Question

Copper is being electrodeposited from a $CuSO_4$ bath into a stainless steel cathode of total surface area of $2m^2$ in an electrolytic cell operated at a current density of $200 Am^{-2}$ with a current efficiency of 90%. If current density of an electrolytic cell is the current per unit area of electrode then mass of copper deposited (in kg) in 24 hours is: [Given: Faraday's constant = 96500 $C mol^{-1}$, atomic mass of copper = 63.5$gmd^{-1}$]

Answer 1

10.23 kg

Answer 2

To calculate the mass of copper deposited, we will follow these steps:

1. Calculate the total current (I): Current density (J) is given as $200 Am^{-2}$ and the total surface area (A) of the cathode is $2m^2$. $I = J \times A$ $I = 200 Am^{-2} \times 2 m^2 = 400 A$

2. Convert time (t) to seconds: The deposition time is 24 hours. $t = 24 \text{ hours} \times 60 \text{ minutes/hour} \times 60 \text{ seconds/minute} = 86400 \text{ seconds}$

3. Determine the number of electrons (n) involved in the deposition of copper: Copper is being electrodeposited from $CuSO_4$, which means $Cu^{2+}$ ions are reduced to Cu metal. The half-reaction is: $Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$ Therefore, $n = 2$ electrons per mole of copper.

4. Apply Faraday's First Law of Electrolysis, considering current efficiency: The mass of substance (w) deposited can be calculated using the formula: $w = \frac{M \times I \times t \times \eta}{n \times F}$ Where:

   • w = mass of copper deposited (in grams)
   • M = atomic mass of copper = 63.5 $g mol^{-1}$
   • I = total current = 400 A
   • t = time in seconds = 86400 s
   • $\eta$ = current efficiency = 90% = 0.90
   • n = number of electrons = 2
   • F = Faraday's constant = 96500 $C mol^{-1}$

   Substitute the values into the formula: $w = \frac{63.5 \text{ g/mol} \times 400 \text{ A} \times 86400 \text{ s} \times 0.90}{2 \times 96500 \text{ C/mol}}$ $w = \frac{63.5 \times (400 \times 86400) \times 0.90}{193000}$ $w = \frac{63.5 \times 34560000 \times 0.90}{193000}$ $w = \frac{63.5 \times 31104000}{193000}$ $w = \frac{1975404000}{193000}$ $w = 10235.253367... \text{ g}$

5. Convert the mass from grams to kilograms: $w_{kg} = \frac{10235.253367... \text{ g}}{1000 \text{ g/kg}} = 10.235253367... \text{ kg}$

Rounding to two decimal places, the mass of copper deposited is approximately 10.24 kg. Among the given options, 10.23 kg is the closest value.

The final answer is $\boxed{\text{10.23 kg}}$.