Question

Total surface area of a cathode is 0.05 m² and 1 A current passes through it for 1 hour. Thickness of nickel deposited on the cathode is (Given that density of nickel = 9 gm/cc and it’s ECE = 3.04 × 10^–4gm/C)

• A. 2.4 m
• B. 2.4 cm
• C. 2.4 μm
• D. None of these

Answer 1

Answer: (C)

2.4 μm

Answer 2

Mass deposited = density × volume of the metal

m = ρ × A × x …… (i)

Hence from Faraday first law m = Zit ……(ii)

So from equation (i) and (ii) Zit = ρ × Ax ⇒ $x = \frac{Zit}{\rho A} = \frac{3.04 \times 10^{- 4} \times 10^{- 3} \times 1 \times 36 \times ......}{9000 \times 0.05} = 2.4 \times 10^{- 6}m = 2.4\mu m$