How do I find the K of O2(g)+2H2O(l)+2Cu(s)→4OH−(aq)+2Cu2+(aq)

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26-04-2018
Chemistry

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How do I find the K of O2(g)+2H2O(l)+2Cu(s)→4OH−(aq)+2Cu2+(aq)

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JABARBS JABARBS · Answer 1 · 2018-06-01T15:41:00+02:00

JABARBS JABARBS

01-06-2018

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

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