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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.