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The equation for the anode space (pH=14) for electrolysis is:
4·OH– → O2 + 2·H2O
Describe the same situation (including the electrons) for total for 6·H2O !
Solution: - 4·OH– → O2 + 2·H2O or more precisely:
- 4·OH– → O2 + 2·H2O + 4·e– // [translation unavailable] 4H+
- 4·OH– + 4H+ → O2 + 2·H2O + 2H+ 2H+ + 4·e–
- 4·H2O → O2 + 2·H3O+ + 2H+ + 4·e– // [translation unavailable] 2·H2O
- 6·H2O → O2 + 2·H3O+ + (2H+ + 2·H2O) + 4·e–
- 6·H2O → O2 + 2·H3O+ + ( 2·H3O+ ) + 4·e–
- 6·H2O → O2 + 4·H3O+ + 4·e–
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