Question

A rod measuring 13.870000 x 3.640000 x 5.980000 cm was plated by means of a current of 84.780000 milliamps for 3.670000 hours. What is the thickness in millimeters of the silver deposit on the rod, given that the density of silver is 10.650000 g / cm3

Answer 1

0.024 mm

Step-by-step explanation:

The quantity of charge deposited Q = It where I = current = 84.780000 mA = 0.084780000 A and t = time = 3.670000 hours = 3.670000 × 3600 s = 132120000 s.

Also Q = nF where n = number of moles of electrons silver deposited and F = Faraday's constant = 96500 C/mol

So, It = nF

n = It/F = 0.084780000 A × 132120000 s/96500 C/mol = 1120.1134 C/96500 C/mol = 0.012 mol

So, we have 0.012 mol of electrons

Our chemical equation is

Ag⁺ + e⁻ → Ag

Since 1 mol of electrons deposits 1 mol of silver atoms, then, 0.012 mol of electrons deposits 0.012 mol of silver atoms.

Since number of moles of silver atoms, n' = m/M where m = mass of silver atoms deposited and M = molar mass of silver = 107.868 g/mol

So, m = n'M

since n' = 0.012 mol,

m = 0.012 mol × 107.868 g/mol = 1.294 g

Since density of silver ρ = m/V where m = mass of silver deposited = 1.294 g and V = volume of silver deposited

V = m/ρ

Since, ρ = 10.650000 g/cm³

V = 1.294 g/10.650000 g/cm3 =

V = 0.122 cm³

Since the dimensions of the measuring rod are 13.870000 x 3.640000 x 5.980000 cm which represent its length, l, width, w and height, h respectively, the volume of silver deposited V = Ah' where A = area of the rod, lw and h' = thickness of silver deposited

So, V = Ah

V = lwh'

h' = V/lw

= 0.122 cm³/13.870000 cm x 3.640000 cm

= 0.122 cm³/50.4868 cm²

= 0.0024 cm

= 0.024 mm