What will be the emf of the following cell at 25^ C ? Fe / Fe ²⁺(0.001 M )|| H ⁺(0.01 M ) H ₂( ~g ) (1 Bar) Pt (s)E_ ( Fe ²⁺ / Fe ) ^ =-0.44 ~V ; E_ ( H ⁺ / H ₂ ) ^ =0.00 ~V

Step-by-step solution

The cell reaction is given by: Anode: Fe (s) Fe ²⁺(aq) + 2e⁻ Cathode: 2 H ⁺(aq) + 2e⁻ H ₂(g) Overall reaction: Fe (s) + 2 H ⁺(aq) Fe ²⁺(aq) + H ₂(g) The standard cell potential E^ _ cell is calculated as: E^ _ cell = E^ _ cathode - E^ _ anode E^ _ cell = 0.00 V - (-0.44 V ) = 0.44 V Using the Nernst equation at 25^ C : E_ cell = E^ _ cell - 0.0591 n [ Fe ²⁺] P_ H ₂ [ H ⁺]² Here n = 2 , [ Fe ²⁺] = 0.001 M , [ H ⁺] = 0.01 M , and P_ H ₂ = 1 bar . E_ cell = 0.44 - 0.0591 2 0.001 1 (0.01)² E_ cell = 0.44 - 0.02955 10⁻³