Silver Electroplating: Efficiency, Side Reactions and Current Increase
Theoretical classroom model only. Use the supplied constants and idealizations; this is not authorization to handle unknown samples, run a combustion experiment or alter an operating chemical plant. A workshop electroplates steel fittings with silver from a nitrate bath. The fitting is the cathode and a pure silver anode is used. A run is carried out at 2.50 A for 45.0 min. The rectifier ammeter is accurate to ±0.05 A and the timer to ±1 s. A coulomb meter in series records 6.20 × 10^3 C. The mass of silver deposited is found to be 6.00 g. The specification requires at least 6.00 g per fitting. The operator notices gas bubbles at the cathode and a dark film on the anode. Standard reduction potentials: Ag+/Ag = +0.80 V; 2H2O + 2e− → H2 + 2OH−, E° = −0.83 V. The bath pH is 9.5. The workshop proposes doubling the current to 5.00 A to increase output.
1. Calculate the theoretical mass of silver corresponding to the coulomb-meter reading and the current efficiency.
Read the worked answer
n(e−) = Q/F = 6200 / 96 485 = 0.06426 mol. Ag+ + e− → Ag, so n(Ag) = 0.06426 mol. m(theoretical) = 0.06426 × 107.87 = 6.93 g. Current efficiency = actual/theoretical × 100 = 6.00/6.93 × 100 = 86.6%. If instead the nominal charge I×t = 2.50 × 2700 = 6750 C is used, the theoretical mass is 7.55 g and the efficiency is 79.5%; the coulomb meter is the more direct measure, but the 8% difference between 6750 C and 6200 C is larger than the ammeter's ±2% and should be investigated (current fluctuation or meter calibration).
2. Explain the gas bubbles at the cathode and the dark film on the anode, and state what each implies for the plating process.
Read the worked answer
Gas evolution is consistent with a competing process; hydrogen from water reduction is one hypothesis to test, not an identification established by visible bubbles alone. In that hypothesis the reaction is 2H2O + 2e− → H2 + 2OH−, because some of the current is used to make hydrogen instead of depositing silver. This lowers the current efficiency and can make the deposit porous. At pH 9.5 the water reduction potential is about −0.56 V, but hydrogen evolution on silver has a significant overpotential; silver-ion depletion near the cathode can also shift the local potential. The dark film on the anode is probably Ag2O or a passivating layer formed by local alkalinity or by impurities; it can slow anodic dissolution and cause the Ag+ concentration to fall. Both observations point to mass-transport or current-density problems, so stirring and checking the silver concentration are indicated.
3. Evaluate the proposal to double the current, using the data and at least one quantitative estimate.
Read the worked answer
Doubling the current doubles the nominal charge: 5.00 A × 2700 s = 13 500 C, which would give 15.1 g of silver at 100% efficiency. Even at 70% efficiency that is 10.6 g, above the 6.00 g specification. At fixed area and otherwise comparable conditions, side reactions or passivation may worsen; the supplied data do not establish that outcome with certainty: higher current density increases the cathodic overpotential, favouring H2 evolution, and can produce dendritic or powdery deposits. The anode may passivate faster. A better approach is to improve agitation and maintain the Ag+ concentration, or to increase the current moderately and monitor efficiency, rather than simply doubling it.