What is discharged at the cathode during the electrolysis of copper (ll) tetraoxosulphate (VI) solution?
Cu 2+ only
H + only
Cu 2+ and H
Cu 2+ and SO 2-
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So...During electrolysis of copper sulfate solutions, the negative cathode electrode attracts `Cu^(2+)` ions and `H^+` ions. Only copper ion is discharged hence reduce to copper ....Gracia's🙂...And hence why Cu2+ is discharged and not H+, Because Cu2+ is the Lowest in electrochemical series 👀 🚶🚶🚶
During the electrolysis of copper(II) tetraoxosulphate(VI) solution (CuSO₄), the cathode attracts positively charged ions from the solution. In this case, the positively charged copper ions (Cu²⁺) are attracted to the cathode. When the copper ions reach the cathode, they gain electrons and are reduced to form solid copper metal (Cu).
The half-reaction at the cathode during the electrolysis of CuSO₄ can be represented as:
Cu²⁺(aq) + 2e⁻ → Cu(s)
This equation shows that only copper ions (Cu²⁺) are discharged at the cathode to form solid copper. Hydrogen ions (H⁺) and sulphate ions (SO₄²⁻) are not discharged at the cathode during the electrolysis of CuSO₄.
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