Practice question

Given: What is the EMF of a ll with Zn(s) | Zn²⁺(0.05 M) || H⁺(0.1 M) | H₂(g, 1 bar) | Pt(s) at 298 K, given E_{Zn^{2+/Zn⁰ = -0.76 V and E_{H^{+/H_2⁰ = 0 V ? These values define the system as per NCERT data. Formula: E_{ll⁰ = 0 - (-0.76) = 0.76 V, n = 2. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: E_{ll = E_{ll⁰ - 0.059/2 log frac[Zn^{2+][H^{+]² . Q = 0.05/(0.1)² = 5, log Q = 0.699 . E_{ll = 0.76 - 0.059/2 × 0.699 = 0.76 - 0.0206 = 0.7394 V approx 0.74 V . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.