Q1(k). Electrolysis of copper sulfate

Answer in brief

For inert electrodes and quantitative current efficiency: 0.316 g Cu at the cathode and 0.0796 g O₂ at the anode.

Key points

• The charge is Q=It=0.800×1200=960 C; use F=96485 C mol⁻¹.
• Cathode: Cu²⁺ + 2e⁻ → Cu. Thus n_(Cu)=Q/(2F)=4.975×10⁻³ mol and m=0.316 g.
• Inert anode: 2H₂O → O₂ + 4H⁺ + 4e⁻. Thus n_(O₂)=Q/(4F)=2.487×10⁻³ mol, or 0.0796 g.
• Electrode material is unspecified: a copper anode instead dissolves by Cu → Cu²⁺ + 2e⁻, losing about 0.316 g, rather than evolving oxygen under ordinary…