In the reaction, F e_ (a q) ²⁺+A g_ (a q) ⁺ F e_ (a q) ³⁺+A g_ (s) The standard emf ( E ^ ) of the cell is: [ . Given: .E_ Ag ⁺ / Ag ^o=a V ; E_ Fe ²⁺ / Fe ^o=b V ; E_ Fe ³⁺ / Fe ^o=c V ]

Step-by-step solution

Oxidation (Anode): Fe²⁺ Fe³⁺ + e^- Reduction (Cathode): Ag^+ + e^- Ag E°_ cell = E°_ cathode - E°_ anode = E°_ Ag^+/Ag - E°_ Fe³⁺/Fe²⁺ Finding E°_ Fe³⁺/Fe²⁺ using Gibbs free energy: Reaction 1: Fe²⁺ + 2e^- Fe , E°₁ = b , G°₁ = -2Fb Reaction 2: Fe³⁺ + 3e^- Fe , E°₂ = c , G°₂ = -3Fc Target (Reaction 2 - Reaction 1): Fe³⁺ + e^- Fe²⁺ G°₃ = G°₂ - G°₁ = -3Fc - (-2Fb) = -3Fc + 2Fb -1 F E°₃ = -3Fc + 2Fb E°_ Fe³⁺/Fe²⁺ = 3c - 2b E°_ cell = a - (3c - 2b) = a + 2b - 3c