Given E°(Cu²⁺/Cu) = +0.34 V and E°(Zn²⁺/Zn) = −0.76 V, E°_cell for Zn + Cu²⁺ is
- A
+0.42 V
- B
+0.76 V
- Ccheck_circle
+1.10 V
- D
−1.10 V
Explanation
E°_cell = E°_cathode − E°_anode = 0.34 − (−0.76) = +1.10 V.
Cell Potential and Free Energy
AP Chemistry· difficulty 4/5
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