁻ are so fast
They do not travel as whole ions. A proton passes from H₃O⁺ to a neighbouring water molecule along the hydrogen-bond network, and the charge moves by a relay of such hops (the Grotthuss mechanism); OH⁻ moves by the reverse relay. So conductance changes sharply whenever H⁺ or OH⁻ is consumed or produced.
  • Degree of dissociation and Ka of a weak electrolyte: α = Λm/Λ°m and Ka = cα²/(1 − α).
  • Solubility of a sparingly soluble salt: subtract the water's conductivity, κ(salt) = κ(solution) − κ(water). The saturated solution is so dilute that Λm ≈ Λ°m, so s = 1000κ(salt)/Λ°m in mol L⁻¹, and Ksp follows (for AgCl, Ksp = s²).
  • Conductometric titrations: with HCl titrated by NaOH the conductance falls as fast H⁺ is replaced by slow Na⁺, reaches a minimum at the equivalence point, then rises as excess OH⁻ builds up (a V-shaped plot). With a weak acid such as acetic acid it starts low, rises gently as Na⁺ and acetate replace the barely ionised acid, and rises steeply after the equivalence point. The end point is the intersection of straight branches, so coloured or turbid solutions pose no problem.
  • Transport number: the fraction of the current carried by an ion, t₊ = λ°₊/Λ°m for a 1:1 electrolyte; in dilute HCl H⁺ carries about 0.82 of the current.

3. Galvanic cells, electrode potentials and EMF

In a galvanic cell a spontaneous redox reaction drives electrons through an external circuit; in an electrolytic cell an external source drives a non-spontaneous one. In both, oxidation occurs at the anode and reduction at the cathode; in a galvanic cell the anode is the negative terminal. The Daniell cell is written Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s): anode on the left, a single bar for a phase boundary, a double bar for the salt bridge (KCl or KNO₃ in agar), which completes the circuit and keeps the liquid-junction potential small.

The EMF is the potential difference between the electrodes when no current flows, measured with a potentiometer or a high-impedance voltmeter. A single electrode potential cannot be measured alone, so each is quoted against the standard hydrogen electrode (SHE: Pt, H₂ at 1 bar, H⁺ at unit activity), assigned 0 V. Standard reduction potentials E° (unit activities, 298 K) then give E°cell = E°(cathode) − E°(anode), both written as reductions. For the Daniell cell, E° = +0.34 − (−0.76) = 1.10 V. In practice a secondary reference, the saturated calomel electrode (about +0.24 V against SHE) or the silver/silver chloride electrode, replaces the awkward SHE.

The electrochemical series ranks couples by E°. The more positive E°, the stronger the oxidised form as an oxidising agent (F₂, Ag⁺, Cu²⁺); the more negative, the stronger the reduced form as a reducing agent (Li, Zn). The oxidised form of a couple can oxidise the reduced form of any couple below it, so metals with negative E° displace H₂ from 1 M acid while copper and silver do not.

⚠️ E° is intensive; ΔG° is not
Writing Ag⁺ + e⁻ → Ag twice does not double +0.80 V, so E° for Cu | Cu²⁺ || Ag⁺ | Ag is 0.80 − 0.34 = 0.46 V. But when two half-reactions are combined into a third half-reaction, the electrons do not cancel and potentials do not add: add ΔG° = −nFE° instead. From Cu²⁺ + e⁻ → Cu⁺ (+0.153 V) and Cu⁺ + e⁻ → Cu (+0.521 V), E°(Cu²⁺/Cu) = (1 × 0.153 + 1 × 0.521)/2 = +0.337 V.

4. The Nernst equation

Away from standard conditions, E = E° − (RT/nF) ln Q, where Q is the reaction quotient of the cell reaction and n the number of electrons transferred. At 298 K, 2.303RT/F = 0.05916 V, so E = E° − (0.05916/n) log Q (at 310 K the factor is 0.0615 V). The same form applies to one electrode written as a reduction: for Mⁿ⁺ + ne⁻ → M, E = E° − (0.05916/n) log(1/[Mⁿ⁺]). As a cell discharges, Q rises and E falls, reaching zero at equilibrium, where Q = K. Worked example: for Zn | Zn²⁺(0.010 M) || Cu²⁺(1.0 M) | Cu at 298 K, Q = [Zn²⁺]/[Cu²⁺] = 0.010, so E = 1.10 − (0.05916/2) log 0.010 = 1.10 + 0.059 = 1.16 V: lowering the product ion raises the EMF.

  • Concentration cells: the same couple on both sides, so E° = 0 and E = (0.05916/n) log(c₂/c₁), with the more concentrated solution at the cathode for a metal-ion electrode. A tenfold ratio gives 59 mV when n = 1. The cell runs until the concentrations become equal.
  • pH measurement: for H⁺ + e⁻ → ½H₂ at 1 bar, E = −0.05916 pH at 298 K. A glass electrode responds the same way, about 59 mV per pH unit at 25 °C; it is calibrated with two standard buffers and read against a reference electrode.
  • Membrane (equilibrium) potentials: a membrane permeable to one ion settles at E = (RT/zF) ln([ion]out/[ion]in), the inside relative to the outside. For Na⁺ at 310 K with 145 mM outside and 12 mM inside, E = 0.02671 × ln(145/12) = +66.6 mV: the inside would have to be about 67 mV positive to stop Na⁺ entering.
  • Biological couples: with NAD⁺ + H⁺ + 2e⁻ → NADH at pH 7, E′ = E°′ − (0.05916/2) log([NADH]/[NAD⁺]) at 298 K, so the cell's NAD⁺/NADH ratio, not E°′ alone, sets the actual driving force.
⚠️ What the Nernst equation does not change
E° is a fixed constant for a reaction at a given temperature; only E changes with concentration. The amount of solid electrode does not enter Q, and n is the number of electrons in the balanced cell reaction, not the number per atom of the anode metal.

5. Thermodynamics of cells: ΔG, K, ΔS and ΔH

The electrical work a cell can deliver reversibly equals the fall in Gibbs energy: ΔG = −nFE and ΔG° = −nFE°, with F = 96485 C mol⁻¹. Since ΔG° = −RT ln K, ln K = nFE°/RT, or log K = nE°/0.05916 at 298 K. A positive E° means ΔG° < 0 and K > 1, and each 0.059 V of E° is worth a factor of ten in K when n = 1. For the Daniell cell, ΔG° = −2 × 96485 × 1.10 = −212 kJ mol⁻¹ and log K = 2 × 1.10/0.05916 = 37.2, so K ≈ 1.5 × 10³⁷: the reaction goes to completion.

Temperature coefficient. Because (∂ΔG/∂T)p = −ΔS, measuring E at several temperatures gives ΔS = nF(∂E/∂T)p and ΔH = −nFE + TΔS = −nF[E − T(∂E/∂T)p]. A cell whose EMF falls as temperature rises has a negative ΔS, and its reaction releases more heat than the electrical work it delivers. Electrochemical measurement thus gives ΔG, ΔH, ΔS and K without calorimetry.

Equilibrium constants from E°. Any equilibrium that is the sum of two half-reactions has a K obtainable from E° values: the solubility product of AgCl from the Ag⁺/Ag and AgCl/Ag couples, the ionic product of water, the stability constants of complexes and disproportionation constants. Cu⁺ disproportionates in water (2Cu⁺ → Cu²⁺ + Cu) because E° = 0.521 − 0.153 = +0.368 V is positive, giving K ≈ 1.7 × 10⁶.

6. Applications: potentiometric titrations and biological redox

Potentiometric titrations follow the EMF of an indicator electrode against a reference as titrant is added; the end point is the steepest part of the curve (maximum dE/dV). In an acid–base titration with a glass electrode the pH at half-neutralisation equals the pKa of the weak acid. In a redox titration of Fe²⁺ by Ce⁴⁺ the potential at half-equivalence equals E°(Fe³⁺/Fe²⁺), because [Fe³⁺] = [Fe²⁺] there.

Biological electron transfer runs from couples of lower E°′ to couples of higher E°′, where E°′ is the reduction potential at pH 7. For n electrons moving from donor to acceptor, ΔG°′ = −nFΔE°′, with ΔE°′ = E°′(acceptor) − E°′(donor). Electrons from NADH fall through about 1.14 V to O₂, releasing over 200 kJ per mole of NADH, which the respiratory chain conserves in steps as a proton gradient.

Standard reduction potentials of respiratory carriers (pH 7, 25 °C)
CoupleE°′ (V)
NAD⁺ + H⁺ + 2e⁻ → NADH−0.320
Ubiquinone + 2H⁺ + 2e⁻ → ubiquinol+0.045
Cytochrome c Fe(III) + e⁻ → cytochrome c Fe(II)+0.254
½O₂ + 2H⁺ + 2e⁻ → H₂O+0.816

Key takeaways

  • G = 1/R, κ = G × cell constant (found with standard KCl) and Λm = 1000κ/c in S cm² mol⁻¹. On dilution κ falls and Λm rises.
  • Kohlrausch: Λm = Λ°m − A√c for strong electrolytes; Λ°m = ν₊λ°₊ + ν₋λ°₋ for all, which gives Λ°m of weak electrolytes, α = Λm/Λ°m, Ka, solubilities (s = 1000κ/Λ°m) and conductometric end points. H⁺ and OH⁻ are fast by the Grotthuss relay.
  • Oxidation at the anode (negative in a galvanic cell); E°cell = E°(cathode) − E°(anode). E° is intensive, but combining half-reactions into a new half-reaction needs ΔG° = −nFE°, not added potentials.
  • Nernst: E = E° − (0.05916/n) log Q at 298 K, zero at equilibrium. It explains concentration cells, pH electrodes (59 mV per pH unit) and membrane potentials, E = (RT/zF) ln([out]/[in]).
  • ΔG° = −nFE° = −RT ln K, so log K = nE°/0.05916 at 298 K; ΔS = nF(∂E/∂T)p and ΔH = −nF[E − T(∂E/∂T)p]. E° data give Ksp, Kw, stability and disproportionation constants.
  • Biological electrons flow from low to high E°′ (pH 7): NADH (−0.320 V) to O₂ (+0.816 V), with ΔG°′ = −nFΔE°′. In potentiometric titrations the half-equivalence point gives pKa or E°.

Practice questions (18)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. When an aqueous solution of KCl is diluted at constant temperature, how do its conductivity κ and its molar conductivity Λm change?

    1. κ increases and Λm increases, because the ions move more freely in the more dilute solution
    2. κ increases and Λm decreases, because fewer ions are left to share the current at lower concentration
    3. κ decreases and Λm increases: there are fewer ions per cm³, but less interionic drag on each mole of salt
    4. κ decreases and Λm decreases, because both quantities are directly proportional to the concentration
    Show answer

    Answer: C — κ decreases and Λm increases: there are fewer ions per cm³, but less interionic drag on each mole of salt

    κ measures the conductance of a fixed volume, so it falls as each cm³ holds fewer ions. Λm = 1000κ/c refers to one mole of electrolyte, and with the ions farther apart the interionic attraction that slows them is weaker, so Λm rises towards Λ°m. Saying both fall treats Λm as if it were proportional to c, which is the common slip.

  2. A conductivity cell filled with 0.0100 M KCl (κ = 1.413 × 10⁻³ S cm⁻¹ at 25 °C) has a resistance of 125.0 Ω. Filled with a 0.0050 M solution of an electrolyte X at 25 °C, it has a resistance of 400.0 Ω. The molar conductivity of X is closest to:

    1. 88.3 S cm² mol⁻¹
    2. 0.0883 S cm² mol⁻¹
    3. 904 S cm² mol⁻¹
    4. 44.2 S cm² mol⁻¹
    Show answer

    Answer: A — 88.3 S cm² mol⁻¹

    Cell constant = κ(KCl) × R(KCl) = 1.413 × 10⁻³ × 125.0 = 0.1766 cm⁻¹. κ(X) = 0.1766/400.0 = 4.416 × 10⁻⁴ S cm⁻¹. Λm = 1000κ/c = 1000 × 4.416 × 10⁻⁴/0.0050 = 88.3 S cm² mol⁻¹. Omitting the factor 1000 gives 0.0883; scaling κ by 400/125 instead of 125/400 gives 904; using 0.0100 M in place of 0.0050 M gives 44.2.

  3. At 25 °C a 0.0100 M solution of acetic acid has a conductivity of 1.63 × 10⁻⁴ S cm⁻¹ (after subtracting the conductivity of the water). The limiting ionic molar conductivities are λ°(H⁺) = 349.8 and λ°(CH₃COO⁻) = 40.9 S cm² mol⁻¹. Using α = Λm/Λ°m and Ka = cα²/(1 − α), the value of Ka × 10⁵, to one decimal place, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 1.8

    Λm = 1000 × 1.63 × 10⁻⁴/0.0100 = 16.3 S cm² mol⁻¹. By Kohlrausch's law Λ°m = 349.8 + 40.9 = 390.7 S cm² mol⁻¹. α = 16.3/390.7 = 0.0417. Ka = 0.0100 × (0.0417)²/(1 − 0.0417) = 1.82 × 10⁻⁵, so Ka × 10⁵ = 1.8. Dropping the (1 − α) term gives 1.74 × 10⁻⁵, which rounds to 1.7.

  4. The limiting molar conductivities of H⁺ and OH⁻ in water are several times larger than those of Na⁺ and Cl⁻ because:

    1. H⁺ and OH⁻ carry larger charges than Na⁺ and Cl⁻, so the same field pulls on them harder
    2. H⁺ and OH⁻ are not hydrated in water, so they move as bare ions that meet very little drag
    3. H⁺ and OH⁻ are the lightest ions, and an ion's speed in solution depends on its mass alone
    4. their charge is relayed by proton transfers along the hydrogen-bonded network of water
    Show answer

    Answer: D — their charge is relayed by proton transfers along the hydrogen-bonded network of water

    In the Grotthuss mechanism a proton passes from H₃O⁺ to a neighbouring water molecule and the charge travels by a relay of such hops, much faster than any ion can diffuse; OH⁻ conducts by the reverse relay. All four ions carry a single charge, and H⁺ is in fact strongly hydrated (as H₃O⁺ and larger clusters), so the bare-ion explanation is wrong; speed in solution is governed by charge, size and hydration, not mass alone.

  5. 50 mL of 0.10 M HCl is titrated conductometrically with 1.0 M aqueous NH₃, a weak base. Which description of the conductance plotted against the volume of NH₃ added is correct?

    1. It rises steadily from the start and rises still more steeply after the equivalence point, as NH₄⁺ ions keep accumulating
    2. It falls until the equivalence point, as H⁺ is replaced by slower NH₄⁺, and then stays nearly constant as excess NH₃ barely ionises
    3. It falls until the equivalence point and then rises as steeply as it fell, because excess NH₃ supplies OH⁻ just as excess NaOH would
    4. It stays constant until the equivalence point, because one ion replaces another, and then falls as excess NH₃ dilutes the ions
    Show answer

    Answer: B — It falls until the equivalence point, as H⁺ is replaced by slower NH₄⁺, and then stays nearly constant as excess NH₃ barely ionises

    Before the equivalence point H⁺ (λ° ≈ 350) is replaced by NH₄⁺ (λ° ≈ 73.5), so the conductance falls. After it, the added NH₃ is a weak base whose ionisation is further suppressed by the NH₄⁺ already present (common-ion effect), so very few ions are added and the line is nearly flat; the titrant is concentrated, so dilution is negligible. A steep rise after the end point is what NaOH, a strong base supplying OH⁻ fully, would give.

  6. Which of the following statements about Kohlrausch's laws are correct?

    1. Λ°m of acetic acid can be found accurately by extrapolating its plot of Λm against √c to zero concentration
    2. Λ°m of acetic acid equals Λ°m(CH₃COONa) + Λ°m(HCl) − Λ°m(NaCl)
    3. For a strong electrolyte at low concentration, Λm decreases linearly with √c
    4. At infinite dilution λ° of Cl⁻ is larger in HCl than in KCl, because Cl⁻ is carried along by the faster cation
    Show answer

    Answer: B — Λ°m of acetic acid equals Λ°m(CH₃COONa) + Λ°m(HCl) − Λ°m(NaCl); C — For a strong electrolyte at low concentration, Λm decreases linearly with √c

    Strong electrolytes obey Λm = Λ°m − A√c, and the law of independent migration lets Λ°m of a weak acid be built from strong electrolytes: the Na⁺ and Cl⁻ terms cancel to leave λ°(H⁺) + λ°(CH₃COO⁻). Acetic acid is a weak electrolyte whose Λm rises steeply near zero concentration, so it cannot be extrapolated. Independence means each ion's λ° at infinite dilution does not depend on its partner.

  7. In a galvanic cell such as Zn | Zn²⁺ || Cu²⁺ | Cu:

    1. oxidation occurs at the anode, which is the negative terminal from which electrons enter the external circuit
    2. oxidation occurs at the cathode, which is the positive terminal into which electrons flow from the circuit
    3. reduction occurs at the anode, which is the negative terminal, because electrons collect on that electrode
    4. the anode is the positive terminal, as in electrolysis, and reduction occurs at the zinc electrode
    Show answer

    Answer: A — oxidation occurs at the anode, which is the negative terminal from which electrons enter the external circuit

    By definition oxidation occurs at the anode in every cell. In a galvanic cell zinc is oxidised (Zn → Zn²⁺ + 2e⁻), leaving electrons on the zinc, which is therefore the negative terminal; they flow through the external circuit to the copper cathode, where Cu²⁺ is reduced. The anode is positive only in an electrolytic cell, where an external source forces the reaction, which is why that choice is tempting.

  8. Standard reduction potentials at 298 K are Ag⁺/Ag +0.80 V, Cu²⁺/Cu +0.34 V, Fe²⁺/Fe −0.44 V and Zn²⁺/Zn −0.76 V. Which reaction is spontaneous with all species in their standard states?

    1. Copper reduces zinc ions: Cu(s) + Zn²⁺(aq) → Cu²⁺(aq) + Zn(s)
    2. Copper reduces iron(II) ions: Cu(s) + Fe²⁺(aq) → Cu²⁺(aq) + Fe(s)
    3. Copper reduces silver ions: Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s)
    4. Silver reduces zinc ions: 2Ag(s) + Zn²⁺(aq) → 2Ag⁺(aq) + Zn(s)
    Show answer

    Answer: C — Copper reduces silver ions: Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s)

    E° = E°(couple reduced) − E°(couple oxidised). For Ag⁺ oxidising Cu: 0.80 − 0.34 = +0.46 V, positive, so ΔG° < 0 and the reaction is spontaneous. The others give −0.76 − 0.34 (−1.10 V), −0.44 − 0.34 (−0.78 V) and −0.76 − 0.80 (−1.56 V), all negative: copper cannot reduce Zn²⁺ or Fe²⁺, and silver cannot reduce Zn²⁺.

  9. At 298 K, Fe³⁺ + e⁻ → Fe²⁺ has E° = +0.771 V and Fe²⁺ + 2e⁻ → Fe has E° = −0.447 V. Using ΔG° = −nFE°, the standard reduction potential of Fe³⁺ + 3e⁻ → Fe, in volts to three decimal places, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: -0.041

    ΔG°₁ = −1 × F × 0.771 and ΔG°₂ = −2 × F × (−0.447) = +0.894F. Sum: ΔG° = −0.771F + 0.894F = +0.123F for the three-electron reduction. E° = −ΔG°/(3F) = −0.123/3 = −0.041 V. Simply adding the potentials (0.771 − 0.447 = 0.324 V) is wrong because the electrons do not cancel in a sum of half-reactions.

  10. Given E°(Au³⁺/Au) = +1.50 V and E°(Ni²⁺/Ni) = −0.25 V at 298 K, the standard EMF of the cell Ni | Ni²⁺ || Au³⁺ | Au, whose reaction is 3Ni + 2Au³⁺ → 3Ni²⁺ + 2Au, is:

    1. +3.75 V, since the gold potential is doubled and the nickel potential tripled
    2. +1.75 V, since E° = E°(cathode) − E°(anode) whatever the coefficients
    3. +1.25 V, since the two reduction potentials are added to each other
    4. −1.75 V, since nickel is written on the left and is oxidised
    Show answer

    Answer: B — +1.75 V, since E° = E°(cathode) − E°(anode) whatever the coefficients

    E° is an intensive property: the coefficients that balance the electrons (six here) multiply ΔG° but not E°. E° = +1.50 − (−0.25) = +1.75 V. Multiplying each potential by its coefficient (2 × 1.50 + 3 × 0.25 = 3.75 V) is the classic trap; adding the reduction potentials gives 1.25 V, and the cell as written has the anode on the left, so its EMF is positive.

  11. A nerve cell has [K⁺] = 5.0 mM outside and 140 mM inside. Using E = (RT/zF) ln([K⁺]out/[K⁺]in) for the potential of the inside relative to the outside, with R = 8.314 J mol⁻¹ K⁻¹, F = 96485 C mol⁻¹ and T = 310 K, the K⁺ equilibrium potential in millivolts, to one decimal place, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: -89.0

    RT/F = 8.314 × 310/96485 = 0.02671 V and z = +1. ln(5.0/140) = −3.332. E = 0.02671 × (−3.332) = −0.0890 V = −89.0 mV: the inside must be negative to hold K⁺ in against its outward gradient. Using 298 K instead of 310 K gives −85.6 mV, and inverting the ratio gives +89.0 mV.

  12. The cell Cu | Cu²⁺(0.0010 M) || Cu²⁺(0.10 M) | Cu is set up at 298 K. Taking 2.303RT/F = 0.0592 V and ignoring activity coefficients, its EMF is:

    1. 0 V, because both electrodes are the same Cu²⁺/Cu couple with E° = +0.34 V
    2. 0.118 V, since a hundredfold ratio gives 0.0592 V for each factor of ten
    3. 0.0296 V, since n = 2 halves the slope and the ratio contributes one decade
    4. 0.0592 V, since E = (0.0592/2) log(0.10/0.0010) for the two-electron couple
    Show answer

    Answer: D — 0.0592 V, since E = (0.0592/2) log(0.10/0.0010) for the two-electron couple

    For a concentration cell E° = 0, and the Nernst equation gives E = (0.0592/n) log(c_cathode/c_anode) = (0.0592/2) × log 100 = 0.0296 × 2 = 0.0592 V. Using n = 1 gives 0.118 V; treating log 100 as 1 gives 0.0296 V. E° is zero, but E is not, because the concentrations differ.

  13. The cell Pt | H₂(g, 1 bar) | H⁺(test solution) || saturated calomel electrode has an EMF of 0.535 V at 298 K. Take the potential of the saturated calomel electrode as +0.241 V against the standard hydrogen electrode and 2.303RT/F = 0.05916 V. The pH of the test solution is closest to:

    1. 4.97
    2. 9.04
    3. 2.48
    4. 13.12
    Show answer

    Answer: A — 4.97

    The hydrogen electrode is the anode: E(H⁺/H₂) = −0.05916 pH. E(cell) = E(calomel) − E(H⁺/H₂) = 0.241 + 0.05916 pH = 0.535, so pH = (0.535 − 0.241)/0.05916 = 4.97. Ignoring the reference potential gives 9.04, adding it gives 13.12, and dividing by 2 × 0.05916 gives 2.48, but the slope is 0.05916 V per pH unit because one electron is transferred per H⁺.

  14. Which of the following statements about the Nernst equation and cell EMF are correct?

    1. When a galvanic cell reaction reaches equilibrium, E = 0 and the reaction quotient Q equals K
    2. Raising [Cu²⁺] in the cathode compartment of a Daniell cell raises its EMF
    3. The standard EMF E° of a cell changes as the concentrations in the cell change during discharge
    4. Doubling the mass of zinc metal in the anode compartment of a Daniell cell raises its EMF
    Show answer

    Answer: A — When a galvanic cell reaction reaches equilibrium, E = 0 and the reaction quotient Q equals K; B — Raising [Cu²⁺] in the cathode compartment of a Daniell cell raises its EMF

    At equilibrium ΔG = 0, so E = 0 and Q = K. For Zn + Cu²⁺ → Zn²⁺ + Cu, Q = [Zn²⁺]/[Cu²⁺], so more Cu²⁺ lowers Q and raises E. E° is a constant at a given temperature; only E changes with concentration. A pure solid has unit activity, so the amount of zinc does not enter Q or E.

  15. In the respiratory chain, electrons pass from NADH to O₂: NADH + H⁺ + ½O₂ → NAD⁺ + H₂O. At pH 7 and 298 K, E°′(NAD⁺/NADH) = −0.320 V and E°′(½O₂/H₂O) = +0.816 V. Using ΔG°′ = −nFΔE°′ with n = 2 and F = 96485 C mol⁻¹, ΔG°′ in kJ mol⁻¹, to one decimal place, is ____.

    Numerical answer — type the value.

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    Answer: -219.2

    ΔE°′ = E°′(acceptor) − E°′(donor) = 0.816 − (−0.320) = 1.136 V. ΔG°′ = −2 × 96485 × 1.136 = −219,214 J mol⁻¹ = −219.2 kJ mol⁻¹. Taking n = 1 halves the answer, and subtracting the potentials the wrong way round gives a positive ΔG°′ for a reaction that is strongly downhill.

  16. At 298 K, E°(Ag⁺/Ag) = +0.7996 V, and E° for AgCl(s) + e⁻ → Ag(s) + Cl⁻(aq) is +0.2223 V. Taking 2.303RT/F = 0.05916 V, the value of pKsp (= −log Ksp) for AgCl, to two decimal places, is ____.

    Numerical answer — type the value.

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    Answer: 9.76

    AgCl(s) → Ag⁺ + Cl⁻ is the sum of AgCl + e⁻ → Ag + Cl⁻ (E° = +0.2223 V) and Ag → Ag⁺ + e⁻ (the reverse of +0.7996 V). With n = 1 in both, E° = 0.2223 − 0.7996 = −0.5773 V. log Ksp = nE°/0.05916 = −0.5773/0.05916 = −9.758, so pKsp = 9.76 and Ksp = 1.7 × 10⁻¹⁰. The negative E° correctly signals that dissolution is not spontaneous under standard conditions.

  17. A galvanic cell with n = 2 has E = 1.0180 V at 298 K, and its temperature coefficient (∂E/∂T)p is −4.0 × 10⁻⁵ V K⁻¹. With F = 96485 C mol⁻¹, which pair of values for the cell reaction is correct?

    1. ΔS = +7.7 J K⁻¹ mol⁻¹ and ΔH = −194.1 kJ mol⁻¹
    2. ΔS = −3.9 J K⁻¹ mol⁻¹ and ΔH = −99.4 kJ mol⁻¹
    3. ΔS = −7.7 J K⁻¹ mol⁻¹ and ΔH = −198.7 kJ mol⁻¹
    4. ΔS = −7.7 J K⁻¹ mol⁻¹ and ΔH = −194.1 kJ mol⁻¹
    Show answer

    Answer: C — ΔS = −7.7 J K⁻¹ mol⁻¹ and ΔH = −198.7 kJ mol⁻¹

    ΔG = −nFE = −2 × 96485 × 1.0180 = −196.4 kJ mol⁻¹. ΔS = nF(∂E/∂T)p = 2 × 96485 × (−4.0 × 10⁻⁵) = −7.72 J K⁻¹ mol⁻¹. ΔH = ΔG + TΔS = −196.4 + 298 × (−7.72 × 10⁻³) = −196.4 − 2.3 = −198.7 kJ mol⁻¹. Subtracting TΔS instead of adding it gives −194.1, the tempting near-miss; using n = 1 gives the −3.9 and −99.4 pair.

  18. Which of the following statements are correct? Where needed, use E°(Cu²⁺/Cu⁺) = +0.153 V and E°(Cu⁺/Cu) = +0.521 V at 298 K.

    1. A cell reaction with a positive E° has a negative ΔG° and an equilibrium constant greater than 1
    2. Multiplying a balanced cell reaction by 2 doubles its ΔG° but leaves its E° unchanged
    3. Cu⁺ is stable in water towards disproportionation into Cu²⁺ and Cu
    4. A cell whose E° is negative can never drive current in the written direction, whatever the concentrations
    Show answer

    Answer: A — A cell reaction with a positive E° has a negative ΔG° and an equilibrium constant greater than 1; B — Multiplying a balanced cell reaction by 2 doubles its ΔG° but leaves its E° unchanged

    ΔG° = −nFE° = −RT ln K, so E° > 0 means ΔG° < 0 and K > 1; doubling the equation doubles n and hence ΔG°, while E° = −ΔG°/(nF) stays the same. For 2Cu⁺ → Cu²⁺ + Cu, E° = 0.521 − 0.153 = +0.368 V, so Cu⁺ does disproportionate. A negative E° can still give a positive E if Q is made small enough, as the Nernst equation shows.