Electrolysis of CuSO₄
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flagWhat you'll discover
- arrow_forwardDescribe the electrode reactions at copper anode and cathode
- arrow_forwardExplain why the solution's blue colour stays constant with copper electrodes
- arrow_forwardMeasure mass deposited and verify m = ZIt quantitatively
- arrow_forwardDefine the electrochemical equivalent Z and calculate it from M and F
- arrow_forwardPredict the effect of doubling current or time on the deposit
What happens at each electrode
Copper sulphate solution contains mobile Cu²⁺ and SO₄²⁻ ions. Connect copper electrodes to a DC supply and the ions drift: Cu²⁺ toward the negative cathode, SO₄²⁻ toward the positive anode. At the cathode, copper ions gain electrons and deposit as fresh pink-brown metal: Cu²⁺ + 2e⁻ → Cu. At the copper anode, the easiest oxidation is the copper of the electrode itself dissolving: Cu → Cu²⁺ + 2e⁻.
The net effect is elegant: copper transfers from anode to cathode while the solution composition stays exactly constant — every Cu²⁺ removed at the cathode is replaced at the anode, which is why the blue colour never fades. (With inert platinum or carbon electrodes the story changes: oxygen gas evolves at the anode instead, Cu²⁺ is genuinely consumed, and the blue fades.) This anode-dissolves arrangement is exactly how industrial copper refining and all electroplating work.
Faraday's first law
Faraday's first law of electrolysis (1833): the mass liberated at an electrode is directly proportional to the quantity of charge passed. Charge Q = I × t (amperes × seconds), so m = Z·I·t, where Z is the electrochemical equivalent — the mass deposited per coulomb.
Z follows from atomic bookkeeping. One mole of electrons carries F = 96,500 C (the Faraday constant). Depositing one Cu atom needs 2 electrons, so one mole of copper (63.5 g) needs 2 × 96,500 C. Hence Z(Cu) = 63.5 / (2 × 96,500) = 0.000329 g/C. Run 1.0 A for 30 minutes: m = 0.000329 × 1.0 × 1800 = 0.59 g — a gain you can weigh easily. The law is beautifully linear: double the current or double the time, and exactly double the copper lands on the cathode. Test it in the simulation.
Doing it accurately
The measurement is a sandwich: weigh the clean, dry cathode; electrolyse at a steady, gentle current for a known time; rinse the cathode with distilled water, dry it carefully, and reweigh. The mass gain is your m. Keep current density low — around 1 A with small electrodes — because forcing current too fast deposits copper as a spongy, poorly-adherent layer that flakes off during rinsing and undercuts your result.
Other error sources: a fluctuating current (use a rheostat and watch the ammeter — the I in ZIt must be the average), timing sloppiness, weighing the cathode wet, and rubbing the deposit off while drying (blot, don't wipe). A neat check on your whole experiment: the anode's mass loss should equal the cathode's gain. If they differ much, some copper ended up as sludge instead of plate. As a bonus check of Faraday's second law, the same charge would deposit silver in the ratio of its equivalent weight — 108/1 versus copper's 63.5/2.