Practice question

Given: What is the EMF of the ll Zn(s) | Zn²⁺(0.1 M) || Cu²⁺(0.01 M) | Cu(s) at 298 K, given E_{Zn^{2+/Zn⁰ = -0.76 V and E_{Cu^{2+/Cu⁰ = 0.34 V ? These values define the system as per NCERT data. Formula: E_{ll⁰ = 0.34 - (-0.76) = 1.10 V, n = 2. This is standard NCERT relation. Substitution & Calculation: E_{ll = E_{ll⁰ - 0.059/2 log frac[Zn^{2+][Cu^{2+] = 1.10 - 0.059/2 log 0.1/0.01 . Q = 10, E_{ll = 1.10 - 0.0295 × 1 = 1.0705 V approx 1.07 V . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.