Electrochemistry

Question 1

The standard electrode potential (E°) for the half-cell reaction Fe³⁺ + e⁻ → Fe²⁺ at 298 K is (Given: E°(Fe³⁺/Fe) = −0.04 V and E°(Fe²⁺/Fe) = −0.44 V at 298 K)

Accepted Answer

+0.76 V. Electrode potentials are not additive, but Gibbs energies are — that is the key. Writing ΔG for the three-electron, one-electron and two-electron steps gives −3F(−0.04) = −1F·E° + −2F(−0.44). Solving for the one-electron couple gives E° = +0.76 V.

Answer

+0.40 V

Answer

−0.48 V

Answer

+0.92 V

Question 2

For a salt XY, which is a strong electrolyte, the plot of Λm versus √c has a slope of −90.0 S cm² mol⁻³ᐟ² L¹ᐟ² at 298 K. At 0.01 M concentration of XY, the value of Λm is 145.0 S cm² mol⁻¹. The limiting molar conductivity of the Y⁻ ion (Λ°(Y⁻), in S cm² mol⁻¹) at 298 K will be (Given: Λ°(X⁺) = 74.0 S cm² mol⁻¹)

Accepted Answer

80.0. Debye-Hückel-Onsager gives Λm = Λ°m − A√c, so Λ°m = 145.0 + 90.0 × √0.01 = 145.0 + 9.0 = 154.0 S cm² mol⁻¹. Kohlrausch's law then splits that between the ions: Λ°(Y⁻) = 154.0 − 74.0 = 80.0 S cm² mol⁻¹.

Answer

100.0

Answer

90.0

Answer

76.0

Question 3

If the molar conductivity (Λₘ) of a 0.050 mol L⁻¹ solution of a monobasic weak acid is 90 S cm² mol⁻¹, its extent (degree) of dissociation will be [Assume λ°₊ = 349.6 S cm² mol⁻¹ and λ°₋ = 50.4 S cm² mol⁻¹]

Accepted Answer

0.225. The degree of dissociation of a weak electrolyte is α = Λₘ/Λ°ₘ, where Λ°ₘ is the limiting molar conductivity. By Kohlrausch’s law of independent migration of ions, Λ°ₘ = λ°₊ + λ°₋ = 349.6 + 50.4 = 400 S cm² mol⁻¹. Hence α = 90/400 = 0.225. The concentration is not needed for this part — it would only matter if the dissociation constant were asked.

Answer

0.115

Answer

0.125

Answer

0.215

Q1.The standard electrode potential (E°) for the half-cell reaction Fe³⁺ + e⁻ → Fe²⁺ at 298 K is (Given: E°(Fe³⁺/Fe) = −0.04 V and E°(Fe²⁺/Fe) = −0.44 V at 298 K)

Q3.If the molar conductivity (Λₘ) of a 0.050 mol L⁻¹ solution of a monobasic weak acid is 90 S cm² mol⁻¹, its extent (degree) of dissociation will be [Assume λ°₊ = 349.6 S cm² mol⁻¹ and λ°₋ = 50.4 S cm² mol⁻¹]