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QJEEM23Che573: JEE Main 2023 Chemistry - Class 12 JEEM23Che573 Solution: A metal surface of area has to be coated with nickel layer of…

A metal surface of 100 cm2100 \mathrm{~cm}^{2} area has to be coated with nickel layer of thickness 0.001 mm0.001 \mathrm{~mm} . A current of 2 A2 \mathrm{~A} was passed through a solution of Ni(NO3)2\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2} for ' x\mathrm{x} ' seconds to coat the desired layer. The value of x\mathrm{x} is __________. (Nearest integer) ( ρNi\rho_{\mathrm{Ni}} (density of Nickel) is 10 g mL10 \mathrm{~g} \mathrm{~mL} , Molar mass of Nickel is 60 g mol−160 \mathrm{~g} \mathrm{~mol}^{-1} F=96500 C mol−1)\left.\mathrm{F}=96500 ~\mathrm{C} ~\mathrm{mol}^{-1}\right)

Step-by-Step Verified Solution

Using the Faraday's law of electrolysis, we can directly relate the amount of substance deposited (in this case, the nickel layer) with the electric charge passed through the electrolyte.

The formula for Faraday's law of electrolysis is:

W=z×i×tW = z \times i \times t

where W is the amount of substance deposited (in grams), z is the electrochemical equivalent (grams per coulomb), i is the current (in amperes), and t is the time (in seconds).

By relating the density and volume of the nickel layer to the electric charge passed through the electrolyte, we can calculate the time needed for the deposition:

10×100×0.0001=( atomic wt.  v.f )×2×x9650010 \times 100 \times 0.0001 = \frac{\left(\frac{\text { atomic wt. }}{\text { v.f }}\right) \times 2 \times x}{96500}

where v.f is the valence factor for the reaction (in this case, 2).

Solving for x, we get:

x=161 secx = 161 \, \mathrm{sec}

So, the value of x is 161 seconds.

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