Trina's answer to Rosie's Secondary 4 E Maths Singapore question.
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positive electrode: 4OH-(aq) -> O2(g) + 2H2O(l) + 4e-
negative electrode: Cu2+(aq) + 2e- -> Cu(s)
negative electrode: Cu2+(aq) + 2e- -> Cu(s)
Date Posted:
8 months ago
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