14.11

[Emily Glaser 1F]

can someone do step by step of part d. I don't understand why, in the solutions manual, it says:

O2 +2H2O + 4e- --> 4OH- for the cathode (positive E value). OH- is listed before O2, so shouldn't the O2 be on the right side and OH on the left with a positive E value

[Clara Rehmann 1K]

Yes, but the catch with this problem is that the battery does not have a positive voltage, as the Eº of the anode is greater than that of the cathode.

Pt(s) | O₂(g) | H⁺(aq) || OH^-(aq) | O₂ (g) | Pt(s)

Overall reaction: 4H₂O(l) → 4H⁺(aq) + 4OH^-(aq)

Anode: O₂(g) + 4H⁺(aq) → 2H₂O(l) +1.23 V
Cathode: O₂(g) + 2H₂O(l) + 4e^- → 4OH^-(aq) +.4 V

Eº = Eº_cathode - Eº_anode = .4 V - 1.23 V = -.83 V

[904676178]

Yes, but the catch with this problem is that the battery does not have a positive voltage, as the Eº of the anode is greater than that of the cathode.

Pt(s) | O₂(g) | H⁺(aq) || OH^-(aq) | O₂ (g) | Pt(s)

Overall reaction: 4H₂O(l) → 4H⁺(aq) + 4OH^-(aq)

Anode: O₂(g) + 4H⁺(aq) → 2H₂O(l) +1.23 V
Cathode: O₂(g) + 2H₂O(l) + 4e^- → 4OH^-(aq) +.4 V

Eº = Eº_cathode - Eº_anode = .4 V - 1.23 V = -.83 V

How did you know what the anode is? Isn't the cathode supposed to be the one with the greater positive Eº?