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- Aufgabe: Redox-24
Dans le L'espace cathodique (pH=0) est Les équation de réaction
2·H3O+ + 2·e– → H2 + 2·H2O
Formes ainsi à, que à Le Droite La page 4·OH– apparaître.
Lösung: - 2·H3O+ + 2·e– → H2 + 2·H2O // – 2·H+
- 2·H2O + 2·e– → H2 + 2·OH– // ·2
- 4·H2O + 4·e– → 2·H2 + 4·OH–
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