Calculate the reduction potential of the half-cell represented below at 298 K: Pt (s) | Cl ₂ (g, 0.1 atm) | Cl ⁻ (aq, 0.01 M) Given: E^ _ Cl ₂/ Cl ⁻ = 1.36 V, 2.303 RT F = 0.059 V

Step-by-step solution

The corresponding reduction half-cell reaction is: Cl ₂ (g) + 2 e ⁻ 2 Cl ⁻ (aq) Using the Nernst equation for the reduction potential: E = E^ _ Cl ₂/ Cl ⁻ - 0.059 n [ Cl ⁻]^2 P_ Cl ₂ Given values: E^ _ Cl ₂/ Cl ⁻ = 1.36 V n = 2 [ Cl ⁻] = 0.01 M = 10⁻² M P_ Cl ₂ = 0.1 atm = 10⁻¹ atm Substituting the values into the Nernst equation: E = 1.36 - 0.059 2 (10⁻²)^2 10⁻¹ E = 1.36 - 0.0295 10⁻⁴ 10⁻¹ E = 1.36 - 0.0295 (10⁻³) E = 1.36 - 0.0295 (-3) E = 1.36 + 0.0885 E = 1.4485 V Answer: 1.4485 V