What is the standard emf of a cell with the reaction Pb(s) + Hg₂²⁺(aq) → Pb²⁺(aq) + 2Hg(l) ? (Given: E°Pb²⁺/Pb = -0.13 V
E°cell = E°cathode - E°anode = 0.79 - (-0.13) = 0.92 V .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Corrosion and Applications of Electrochemistry