Copper Plating: A Mass Gain Above 100% Efficiency
A student electroplates copper onto a brass key using copper electrodes and copper(II) sulfate solution. The key is weighed at 4.86 g, immersed as the cathode, and a current of 0.50 A is passed for 32.0 minutes (1920 s, so 960 C). At the end the key is rinsed in water, blotted dry and reweighed at 5.20 g. Taking the Faraday constant as 96 500 C mol⁻¹ and Ar(Cu) = 63.5, the student calculates the expected mass of copper and then claims a current efficiency of 108%. She also notes that the blue colour of the electrolyte did not fade during the experiment. Her partner argues that the efficiency above 100% is genuine 'because copper is also coming from the anode and adding to the plate'.
1. Calculate the theoretical maximum mass of copper that 960 C could deposit, and show why the 108% claim cannot be correct.
Read the worked answer
Charge passed: Q = I t = 0.50 A × 1920 s = 960 C
Moles of electrons: n(e⁻) = Q / F = 960 C / 96 500 C mol⁻¹ = 9.948 × 10⁻³ mol
Cathode half-equation: Cu²⁺ + 2e⁻ → Cu, so 2 mol electrons deposit 1 mol copper. n(Cu) = 9.948 × 10⁻³ / 2 = 4.974 × 10⁻³ mol
Mass: m = 4.974 × 10⁻³ mol × 63.5 g mol⁻¹ = 0.316 g
Apparent efficiency: 0.34 g / 0.316 g × 100 = 108%
Faraday's laws give the maximum mass that can be deposited for a given charge, assuming every electron reduces a Cu²⁺ ion. Mass deposited cannot exceed that value, so an apparent efficiency above 100% is always a sign of a measurement or procedural error, not a real efficiency.
2. Explain the most likely reason for the mass gain being too high, and evaluate the partner's claim that copper from the anode adds to the plate.
Read the worked answer
Likely cause: the key was not properly dried. It was rinsed and only blotted, so residual water (and any loosely adhering solution) is weighed as if it were copper. Other possibilities include a greasy or oxidised key surface trapping solution, a loose/spongy deposit holding moisture, or an inconsistent balance reading.
The copper deposited is fixed by the charge: each Cu²⁺ reduced at the cathode requires two electrons, so a copper atom cannot arrive 'for free' from the anode. The partner has confused the source of the ions with the cause of deposition. The anode does dissolve (Cu → Cu²⁺ + 2e⁻), but those ions must still be reduced at the cathode using electrons from the circuit, so the anode cannot raise the mass above the Faraday maximum - it only maintains the ion concentration.
A correct procedure is to dry the key thoroughly (e.g. rinse with distilled water then propan-2-ol, or dry in a desiccator) and reweigh to constant mass. A realistic efficiency is slightly under 100%, because some charge drives side reactions.
3. Explain why the blue colour of the electrolyte did not fade, and describe what would be observed if graphite electrodes were used instead.
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 ions are removed from solution at the cathode at essentially the same rate as they are replaced by the dissolving anode, so the concentration of Cu²⁺ stays roughly constant and the blue colour persists. (The anode loses mass; the cathode gains it.)
With graphite (inert) electrodes: there is no source of new Cu²⁺, so the cathode reaction depletes copper ions and the blue colour fades. At the anode, water (or hydroxide) is oxidised instead of copper: 2H₂O → O₂ + 4H⁺ + 4e⁻ so oxygen gas is seen at the anode and the solution becomes acidic. Once the Cu²⁺ is exhausted, hydrogen is produced at the cathode (2H⁺ + 2e⁻ → H₂). The key observation is that with an inert anode the electrolyte changes, whereas with an active copper anode it is regenerated.