Question

A teapot with a surface area of 700 cm2 is to be silver plated. It is attached to the negative electrode of an electrolytic cell containing silver nitrate (Ag+ NO3 −). If the cell is powered by a 12.0 V battery and has a resistance of 1.80 Ω, how long does it take for a 0.133 mm layer of silver to build up on the teapot? (The density of silver is 10.5 × 103 kg m3 ⁄ ).

Answer 1

Time taken = 13095 seconds

Step-by-step explanation:

Mass of silver required = density of silver * volume of silver

volume of silver = surface area of teapot * height of silver layer

volume = 700 cm² * 1 m²/ 10⁴ cm² * 0.133 * 1 m/1000 mm

volume = 9.31 * 10⁻⁶ m³

mass of silver = 10.5 * 10³ kg/m³ * 9.31 * 10⁻⁶ m³

mass of silver = 9.7755 * 10⁻² kg * 1000 g/ 1 kg = 97.755 g

1 mole of Ag⁺ will be discharged by 96500 C of charge

molar mass of silver 108 g/mol

97.755 g of silver will be discharged by = 96500 C * 97.755/108 = 87345.9 C of charge

From Q = I * t

I = V/R = 12/1.8 = 6.67 A

Therefore, t = Q/I

t = 87345.9/6.67

t = 13095 seconds