Question

How many grams of silver could be plated out on a serving tray by electrolysis of a solution containing silver in +1 oxidation state for a period of 8.0 hours at a current of 8.46 amperes? What is the area of the tray if the thickness of the silver plating is 0.00254 cm? Density of silver is 10.5 g cm⁻³.

Answer 1

The electrode reaction is:

\[\ce{Ag{^{+}_{(aq)}} + e- -> Ag_{(s)}}\]

Equivalent mass of Ag = \[\ce{\frac{108}{1}}\]

= 108

∴ Electrochemical equivalent of Ag = \[\ce{\frac{Equivalent mass}{F}}\]

= \[\ce{\frac{108}{96500}}\]

= 1.12 × 10⁻³ g C⁻¹

According to Faraday's first law of electrolysis,

W = Z × I × t

= 1.12 × 10⁻³ × 8.46 × 8 × 60 × 60

= 272.88 g

Suppose the area of the tray is A cm².

∴ Volume of coated silver = Area × Thickness

= A × 0.00254 cm³

Mass of coated silver = Volume × Density

= A × 0.00254 × 10.5 g

Hence, A × 0.00254 × 10.5 = 272.88

or, A = \[\ce{\frac{272.88}{0.00254 × 10.5}}\]

= 10231.7 cm²

= 1.02 × 10⁴ cm²