Practice question

Given: What is the EMF of the ll Pb(s) | Pb²⁺(0.05 M) || Ag⁺(0.002 M) | Ag(s) at 298 K, given E_{Pb^{2+/Pb⁰ = -0.13 V and E_{Ag^{+/Ag⁰ = 0.80 V ? These values define the system as per NCERT data. Formula: E_{ll⁰ = 0.80 - (-0.13) = 0.93 V, n = 2. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: E_{ll = E_{ll⁰ - 0.059/2 log frac[Pb^{2+][Ag^{+]² . Q = 0.05/(0.002)² = 12500, log Q = 4.0969 . E_{ll = 0.93 - 0.059/2 × 4.0969 = 0.93 - 0.1209 = 0.8091 V approx 0.81 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.