Copper is being electrodeposited from a CuSO\(_4\) bath onto a stainless steel cathode of total surface area of 2 m\(^2\) in an electrolytic cell operated at a current density of 200 A m\(^{-2}\) with a current efficiency of 90%. The mass of copper deposited in 24 h is _________ kg (rounded off to two decimal places). Given: Faraday's constant = 96500 C mol\(^{-1}\), Atomic mass of copper = 63.5 g mol\(^{-1}\).
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When calculating the mass of an element deposited by electrolysis, use Faraday's law: \( {mass} = \frac{Q}{nF} \times M \), where \( M \) is the molar mass and \( n \) is the number of electrons involved in the reaction. Don't forget to apply the current efficiency to get the actual mass deposited.
First, we need to calculate the total charge passed through the electrolyte:
\[
{Current density} = 200 \, {A/m}^2, \quad {Area} = 2 \, {m}^2
\]
The total current \( I \) is:
\[
I = 200 \times 2 = 400 \, {A}
\]
Next, we calculate the total charge passed in 24 hours:
\[
Q = I \times t = 400 \times (24 \times 3600) = 34,560,000 \, {C}
\]
Now, using Faraday's law, the amount of substance deposited can be calculated. For copper, the number of moles of copper deposited is given by:
\[
{moles of Cu} = \frac{Q}{nF}
\]
where \( n = 2 \) (since copper undergoes a 2-electron transfer), and \( F = 96500 \, {C/mol} \). So:
\[
{moles of Cu} = \frac{34,560,000}{2 \times 96500} = 179.8 \, {mol}
\]
The mass of copper deposited is:
\[
{mass of Cu} = 179.8 \times 63.5 = 11,419.3 \, {g} = 11.42 \, {kg}
\]
Since the current efficiency is 90%, we adjust the mass accordingly:
\[
{mass deposited} = 11.42 \times 0.90 = 10.28 \, {kg}
\]
Thus, the mass of copper deposited is approximately 10.20 kg to 10.30 kg.
