Practice question

Given: What is the EMF of a ll with Fe(s) | Fe²⁺(0.01 M) || H⁺(1 M) | H₂(g, 1 bar) | Pt(s) at 298 K, given E_{Fe^{2+/Fe⁰ = -0.44 V and E_{H^{+/H_2⁰ = 0 V ? Formula: E_{ll⁰ = 0 - (-0.44) = 0.44 V, n = 2. Substitution & Calculation: E_{ll = E_{ll⁰ - 0.059/2 log frac[Fe^{2+][H^{+]² . Q = 0.01/1² = 0.01, log Q = -2 . E_{ll = 0.44 - 0.059/2 × (-2) = 0.44 + 0.059 = 0.499 V approx 0.50 V . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.