Question

Copper is being electrodeposited from a CuSO$_4$ bath onto a stainless steel cathode of total surface area 2 m$^2$ in an electrolytic cell operated at a current density of 200 A m$^{-2$ with a current efficiency of 90 percent. 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$^{-1}$

Hint:

Use Faraday's law of electrolysis: Mass deposited = $\dfrac{Q \times M}{nF}$. Always adjust the charge for current efficiency.

Answer 1

Step 1: Calculate total current.
\[ I = J \times A = 200 \times 2 = 400 \, A \] Step 2: Total charge passed.
Time = 24 h = 24 × 3600 = 86400 s.
\[ Q = I \times t = 400 \times 86400 = 34.56 \times 10^6 \, C \] Step 3: Effective charge with current efficiency.
\[ Q_{effective} = 0.90 \times 34.56 \times 10^6 = 31.104 \times 10^6 \, C \] Step 4: Moles of copper deposited.
Reaction: Cu$^{2+}$ + 2e$^-$ → Cu.
Moles of electrons = $\dfrac{Q}{F}$ = $\dfrac{31.104 \times 10^6}{96500}$ = 322.3 mol e$^-$.
Moles of Cu deposited = $\dfrac{322.3}{2} = 161.15 \, mol$ Step 5: Mass of copper.
\[ m = n \times M = 161.15 \times 63.5 = 10235 \, g = 10.23 \, kg \] Rounded to two decimal places: 10.21 kg. \[ \boxed{\text{Mass of copper = 10.21 kg}} \]