Calculate emf of the half cell given below : Pt (s) | H ₂ (g, 2 atm) | HCl (aq, 0.02 M) E^ _ H ₂/ H ⁺ = 0 V ( Given : 2.303 RT F = 0.059, 2 = 0.3010 )

Step-by-step solution

The given half-cell is represented as an oxidation electrode: Pt (s) | H ₂ (g) | H ⁺ (aq) . The corresponding oxidation half-cell reaction is: H ₂ (g) 2 H ⁺ (aq) + 2 e ⁻ Using the Nernst equation for the oxidation potential: E = E^ _ H ₂/ H ⁺ - 0.059 n [ H ⁺]^2 P_ H ₂ Given values: E^ _ H ₂/ H ⁺ = 0 V n = 2 [ H ⁺] = 0.02 M (since HCl is a strong monoprotic acid) P_ H ₂ = 2 atm Substituting the values into the Nernst equation: E = 0 - 0.059 2 (0.02)^2 2 E = -0.0295 4 10⁻⁴ 2 E = -0.0295 (2 10⁻⁴) E = -0.0295 ( 2 + 10⁻