A metal surface of 100 cm² area has to be coated with nickel layer of thickness 0.001 mm. A current of 2 A was passed through a solution of Ni(NO₃)₂ for 'x' seconds to coat the desired layer. The value of x is __________. (Nearest integer) ( ρ_(Ni) (density of Nickel) is 10 g mL, Molar mass of Nickel is 60 g mol⁻¹ F=96500 C mol⁻¹)

Solution

Chemistry ma’am · 4 steps

Turn the coating into a mass first, then let Faraday convert that mass into charge and time.

1. Volume of nickel. Put the thickness in cm: 0.001 mm = 10⁻⁴ cm. The volume of the layer is 100 × 10⁻⁴ = 0.01 cm³, which is 0.01 mL.
2. Mass and moles. Mass = 10 × 0.01 = 0.1 g. Moles of Ni = 0.1/60 = 1/600 mol.
3. Charge needed. Each nickel ion takes two electrons, Ni²⁺ + 2e⁻ → Ni. So the charge is 1/600 × 2 × 96500 = 321.7 C.

   x = Q/I = (321.7 C)/(2 A) = 160.8 s

4. So. The current must flow for about 160.8 s, which rounds to 161.

The common slip: 1608. Treating 0.001 mm as 0.001 cm makes the layer ten times thicker and gives 1608 s, not 161.

The idea to keep. For plating questions, go from the layer's volume to its mass, then to moles. Multiply by the electrons per ion and by F to get the charge, then divide by the current for the time.

Answer from the NTA final answer key. Solution by PrepWiser.