Determine E°_cell for the reaction: Pb²⁺(aq) + Zn(s) → Pb(s) + Zn²⁺(aq) .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: Pb²⁺(aq)+ Zn(s)→ Pb(s)+ Zn²⁺(aq).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

The E°cell can be calculated using the formula E°cell = E°cathode - E°anode. 
In this case, since Pb is reduced (gains electrons) and Zn is oxidized (loses electrons), the half reaction with the more positive E° value (Zn) is the anode and the one with the more negative E° value (Pb) is the cathode. 
Therefore, E°cell = E°cathode - E°anode = -0.125 V - (-0.763 V) = 0.638 V. 
Note that the value is positive, indicating that the reaction is spontaneous in the forward direction (as written).

Determine E°cell for the Reaction: Pb2+(aq)+ Zn(s)→ Pb(s)+ Zn2+(aq)

Determine E°_cell for the Reaction: Pb²⁺(aq)+ Zn(s)→ Pb(s)+ Zn²⁺(aq)