JEE Main202330 Jan 2023Morning ShiftChemistryElectrochemistryActual
Consider the cell Pt s H 2 g , 1 atm H + aq , 1 M | Fe 3 + aq , Fe 2 + aq Pt s When the potential of the cell is 0 . 712 V at 298 K , the ratio Fe 2 + / Fe 3 + is (Nearest integer) Given: Fe 3 + + e - = Fe 2 + , E ° Fe 3 + , Fe 2 + Pt = 0 . 771 2 . 303 RT F = 0 . 06 V
Correct answer
0
Step-by-step solution
Given cell reaction: Pt s H 2 g , 1 atm H + aq , 1 M | | Fe 3 + aq , Fe 2 + aq Pt s at anode(oxidation) H 2 ⟶ 2 H + + 2 e - At cathode(reduction) Fe aq 3 + + e - ⟶ Fe aq 2 + E ° cell = E cathode o   -   E anode o = E o H 2 | H +   + E o Fe 3 + ∣ Fe 2 + = 0 . 771   V Thus, using Nernst equation, ⇒ E = E ° - 0 · 06 1 log Fe 2 + Fe 3 + ⇒ 0 . 712 = 0 + 0 . 771 - 0 . 06 1 log Fe 2 + Fe 3 + ⇒ log Fe 2 + Fe 3 + = 0 . 059 0 . 06 ≈ 1 Fe 2 + Fe 3