Zn2+(aq) + 2 e-→ Zn(s) E° = -0.76 V ↓ E° → oxidation → anode
Cu2+(aq) + 2 e-→ Cu(s) E° = 0.34 V ↑ E° → reduction → cathode
E°cell = 1.10 V
Zn2+(aq) + 2 e-→ Zn(s)
Cu(s) → Cu2+(aq) + 2 e-
Cu2+ (aq) + Zn (s) → Cu (s) + Zn2+ (aq)
n = 2
Consider the cell described below:
Zn | Zn 2+(1.00 M) ‖ Cu 2+(1.00 M) | Cu
Calculate the cell potential after the reaction has operated long enough for the [Zn 2+] to have changed by 0.20 mol/L. (Assume T = 25°C.)
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