Calculate the reduction potential of the half-cell given below at 298 K : Pt(s) | Cl ₂ (g, 9 atm) | Cl ^- (aq, 0.03 M) Given: E^ _ Cl ₂/ Cl ^- = 1.36 V ( Assume 2.303 RT F = 0.059 V )

Step-by-step solution

The reduction half-reaction for the given chlorine electrode is: Cl ₂ (g) + 2e^- 2 Cl ^- (aq) The Nernst equation for the reduction potential is given by: E = E^ - 0.059 n [ Cl ^-]^2 P_ Cl ₂ Here, the number of electrons transferred is n = 2 . Substituting the given values: E = 1.36 - 0.059 2 (0.03)^2 9 E = 1.36 - 0.059 2 9 10⁻⁴ 9 E = 1.36 - 0.059 2 (10⁻⁴) E = 1.36 - 0.059 2 (-4) E = 1.36 + 0.118 E = 1.478 V Answer: 1.478 V