Cd2+ (aq) + Zn (s) → Cd (s) + Zn2+
↓ E° → oxidation → anode
↑ E° → reduction → cathode
Anode: Zn2+ (aq) + 2 e- → Zn (s) E = -0.763 V
Cathode: Cd2+ (aq) + 2e- → Cd (s) E = -0.403 V
E°cell = -0.403 - 0.763 = 0.36 V
Compute the equilibrium constant for the spontaneous reaction between Cd2+(aq) and Zn(s). Express your answer using two significant figures.
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