Determine E°cell at 25 °C for the following reaction: Pb2+(aq) + Cu(s) → Cu2+(aq) + Pb(s)
The half-reactions are:
Pb2+(aq) + 2 e- → Pb(s) E° = -0.125 V
Cu2+(aq) + 2 e- → Cu(s) E° = +0.337 V

Question

Quizplus · Accepted Answer

E°cell = E°cathode - E°anode = 0.337 V - (-0.125 V) = 0.462 V. However, since the reaction is reversed, the sign changes, resulting in -0.462 V.

Determine E°cell at 25 °C for the Following Reaction: Pb2+(aq)