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OpenStudy (anonymous):
ELECTROLYSIS: How do you find the product of Copper sulphate(CuSO4) at the anode and cathode .And the half equations at the anode and cathode
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OpenStudy (anonymous):
At Anode: 4OH- (aq) → 2H2O (l) + O2 (g) + 4e- At Cathode: Cathode: Cu2+ (aq) + 2e- → Cu (s) Overall reaction: 2CuSO4 (aq) + 2H2O (l) → 2Cu (s) + 2H2SO4 (aq) + O2 (g)
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