Question

# The density of copper is 8.94$m{{L}^{-1}}$. Find out the number of coulombs needed to plate an area 10 x 10 $c{{m}^{2}}$ to a thickness of ${{10}^{-2}}$ cm using $CuS{{O}_{4}}$ solution as electrolyte.(At. Mass of Cu = 63.6).

Hint: Calculate the volume of Cu to be deposited by multiplying the area and the thickness mentioned in the question above. Now find the mass of Cu needed by substituting the calculated volume in the formula for density. Now apply Faraday's first law of electrolysis to find the number of coulombs of electrons required.

Let us calculate the volume of Cu to be plated:
Volume = Area x Thickness
\begin{align} & {{V}_{{}}}{{=}_{{}}}{{100}_{{}}}{{\text{x}}_{{}}}{{10}^{-2}}_{{}}c{{m}^{3}} \\ & {{V}_{{}}}{{=}_{{}}}{{1}_{{}}}c{{m}^{3}} \\ \end{align}
Weight of Cu to be plated is given by:
Density =$\dfrac{\text{Mass}}{\text{Volume}}$
Mass = Density x Volume
Mass = 8.94 x 1 g = 8.94 g
Number of moles of Cu is n = $\dfrac{8.94}{63.5}$= 0.14
Faraday’s First Law of Electrolysis states that the chemical deposition of a substance due to the flow of current through an electrolyte is directly proportional to the quantity of electricity passed through it.
\begin{align} & {{\text{m}}_{{}}}{{\alpha }_{{}}}Q \\ & {{m}_{{}}}{{=}_{{}}}Z.Q \\ \end{align}
Where,
m is the mass of electrolyte deposited,
Q is the quantity of electricity
Z is the constant of proportionality and is known as the electro-chemical equivalent.
$\text{C}{{\text{u}}^{2+}}_{{}}{{+}_{{}}}2{{e}^{-}}_{{}}{{\to }_{{}}}Cu$
Hence 2 moles of electrons or 2F(Faraday) are required to deposit 1 mole of Cu.
To deposit 0.14 mole of Cu the moles of electrons required :
\begin{align} & \text{Q = 0}\text{.14 x 2 F} \\ & \text{Q = 0}\text{.28 F} \\ \end{align}
1F is equal to 96500 C, hence the number of coulombs required to plate Cu is :
C = 96500 x 0.28 coulombs
C = 27172 coulombs.
Therefore, the number of coulombs required for plating of Cu is 27172 C.

Note: The term faraday is used in chemistry and farad used in physics. One Farad is defined as the capacitance across two plates when charged to 1 coulomb and the potential difference is 1 volt. On the other hand, one faraday is the magnitude of charge of 1 mole of electrons. Although both terms are in honor of the same scientist, they are two completely different terms.