Determine E°_cell for the reaction:
Pb(s) + Zn²⁺(aq) → Pb²⁺(aq) + Zn(s) .
The half reactions are:
Pb²⁺(aq) + 2 e^- → Pb(s) E° = -0.125 V
Zn²+(aq) + 2 e^- → Zn(s) E° = -0.763 V

Question

Determine E°_cell for the reaction: Pb(s)+ Zn²⁺(aq)→ Pb²⁺(aq)+ Zn(s). The half reactions are: Pb²⁺(aq)+ 2 e^- → Pb(s)E° = -0.125 V Zn²+(aq)+ 2 e^- → Zn(s)E° = -0.763 V

Quizplus · Accepted Answer

To determine the standard cell potential (E°cell), we need to subtract the reduction potential of the anode (Pb) from the reduction potential of the cathode (Zn): E°cell = E°reduction (Zn) - E°reduction (Pb) E°cell = (-0.763 V) - (-0.125 V) E°cell = -0.638 V Since the question asks for E°_cell (where _{represents solid, liquid, or gas and} represents the standard state aqueous solution), the final answer is -0.638 V, which is choice B.

Determine E°cell for the Reaction

Determine E°_cell for the Reaction