Silver Plating Brass Handles: Faraday Law and Anode Choice
A small heritage-metal workshop has an order for 40 brass spoon handles, each with a total surface area of 35 cm², to be silver-plated to a nominal thickness of 12 µm. The workshop uses a cyanide-free silver bath, a 0.80 A current per item, and a quoted cathode efficiency of 94%. The bath is a 2.0 L tank containing 45 g of silver nitrate (M = 169.87 g mol⁻¹; Ar(Ag) = 107.87; density of silver = 10.49 g cm⁻³). The technician plans to hang all 40 items in parallel from the same rectifier, use a stainless-steel sheet as the anode, and stop when the calculated time has elapsed. A conservator warns that thickness on complex handles may vary and that the bath chemistry will drift. The workshop needs to know the plating time, whether the dissolved silver is sufficient, and whether the anode and parallel arrangement are appropriate.
1. Calculate the mass of silver required per item and for all 40 items, the plating time for one item, and the silver available in the bath. Hence judge whether the bath contains enough silver for the order.
Read the worked answer
Volume of silver per item
12 µm = 12 × 10⁻⁴ cm = 0.0012 cm.
Volume = 35 cm² × 0.0012 cm = 0.042 cm³.
Mass per item = 0.042 cm³ × 10.49 g cm⁻³ = 0.4406 g.
For 40 items: 40 × 0.4406 = 17.62 g Ag.
Charge and time for one item
The electrode reaction is Ag⁺ + e⁻ → Ag, so n = 1.
Theoretical charge:
Q = nF = (0.4406/107.87) × 96485 = 394.1 C.
At 94% cathode efficiency, the actual charge required is:
Q(actual) = 394.1/0.94 = 419.3 C.
At 0.80 A per item:
t = Q/I = 419.3/0.80 = 524 s ≈ 8.7 min per item.
If all 40 items are plated in parallel at 0.80 A each, the total current is 32 A and the time is still about 8.7 min. If the rectifier supplies only 0.80 A total, the time would be 40 times longer, about 5.8 h.
Silver in the bath
Mass of Ag in 45 g AgNO₃:
45 × (107.87/169.87) = 28.58 g Ag.
The order needs 17.62 g, so the bath contains enough dissolved silver for the order even if no anode replenishment occurs. However, the dissolved silver concentration would fall from about 14.3 g dm⁻³ to about 5.5 g dm⁻³ if all of it were used, which would change the plating kinetics and deposit quality. A soluble silver anode is therefore needed to maintain the bath, not just to supply the initial silver.
2. Evaluate the use of a stainless-steel sheet as the anode and the plan to hang all 40 items in parallel from one rectifier. Recommend improvements.
Read the worked answer
A stainless-steel sheet is an inert anode, not a silver source. At the anode, water would be oxidised:
2H₂O → O₂ + 4H⁺ + 4e⁻.
This generates oxygen, acidifies the bath, and does not replenish Ag⁺. It may also introduce chromium, nickel, or iron species if the stainless steel corrodes, contaminating the deposit. The correct anode is high-purity silver, preferably in an anode bag, with an anode area at least as large as the total cathode area. The anode dissolves as Ag → Ag⁺ + e⁻, maintaining the silver concentration.
Hanging 40 items in parallel from one rectifier is also problematic. In parallel, each item sees the same applied voltage, but the current through each branch depends on contact resistance, racking, and the local bath conductivity. The technician cannot assume that each item draws exactly 0.80 A. Some items may receive too little current and be under-plated; others may receive too much and produce rough or dendritic deposits. Poor contact at one hanger can also cause local overheating or no deposit at all.
Better practice is to use a proper plating rack with individual, cleaned contacts, control the total current and voltage, and monitor the current per rack. Agitation or cathode movement improves throwing power into recesses, and a lower current density with longer time usually gives a more uniform thickness on complex handles. The workshop should verify thickness on a test coupon or by X-ray fluorescence rather than relying only on Faraday's law.
3. If the cathode efficiency falls to 85%, calculate the actual silver thickness after the originally calculated time and the time needed to reach 12 µm. Explain why thickness may still vary across a complex handle.
Read the worked answer
The original time was 524 s at 0.80 A, corresponding to a charge of:
Q = 0.80 × 524 = 419.2 C.
At 85% efficiency, the silver deposited is:
m = (Q × M × efficiency)/(nF)
= (419.2 × 107.87 × 0.85)/96485
= 0.3985 g.
Thickness = m/(area × density)
= 0.3985/(35 × 10.49)
= 0.001085 cm = 10.85 µm.
So the deposit would be about 10.9 µm, roughly 9% below the 12 µm target.
To reach 12 µm at 85% efficiency, the required charge is:
Q = (12 × 10⁻⁴ cm × 35 cm² × 10.49 g cm⁻³ × 96485)/(107.87 × 0.85)
= 463.6 C.
At 0.80 A, t = 463.6/0.80 = 580 s ≈ 9.7 min.
Thickness can still vary across a complex handle because current density is not uniform. Edges, corners, and areas facing the anode receive a higher current density and plate thicker; recesses and areas shielded by the rack receive less. Gas evolution, poor agitation, and variations in surface preparation also affect the deposit. The workshop should use agitation, a well-spaced anode arrangement, possibly auxiliary anodes or shielding, and measure thickness at several points rather than assuming a single average.