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Cathode (reduction): 2 H2O(l) + 2e− → H2(g) + 2 OH-(aq)Anode (oxidation): 4 OH-(aq) → O2(g) + 2 H2O(l) + 4 e−Overall reaction: 2 H2O(l) → 2 H2(g) + O2(g)

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Q: What is the product formed at cathode and anode on electrolysis of acidified water?
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