Given:
Current, i = 15 A
Surface area of the plate = 200 cm2,
Thickness of silver deposited= 0.1 mm = 0.01 cm
Volume of Ag deposited on one side = 200 × 0.01 cm3 = 2 cm3
∴ Volume of Ag deposited on both side = 4 cm3
Mass of silver deposited,
m = Volume × Specific gravity × 1000 = 4 × 10-3 × 10.5 ×1000 = 42 kg
Using the formula, m = Zit, we get:
42 = ZAg × 15 × t