What is the emf of the cell Pb(s) | Pb²⁺(0.005 M) || Ag⁺(0.05 M) | Ag(s) at 298 K? (Given: E°Pb²⁺/Pb = -0.13 V , E°Ag⁺/A
E°cell = 0.80 - (-0.13) = 0.93 V . Ecell = 0.93 - (0.059/2) log (0.005/0.05²) = 0.93 - 0.0295 = 0.9005 V .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Corrosion and Applications of Electrochemistry