Answer:
0.024 mm
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
