Question

A teapot with a surface area of 450 cm2 is to be plated with silver. It is attached to the negative electrode of an electrolytic cell containing silver nitrate (Ag+ NO3−). The cell is powered by a 12.0-V battery and has a resistance of 1.40 Ω. If the density of silver is 1.05 104 kg/m3, over what time interval does a 0.133-mm layer of silver build up on the teapot?

Answer 1

Given :-

A = 450 cm^2 = 450 x 10^-4 m^2 = 0.0450 m^2

V = 12 V

R = 1.40 ohm

L = 0.133 m = 0.133 x 10^-3 m

= 1.05 x 10^4 kg/m^3

mass of silver plated,

m = V = AL

m = 1.05 x 10^4 kg/m^3 x 0.0450 m^2 x 0.133 x 10^-3 m

m = 0.0628 kg = 62.8 g

no of atom plated, n = m/M x avogadro number

n =  62.8 g / 107.87 g x 6.022 x 10^23

n = 3.5 x 10^23

It = Q = ne

(V/R)t = ne

t = Rne / V

t = (1.40 x 3.5 x 10^23 x 1.6 x 10^-19) / 12

t = 6533.33 sec

t =6533.33 / 3600 = 1.815 hours..