Electrolysis
Faraday’s first lawQ = I × tm = (Q × M) ÷ (n × F)For copper: M = 63.5 g mol⁻¹, n = 2 and F = 96,500 C mol⁻¹.
CATHODE · REDUCTIONCu²⁺ + 2e⁻ → Cu
ANODE · OXIDATIONCu → Cu²⁺ + 2e⁻
Predicted charge2400 C
Predicted electron amount0.0249 mol
Predicted copper deposit0.790 g
Copper transfers from the anode to the cathode. With copper electrodes, no gas is expected at either electrode. The selected time instantly shows Faraday’s-law prediction for the final deposit; moving ions show direction, not elapsed-time progress.
- Pass at least 1000 C of charge through the cell
- Deposit at least 0.20 g of copper on the cathode
- Deposit about 1.0 g of copper
- Pass 0.05 mol of electrons through the cell
- Now deposit about 2.0 g — double the copper needs double the charge