Determine E°_cell for the reaction: MnO₄^-(aq) + 8 H⁺(aq) + 5 Fe²⁺(aq) → Mn²⁺(aq) + 4 H₂O(l) + 5 Fe³⁺(aq) .The half reactions are:
Fe³⁺(aq) + e^- → Fe²⁺(aq) E° = +0.771 V
MnO₄^-(aq) + 8 H⁺(aq) + 5 e^- → Mn²⁺(aq) + 4 H₂O(l) E° = 1.507 V

Question

Quizplus · Accepted Answer

To find the standard cell potential (E°cell), subtract the standard reduction potential of the anode (more negative or less positive) from the standard reduction potential of the cathode (more positive). Here, the MnO4- reduction has a potential of +1.507 V, and the Fe3+ reduction has a potential of +0.771 V. Since the Fe3+/Fe2+ couple acts as the anode (it is being oxidized), and the MnO4-/Mn2+ couple acts as the cathode (it is being reduced), the E°cell is calculated as follows: E°cell = E°cathode - E°anode = 1.507 V - 0.771 V = 0.736 V, which rounds to 0.74 V.

Determine E°cell for the Reaction: MnO4-(aq)+ 8 H+(aq)+ 5 Fe2+(aq)→

Determine E°_cell for the Reaction: MnO₄^-(aq)+ 8 H⁺(aq)+ 5 Fe²⁺(aq)→