Respuesta :

O₂(g) + 2H₂O(l) + 2CU(s) → 4OH⁻(aq) +2CU⁴(aq)
Anode: 2CU(s) →2CU²⁺(aq) + 4C⁻ 
E⁰ = 0.34V
Cathode: O₂(g) + 2H₂O(l) + 4e⁻ → 40⁴⁻(aq)
E⁰ = 0.40
E⁰ cell = E⁰ cathode - E⁰ anode = 0.40 -(.34V) = 0.06V
∴ 0.06 = 0.059L/4 log K
n = 4e⁻
K = 10⁴ 
= 1.0 × 10₄