Practice question

Given: What is the EMF of a ll with Ni(s) | Ni²⁺(0.005 M) || Ag⁺(0.001 M) | Ag(s) at 298 K, given E_{Ni^{2+/Ni⁰ = -0.25 V and E_{Ag^{+/Ag⁰ = 0.80 V ? These values define the system as per NCERT data. Formula: E_{ll⁰ = 0.80 - (-0.25) = 1.05 V, n = 2. This is standard NCERT relation. Substitution & Calculation: E_{ll = E_{ll⁰ - 0.059/2 log frac[Ni^{2+][Ag^{+]² . Q = 0.005/(0.001)² = 5000, log Q = 3.699 . E_{ll = 1.05 - 0.059/2 × 3.699 = 1.05 - 0.109 = 0.941 V approx 0.94 V . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.