Copper electrodes versus platinum electrodes in copper(II) sulfate electrolysis
A student electrolyses 100 cm3 of 0.500 mol dm−3 copper(II) sulfate solution using two copper electrodes. The current is held at an average of 0.500 A for 30.0 minutes; the ammeter fluctuates by about ±0.02 A. The cathode gains a salmon-pink deposit and the anode slowly becomes thinner. The student repeats the experiment with inert platinum electrodes under the same conditions and observes gas bubbles at the anode. In the copper-electrode cell the blue colour stays almost unchanged, but in the platinum cell it fades. The teacher asks the student to calculate the expected mass change at one electrode, explain why the copper-ion concentration behaves differently in the two cells, and decide which cell would be suitable for electroplating a steel spoon without changing the electrolyte concentration significantly.
1. Calculate the expected mass change at the cathode in the copper-electrode cell and the approximate uncertainty caused by the current fluctuation.
Read the worked answer
- Charge passed: Q = It = 0.500 × (30.0 × 60) = 900 C.
- Amount of electrons: n(e−) = Q/F = 900/96 500 = 9.33 × 10−3 mol.
- Cathode reaction: Cu2+(aq) + 2e− → Cu(s), so n(Cu) = n(e−)/2 = 4.66 × 10−3 mol.
- Mass of copper deposited = 4.66 × 10−3 × 63.5 = 0.296 g, about 0.30 g.
- The anode loses the same mass if the current efficiency is 100% and the copper anode dissolves as Cu2+.
- The current uncertainty is ±0.02/0.500 = ±4.0%. The mass uncertainty is therefore about ±0.012 g, so the expected cathode gain is roughly 0.296 ± 0.012 g. The judgement is that the calculated value is an average, not an exact prediction, because the current fluctuated and some charge may be used in side reactions.
2. Explain why the copper(II) ion concentration remains almost constant with copper electrodes but falls with platinum electrodes. Include electrode equations.
Read the worked answer
With copper electrodes, the anode reaction is Cu(s) → Cu2+(aq) + 2e− and the cathode reaction is Cu2+(aq) + 2e− → Cu(s). Copper dissolves at the anode at the same rate as copper is deposited at the cathode, so Cu2+ ions are removed and replaced in equal amounts. The concentration of copper(II) ions therefore stays almost constant, although the anode loses mass and the cathode gains mass.
With platinum electrodes, platinum is inert. The cathode reaction is still Cu2+(aq) + 2e− → Cu(s), so copper ions are removed from solution. At the anode, water is oxidised: 2H2O(l) → O2(g) + 4H+(aq) + 4e−. No copper ions are produced to replace those deposited, so [Cu2+] falls and the blue colour fades. The solution also becomes more acidic as H+ ions are generated. This explains the gas bubbles at the platinum anode and the difference in colour change between the two cells.
3. Which cell should be used to electroplate a steel spoon, and what practical conditions would give an even coating?
Read the worked answer
The copper-electrode cell should be used. The steel spoon must be made the cathode so that Cu2+ ions are reduced and deposited onto it, and the anode should be pure copper. The electrolyte should be copper(II) sulfate solution. Because the copper anode dissolves as the cathode plates, [Cu2+] remains roughly constant, so the electrolyte does not need frequent replacement and the coating thickness is easier to control.
For an even coating, clean and degrease the spoon first, because grease or oxide prevents adhesion. Use a moderate current rather than a very high one, since too high a current gives a rough, powdery deposit. Stir or gently move the electrolyte to keep the ion concentration uniform, and keep the spoon fully immersed with a suitable shape and spacing of anode. A platinum anode would not be suitable: it would produce oxygen gas at the anode and steadily remove Cu2+ from the electrolyte, so the plating rate would change and the solution would become acidic.